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1. ## Test Test for Dementia Below are four ( 4 ) questions and a bonus question. You have to answer them instantly. You can't take your time, answer all of them immediately . OK? Let's find out just how clever you really are.... First Question: You are participating in a race. You overtake the second person. What position are you in? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~ absolutely wrong! If you overtake the second person and you take his place, you are second! Try not to screw up next time. but don't take as much time as you too! k for the first question, OK ? Second Question: I f you overtake the last person, then you are...? (scroll down) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~ Answer: If you answered that you are second to last, then you are wrong again. Tell me, how can you overtake the LAST Person? You're not very good at this, are you? Third Question: Very tricky arithmetic! Note: This must be done in your head only . Do NOT use paper and pencil or a calculator. Try it. Now add 10 . What is the total? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~ Did you get 5000 ? The correct answer is actually 4100. If you don't be! lieve it, check it with a calculator! Today is definitely not your day, is it? Maybe you'll get the last question right.... ....Maybe. Fourth Question: Mary's father has five daughters: 1. Nana, 2. Nene, 3. Nini, 4. Nono. What is the name of the fifth daughter? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~ NO! Of course it isn't. Her name is Mary. Read the question again! Okay, now the bonus round: A mute person goes into a shop and wants to buy a toothbrush. By imitating the action of brushing his teeth he successfully expresses himself to the shopkeeper and! the purchase is done. Next, a blind man comes into the shop who wants to buy a pair of sunglasses; how does HE indicate what he wants? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~ He just has to open his mouth and ask... It's really very simple.... 2. ## Thanks given for this post: IanF (28-Apr-11) 3. Brilliant 4. If this was a job interview... I got all the questions right but... the comments after each question would have provoked me to the point where the interviewer would have needed medical assistance and I would have needed a lawyer...

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## Intermediate Algebra for College Students (7th Edition) $x^2+4x+4=x^2+2\cdot x\cdot2+2^2=(x+2)^2$, thus the answer is f

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# Contractive Sequence • Oct 24th 2006, 09:07 PM OntarioStud Contractive Sequence Let x1 be greater than zero. x(n+1)=1/(2+xn) for all n greater than or equal to 2. Show that (xn) is a contractive sequence and find its limit. Stuck on the proof. Pretty sure the limit would just be the roots of (x^2)+2x-1=0. Any help appreciated. • Oct 25th 2006, 09:08 AM OntarioStud I just need a little help. I have that: abs(x(n+1)-x(n))=abs(1/(2+xn)-1(2+x(n-1))). I just don't know how to simplify this in terms of abs(xn-x(n-1)) and find what C in the contractive sequence is. Could anyone just steer me in the right direction? • Oct 25th 2006, 10:04 AM ThePerfectHacker We need to show, $\displaystyle |x_{n+1}-x_n|\leq k|x_n-x_{n-1}|$ For some $\displaystyle k\in (0,1)$ Now, Let $\displaystyle a=x_{n-1}$ then we have, $\displaystyle x_{n+1}=\frac{1}{2+x_n}=\frac{1}{2+\frac{1}{2+a}}= \frac{a+2}{2a+5}$ $\displaystyle x_n=\frac{1}{a+2}$ Thus, $\displaystyle \left| \frac{a+2}{2a+5}-\frac{1}{a+2} \right| \leq k \left| \frac{1}{a+2}-a\right|$ If und only if, $\displaystyle \left| \frac{a^2+2a-1}{(2a+5)(a+2)}\right| \leq k \left|\frac{-a^2-2a+1}{2a+5} \right|$ If und only if, $\displaystyle 0<\frac{1}{a+2}\leq k<1$ Thus, by choosing this konstant to be that value we can form a contractive sequence.

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# Whittle’s Stabilizer Theorem Representable matroids are an attractive subclass of matroids, because in their study you have access to an extra tool: a matrix representing this matroid. This is a concise way to describe a matroid: $O(n^2)$ numbers as opposed to $O(2^{2^n})$ bits declaring which subsets are (in)dependent. Let $M$ be a matroid, and $A$ a representation matrix of $M$. The following operations do not change the matroid: • Add a multiple of a row of $A$ to another row of $A$; • Scale a row of $A$ by a nonzero constant; • Scale a column of $A$ by a nonzero constant; • Add or remove all-zero rows; • Apply a field automorphism to each entry of $A$. If a matrix $A_1$ can be turned into a matrix $A_2$ through such operations, then we say $A_1$ and $A_2$ are equivalent. If we don’t use any field automorphisms, then we say they are projectively equivalent. Generally, a matroid can have multiple inequivalent representations over a field. The exceptions are the finite fields $\textrm{GF}(2)$ and $\textrm{GF}(3)$ (shown in [BL]). When we try to prove some theorem about a matroid or class of matroids, inequivalent representations can be a major complicating factor. For instance, the excluded-minor characterization of ternary matroids can be proven in under five pages [Oxley, pp. 380-385], whereas the excluded-minor characterization of quaternary matroids takes over fifty [GGK]. It is not surprising, then, that significant efforts have been made to get a handle on inequivalent representations. In this post I will focus on one such effort, namely a very attractive theorem by Geoff Whittle, who recently celebrated his 65th birthday with a wonderful workshop. First, a definition. Definition. Let $\mathbb{F}$ be a field, and $\mathcal{M}$ a minor-closed class of $\mathbb{F}$-representable matroids. Let $N \in \mathcal{M}$. We say $N$ is a stabilizer for $\mathcal{M}$ if, for every 3-connected matroid $M \in \mathcal{M}$ that has $N$ as a minor, each representation of $N$ (over $\mathbb{F}$) extends to at most one representation of $M$ (up to the equivalence defined above). In other words, once we select a representation for $N$, we have uniquely determined a representation of $M$. A small example: let $\mathbb{F} = \textrm{GF}(5)$, let $\mathcal{M}$ be the set of all minors of the non-Fano matroid $F_7^-$, and let $N$ be the rank-3 wheel. Now $N$ has the following representation: $$\begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 1\\ 0 & 1 & 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 1 & a \end{bmatrix}$$ where $a \in \{1, 2, 3\}$. The only 3-connected matroids in $\mathcal{M}$ that have $N$ as a minor are $N$ itself and $F_7^-$. We need to check that each representation of $N$ extends to at most one representation of $F_7^-$. Up to equivalence, the latter representation must look like $$\begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 1 & 1\\ 0 & 1 & 0 & 1 & 1 & 0 & b\\ 0 & 0 & 1 & 0 & 1 & a & c \end{bmatrix}$$ and it is readily checked that we must have $b = c = a = 1$. Hence two of the representations of $N$ do not extend to a representation of $M$, whereas one extends uniquely to a representation of $M$. So $N$ is a stabilizer for $\mathcal{M}$. If $\mathcal{M}$ is an infinite class, we cannot do an exhaustive check as in the example to verify a stabilizer. But Geoff Whittle managed to prove that a finite check still suffices: Whittle’s Stabilizer Theorem [Whi]. Let $\mathcal{M}$ be a minor-closed class of $\mathbb{F}$-representable matroids, and $N \in \mathcal{M}$ a 3-connected matroid. Exactly one of the following holds: • $N$ is a stabilizer for $\mathcal{M}$ over $\mathbb{F}$; • There is a 3-connected matroid $M \in \mathcal{M}$ such that either: • $N = M\backslash e$ and some representation of $N$ extends to more than one representation of $M$; • $N = M / e$ and some representation of $N$ extends to more than one representation of $M$; • $N = M / e \backslash f$, $M / e$ and $M \backslash f$ are 3-connected, and some representation of $N$ extends to more than one representation of $M$. I will conclude this post with two applications. I will leave the finite case checks to the reader. Lemma. The matroid $U_{2,4}$ is a stabilizer for the class of quaternary matroids. Corollary [Kah]. A 3-connected, quaternary, non-binary matroid has a unique representation over $\textrm{GF}(4)$. Proof. A non-binary matroid $M$ has a $U_{2,4}$-minor. The matroid $U_{2,4}$ has the following representation: $$\begin{bmatrix} 1 & 0 & 1 & 1\\ 0 & 1 & 1 & a \end{bmatrix}$$ where $a \not\in \{0,1\}$. This leaves two choices for $a$, that are related through a field automorphism. Hence $U_{2,4}$ has (up to equivalence) a unique representation over $\textrm{GF}(4)$. But $U_{2,4}$ is a stabilizer, so $M$ is uniquely representable over $\textrm{GF}(4)$ as well. $\square$ Lemma. The matroids $U_{2,5}$ and $U_{3,5}$ are stabilizers for the class of quinary matroids. Lemma. The matroid $U_{2,4}$ is a stabilizer for the class of quinary matroids with no minor isomorphic to $U_{2,5}$ and $U_{3,5}$. Corollary [OVW]. A 3-connected, quinary matroid has at most six inequivalent representations over $\textrm{GF}(5)$. Proof. $U_{2,5}$ and $U_{3,5}$ have six inequivalent representations and are stabilizers. If $M$ does not have such a minor, then either $M$ is regular (and thus uniquely representable over any field) or $M$ has a $U_{2,4}$-minor, which is has three inequivalent representations. $\square$ ### References • [BL] Tom Brylawski and Dean Lucas, Uniquely representable combinatorial geometries. In Teorie Combinatorie (proc. 1973 internat. colloq.) pp. 83-104 (1976). • [GGK] Jim Geelen, Bert Gerards, Ajai Kapoor, The excluded minors for $\textrm{GF}(4)$-representable matroids. J. Combin. Th. Ser. B, Vol. 79, pp. 247-299 (2000). • [Kah] Jeff Kahn, On the uniqueness of matroid representations over $\textrm{GF}(4)$. Bull. London Math. Soc. Vol. 20, pp. 5–10 (1988). • [OVW] James Oxley, Dirk Vertigan, Geoff Whittle, On inequivalent representations of matroids over finite fields.J. Combin. Theory Ser. B. Vol. 67, pp. 325–343 (1996). • [Oxley] James Oxley, Matroid Theory, 2nd edition. Oxford University Press (2011). • [Whi] Geoff Whittle, Stabilizers of classes of representable matroids. J. Combin. Theory Ser. B, Vol. 77, pp. 39–72 (1999). This site uses Akismet to reduce spam. Learn how your comment data is processed.

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Maximum Loan Amount Formula A loan, by definition, is an annuity, in that it consists of a series of future periodic payments. The PV, or present value, portion of the loan payment formula uses the original loan amount. The original loan amount is essentially the present value of the future payments on the loan, much like the present value of an annuity. VA uses the one-unit loan limit to determine maximum VA loan amounts. Calculate Maximum VA Loan Amount & Tier 2 VA entitlement. includes 2019 county loan limits. Yes, you can have more than one VA loan. You can also use a VA loan to buy a home priced above the VA county loan limit. Retail Mortgage Lending It’s no secret the mortgage business has been tough lately, but bankers who have downsized in home lending say it’s no fad – they. line will be substantially smaller, focused on our retail deposit. · How to calculate gratuity amount using gratuity formula in 2019? The latest amendment bill increased the new limit of gratuity to Rs. 20 lakhs . This article covers Gratuity Formula – Last drawn salary X years of service X 15/26. Gratuity New Limit – Rs. 20 Lakhs Gratuity Eligibility – 5 Years of Services Rules – Gratuity is Mandatory Gratuity Calculations What is gratuity in salary slip in India? Mortgage Calculator for Singapore Property. Estimate your maximum loan amount with our mortgage calculator for your next home loan. Get a AIP (Approval In Principle) with our partner banks to confirm the loan amount and give you more clarity when budgeting your purchase. Calculate Loan Payments and Costs: Formulas and Tools. Formula for Amortizing Loan Payment . This formula works for most amortizing loans, which covers most loans-except credit cards and interest-only loans.. Your monthly payment is just a result of the loan amount, interest rate, and. A maximum loan amount describes the total amount that a borrower is authorized to borrow. maximum loan amounts are used in standard loans, credit cards and line of credit accounts. BREAKING DOWN ‘Maximum Loan Amount’. The maximum loan amount for a borrower is based on various factors and determined by the loan underwriter. Formula Used: A = (P / r) * [ 1 – (1+r) -N ] Where, A = Loan Amount P = Payment Amount r = Rate of Interest (compounded) N = Number of Payments Rate of Interest Compounded is, If Monthly, r = i / 1200 and N = n * 12 If Quarterly, r = i / 400 and N = n * 4 If Half yearly, r = i / 200 and N = n * 2 If Yearly, r = i / 100 and N = n The base limit will remain at \$417,000, but the formula for establishing limits. and for counties whose housing is priced somewhere between that amount and the lowest ceiling of \$271,050, the FHA. Loan Calculator Bankrate Com Check out the web’s best free mortgage calculator to save money on your home loan today. Estimate your monthly payments with PMI, taxes, homeowner’s insurance, HOA fees, current loan rates & more. Also offers loan performance graphs, biweekly savings comparisons and easy to print amortization schedules.Balloon Payment Excel for \$175 million in cash up-front and a maximum of \$125 million in contingent payments over the period of 2018-2020, based on achievements of clinical and regulatory milestones. Apama Medical has.

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1/57 # 2016年英语二轮提分策略3.29.ppt 0/100 2016年英语二轮提分策略3.29.ppt 20162016高考高考二轮提分策略二轮提分策略Lyyg 赵永茂年份平均分数客观平均主观平均难度系数语法填空短文改错书面表达201166.2350.0516.180.594.1811.99201264.0649.3714.690.534.7412.31201361.2546.7614.500.512.9111.37201476.3442.4924.770.649.294.3711.21201573.3142.0322.640.617.654.6610.00近几年高考成绩数据统计?一轮(基础)复习回顾?二轮(专题)策略方法?试卷结构及用时参考?阅读模块(真题研究,命题规律,方法策略)?语篇填空(真题研究,命题规律,方法策略)?完形模块(真题研究,命题规律,方法策略)?短文改错(真题研究,命题规律,方法策略)?书面表达(真题研究,命题规律,方法策略)一轮(基础)复习抓基础重巩固强规范保提升扎实积累、科学高效?早读“朗读”要求:?Sit Straight?Book in hand?Read aloud?Keep in mind?背诵内容→具体化。内容:依纲扣本考纲词汇?课本中含高考“考点”句子(解题);?课本中可“借用”到“书面表达”中的句子;?模拟练习好篇章?高考真题:完形填空、书面表达。?不良习惯:?①“早读“变成“早自习”?②“朗读”变成“阅读”?③“早读”做“书面习题”“早读”(一轮适用、二轮调整)★态度、★方法、★努力、★时间阅读理解阅读理解?我们知道賢鏛是在生活中很重要的。比如在鼙蠻和贎鬍里,有彃燊在罅鷄那里蘩墝,之前他们鏈鴊恆闳嘑傡彚槩滼鞷蕻賤鬡艐倏雫寬褲灣。?1.鞷在文中是什么意思??2.这篇文章的最佳标题是什么??3.从这篇文章我们可以推断出什么??4.作者为什么说“恆闳嘑傡彚槩”??阅读短文,回答下面问题:?What is the most important in learning English??Do you have the confidence in Jumping over theDragonGate??These people eeded because they understood that you can’t let your failuresdefineyou -- you have to let your failures teach you. You have to let them show you what to do differently the next time. So if you get into trouble, that doesn’t mean you’re a troublemaker, it means you need to try harder to act right. If you get a bad grade, that doesn’t mean you’re stupid, it just means you need to spend more time studying.得阅读者得天下 1

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# Counting Problem Using Product rule and Subtraction rule (plz check my answer) #### HeartyBowl Problem: There are 10 people in a line, where each person is either male or female. How many different lineups are there, where there are either 5 consecutive men, or 6 consecutive women? For the case of 5 consecutive men: say you have the lineup M M M M M _ _ _ _ _ then for the other 5 spaces to the right you can either choose a man or a woman. Using the product rule, there are 2^5 different ways to create a lineup with 5 consecutive men in the first 5 spots. Now if you shift the 5 consecutive men 6 times to the right, you get a total of 6 * 2^5 different ways to create a lineup with 5 consecutive men. For the case of 5 consecutive women: This would be the same situation as above and we would be able to create a total of 6 * 2^5 different ways to create a lineup with 5 consecutive women. But there are 2 cases where there are both 5 consecutive men and 5 consecutive women such as: M M M M M W W W W W and W W W W W M M M M M So the answer for the amount of different lineups is 12*2^5 - 2 #### chiro MHF Helper Hey HeartyBowl. If you have five consecutive men (or women) then the next person must be the opposite gender. In light of this you have to slightly adjust your analyses. You will then have to shift this and take into account that the tail (as opposed to the head of the line) just before (and after) must also be the opposite gender. See if you can take the above into account to get a new answer. Also if you meant to say "at least" so many people then disregard my response. #### HeartyBowl I get what you're saying and now that i realize it, the problem isn't being clear enough. I don't know if it means "at least" or if it means "at most"....

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Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs! #### Online Quiz (WorksheetABCD) Questions Per Quiz = 2 4 6 8 10 ### MEAP Preparation - Grade 5 Mathematics2.21 Changing Units in the Metric System Example: How many kilograms are there in 5.5 grams? 1 kg = 1000 g 5.5 gms = 5.5/1000 Answer: 0.0055 kg Directions: Answer the following questions. Also write at least ten examples of your own. Q 1: How many milligrams are in a gram?Answer: Question 2: This question is available to subscribers only!

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#### Overview of this book IPython Notebook Essentials Credits www.PacktPub.com Preface Free Chapter A Tour of the IPython Notebook The Notebook Interface Graphics with matplotlib Handling Data with pandas Advanced Computing with SciPy, Numba, and NumbaPro IPython Notebook Reference Card A Brief Review of Python NumPy Arrays Index ## Basic types, expressions, and variables and their assignment Any data that can be referred to in a Python code is considered an object. Objects are used to represent everything from atomic data, such as numbers, to very complex data structures, such as multidimensional arrays, database connections, and documents in several formats. At the root of the object hierarchy are the numeric data types. These include the following: • Integers: There are three types of integers in Python. • Plain integers: They are represented in the native architecture, which, in most systems, will be either 32- or 64-bit signed values. • Long integers: They are integers with unlimited range, subject to available memory. Most of the time, the programmer does not need to be concerned with the distinction between plain and long integers. Python deals with conversions between the types in a transparent way. • Booleans: They represent the values `False` and `True`. In most situations, they are equivalent to `0` and `1`, respectively. • Floats: They represent the native double-precision floating-point numbers. • Complex: They represent complex numbers, represented as a pair of double-precision floating-point numbers. The following table has examples of literals (that is, constants) for each data type: Data type Literals Integers `0`, `2`, `4`, …, `43882838388` `5L`, `5l` (long integer) `0xFE4` (hexadecimal) `03241` (octal) Real numbers (float) `5.34`, `1.2`, `3.`, `0` `1.4e-32` (scientific notation) Complex `1.0+3.4j`, `1+2j`, `1j`, `0j`, `complex(4.3, 2.5)` The imaginary unit is represented by `j`, but only if it follows a number literal (otherwise, it represents the variable named `j`). So, to represent the imaginary unit we must use `1j` and the complex zero is `0j`. The real and imaginary part of a complex number are always stored as double-precision floating-point values. ### Note Note that the set of numeric types is greatly extended by `NumPy` to allow efficient numeric computations. The assignment statement is used to store values in variables, as follows: ```a = 3 b = 2.28 c = 12 d = 1+2j ``` Python supports multiple simultaneous assignments of values, so the previous four lines of code could be equivalently written in a single line as follows: ```a, b, c, d = 3, 2.28, 12, 1+2j ``` In a multiple assignment, all expressions in the right-hand side are evaluated before the assignments are made. For example, a common idiom to exchange the values of two variables is as follows: ```v, w = w, v ``` As an exercise, the reader can try to predict the result of the following statement, given the preceding variable assignments: ```a, b, c = a + b, c + d, a * b * c * d print a, b, c, d ``` The following example shows how to compute the two solutions of a quadratic equation: ```a, b, c = 2., -1., -4. x1, x2 = .5 * (-b - (b ** 2 - 4 * a * c) ** 0.5), .5 * (-b + (b ** 2 - 4 * a * c) ** 0.5) print x1, x2 ``` Note that we force the variables `a`, `b`, and `c` to be floating-point values by using a decimal point. This is good practice when performing numerical computations. The following table contains a partial list of Python operators: Operators Python operators Arithmetic `+` (Addition) `-` (Subtraction, unary minus) `*` (Multiplication) `/` (Division, see the note below the table) `//` (Integer division) `%` (Remainder) Comparison `==` (Equal to) `>` (Greater than) `<` (Less than) `>=` (Greater than or equal to) `<=` (Less than or equal to) `!=` (Not equal to) Boolean `and` `or` `not` Bitwise Boolean `&` (AND) `|` (OR) `^` (XOR) `~` (NOT) Bitwise shift `<<` (Left shift) `>>` (Right shift) ### Note Care should be taken with the division operator (`/`). If the operands are integers, the result of this operation is the integer quotient. For example, `34/12` results `2`. To get the floating point result, we must either enter floating point operands, as in `34./12.`, or add the following statement: `from __future__ import division` The `//` operator always represents integer division. Arithmetic operators follow the rules for the order of operations that may be altered with the use of parenthesis. Comparison operators have lower precedence than arithmetic operators, and the `or`, `and`, and `not` operators have even lower precedence. So, an expression like the following one produces the expected result: ```2 + 3 < 5 ** 2 and 4 * 3 != 13 ``` In other words, the preceding command line is parsed as follows: ```(((2 + 3) < (5 ** 2)) and ((4 * 3) != 13)) ``` The logical operators `and` and `or` short circuit, so, for example, the second comparison is never evaluated in the command: ```2 < 3 or 4 > 5 ``` The precedence rules for the bitwise and shift operators may not be as intuitive, so it is recommended to always use parenthesis to specify the order of operations, which also adds clarity to the code. Python also supports augmented assignments. For example, the following command lines first assign the value `5` to `a`, and then increment the value of `a` by one: ```a = 5 a += 1 ``` ### Note Python does not have increment/decrement operators, such as a++ and ++a, as in the C language. All Python operators have a corresponding augmented assignment statement. The general semantic for any operator `\$` is the following statement: ```v \$= <expression> ``` The preceding statement is equivalent to the following: ```v = v \$ (<expression>) ``` ### Note Note that `\$` is not a valid Python operator, it is just being used as a placeholder for a generic operator.

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# Questions on Algebra: Polynomials, rational expressions and equations answered by real tutors! Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> Questions on Algebra: Polynomials, rational expressions and equations answered by real tutors!      Log On Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Algebra: Polynomials, rational expressions and equations Solvers Lessons Answers archive Quiz In Depth Question 121027: Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smit, and his trip took one-half hour longer than Smith's. How fast was each one treveling? Click here to see answer by ankor@dixie-net.com(15645) Question 121024: The acient Greeks thought that the most pleasing shape for a rectangle was one for which the ratio of the length to the width was 8 to 5, the golden ratio. If the length of the painting is 2 ft longer than its width, then for what dimensions would the length and width have the golden ratio? Click here to see answer by checkley71(8403) Question 121161: To estimate the size of the bear population on the Keweenaw Peninsula, conservasionists captured, tagged, and released 50 bears. One year later, a random sample of 100 bears included only 2 tagged bears. What is the conservasionist's estimate of the size of the bear population? Click here to see answer by checkley71(8403) Question 121192: A guy wire of length 50 feet is attached to the ground and to the top of an antenna. The height of the antenna is 10 feet larger than the distance from the base of the antenna to the point where the guy wire is attached to the ground. What is the height of the antenna? Click here to see answer by checkley71(8403) Question 121203: factor the polynomial by factoring out the GCF. 2a^2-4ab^2-ab Click here to see answer by jim_thompson5910(28536) Question 121160: Use the fomula 1/f=1/a+1/i to find the image distance i for an object that is 2,000,000 mm from a 250mm telephoto lens. Click here to see answer by MathLover1(6625) Question 121423: Classify the given numbers as real and rational, real and irrational, or complex. a.) (2)1/2+2 b.) i c.) 0 d.) 3-(-5)1/2i e.) (100)1/5 f.) 30007 g.) (8)1/3 h.) 4+0i Click here to see answer by stanbon(57282) Question 121548: Can you please help me with this application problem? The equation V= -3000t + 22,000 describes the value in dollars of a certain model of car after it is t years old. If a car is worth \$13,000 substitute 13,000 into the equation to find the age of the car. Show equation and solve. Click here to see answer by algebrapro18(206) Question 121553: can you please help me solve this one. One number exceeds another by 10. The sum of the numbers is 8. What are the numbers? What I had for the equation was x+x+10=8 Click here to see answer by algebrapro18(206) Question 121546: I can't figure this one out, and you please help me? A train ticket in a certain city is \$1.50. People who use the train also have the option of purchasing a frequent rider pass for \$18.75 each month. With the pass, each ticket costs only \$.75. Determine the number of times in a month the train must be used so that the total monthly cost without the pass is the same as the total monthly cost with the pass. Show the equation and solve. Click here to see answer by checkley71(8403) Question 121543: I have tried to figure this one out and can't seem to do it. Can you please help me? The height of the bookcase is 2ft longer than the length of a shelf. If 22ft. of lumber is available for the entire unit(including the shelves, but not the back of the bookcase), find the length and height of the unit. Show your equation and solve. Click here to see answer by solver91311(16872) Question 121615: how do i divide these two problems: y^2-12y+20 /y-2 and 2x^2+7x+3 / 2x+1 i need the steps for these two problems by today january 21,2008 at 8:00pm please its urgent Click here to see answer by stanbon(57282) Question 121678: i pretty much understand how to muliply rational expressions but im having trouble factoring this problem out. its been a while since ive done this 4x^2/9x^2 * 12/21x^3 Click here to see answer by checkley71(8403) Question 121678: i pretty much understand how to muliply rational expressions but im having trouble factoring this problem out. its been a while since ive done this 4x^2/9x^2 * 12/21x^3 Click here to see answer by edjones(7569) Question 121727: Can you please help me with this problem? y1= 1/x+2, y2= 1/x-2, y3= 1/x^2-4, and the difference between 4 times y1 and 7 times y2 is the product of 11 and y3. Show equation and solve Click here to see answer by stanbon(57282) Question 121724: Can you please help me figure this one out? I have tried doing it but doesn't come out right. y1=x, y2=6+x, y3= 3(x-4)/10x, and the sum of 8 times y1 and 4 times y2 equals y3. Show equation and solve. Click here to see answer by stanbon(57282) Question 121733: f(x) = 3x-5 / x+2 f ^ -1 (x) = a) 2x+5 / x-3 b) -2x-5 / x-3 c) -2x-5 / x+5 d) 2x+3 / x-5 e) None of these Click here to see answer by jim_thompson5910(28536) Question 121737: 2) e^x + 2x = 7 a) 1.424 b) 7 c) -1.621 d) 3.814 e) None of these Click here to see answer by jim_thompson5910(28536) Question 121738: 3) In decomposing 8x-1 / (x+1)(x-2) one of the fractions obtained is: a) 1 / x+1 b) 1 / x-2 c) 5 / x+1 d) 3 / x+1 e) 3 / x-2 Click here to see answer by jim_thompson5910(28536) Question 121739: 4) How long will it take a sum of money to double at 5% per annum compounded continuously? a) 5 years b) 13.9 years c) 16.4 years d) 19.1 years e) 12.5 years Click here to see answer by jim_thompson5910(28536)

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# MATLAB代做|FPGA代做--基于脉冲耦合神经网络的图像分割 ### 时间：2018-10-24 15:10:50 点击： 核心提示：基于脉冲耦合神经网络的图像分割... function [Edge,Numberofaera]=pcnn(X) % 功能：采用PCNN算法进行边缘检测 % 输入：X—输入的灰度图像 % 输出：Edge—检测到的       Numberofaera—表明了在各次迭代时激活的块区域 figure(1); imshow(X); X=double(X); % 设定权值 Weight=[0.07 0.1 0.07;0.1 0 0.1;0.07 0.1 0.07]; WeightLI2=[-0.03 -0.03 -0.03;-0.03 0 -0.03;-0.03 -0.03 -0.03]; d=1/(1+sum(sum(WeightLI2))); %%%%%%测试权值%%%%%% WeightLI=[-0.03 -0.03 -0.03;-0.03 0.5 -0.03;-0.03 -0.03 -0.03]; d1=1/(sum(sum(WeightLI))); %%%%%%%%%%%%%%%%%% Beta=0.4; Yuzhi=245; %衰减系数 Decay=0.3; [a,b]=size(X); V_T=0.2; %门限值 Threshold=zeros(a,b); S=zeros(a+2,b+2); Y=zeros(a,b); %点火频率 Firate=zeros(a,b); n=1; %统计循环次数 count=0; Tempu1=zeros(a,b); Tempu2=zeros(a+2,b+2); %%%%%%图像增强部分%%%%%% Out=zeros(a,b); Out=uint8(Out); for i=1:a for j=1:b if(i==1|j==1|i==a|j==b) Out(i,j)=X(i,j); else H=[X(i-1,j-1)  X(i-1,j) X(i-1,j+1); X(i,j-1)   X(i,j)   X(i,j+1); X(i+1,j-1) X(i+1,j) X(i+1,j+1)]; temp=d1*sum(sum(H.*WeightLI)); Out(i,j)=temp; end end end figure(2); imshow(Out); %%%%%%%%%%%%%%%%%%% for count=1:30 for i0=2:a+1 for i1=2:b+1 V=[S(i0-1,i1-1)  S(i0-1,i1) S(i0-1,i1+1); S(i0,i1-1)   S(i0,i1)   S(i0,i1+1); S(i0+1,i1-1) S(i0+1,i1) S(i0+1,i1+1)]; L=sum(sum(V.*Weight)); V2=[Tempu2(i0-1,i1-1)  Tempu2(i0-1,i1) Tempu2(i0-1,i1+1); Tempu2(i0,i1-1)   Tempu2(i0,i1)   Tempu2(i0,i1+1); Tempu2(i0+1,i1-1) Tempu2(i0+1,i1) Tempu2(i0+1,i1+1)];        F=X(i0-1,i1-1)+sum(sum(V2.*WeightLI2)); %保证侧抑制图像无能量损失 F=d*F; U=double(F)*(1+Beta*double(L)); Tempu1(i0-1,i1-1)=U; if U>=Threshold(i0-1,i1-1)|Threshold(i0-1,i1-1)<60 T(i0-1,i1-1)=1; Threshold(i0-1,i1-1)=Yuzhi; %点火后一直置为1 Y(i0-1,i1-1)=1; else T(i0-1,i1-1)=0; Y(i0-1,i1-1)=0; end end end Threshold=exp(-Decay)*Threshold+V_T*Y; %被激活过的像素不再参与迭代过程 if n==1 S=zeros(a+2,b+2); else S=Bianhuan(T); end n=n+1; count=count+1; Firate=Firate+Y; figure(3); imshow(Y); Tempu2=Bianhuan(Tempu1); end Firate(find(Firate<10))=0; Firate(find(Firate>=10))=10; figure(4); imshow(Firate); %%%%%%子函数 %%%%%%% function Y=Jiabian(X) [m,n]=size(X); Y=zeros(m+2,n+2); for i=1:m+2 for j=1:n+2 if i==1&j~=1&j~=n+2 Y(i,j)=X(1,j-1); elseif j==1&i~=1&i~=m+2 Y(i,j)=X(i-1,1); elseif i~=1&j==n+2&i~=m+2 Y(i,j)=X(i-1,n); elseif i==m+2&j~=1&j~=n+2 Y(i,j)=X(m,j-1); elseif i==1&j==1 Y(i,j)=X(i,j); elseif i==1&j==n+2 Y(i,j)=X(1,n); elseif i==(m+2)&j==1 Y(i,j)=X(m,1); elseif i==m+2&j==n+2 Y(i,j)=X(m,n); else Y(i,j)=X(i-1,j-1); end end end %%%%%%子函数%%%%%% function Y=Bianhuan(X) [m,n]=size(X); Y=zeros(m+2,n+2); for i=1:m+2 for j=1:n+2 if i==1|j==1|i==m+2|j==n+2 Y(i,j)=0; else Y(i,j)=X(i-1,j-1); end end end %%%%%%子函数%%%%%% function Y=judge_edge(X,n) %X:每次迭代后PCNN输出的二值图像，如何准确判断边界点是关键 [a,b]=size(X); T=Jiabian(X); Y=zeros(a,b); W=zeros(a,b); for i=2:a+1 for j=2:b+1 if (T(i,j)==1)&((T(i-1,j)==0&T(i+1,j)==0)|(T(i,j-1)==0&T(i,j+1)==0)|(T(i-1,j-1)==0&T(i+1,j+1)==0)|(T(i+1,j-1)==0&T(i-1,j+1)==0)) Y(i-1,j-1)=-n; end end end QQ ：1224848052 • 百度搜索 • 查阅资料过程中 • 论坛发现 • 百度贴吧发现 • 朋友介绍 • 没有相关文章 • 大名： • 内容：

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# AREA OF CIRCLES. REVIEW-CIRCUMFERENCE WORDS: The circumference of a circle is equal to pi times its diameter or pi times twice its radius. Symbols: or. ## Presentation on theme: "AREA OF CIRCLES. REVIEW-CIRCUMFERENCE WORDS: The circumference of a circle is equal to pi times its diameter or pi times twice its radius. Symbols: or."— Presentation transcript: AREA OF CIRCLES REVIEW-CIRCUMFERENCE WORDS: The circumference of a circle is equal to pi times its diameter or pi times twice its radius. Symbols: or Find the circumference. Use. 8cm Find the circumference. Use. mm 6 x 7 AREA WORDS: The area of a circle equals the product of pi and the square of the radius. Symbols: Find the area. Use. 6 cm Find the area. Use. 8cm d = 8 r = 4 A circle has a radius of 11 feet. Find the area of the circle. Use. The world’s largest cylindrical sundial is at Walt Disney World. The face of the sundial has a diameter of 122 feet. What is the area of the face? d = 122 ft r = 122÷2 r = 61 ft Use. Download ppt "AREA OF CIRCLES. REVIEW-CIRCUMFERENCE WORDS: The circumference of a circle is equal to pi times its diameter or pi times twice its radius. Symbols: or." Similar presentations

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Search by Topic Resources tagged with Volume and capacity similar to Food Chains: Filter by: Content type: Stage: Challenge level: There are 32 results Broad Topics > Measures and Mensuration > Volume and capacity Fill Me Up Stage: 3 Challenge Level: Can you sketch graphs to show how the height of water changes in different containers as they are filled? Chocolate Cake Stage: 3 Challenge Level: If I don't have the size of cake tin specified in my recipe, will the size I do have be OK? Concrete Calculation Stage: 4 Challenge Level: The builders have dug a hole in the ground to be filled with concrete for the foundations of our garage. How many cubic metres of ready-mix concrete should the builders order to fill this hole to. . . . Biology Measurement Challenge Stage: 4 Challenge Level: Analyse these beautiful biological images and attempt to rank them in size order. Immersion Stage: 4 Challenge Level: Various solids are lowered into a beaker of water. How does the water level rise in each case? Scientific Measurement Stage: 4 Challenge Level: Practice your skills of measurement and estimation using this interactive measurement tool based around fascinating images from biology. Maths Filler 2 Stage: 4 Challenge Level: Can you draw the height-time chart as this complicated vessel fills with water? Maths Filler Stage: 3 Challenge Level: Imagine different shaped vessels being filled. Can you work out what the graphs of the water level should look like? Stage: 3 Challenge Level: Can you rank these sets of quantities in order, from smallest to largest? Can you provide convincing evidence for your rankings? Conical Bottle Stage: 4 Challenge Level: A right circular cone is filled with liquid to a depth of half its vertical height. The cone is inverted. How high up the vertical height of the cone will the liquid rise? Uniform Units Stage: 4 Challenge Level: Can you choose your units so that a cube has the same numerical value for it volume, surface area and total edge length? Changing Areas, Changing Volumes Stage: 3 Challenge Level: How can you change the surface area of a cuboid but keep its volume the same? How can you change the volume but keep the surface area the same? Sliced Stage: 4 Challenge Level: An irregular tetrahedron has two opposite sides the same length a and the line joining their midpoints is perpendicular to these two edges and is of length b. What is the volume of the tetrahedron? More Pebbles Stage: 2 and 3 Challenge Level: Have a go at this 3D extension to the Pebbles problem. Cola Can Stage: 3 Challenge Level: An aluminium can contains 330 ml of cola. If the can's diameter is 6 cm what is the can's height? Cuboid Challenge Stage: 3 Challenge Level: What size square corners should be cut from a square piece of paper to make a box with the largest possible volume? Growing Rectangles Stage: 3 Challenge Level: What happens to the area and volume of 2D and 3D shapes when you enlarge them? Sending a Parcel Stage: 3 Challenge Level: What is the greatest volume you can get for a rectangular (cuboid) parcel if the maximum combined length and girth are 2 metres? The Genie in the Jar Stage: 3 Challenge Level: This jar used to hold perfumed oil. It contained enough oil to fill granid silver bottles. Each bottle held enough to fill ozvik golden goblets and each goblet held enough to fill vaswik crystal. . . . Three Cubes Stage: 4 Challenge Level: Can you work out the dimensions of the three cubes? Mouhefanggai Stage: 4 Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai. Volume of a Pyramid and a Cone Stage: 3 These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts. Boxed In Stage: 3 Challenge Level: A box has faces with areas 3, 12 and 25 square centimetres. What is the volume of the box? Thousands and Millions Stage: 3 Challenge Level: Here's a chance to work with large numbers... Plutarch's Boxes Stage: 3 Challenge Level: According to Plutarch, the Greeks found all the rectangles with integer sides, whose areas are equal to their perimeters. Can you find them? What rectangular boxes, with integer sides, have. . . . All in a Jumble Stage: 3 Challenge Level: My measurements have got all jumbled up! Swap them around and see if you can find a combination where every measurement is valid. Zin Obelisk Stage: 3 Challenge Level: In the ancient city of Atlantis a solid rectangular object called a Zin was built in honour of the goddess Tina. Your task is to determine on which day of the week the obelisk was completed. Efficient Cutting Stage: 3 Challenge Level: Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end. Tubular Stand Stage: 4 Challenge Level: If the radius of the tubing used to make this stand is r cm, what is the volume of tubing used? Cylinder Cutting Stage: 2 and 3 Challenge Level: An activity for high-attaining learners which involves making a new cylinder from a cardboard tube. In a Spin Stage: 4 Challenge Level: What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse? Funnel Stage: 4 Challenge Level: A plastic funnel is used to pour liquids through narrow apertures. What shape funnel would use the least amount of plastic to manufacture for any specific volume ?

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ღSupperღ Jiangly Fan Club 1458 Foraino0267 nixnix ShawnXIAOGNAH chouhui Reminiscence ycf_gogogo bz2021 Wawa机 RyanMoriarty ღSupperღ 40分钟前 #include <bits/stdc++.h> using namespace std; using i64 = long long; const int N = 1e5+10,M = 110; int a[N]; int f[N][M][2]; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n,m; cin >> n >> m; for(int i = 1; i <= n; i ++ ) cin >> a[i]; memset(f,-0x3f,sizeof f); for(int i = 0; i <= n; i ++ ) f[i][0][0] = 0; for(int i = 1; i <= n; i ++ ) { for(int j = 1; j <= m; j ++ ) { f[i][j][0] = max(f[i - 1][j][0], f[i - 1][j][1] + a[i]); f[i][j][1] = max(f[i - 1][j][1], f[i - 1][j - 1][0] - a[i]); } } int res = 0; for(int i = 0; i <= m; i ++ ) res = max(res,f[n][i][0]); cout << res << "\n"; return 0; } ღSupperღ 1小时前 #include <bits/stdc++.h> using namespace std; using i64 = long long; const int N = 40; int n; int w[N]; vector<int> level[N]; int getmin(int l, int r) { int res = l; for (int i = l; i <= r; i ++ ) if (w[res] > w[i]) res = i; return res; } void dfs(int l, int r, int d) { if (l > r) return; int root = getmin(l, r); level[d].push_back(w[root]); dfs(l, root - 1, d + 1); dfs(root + 1, r, d + 1); } int main() { cin >> n; for (int i = 0; i < n; i ++ ) cin >> w[i]; dfs(0, n - 1, 0); for (int i = 0; level[i].size(); i ++ ) for (auto x: level[i]) cout << x << ' '; return 0; } ღSupperღ 1小时前 #include <bits/stdc++.h> using namespace std; using i64 = long long; const int N = 1010; int f[N][N]; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n,m; while(cin >> m >> n) { for(int i = 0; i <= n; i ++ ) f[i][0] = 1; for(int i = 0; i <= m; i ++ ) f[1][i] = 1; for(int i = 1; i <= n; i ++ ) { for(int j = 1; j <= m; j ++ ) { if(j < i) f[i][j] = f[j][j]; else { f[i][j] = f[i-1][j] + f[i][j - i]; } } } cout << f[n][m] << "\n"; } return 0; } ღSupperღ 2小时前 #include <bits/stdc++.h> using namespace std; using i64 = long long; const int N = 1010; int f[N][N],g[N][N]; int dx[] = {-1,0,1,0},dy[] = {0,1,0,-1}; int n; int dfs(int x,int y) { if(f[x][y] != -1) return f[x][y]; f[x][y] = 1; for(int i = 0; i < 4; i ++ ) { int a = x + dx[i],b = y + dy[i]; if(a >= 1 and a <= n and b >= 1 and b <= n and g[x][y] + 1 == g[a][b]) { f[x][y] = max(f[x][y], dfs(a,b) + 1); } } return f[x][y]; } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int T; cin >> T; for(int i = 1; i <= T; i ++ ) { cin >> n; for(int i = 1; i <= n; i ++ ) for(int j = 1; j <= n; j ++ ) cin >> g[i][j]; memset(f,-1,sizeof f); int res = 0,cnt = 0; for(int i = 1; i <= n; i ++ ) { for(int j = 1; j <= n; j ++ ) { int t = dfs(i,j); if(t > cnt or g[i][j] < res and t == cnt) { res = g[i][j];//当前几号点 cnt = t;//走了几步 } } } cout << "Case #" << i << ": " << res << " " << cnt << "\n"; } return 0; } ღSupperღ 2小时前 #include <bits/stdc++.h> using namespace std; using i64 = long long; const int N = 200100; int a[N]; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int T; cin >> T; for(int i = 1; i <= T; i ++ ) { int n; cin >> n; for(int i = 0; i < n; i ++ ) cin >> a[i]; int res = 0; for(int i = 0; i < n; i ++ ) { int j = i + 2; while(j < n and a[j] - a[j - 1] == a[j - 1] - a[j - 2]) j ++; res = max(res,j - i); i = j - 2; } cout << "Case #" << i << ": "; cout << res << "\n"; } return 0; } ღSupperღ 3小时前 #include <bits/stdc++.h> using namespace std; using i64 = long long; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int T; cin >> T; for(int i = 1; i <= T; i ++ ) { int n; cin >> n; unordered_map<string,int> m1; unordered_map<string,string> m2; while(n -- ) { string a,b; cin >> a >> b; m1[a] ++,m1[b] ++,m2[a] = b; } string r; for(auto t : m1) { if(t.second == 1 and m2.count(t.first)) r = t.first; } cout << "Case #" << i << ": "; for(string i = r; m2.count(i); i = m2[i]) cout << i << "-" << m2[i] << " "; cout << "\n"; } return 0; } ღSupperღ 3小时前 #include <bits/stdc++.h> using namespace std; using i64 = long long; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n; cin >> n; int a = 0,b = 0; for(int i = 0; i < n; i ++ ) { int x; cin >> x; a += x; } for(int i = 0; i < n; i ++ ) { int x; cin >> x; b += x; } if(a >= b) cout << "Yes\n"; else cout << "No\n"; return 0; } ღSupperღ 6小时前 ### y总 #include <cstring> #include <iostream> using namespace std; const int N = 1010, mod = 1e9 + 7; int n, m; int f[N], g[N]; int main() { cin >> n >> m; memset(f, -0x3f, sizeof f); f[0] = 0; g[0] = 1; for (int i = 0; i < n; i ++ ) { int v, w; cin >> v >> w; for (int j = m; j >= v; j -- ) { int maxv = max(f[j], f[j - v] + w); int s = 0; if (f[j] == maxv) s = g[j]; if (f[j - v] + w == maxv) s = (s + g[j - v]) % mod; f[j] = maxv, g[j] = s; } } int res = 0; for (int i = 1; i <= m; i ++ ) if (f[i] > f[res]) res = i; int sum = 0; for (int i = 0; i <= m; i ++ ) if (f[i] == f[res]) sum = (sum + g[i]) % mod; cout << sum << endl; return 0; } ### 稽宝 #include <bits/stdc++.h> using namespace std; using i64 = long long; const int N = 1010, mod = 1e9 + 7; int f[N],cnt[N]; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n,m; cin >> n >> m; for(int i = 0; i <= m; i ++ ) cnt[i] = 1; for(int i = 0; i < n; i ++ ) { int v,w; cin >> v >> w; for(int j = m; j >= v; j -- ) { int maxx = f[j - v] + w; if(maxx > f[j]) //价值最大,选其求方案数 { f[j] = maxx; cnt[j] = cnt[j - v] % mod; } else if (maxx == f[j]) //相等, 更新两个的总和为方案数 { cnt[j] = (cnt[j] + cnt[j - v]) % mod; } } } cout << cnt[m] << "\n"; return 0; } ღSupperღ 6小时前 #include <bits/stdc++.h> using namespace std; using i64 = long long; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); string s, s1, s2; char c; while (cin >> c, c != ',') s += c; while (cin >> c, c != ',') s1 += c; while (cin >> c) s2 += c; if (s.size() < s1.size() or s.size() < s2.size()) puts("-1"); else { int l = 0; while (l + s1.size() <= s.size()) { int k = 0; while (k < s1.size()) { if (s[l + k] != s1[k]) break; k ++ ; } if (k == s1.size()) break; l ++ ; } int r = s.size() - s2.size(); while (r >= 0) { int k = 0; while (k < s2.size()) { if (s[r + k] != s2[k]) break; k ++ ; } if (k == s2.size()) break; r -- ; } l += s1.size() - 1; if (l >= r) puts("-1"); else printf("%d\n", r - l - 1); } return 0; } ღSupperღ 6小时前 #include <bits/stdc++.h> using namespace std; using i64 = long long; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); string str; while (cin >> str, str != ".") { int len = str.size(); for (int n = len; n; n -- ) if (len % n == 0) { int m = len / n; string s = str.substr(0, m); string r; for (int i = 0; i < n; i ++ ) r += s; if (r == str) { cout << n << endl; break; } } } return 0; }

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This site is supported by donations to The OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A063881 Number of oriented trees rooted at an arc. 1 1, 4, 18, 80, 367, 1708, 8122, 39204, 191963, 950984, 4759626, 24030736, 122258314, 626162464, 3225926450, 16706775984, 86928097451, 454203897192, 2382255252398, 12537764465072, 66193294753768, 350472816969976, 1860542261745782, 9901018433270812 (list; graph; refs; listen; history; text; internal format) OFFSET 2,2 REFERENCES F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 61, (3.3.7). LINKS Vincenzo Librandi, Table of n, a(n) for n = 2..100 FORMULA a(n) = A000151(n)- A000238(n). G.f.: A(x) = B(x)^2, where B(x) is g.f. for A000151. MAPLE B:= proc(n) option remember; if n<=1 then unapply(x, x) else unapply(convert(series(x*exp(2*sum(B(n-1)(x^k)/k, k=1..n-1)), x, n+1), polynom), x) fi end: a:= proc(n) local T; T:=B(n-1)(x); add(coeff(T, x, k)* coeff(T, x, n-k), k=1..n-1) end: seq(a(n), n=2..23); # Alois P. Heinz, Aug 23 2008 MATHEMATICA B[n_ /; n <= 1] = Identity; B[n_] := B[n] = Function[x, Evaluate[Normal[Series[x*Exp[2*Sum[B[n-1][x^k]/k, {k, 1, n-1}]], {x, 0, n+1}]]]]; a[n_] := Module[{T}, T = B[n-1][x]; Sum[Coefficient[T, x, k]*Coefficient[T, x, n-k], {k, 1, n-1}]]; Table[a[n], {n, 2, 23}] (* Jean-François Alcover, Feb 17 2014, after Alois P. Heinz *) CROSSREFS Cf. A000151, A000238. Sequence in context: A257390 A104631 A106391 * A264004 A282708 A252823 Adjacent sequences:  A063878 A063879 A063880 * A063882 A063883 A063884 KEYWORD nonn AUTHOR Vladeta Jovovic, Aug 27 2001 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified October 15 22:25 EDT 2019. Contains 328038 sequences. (Running on oeis4.)

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# Metamath Proof Explorer ## Theorem pm10.42 Description: Theorem *10.42 in WhiteheadRussell p. 155. (Contributed by Andrew Salmon, 17-Jun-2011) Ref Expression Assertion pm10.42 ${⊢}\left(\exists {x}\phantom{\rule{.4em}{0ex}}{\phi }\vee \exists {x}\phantom{\rule{.4em}{0ex}}{\psi }\right)↔\exists {x}\phantom{\rule{.4em}{0ex}}\left({\phi }\vee {\psi }\right)$ ### Proof Step Hyp Ref Expression 1 19.43 ${⊢}\exists {x}\phantom{\rule{.4em}{0ex}}\left({\phi }\vee {\psi }\right)↔\left(\exists {x}\phantom{\rule{.4em}{0ex}}{\phi }\vee \exists {x}\phantom{\rule{.4em}{0ex}}{\psi }\right)$ 2 1 bicomi ${⊢}\left(\exists {x}\phantom{\rule{.4em}{0ex}}{\phi }\vee \exists {x}\phantom{\rule{.4em}{0ex}}{\psi }\right)↔\exists {x}\phantom{\rule{.4em}{0ex}}\left({\phi }\vee {\psi }\right)$

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# Re: st: Calculating Percent Change In Regression Coeffecients From "Tim Wade" To statalist@hsphsun2.harvard.edu Subject Re: st: Calculating Percent Change In Regression Coeffecients Date Tue, 30 May 2006 16:06:13 -0400 Nick, One use for this type of metric could be used in a "backward deletion" or "change in estimate" procedure for selecting important covariates in a regression model. In this process a full model with an exposure of interest and all other potentially important covariates is fit, then the relative importance of each covariate is judged by how much of an effect its removal has on the coefficient of the exposure of interest. Variables that, when removed from the model, have little effect on the exposure coefficient are rdeleted from the final model. The result is a model where the effect of the exposure has been adjusted for confounding but with unimportant variables removed. It is sometimes advocated as an alternative to stepwise methods that rely on statistical significance. Tim On 5/30/06, Nick Cox <n.j.cox@durham.ac.uk> wrote: On a different note, why this interest in percent change in coefficient as a metric? 1. The behaviour of ratios can be complicated already. This measure is a ratio calculated from ratios. 2. Specifically, is the behaviour as the denominator goes from small positive through zero to small negative regarded as a feature? 3. There is a lack of symmetry in the calculation. I can imagine a practical argument that (1) and (2) do not matter for the application, and (3) might be irrelevant given a time order, but I wouldn't put much weight on this measure. Nick n.j.cox@durham.ac.uk > Hi Raphael, I don't know how to do this in Mata, but here is a brute > force solution using macros and for loops: > > > . regress price headroom rep78 gear_ratio > > Source | SS df MS Number > of obs = 69 > -------------+------------------------------ F( 3, > 65) = 4.68 > Model | 102521828 3 34173942.7 Prob > > F = 0.0051 > Residual | 474275131 65 7296540.47 > R-squared = 0.1777 > R-squared = 0.1398 > Total | 576796959 68 8482308.22 Root > MSE = 2701.2 > > -------------------------------------------------------------- > ---------------- > price | Coef. Std. Err. t P>|t| > [95% Conf. Interval] > -------------+------------------------------------------------ > ---------------- > headroom | -136.9778 414.9107 -0.33 0.742 > -965.6117 691.6561 > rep78 | 576.2363 362.8717 1.59 0.117 > -148.4686 1300.941 > gear_ratio | -2995.126 829.7523 -3.61 0.001 > -4652.256 -1337.996 > _cons | 13577.64 3025.567 4.49 0.000 > 7535.166 19620.12 > -------------------------------------------------------------- > ---------------- > > /*only include coefficients you want to compare*/ > > . foreach var of varlist headroom rep78 { > 2. local `var'1=_b[`var'] > 3. } > > . regress price headroom rep78 > > Source | SS df MS Number > of obs = 69 > -------------+------------------------------ F( 2, > 66) = 0.43 > Model | 7450346.06 2 3725173.03 Prob > > F = 0.6511 > Residual | 569346613 66 8626463.83 > R-squared = 0.0129 > R-squared = -0.0170 > Total | 576796959 68 8482308.22 Root > MSE = 2937.1 > > -------------------------------------------------------------- > ---------------- > price | Coef. Std. Err. t P>|t| > [95% Conf. Interval] > -------------+------------------------------------------------ > ---------------- > headroom | 391.6261 422.1074 0.93 0.357 > -451.1385 1234.391 > rep78 | 69.23416 363.8024 0.19 0.850 > -657.1208 795.5892 > _cons | 4735.368 1930.863 2.45 0.017 > 880.276 8590.459 > -------------------------------------------------------------- > ---------------- > > . foreach var of varlist headroom rep78 { > 2. local `var'2=_b[`var'] > 3. } > > . foreach var of varlist headroom rep78 { > 2. di as result "percent change for > `var'="((``var'2'-``var'1')/``var'1')*100 > 3. } > percent change for rep78=-87.985109 Raphael Fraser > > I would like to calculate the percentage change in the regression > > coeffecients of model 1 and model 2. Can any one help? I tried using > > Mata but I did not know how to divide each element in a matrix with > > different scalars. > > > > sysuse auto, clear > > stset mpg, failure(foreign) > > stcox mpg price weight rep78, nohr nolog /*Model 1*/ > > stcox mpg weight rep78, nohr nolog /*Model 2*/ > > > > For example % change = (rep78_m2 - rep78_m1) / rep78_m1 * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

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# How to Calculate a Part-time Employee Salary Employee status may change during the course of a career, and sometimes, it changes from full-time to part-time employment. When that happens, the employee's salary also changes from an annual salary to a part-time wage. This is typically the hourly rate multiplied by the number of hours an employee works each pay period. ## Does the Job Classification Change? Along with the status change from full to part-time employee, there may also be changes pursuant to the Fair Labor Standards Act, if the employee's wages drop below the threshold regulation from exempt to non-exempt. The regulations concerning exempt vs. non-exempt pay come with other rules that should be factored into the employee's status. Those rules pertain to job classification, which the change to pay might affect. Prudent employers will determine whether changing status and pay from full-time to part-time will affect employee classification. ## Determine the Employee's Current Annual Pay Verify the employee's current pay rate, which should be on an annual basis. Employees who are paid on a salary basis can have their pay converted to an hourly rate; however, their classification may also change. This means that you should check the U.S. Department of Labor, Fair Labor Standards Act to ensure that the status change is appropriate, and to confirm that the employee's classification will remain exempt vs. non-exempt. When an employee is paid on a salary basis, typically, it means that she has a level of autonomy that exempts her from overtime pay. Employees who are part-time workers may be non-exempt and, therefore, they are entitled to overtime pay when they work more than 40 hours per week. ## Calculate Employee's Equivalent Hourly Wage If you are converting the employee's annual pay to an hourly wage, determine the number of hours she works on a salary basis - generally, it's 2,080 hours in most industries. Assume her annual salary is \$80,000 and if she works 2,080 per year, her equivalent hourly rate is \$38.46. Based on this calculation, if the employee converts from 40 hours per week to 20 hours per week and she's paid semi-monthly, her pay is reduced from \$3,333.33 (before taxes and other deductions) per pay period to approximately \$1,538.40 for each semi-monthly paycheck. ## Explain Difference Between Annual Pay vs. Pay Per Hours Worked What's important to note is that this is not precisely one-half of her pay, because the semi-monthly pay is based on 24 pay periods, which is \$3,333.33 per pay period. If she works approximately 40 hours each pay period, she is paid for 20 hours per week, which is slightly less than the full-time hours. Her pay for 40 hours for roughly two weeks is \$1,538.40, or \$38.46 multiplied by 40 hours. The approximate number of hours that a full-time employee works on a semi-monthly pay schedule is 86.67 hours, calculated twice monthly. This explains that the difference is that the employee is expecting exactly half the pay she was making as a full-time employee.

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{[ promptMessage ]} Bookmark it {[ promptMessage ]} # GE330_lect10 - Lecture 10 Post-Optimal Analysis Lecture 10... This preview shows pages 1–7. Sign up to view the full content. Lecture 10: Post-Optimal Analysis February 24, 2009 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Lecture 10 Today Primal-Dual Relationship Economic Interpretation of Duality Post-Optimal Analysis Chapters 4.2, 4.3, and 4.5. GE330 1 Lecture 10 More on Primal-Dual Relations Primal maximize z = 5 x 1 + 12 x 2 + 4 x 3 subject to x 1 + 2 x 2 + x 3 + x 4 = 10 2 x 1 - x 2 + 3 x 3 = 8 x 1 , x 2 , x 3 0 Its dual minimize w = 10 y 1 + 8 y 2 subject to y 1 + 2 y 2 5 2 y 1 - y 2 12 y 1 + 3 y 2 4 y 1 0 , y 2 is unrestricted GE330 2 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Lecture 10 Optimal Tableau (for primal problem, using M-method) Basic x 1 x 2 x 3 x 4 R Solution z 0 0 3 / 5 29 / 5 - 2 / 5 + M 54 4 5 x 2 0 1 - 1 / 5 2 / 5 - 1 / 5 12 / 5 x 1 1 0 7 / 5 1 / 5 2 / 5 26 / 5 Optimal inverse = 2 / 5 - 1 / 5 1 / 5 2 / 5 Optimal dual = original objective coeff. × optimal inverse (order matters), y 1 y 2 = 12 5 × 2 / 5 - 1 / 5 1 / 5 2 / 5 GE330 3 Lecture 10 Relations in Simplex Table Given a simplex table at any iteration i , we have Column Rule x j -column data = inverse at iteration i × original x j -column Here, “column data” does not include the z -row values. The rule applies to “rhs”-column as well. Shadow Prices at iteration i (dual variables) shadow price = original cost of basis at iteration i × inverse at iteration i Here, order in the basis matter. Shadow price is optimal ONLY if the given table is optimal. Reduced Cost ( z -row data) reduced cost of x j = l.h.s. of j -th dual constraint - r.h.s. of j -th dual constraint The left hand-side of dual constraint is evaluated at the shadow price GE330 4 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Lecture 10 Dual Optimal Solutions Important to sensitivity analysis Change in the right-hand side Change in the objective coefficients Help us analyze additional changes Addition of new operation (constraint in primal) Addition of new activity (variable in primal) We discuss this changes in the light of dual optimal solutions Changes affecting feasibility (right-hand side change or a new constraint) How to recover optimal if the perturbation causes the change in basic optimal solution? This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]}

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Generates a radial layout for trees whose nodes are associated with a distance to the root. Project description Generates a radial layout for trees whose nodes are associated with a distance to the root, similar to how it has been done in [1]. You can choose the basic method or a more sophisticated version which makes a more efficient use of space. [1] *The Hidden Geometry of Complex, Network-Driven Contagion Phenomena*, D Brockmann, D Helbing, Science Vol. 342, Issue 6164, pp. 1337-1342 (2013) ## Install $sudo python setup.py install ## Example$ python example.py or look here: #!python import matplotlib.pyplot as pl import networkx as nx paths = [ [ 'a','b','c'] ] paths += [ [ 'a','b','d'] ] paths += [ [ 'a','e','f','g'] ] paths += [ [ 'a','e','f','h'] ] paths += [ [ 'a','e','i'] ] paths += [ [ 'a','j','k'] ] paths += [ [ 'a','j','l'] ] dists = {'a': 0, 'b':1.1, 'e': 1.2, 'j': 1.4, 'c':2.1, 'd': 2.2, 'f': 2.1, 'i': 2.34, 'k':3.8, 'l':2.5, 'g': 3.9, 'h': 3.8} #The Tree has to be a DiGraph! The root is always the one with distance 0. T = nx.DiGraph() for p in paths: keystr = 'dist' nx.set_node_attributes(T,keystr,dists) fig,ax = pl.subplots(1,2,figsize=(15,8)) nx.draw_networkx(T,pos,ax=ax[0]) nx.draw_networkx(T,pos,ax=ax[1]) pl.show() Project details Uploaded Source

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# nLab cartesian closed category ### Context #### Monoidal categories monoidal categories category theory # Contents ## Definition A cartesian closed category (sometimes: ccc) is a category with finite products which is closed with respect to its cartesian monoidal structure. The internal hom $[S,X]$ in a cartesian closed category is often called exponentiation and is denoted $X^S$. A cartesian closed functor between cartesian closed categories $C$, $D$ is a functor $F \colon C \to D$ that is product-preserving and that is also exponential-preserving, meaning that the canonical map $F(a^b) \to F(a)^{F(b)},$ corresponding to the composite $F(a^b) \times F(b) \cong F(a^b \times b) \stackrel{F(eval_{a, b})}{\to} F(a),$ is an isomorphism. ## Some basic consequences A category is cartesian closed if it has finite products and if for any two objects $X$, $Y$, there is an object $Y^X$ (thought of as a “space of maps from $X$ to $Y$”) such that for any object $Z$, there is a bijection between the set of maps $Z \to Y^X$ and the set of maps $Z \times X \to Y$, and this bijection is natural in $Z$. There is an evaluation map $ev_{X, Y}: Y^X \times X \to Y$, which by definition is the map corresponding to the identity map $Y^X \to Y^X$ under the bijection. Using the naturality, it may be shown that the bijection $\hom(Z, Y^X) \to \hom(Z \times X, Y)$ takes a map $\phi: Z \to Y^X$ to the composite $Z \times X \stackrel{\phi \times 1_X}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} Y$ and the universal property of $Y^X$ may be phrased thus: given any $\psi: Z \times X \to Y$, there exists a unique map $\phi: Z \to Y^X$ for which $\psi = ev_{X, Y} \circ (\phi \times 1_X)$. Taking $Z = 1$ (the terminal object), maps $1 \to Y^X$ (or “points” of $Y^X$) are in bijection with maps $X \stackrel{\pi_{2}^{-1}}{\to} 1 \times X \to Y$. So the “underlying set” of $Y^X$, namely $\hom(1, Y^X)$, is identified with the set of maps from $X$ to $Y$. Let us denote the point corresponding to $f: X \to Y$ by $[f]: 1 \to Y^X$. Then, by definition, $(1 \times X \stackrel{[f] \times 1_X}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} Y) = (1 \times X \stackrel{\pi_2}{\to} X \stackrel{f}{\to} Y)$ The following lemma says that the internal evaluation map $ev_{X, Y}$ indeed behaves as an evaluation map at the level of underlying sets. ###### Lemma Given a map $f: X \to Y$ and a point $x: 1 \to X$, the composite $1 \stackrel{\langle [f], x \rangle}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} Y$ is the point $f x: 1 \to Y$. ###### Proof We have $\array{ 1 \stackrel{\langle [f], x \rangle}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} X & = & 1 \stackrel{\delta_1}{\to} 1 \times 1 \stackrel{[f] \times x}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} Y\\ & = & 1 \stackrel{\delta_1}{\to} 1 \times 1 \stackrel{1 \times x}{\to} 1 \times X \stackrel{[f] \times 1_X}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} Y\\ & = & 1 \stackrel{\delta_1}{\to} 1 \times 1 \stackrel{1 \times x}{\to} 1 \times X \stackrel{\pi_2}{\to} X \stackrel{f}{\to} Y\\ & = & 1 \stackrel{\delta_1}{\to} 1 \times 1 \stackrel{\pi_2}{\to} 1 \stackrel{x}{\to} X \stackrel{f}{\to} Y\\ & = & 1 \stackrel{x}{\to} X \stackrel{f}{\to} Y }$ where the penultimate equation uses naturality of the projection map $\pi_2$. ###### Definition Internal composition $c_{X, Y, Z}: Z^Y \times Y^X \to Z^X$ is the unique map such that $(Z^Y \times Y^X \times X \stackrel{c \times 1_X}{\to} Z^X \times X \stackrel{ev_{X, Z}}{\to} Z) = (Z^Y \times Y^X \times X \stackrel{1 \times ev_{X, Y}}{\to} Z^Y \times Y \stackrel{ev_{Y, Z}}{\to} Z)$ One may show that internal composition behaves as the usual composition at the underlying set level, in that given maps $g: Y \to Z$, $f: X \to Y$, we have $(1 \stackrel{\langle [g], [f] \rangle}{\to} Z^Y \times Y^X \stackrel{c_{X, Y, Z}}{\to} Z^X) = [g f]: 1 \to Z^X$ ## Properties ### Inheritance by reflective subcategories In showing that a given category is cartesian closed, the following theorem is often useful (cf. A4.3.1 in the Elephant): ###### Theorem If $C$ is cartesian closed, and $D\subseteq C$ is a reflective subcategory, then the reflector $L\colon C\to D$ preserves finite products if and only if $D$ is an exponential ideal (i.e. $Y\in D$ implies $Y^X\in D$ for any $X\in C$). In particular, if $L$ preserves finite products, then $D$ is cartesian closed. ### Exponentials of cartesian closed categories The following observation was taken from a post of Mike Shulman at MathOverflow. If $\mathcal{C}$ is small and $\mathcal{D}$ is complete and cartesian closed, then $\mathcal{D}^{\mathcal{C}}$ is also complete and cartesian closed. Completeness is clear since limits in $D^C$ are computed pointwise. As for cartesian closure, we can compute exponentials in essentially the same way as for presheaves, motivated by $\mathcal{D}$-enriched category theory: $G^F(x) = \int_{y\in \mathcal{C}} \prod_{\mathcal{C}(x,y)} G(y)^{F(y)}.$ Then we can compute $\array{ \mathcal{D}^{\mathcal{C}}\left(H,G^F\right) &=& \int_{x\in \mathcal{C}} \mathcal{D}\left(H(x), G^F(x)\right)\\ &=& \int_{x\in \mathcal{C}} \mathcal{D}\left(H(x), \int_{y\in \mathcal{C}} \prod_{\mathcal{C}(x,y)} G(y)^{F(y)}\right)\\ &=& \int_{x\in \mathcal{C}} \int_{y\in \mathcal{C}} \prod_{\mathcal{C}(x,y)} \mathcal{D}\left(H(x), G(y)^{F(y)}\right)\\ &=& \int_{y\in \mathcal{C}} \mathcal{D}\left( \int^{x\in \mathcal{C}} \sum_{\mathcal{C}(x,y)} H(x), G(y)^{F(y)}\right)\\ &=& \int_{y\in \mathcal{C}} \mathcal{D}\left(H(y), G(y)^{F(y)}\right)\\ &=& \int_{y\in \mathcal{C}} \mathcal{D}\left(H(y)\times F(y), G(y)\right)\\ &=& \mathcal{D}^{\mathcal{C}}(H\times F, G). }$ Here the antepenultimate step uses the co-Yoneda lemma. This appears to involve colimits in $\mathcal{D}$ as well, but the existence of these colimits is not actually an extra assumption: the co-Yoneda lemma tells us that $\int^{x\in \mathcal{C}} \sum_{\mathcal{C}(x,y)} H(x)$ exists and is isomorphic to $H(y)$. Similarly, the above argument can be interpreted to say that even if $\mathcal{D}$ is not complete, then the exponential $G^F$ in $\mathcal{D}^{\mathcal{C}}$ exists if and only if the particular limits above exist, and in that case they are isomorphic. A more abstract argument using comonadicity and the adjoint triangle theorem, which also applies to locally cartesian closed categories, can be found in Theorem 2.12 of • Street and Verity, The comprehensive factorization and torsors, TAC and is reproduced in the setting of closed monoidal categories at closed monoidal category. ### Functional completeness theorem ###### Theorem Let $C$ be a cartesian closed category, and $c$ an object of $C$. Then the Kleisli category of the comonad $c \times - \colon C \to C$, denoted $C[c]$, is also a cartesian closed category. ###### Remark The Kleisli category of the comonad $c \times -$ on $C$ is canonically equivalent to the Kleisli category of the monad $(-)^c$ on $C$. ###### Proof of Theorem Let $a$ and $b$ be objects of $C[c]$. Claim: the product $a \times b$ in $C$, considered as an object of $C[c]$, is the product of $a$ and $b$ in $C[c]$, according to the following series of natural bijections: $\array{ \underline{z \to a \qquad z \to b} & C[c] \\ \underline{c \times z \to a, c \times z \to b} & C \\ \underline{c \times z \to a \times b} & C \\ z \to a \times b & C[c] }$ Claim: the exponential $a^b$ in $C$, considered as an object of $C[c]$, is the exponential of $a$ and $b$ in $C[c]$, according to the following series of natural bijections: $\array{ \underline{z \to a^b} & C[c] \\ \underline{c \times z \to a^b} & C \\ \underline{c \times z \times b \to a} & C \\ z \times b \to a & C[c] }$ where in the last step, we used the prior claim that the product of objects in $C$, when viewed as an object of $C[c]$, gives the product in $C[c]$. Observe that the object $c$ in $C[c]$ has an element $e \colon 1 \to c$, corresponding to the canonical isomorphism $c \times 1 \cong c$ in $C$. We call this element the “generic element” of $c$ in $C[c]$, according to the following theorem. This theorem can be viewed as saying that in the doctrine of cartesian closed categories, $C[c]$ is the result of freely adjoining an arrow $1\to c$ to $C$. ###### Theorem (Functional completeness) Let $C$ and $D$ be cartesian closed categories, and let $F \colon C \to D$ be a cartesian closed functor. Let $c$ be an object of $C$, and let $t: F(1) \to F(c)$ be an element in $D$. Then there exists an extension of $F$ to a cartesian closed functor $\tilde{F} \colon C[c] \to D$ that takes the generic element $e \colon 1 \to c$ in $C[c]$ to the element $t$, and this extension is unique up to isomorphism. ###### Proof (sketch) On objects, $\tilde{F}(a) = F(a)$. Let $f \colon a \to b$ be a map of $C[c]$, i.e., let $g \colon c \times a \to b$ be the corresponding map in $C$, and define $\tilde{F}(f)$ to be the composite $F(a) \cong 1 \times F(a) \stackrel{t \times 1}{\to} F(c) \times F(a) \cong F(c \times a) \stackrel{F(g)}{\to} F(b)$ It is straightforward to check that $\tilde{F}$ is a cartesian closed functor and is, up to isomorphism, the unique cartesian closed functor taking $e$ to $t$. ## References Textbook account: Discussion with focus on mapping spaces (compact-open topology) in topological spaces and compactly generated topological spaces:

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## Saturday, October 20, 2018 ### EVERYTHING IS MATTER MOVING THROUGH SPACE My new book is available here. “According to author Palazzo (Are the Laws of Physics Weird?, 2016, etc.), the stubborn resistance of time to definitive elucidation is the function of a long-standing mistake: Motion has long been considered a function of the relationship between distance and time, which makes time a fundamental concept and motion a derivative one. Palazzo, however, provocatively suggests that relationship is reversed: Motion is fundamental, and time is a convenient mental construct devised to measure motion… The author’s efforts are remarkably ambitious, and his command of the material is powerful.” Kirkus Reviews ## Wednesday, July 20, 2016 ### ARE THE LAWS OF PHYSICS WEIRD? See info on my recent book: ARE THE LAWS OF PHYSICS WEIRD? It's available here. ## Friday, September 04, 2015 ### Superposition and Quantum States A lot of confusion in Quantum Mechanics is the result from not being able to differentiate between the real world and the Hilbert Space. Vectors in real space – like velocities, accelerations, forces, etc. – are objects one can actually measure in the real world. On the other hand, quantum states are represented by vectors (more precisely by rays) in a Hilbert space, but these are NOT subjects of measurement. What we measure for a quantum system are probabilities, and those vectors in that Hilbert space are useful mathematical tools to calculate those probabilities. Illustration Suppose we have a beam of electrons flowing from right to left: Notice this is a thought experiment as we really don’t know in what direction the spin of each individual electron points. We can safely say that these directions are at random. Now physicists are interested in measuring these spins. So I need some kind of apparatus, and the good news is that there exists one – a magnetic field. Trouble is that these electrons, with their spin, are tiny magnets, and we know that magnets placed in a magnetic field will align (or anti-align) with the magnetic field. Suppose I place the magnetic field along a certain direction, say the z-axis. Now let’s look at one electron as it approaches the magnetic field. When that electron penetrates the magnetic field, it will align its spin such that its z-component will yield the value of +ℏ/2 along the z-axis, a spin up, which can be represented as: Here’s another electron about to penetrate the magnetic field: This time it will anti-align with the magnetic field, with a spin value of -ℏ/2, a spin down. On the whole, 50% of the electrons will align with the magnetic field (spin =+ℏ/2, or up), and 50% will anti-align (spin = -ℏ/2, or down). Comments (1) Note that before the measurement, the spin of an electron can be in any direction. Passing the electron through the magnetic field forces the electron to change its spin orientation such that it either aligns or anti-aligns with its z-component to be ± ℏ/2. This is what distinguishes quantum physics from classical physics: the act of measuring a quantity will disturb the system. (2) The other components of the spin are indeterminate: if I were to pass these electrons into another magnetic field, say aligned with the x-axis, again I will find that 50% of the electrons will align with the magnetic field (spin = +ℏ/2), and 50% will anti-align (spin = -ℏ/2), this time along the x-axis. I will no longer know what the spin along the z-axis is. (3) One way to mathematically represent this quantum system (read, the wave function) is this: | ψ> = 1/(2½) (| > + | >). Now this is called a superposition of two quantum states, the up and down states. Note that if I want to calculate the probability that the electron has a spin up, I take the product of the vector | > with the wave function | ψ>, and square that. P = |< | ψ >|2 = 1/2 [< | { | > + | > }]2 =1/2 [{ < | > + < | > }]2 Using the orthogonality condition, < | > = 1 and < | > =0, we get, P=1/2, or 50%, which is what is observed in the lab. (4) Now here comes the real crunch. Writing | ψ> = 1/(2½) (| > + | >) is called a superposition but it’s not meant to mean that the electron “lives” simultaneously in two states and can’t make up its “mind” in which one it wants to live. Those states do not represent ordinary vectors of real objects - like velocities, acceleration, forces, which was a crucial point that was mentioned above. If it were the case, then since these two vectors are equal in magnitude and opposite in direction I would be able to claim, | > = (-1) | >, and the orthogonality condition would no longer hold, and P would not equal to 50% - actually it would turn out to be zero!!! What needs to be reminded is that the two vectors, | > and | > represent possible states, and the beauty of it all is that they form a complete set of orthogonal unit vectors, in an abstract space called the Hilbert space, which provides a powerful method of calculating probabilities. Appendix A word on semantics: note that I used the word "apparatus" when that word description is NOT needed. For instance at the LHC, one thinks of two beam interacting (colliding), and not as one beam interacting with an apparatus - the second beam. Similarly, the beam of electrons described above are interacting with a magnetic field (the "apparatus"). Hence, the whole concept of "wave function collapse" is totally unnecessary. The so-called measurement between a microscopic system and a macroscopic system is illusive as it never happens, it is always a microscopic system interacting with another microscopic system. And the wave function cannot collapse as it is not a function of a real wave. Also, there is no need of hidden variables or "beables". QM can do very well without this extra baggage. ## Wednesday, August 26, 2015 ### Entanglement and the Uncertainty Principle Here's your typical experiment. Suppose these particles are electrons/positrons and they are sent one pair at a time. At O, the two particles are released from rest and sent in opposite directions, and of course because of the conservation of angular momentum, they will have opposite spins as they move toward A and B, where observers are stationed. A..............................←●O●→...............................B Let's say those particles are going to be passed through a magnetic field aligned in the direction of the Z-axis, giving each either a spin up (+) or a spin down (-). Alice will record each individual electron as they arrived, along with their spin. Ditto for the positrons on Bob's side. Here's one entry that Alice might have recorded on day 1. It is no surprise that after a considerable number of observations, she will find that 50% will be up, which will be denoted by (+), and 50% down, designated by (-). Case 1. She can verify her results by passing all the electrons along a second magnetic field, also aligned with the Z-axis, and she will find all those which had a spin up(+) will still have spin up(+), and all those with spin down(-) will still have spin down(-). This is what is meant by preparing a quantum system in a given state: the first measurement prepared our quantum system in a given state - it forced all the electrons to have either spin up or down along the Z-axis. The second measurement confirmed that. Mathematically, the system is in a eigenstate. Of course, because the particles are entangled, Bob if he also measures his positrons along the Z-axis, will record for each positron its opposite spin. The law of conservation of angular momentum demands it. Now, there's no mystery here. Case 2. But what if Bob had chosen to measure along a different axis, say the X-axis, by putting a magnetic field along that axis. Now again he will measure along the X-axis either spin up(+) or spin down(-). Say the first electron that Alice actually measured to be up (+) on the Z-axis, and would have been down (-) on the Z-axis for Bob's particle is now up (+) on the X-axis, did Bob overcome the Uncertainty Principle, since he now "knows" the z-component, down (-), and x-components up (+), of that particle? Not really. Should he take that particle into a second magnetic field now aligned with the Z-axis, he will find that there will be a 50% chance that it will be up (+), and a 50% chance down (-). So he doesn't "know" the spin along the Z-axis, only along the X-axis. The act of measuring along the X-axis no longer guarantees that he has a particle with spin down (-) on the Z-axis. And this is at the heart of Quantum Mechanics: we don't know the state of the particle until we make a measurement, and whatever the state of a particle was before the measurement, it can be altered if we pass the particle into a different apparatus(Case 2). ## Wednesday, May 20, 2015 ### Killing Vectors and Hawking Radiation Preliminary We start out with the interval (see equations (6) to (23) in Relativistic Doppler Effect ), (1) ds2 = -dt2 + dx2 + dy2 + dz2 We define the metric as the coefficient of each of the terms in the above: (2) η00 = -1, η11 = 1,η22 = 1,η33 = 1,and ηij = 0 for i≠j We can rewrite equation (1) in the general form, (3) ds2 = ηαβdxα dxβ The proper time τ is, (4) dτ2 = - ds2 This yields, (5) dτ = dt/γ (6) where γ = (1 - v2) We measure the velocity with respect to the proper time τ, not the ordinary time t. (7) uβ =dxβ/dτ This gives the important result, (8) u2 = uu = -1 We define a 4-vector momentum as, (9) pβ =(p0,pi) = (p0,p) This gives the following: (10)p2 = muβmuβ = m2u2 = - m2 And, (11) E2 = m2 + (p)2. Putting c into the equation, (11) E2 = m2c4 + p2c2. Euler-Lagrange Equations for a free particle in motion Consider two timelike separated points A and B, and all the timelike worldlines. In fig 1, two such lines are illustrated - a straight line path and a nearby path. By the variational principle, the world line of a free particle between two timelike separated points extremizes the proper time between them. To see this, each curve will have a value in terms of the proper time, (12) τAB = ∫AB Using equations (1) and (4), (13) τAB = ∫AB {dt2 - dx2 - dy2 - dz2}½ We parametrize this equation by choosing σ such that at point A, σ = 0, and at B, σ =1 (14) τAB = ∫01 dσ {(dt/dσ)2 - (dx/dσ)2 - (dy/dσ)2 - (dz/dσ)2}½ This has the same form as the action of equation (1) in The Essential Quantum Field Theory , repeated below (15) S = ∫ dt L By making the correspondence: the action S → τAB, the time t → σ and the Lagrangian L → {(dt/dσ)2 - (dx/dσ)2 - (dy/dσ)2 - (dz/dσ)2}½ We can rewrite the Lagrangian L in terms of the general form (equations 3 and 4), (16) L = { - ηαβ(dxα/dσ) (dxβ/dσ) }½ Also, another form of the Lagrangian is, (17) L = dτ/dσ The corresponding Euler-Lagrange equation ( see paragraph below equation 1 in The Essential Quantum Field Theory ) (18) Consider a particle freely moving along the x-axis ( x1 = x, x2 = y =0, x3 = z = 0) (19)Equation (18) becomes (see appendix A), (20) Using equation (17), substitute for L in the above, we get, (21) Now multiply both sides by dσ/dτ, we get, In case you haven't recognized, this is the equation of a straight line. Integrate once, (22) dx/dτ = c Integrate a second time, (23) x = cτ + d Hence for the extremal proper time, the world line for a particle freely moving from point A to point B is a straight line path (fig 1). Killing Vectors Generally speaking, conservation laws are connected to symmetries. For instance, if there is a symmetry under displacement in time, energy is conserved; under displacement in space, momentum is conserved; under rotations, angular momentum is conserved. However, in GR, the metric is often time dependent, angle dependent, position dependent, etc. So how does one tell if there is a symmetry? One clue is if the metric is independent of one of its coordinates. For instance, say the metric is independent of x1. That means, it transforms as, (24) x1 → x1 + const. leaving the metric unchanged The vector ξ with components, (25) ξα = (0,1,0,0) lies along the direction the metric doesn't change. This is a Killing vector (in honor of Wilhelm Killing, German mathematician 1847-1923). A Killing vector is a general way of characterizing a symmetry in any coordinate system. For a freely moving particle, one can show, (26) ξu = constant, (see appendix B) (27) Also,ξp = constant, where p is the particle momentum. Schwarzschild Geometry In GR, the Minkowsky metric ηαβ is replaced by a more general metric gαβ so that equation (3) now reads as, (28) ds2 = gαβdxα dxβ Specifically in a Schwarzschild geometry, the metric reads as, (G=c=1) (29) g00 = -(1 - 2M/r), g11 = (1 - 2M/r)-1, g22 = r2,g33 = r2sin2θ,and gij = 0 for i≠j For our purposes, we note that the metric is time-independent, and therefore there is a Killing vector, which has the components, (30)ξα = (1,0,0,0) Hawking Radiation Fig 2 shows a rest-mass zero particle-antiparticle pair which has been created by vacuum fluctuations in such a way that the two particles were created on opposite sides of the horizon of a black hole. The components ξp and ξp' must be equal and opposite so that ξ(p +p') = 0, (value of the vacuum). The particle ( ξp > 0) can propagate and can be seen as radiation by an observer at infinity. This also means that the antiparticle ( ξp' < 0) will be absorbed by the black hole, thus decreasing its mass in the process. This is the basis of Hawking's claim that black holes radiate, and in time, will evaporate. Appendix A (A3) Equation (18) now reads as, First calculate, Putting it altogether, equation (A3) becomes, (A4) Appendix B let α =1, Equation (A3) becomes, (B1) (B2) from (A2), (B3) LHS of (B1) , (B4) therefore, (B5) Now consider, Using equations (A1) and (17) Note that we can write, (B6) η = ηαβ ξα Substituting in the above, (B7) (B8) From (B4), we get , ξu = constant ## Saturday, December 20, 2014 ### The Equivalence Principle: BBT or SUT Einstein made the following reasoning: he took the Doppler Effect and using the Equivalence Principle, turned it into a Gravitational Shift. And then from there, he derived his eponymous Field Equations. See The Essential General Relativity. By reversing this reasoning I've shown the following: from the Gravitational Shift and using the Equivalence Principle, I've turned that into a Doppler Effect, and from there I then derived the Hubble Equation. See The Equivalence Principle and the Big Bang Theory. So one inevitable conclusion is: either the redshift from faraway galaxies is the result of a Doppler Effect and that gives the BBT - or it's a gravitational effect and that gives the SUT. The second inevitable conclusion from the EP is that one interpretation cannot be differentiated from the other, that is, there are no conclusive tests that would favor one over the other. You either accept GR and the EP, or you reject both. Now note that only facts independent of GR/EP can tip the balance in favor of one of those models. ## Friday, December 05, 2014 ### Big Bang Theory Versus Static Universe Theory In this blog, I will compare the Static Universe Theory (SUT) versus the Big Bang Theory (BBT) in regard to their respective assumptions. As it has been mentioned in Another argument against the BBT, it is believed that only a Hot Big Bang scenario can explain the CMB, and any other explanation would have to be contrived. We show here quite the contrary that it is the BBT which is contrived in its assumptions to make all the parts fit in. Static Universe Theory (1) The Equivalence Principle is valid and the redshift from faraway galaxies is gravitational in nature. See The Equivalence Principle and the Big Bang Theory. (2) The universe is eternal and infinite. (3) The CMB can be explained in terms of the surface of infinite redshift. See Olbers' Paradox . Big Bang Theory (1) The Equivalence Principle is valid and the redshift is due to a Doppler Effect – the galaxies are moving away from each other. (2) By extrapolating backward in time, the universe started as a singularity and then expanded. (3) There is a 4th spatial dimension into which our 3-spatial dimensional world is expanding. See Riemannian Geometry and the Big Bang Theory. (4) In order to solve the Einstein Field Equations to get to the Friedman Equations, one must assume the universe is homogeneous and isotropic. (5) To justify (4), one must assume that the universe went through an inflationary period in the early stage. (6) To justify (5), one must assume that quantum fluctuations popped out of the vacuum some 13.7 billion years. (7) Since the universe is accelerating, one must assume that the universe is filled with Dark Energy, which must make up 75% of the universe in order to justify a flat space universe (As of now, the Vacuum Energy from (6) is out of step by 122 orders of magnitude with Dark Energy). (8) To calculate the density of the universe, one must assume the universe is finite in size with its radius equal to its Schwarzschild radius. (9) In order to tie in the CMB with the BBT, one must assume that the universe must behave like a nearly perfect idealized fluid, so that one can tie in the redshift to the scale factor in (4), which itself is tied in with temperature and time. One can then set a chronology of different reactions that would have happened at different temperatures/times, all of these requiring a number of parameters that can be fine-tuned with observation. Conclusions The BBT is a contrived theory which besides the number of assumptions that is needed to support the BBT - a much larger number than the SUT - it nevertheless leaves a certain number of unanswered questions such as: what evidence do we have that a 4th spatial dimension exists? If the universe didn't exist from t = -∞ to t = -13.7 billion years, what caused it to spring out of the vacuum some 13.7 billion years ago? How many more assumptions will the BBT need in order to reconcile the Vacuum Energy with Dark Energy in order to make that fit into the theory?

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Are all of compact support functions of $A(G)$ in its abstract Segal algebras? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T12:26:00Z http://mathoverflow.net/feeds/question/73041 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73041/are-all-of-compact-support-functions-of-ag-in-its-abstract-segal-algebras Are all of compact support functions of $A(G)$ in its abstract Segal algebras? Mahmood Alaghmandan 2011-08-17T09:48:13Z 2011-08-17T09:55:32Z <p>Let $G$ be a locally compact group. We know that if $G$ is abelian and $\cal F$ implies the Fourier transform, for every Segal algebra of $G$ say $S^1(G)$, ${\cal F}S^1(G)$ is an abstract Segal algebra with respect to $A(\widehat{G})$ (since $S^1(G)$ is an abstract Segal algebra of $L^1(G)$). Reiter in [1, Proposition 6.2.5] has shown that $C_c(\widehat{G})\cap A(\widehat{G}) \subseteq {\cal F}S^1(G)$.</p> <p>My Question is: Can we show that for an aribtarary locally compact group $G$ (not necessarily abelian) every abstract Segal algebra of $A(G)$ contains all functions in $A(G)$ that have compact support?</p> <p>[1] H. Reiter, and J. D. Stegeman, Classical harmonic analysis and locally compact groups, 2nd edn, London Mathematical Society Monographs, New series 22, Oxford university press, New York, 2000.</p>

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# Poly1305 variants with bigger output? This is a rather simple question, but answers are nowhere to be found. Are there any variants of Poly-n hashing algorithms which provide bigger outputs (like 32 instead of 16 bytes)? Or, is there any research which discusses the variability of the constant $$2^{130}-5$$, why this number is special or whether there are other good alternatives? I understand that it would be better if it was a prime number, but are all primes good (ignoring the cost of modular arithmetic)? Or, could I use the Poly1305 two times with two differenst pairs of $$(r,s)$$ and concatenate to produce a 32-byte hash? • You could do something like Daence. May 27, 2023 at 17:54 $$\newcommand{FP}{\mathbb F_p}$$ The authenticator built on Poly1305 follows a generic construction to build MAC from a universal hash and a pseudo-random function. So, increasing the field size and hence output size (assumedly for security) is possible, in theory at least. Poly1305 precisely refers to the universal hash part of the construction. Which itself is a polynomial evaluation hash that can be generically defined over any field $$\FP$$. The actual details of Poly1305 are somewhat more complicated. But the point is that this construction yields an excellent Difference-Unpredictable Hash. The “issue” comes from turning such a hash into a MAC, especially when $$p$$ is a prime. In the purest version of the MAC scheme, we would simply have a PRF generating values in $$\FP$$. However, the most practical and efficient PRFs, including AES that is commonly used with Poly1305, produce $$n$$-bit strings instead. Leaving us with two options for how big $$p$$ should be: 1) either slightly larger than $$2^n$$ or 2) slightly smaller. Poly1305 makes the first choice, which offers one big advantage: The message can be broken up and processed in $$n$$-bit blocks. With the second option, we would need to process inputs as field elements. Which is inconvenient if $$p$$ is prime. Again, the actual implementation details are likely to be more involved than this. Finally, increasing the field size would require adjusting the PRF size accordingly. Otherwise, there might not be significant security gains. Which means that we would also need a performant PRF with bigger output space. To add to Marc Ilunga's excellent answer, I would point out that that the efficiency of Poly1305 is platform dependent. It covers the very important software space where chip instructions can be relied on to include highly optimised integer/float multiplication and addition operations. I believe that the choices in Poly1305 are highly tuned to the capabilities of current chip sets. In more constrained environment, it can be better to base the polynomial evaluation hash over a binary field $$\mathbb F_{2^w}$$. This is done in the GHASH function that is used in Galois Counter Mode and although usually described over $$\mathbb F_{2^{128}}$$, the 2004 GHASH paper describes GHASH for any even $$w$$. This also gives an exact match between blocks and field elements and hence provable universality. There have been a few other polynomial evaluation hashes over prime fields tested for performance in the literature these include $$p=2^{127}-1$$ and $$p=2^{64}-59$$. This SoK seems to provide variants of Poly1305 with bigger output and also discuss which other primes could be good alternatives. Regarding concatenating two times Poly1305 with independent keys, this is also another secure option as proven in section 7.5 of this other paper but according to the benchmarks in the SoK (cf. Appendix D), it is less efficient than simply choosing a bigger and good prime.

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# Math Do YOU Like It? I can handle the basics Math wasn’t my favorite subject in school. I mean, I can add, subtract, multiply, divide and do fractions (full disclosure: When I’m ready to cut drapery fabric I mark where I’m going to cut with pins so Mike can double check the measurements BEFORE I cut the fabric.). To be clear, I don’t like math of the algebra, geometry, trigonometry varieties, never mind calculus. ‘Ometries make my brain hurt, and as hard as I worked at it, I never quite ‘got it’ in high school. <sigh> Looking back, I see now what a wonderful math teacher Mrs. La Mastra was. And a good thing too since I had her for three years and two weeks, to be exact. An early graduate of Douglass College in New Jersey, she was patient, committed and consistent in her ability to convey how to work out the problem on the board. The problem, at least as I saw it, is that some people (that would be me) just don’t have math brains. In Algebra I held my own by working extra problems every night; I got a B in the class and was proud of it. During my year of Geometry I had nightmares about drowning in theorems and proofs, really. My brain takes leaps of faith, and trying to make Mrs. La Mastra understand that my brain couldn’t put down all the steps in the proof because it didn’t ‘see’ all the requisite steps was frustrating for both of us; either I tried to slow my brain down to turtle speed and put way too much information on the paper or I used my regular speed brain and hopped to the end like the bunny-quick thinker I am. She would often ask the recalcitrant brain students in our class, ‘Time will pass, but will you?” As a junior I faced Algebra II and I can assure you that I have absolutely no recollection of what I may have learned, nor the grade I received (Okay, I got a B, but the only reason I know is because I have my old report card because I’m weird that way. Oh come on, didn’t you save yours?!). I can tell you, though, that my friend Fletcher sat across from me. He always slunk into class late, buried his head in his army surplus jacket and appeared to sleep until called on for an answer. Whenever Mrs. La Mastra thought she had him (out) cold, he always sighed, straightened up a little, gave a correct answer and then sank back into his coat. I really didn’t like it that he could come up with an answer so easily when he never did his homework and he never really tried that hard and all I did was try and struggle. And boy did it tee Mrs. La Mastra off that she never did catch him with a problem he couldn’t work out. Senior year I walked into my calculus class and guess what? Mrs. La Mastra was my teacher again. What luck, I thought. She spotted me, and I swear she paused, a long pause, before she resumed her walk to her desk. Class commenced, and two weeks later I still didn’t understand even one concept, and I was getting pretty frantic since I wanted to go to college and was afraid calculus would be the reason I didn’t get into even one. After class one day Mrs. La Mastra watched me leave with what I thought was a more than passing interest. Sure enough, she stopped me just outside the door to her classroom. “Do you have plans to attend college?” she asked. Blanching, figuring I was already failing and this was her way of telling me, I replied, “Yes. Why?” “What do you plan to major in?” Mrs. La Mastra persisted. “English. Journalism. Something in that area,” I replied cautiously, wondering why she was asking and feeling the little ‘perks’ starting up deep in my stomach. “Drop/add day is next week, Miss. You could drop out of calculus without any harm done, you know. I would be willing to sign the necessary form for you. As an English major you wouldn’t need the calculus on your transcript. After having you as a student for the past three years, it’s my considered opinion that you do not have a ‘math brain,’ as you like to call it. Nothing wrong with that, but perhaps you’d like to enjoy your senior year and take an additional elective instead?” she asked with a ghost of a smile flitting across her face. I must have turned five shades of pale before assuming the hue of a Jersey tomato. “Mrs.,  Mrs. La Mastra. Really? Do you really think so?” I asked hopefully. “Yes. Yes I do,” she replied as she walked brusquely back into her classroom. As I hurried off to Guidance, I heard her ask “Time will pass, but will you?” Time will pass, but. . .

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practice test 2 practice test 2 - PRACTICE TEST#2 Physics 1021 NAME 1 Two... This preview shows pages 1–2. Sign up to view the full content. PRACTICE TEST #2 Physics 1021 NAME:___________________________ 1. Two horizontal curves on a bobsled run are banked at the same angle, but one has twice the radius of the other. The safe speed (no friction needed to stay on the run) for the smaller radius curve is v. What is the safe speed on the larger radius curve? a. approximately 0.707 v b. 2v c. approximately 1.41v d. 0.5 v 2. A hypothetical planet has a mass of half that the Earth and a radius of twice that of the Earth. What is the acceleration due to gravity on the planet in terms of g, the acceleration due to gravity at the Earth? a. g b. g/2 c. g/4 d. g/8 3. Does the centripetal force acting on an object do work on the object? a. Yes, since a force acts and the object moves, and work is force times distance. b. Yes, since it takes energy to turn the object c. No, because the object has constant speed. d. No, because the force and the displacement of the object are perpendicular. 4. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. This note was uploaded on 04/07/2008 for the course PHYS 1021 taught by Professor Borovitskaya during the Spring '08 term at Temple. Page1 / 3 practice test 2 - PRACTICE TEST#2 Physics 1021 NAME 1 Two... This preview shows document pages 1 - 2. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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Which statement is always true for a square, but not always true for a rhombus? - Two angles are obtuse. - All sides are equal. - Two or more angles are right angles. - More than two sides are parallel. 2 by HoseaPomainville764 2014-08-17T15:37:58-04:00 two or more angles are right angles. Because a square always has 4 right angles to be a square, however a rhombus has no right angles because that would be a lame rhombus if its a square. Um yeah. 2014-08-17T15:44:59-04:00 Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest. - Two angles are obtuse. No.  There are no obtuse angles in a square. - All sides are equal. No.  It's true of squares AND rhombuses. - Two or more angles are right angles. Yes.  Always true for a square, sometimes true for a rhombus. - More than two sides are parallel. Confusing, so call it a 'no'. In both squares and rhombuses, every one of the four sides is parallel to ONE other side. Another important note: Look at the statement that's true always for a square and sometimes for a rhombus: "Two or more angles are right angles". WHEN does a rhombus have right angles in it ?  Fasten your seat belt and hold on tight: A square is a special kind of rhombus.  It's a rhombus that has right angles in it. So a rhombus has right angles in it when it's a square.

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## Helix - not one full turn - HOWTO? The ideal place for beginners to get help using SketchUp ### Re: Helix - not one full turn - HOWTO? Well, the pictures I posted are from the stairmakers website, so they don't correspond with our stair to be. We'd need to cover a height of 3100 mm in 17 treads, so each riser would need to be 182,35 mm. Also, there's something as the 'walkline' (translated literally, maybe it should be line of walk (?)) This line of walk would indicate that each tread be 230 mm across at a radius of 650 mm... If I'm not mistaken you're thinkering of 16 or 18 to be a multiple of a pair? Erwin 0 lexje ### Re: Helix - not one full turn - HOWTO? Well, true that not exactly the same dimensions... Especially the "walkline" - I used completely different measures so this is just an example. The 18 is handy because 360 can be divided by 18. Then you take two of them and have a very similar number as your 324,xxx degrees (exactly 320). When working with segmented curves in SU, it is extremely hard to count with those long decimal digits. 0 Gaieus ### Re: Helix - not one full turn - HOWTO? Had I realized you wanted to make 'spiral stairs' I would have suggested my 'Grow' tool... Unzip this Grow.zip file and put the resultant Grow.rb into the Plugins folder and the growCursor.png into the Plugins/Icons folder [just as they are zipped]. The attached skps have some tutorials on how to make a helical [spiral] stair and also helices in general... Grow.zip GrowHelicalSteps.skp GrowHelix.skp GrowHelicalSteps.png 0 TIG TIG Global Moderator ### Re: Helix - not one full turn - HOWTO? When making spiral steps there are often rules/codes to adhere to.. In the UK on narrower flights <1m [typically domestic or small commercial flights] you must have at least 50mm of plan-tread at the innermost part and the plan-tread on the centerline should be even for every step, and follow the rule 2R+G >=550mm <=700mm - where R is the Rise and G is the Going [plan-tread]. If you draw this in plan it limits the relative angle of adjacent nosings and effectively sets a limit on the taper angle between them = ~15degrees depending on the configuration etc... For wider stairs you measure in from both edges [strings] 270mm and the same max/min pitch applies, 2R+G >=550mm <=700mm to the two locations Any landings shall be the same length as their width measured along their centerline. The minimum/maximum allowable Rise and Going varies with the building type too. 'Private Stair'[one dwelling] R >=155mm <=220mm and G >=223mm <=300mm [vertical pitch is also limited to 42 degrees - usually trapped by 2R+G rules] 'Institutional and Assembly Stair' R >=135mm <=180mm and G >=280mm <=340mm 'Other Stair' R >=150mm <=190mm and G >=250mm <=320mm BUT AD-Part-M [accessible/disabled=all commercial/public building etc] R <= 170mm G >= 250mm It's best to draw some test treads and place rotate a couple to see it they are right before committing to a complex 3D assembly... 0 TIG TIG Global Moderator ### Re: Helix - not one full turn - HOWTO? I have fixed the problems with drawhelix13.rb. It now allows turns that are not whole numbers... Here it is drawhelix13_TIGd.rb 0 TIG TIG Global Moderator ### Re: Helix - not one full turn - HOWTO? Nice information TIG. Did you use another ruby for the stringer? 0 dylan ### Re: Helix - not one full turn - HOWTO? No - it's made as a part of the tread and Grown... 0 TIG TIG Global Moderator ### Re: Helix - not one full turn - HOWTO? Hi TIG, thanks a tremendous lot for your input Greatly appreciated (Gaieus too b.t.w) I did explain the spiral stair stuff though, see herehttp://forums.sketchucation.com/viewtopic.php?f=79&t=26214 I'll give your attachments a whirl soon and post back. Although I must say my last experiments with Goh's toolkit are promising. (I've not found how to change direction of rotation clockwise/counterclockwise though) One advantage I see in using Goh's plugin is that the treads are more detailed i.e. the outside of the tread are 'curved' instead of straight. Nevertheless a couple of questions: I'm native Dutch (Flemish) speaking, our stairmaker is French, and out here we're talking English, certainly with this kind of lingo I need some vocabulary help Let me first paraphrase the layout of our stairs to be: Hellicoïdal with hollow core with R=200 mm Riser = 182,3 mm for V1 Riser = 189,2 mm for V2 Going = 230 mm (Giron in French) LF = Ligne de foulée ~= "walkline" ~= plan-tread (??) Plan-tread @ R = 650 mm -> V1 = 3/4T + 1/8T +107 mm -> V2 = 3/4T - 70 mm I really think, in conjunction with what I learnt from the stairmaker, the best way is to draft a plan and side view (like the picture I posted before), and build your model from there, as I also found in a post from TaffGoh here http://forums.sketchucation.com/viewtopic.php?f=79&t=12511 I must say, although the ruby depot is a tremendous help, it's lacking responsiveness, and certainly for newcomers, clarity and a little more context. E.g. most rb-scripts are already several years old and have often already been improved upon or contained within newer / other ones. Also a minimal documentation including ONE picture would often help to at least give an impression of what it's all about. One request remaining: For clarity's sake and certainly for others researching this 'specialized' knowledge - would you mind drafting a quick SU image adding the lingo 'terms' to exclude misinterpretation? Thanks again for this great and responsive forum and all the help!! Erwin 0 lexje ### Re: Helix - not one full turn - HOWTO? TIG wrote:I have fixed the problems with drawhelix13.rb. It now allows turns that are not whole numbers... Here it is drawhelix13_TIGd.rb Jeez, TIG you fix faster than I can think & write )) Erwin 0 lexje ### Re: Helix - not one full turn - HOWTO? B.t.w How do I get URL's right?? That is: to not show up as the full URL, but as a 'clickable' word? I've had a look [url]http://www.phpbb.com/community/faq.php?mode=bbcode/]here[/url], but this doesn't seem to work.. Allthough the one below DOES work ? phpBB BBCode supports a number of ways of creating URIs (Uniform Resource Indicators) better known as URLs. * The first of these uses the tag, whatever you type after the = sign will cause the contents of that tag to act as a URL. For example to link to phpBB.com you could use: Visit phpBB! This would generate the following link, Visit phpBB! Please notice that the link opens in the same window or a new window depending on the users browser preferences. Erwin 0 lexje ### Re: Helix - not one full turn - HOWTO? I'm native Dutch (Flemish) speaking... Haha... We know that1s not the same (I had a Flemish room-mate when I went to an American univ. for a semester ) Anyway, for staircase building, TIG's Grow is definitely a good tool (and thanks TIG for a revised version or the helix plugin, too). lexje wrote:B.t.w How do I get URL's right?? This is not the correct way: Code: Select all `[url]http://www.phpbb.com/community/faq.php?mode=bbcode/]here[/url]` But do it like this: Code: Select all `[url=http://www.phpbb.com/community/faq.php?mode=bbcode/]here[/url]` And the result will be this: here _______ Edit: wow, does somebody read the FAQ??? You have a bright future here, my friend! (Just look at what our other moderator, solo has in his signature about this) 0 Gaieus ### Re: Helix - not one full turn - HOWTO? Code: Select all `Edit: wow, does somebody read the FAQ???You have a bright future here, my friend! (Just look at what our other moderator, solo has in his signature about this);)` As a matter of fact, I did search the FAQ, and it simply is not there (correct me if I'm wrong)Off course being the better than average user, I searched phpBB's faq for this... Hence.. Solo doesn't seem to have too much of a Stoppelbaard Erwin 0 lexje ### Re: Helix - not one full turn - HOWTO? Hi, I have just tried the plugin about Helix and it works perfectly!. So I want to thank you, guys, for the script and the info. Then I have a question: I would like draw my helix/spiral giving the radius and the pitch information but I would like to decide the height of the spiral. For example In the image I have a fixed height 11590mm and in the whole height I have to build my helix. I have tried with some parameters in terms of radius and pitch (the radius is fixed as well) but it is smaller. Any tips? 0 stefx ### Re: Helix - not one full turn - HOWTO? Once the helix is made with the expected turns etc you can Scale it in the Z, typing in 1590mm [note the units 'mm'] - it then scales to that exact z-size rather than by a factor... 0 TIG TIG Global Moderator ### Re: Helix - not one full turn - HOWTO? Thank you TIG. I have tried and it works. 0 stefx ### Re: Helix - not one full turn - HOWTO? I'm currently working on a plugin that can produce a highly customizable curved or spiral staircase in SketchUp. I hope to release it soon, but in the meantime, take a look at this page: http://www.freesketchupplugins.co.uk/spiral_staircase.php As for a helix, I also have a tool that will make one. Unlike helix.rb or drawhelix13.rb, this will automatically give it a thickness based on a value you can specify. http://www.freesketchupplugins.co.uk/springtool.php 0 PlyrJames791 ### Re: Helix - not one full turn - HOWTO? Seems you have one inside 1001bits 0 Frenchy Pilou Is beautiful that please without concept! Speedy Galerie pilou Top SketchUcator ### Re: Helix - not one full turn - HOWTO? The plugin on 1001 bits is nothing to do with me. This one is freeware, not a trial: http://www.freesketchupplugins.co.uk/springtool.php 0 PlyrJames791

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https://quant.stackexchange.com/questions/40265/two-papers-two-different-solutions-of-the-ornstein-uhlenbeck-process

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# Two papers - two different solutions of the Ornstein-Uhlenbeck process Bernal 2016 says that the solution of $$dr_{t}=\lambda*(\mu-r_{t})*dt+\sigma dW_{t} \qquad (eq.1)$$ equals $$r_{t}=r_0*exp(-\lambda t)+\mu(1-exp(-\lambda t))+\sigma \int_{0}^{t} exp(-\lambda t)dW_{t} \qquad (eq.2)\\$$ which leads to following Euler Maruyana Scheme: $$r_{t+\delta t}=r_t+\lambda (\mu -r_t)\delta t+\sigma \sqrt{\delta t} *\mathcal{N}(0,1) \qquad (eq.3)\\$$ On the other side this paper tells us that the solution of the SDE should be $$S_{i+1}=S_i*exp(-\lambda \delta)+\mu(1-exp(-\lambda t))+\sigma \sqrt{\frac{(1-exp(-2\lambda t)}{2\lambda}}*\mathcal{N}(0,1) \qquad (eq.4)\\$$ Bernal uses $(eq.3)$ for calibration whereas GE and Berg use $(eq.4)$. Why the difference? Bernals method makes completely sense to me. • There seems to be a type in your equation 4 ($S_0$ should replace $S_i$). Other than that, equation 2 is identical to equation 4. Because the distribution of $\int_0^t exp(-\lambda t) dW_t$ is $\mathcal{N}(0, \frac{1 - exp(-2\lambda t)}{2 \lambda})$ Jun 11, 2018 at 12:32 Note that the Ito integral of a deterministic integrand $f: \mathbb{R}_+ \rightarrow \mathbb{R}$ is normally distributed $$\int_0^t f(u) \mathrm{d}W_u \sim \mathcal{N} \left( 0, \int_0^t f^2(u) \mathrm{d}u \right).$$ In your case, we have $f(t) = e^{-\lambda t}$ and thus $$\int_0^t f^2(u) \mathrm{d}u = \frac{1}{2 \lambda} \left( 1 - e^{-2 \lambda t} \right).$$

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https://byjusexamprep.com/gate-2022-power-electronics-quiz-5-i-7fe51700-37ae-11ec-b8eb-c8442d0a0391

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Time Left - 15:00 mins # GATE 2022 : Power Electronics Quiz 5 Attempt now to get your rank among 222 students! Question 1 For 3ϕ full wave rectifier following Relation hold true for which load. [given α > 60°, α : firing angle]. Question 2 A three-phase full converter delivers a ripple free load current of 15 A with firing angle delay of 45o. The current distortion factor is Question 3 A single Phase full wave transistor rectifier feed power to motor load. the source voltage is 250 V, 50 Hz. R = 4 Ω, L = 12 mH, E = 120 V. For a firing angle of 40o, the average value of output current such that the current releases energy till 1600 is? Question 4 A 3-phase fully controlled converter is fed from 400 V, 50 Hz mains for a firing angle of 60°, output voltage is 250 V. Calculate load resistance, source inductance & overlap angle. The load current is 25 A. Question 5 A 3 – ϕ full converter fed from 3-Phase 450 V ,50 Hz source is connected to R= 12Ω, E = 400 V and large inductance so that output current is ripple free. Calculate the input power factor for firing advance angle of 50o ? Question 6 If the converter shown below is triggered at , then rectification efficiency is • 222 attempts • 1 upvote • 0 comments Oct 30ESE & GATE EE GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 help@byjusexamprep.com

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https://vocal.com/noise-reduction/unifying-spectrum-subtraction-with-wiener-filtering/

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## Complete Communications Engineering It may seem that spectrum subtraction noise reduction is only based on an engineering intuition. However, this engineering intuition has a direct mathematics connection to the theoretically optimal Winer filter approach. The basic spectrum subtraction in frequency domain, $\left|\hat{S}\left(f\right)\right|^2=\left|X\left(f\right)\right|^2-\left|N\left(f\right)\right|^2$   (1) if N(f) is available or can be estimated. We can write is as the following, $\left|\hat{S}\left(f\right)\right|^2=\left|X\left(f\right)\right|^2-\left|N\left(f\right)\right|^2$   (2) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\left|X\left(f\right)\right|^2-\left|N\left(f\right)\right|^2}{\left|X\left(f\right)\right|^2}\left|X\left(f\right)\right|^2\$   (3) and we can recognize immediately the spectrum subtraction approach is actually equivalent to the Wiener filtering solution. Therefore, spectrum subtraction is mathematically optimum in the same sense as Wiener filtering! The spectrum subtraction approach can be viewed as zero phase filter with frequency response $H(f)=\frac{\left|X\left(f\right)\right|^2-\left|N\left(f\right)\right|^2}{\left|X\left(f\right)\right|^2}$ and 0< H(f) < 1. It applies a SNR-dependent gain to the noisy signal. When the SNR is high, the gain applied is high while SNR is low, the gain is low. The main conceptual difference between Wiener filtering and spectrum subtraction is the in way the gain factor is represented. In spectrum subtraction, the powers are estimated locally in time, or we may say, instantaneous power, while Wiener uses statistical ensemble.

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https://mathematicsart.com/solved-exercises/

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# Maths Solved Exercises ### Maths Solved Exercises | Enjoy Learning Mathematics #### Choose a problem to access the corresponding solution. Find the volume of the square pyramid as a function of $$a$$ and $$H$$ by slicing method. Prove that $\lim_{x \rightarrow 0}\frac{\sin x}{x}=1$ Prove that #### ↓ ↓ Calculate the half derivative of $$x$$ Prove Wallis Product Using Integration #### ↓ ↓ Calculate the volume of Torus using cylindrical shells Find the derivative of exponential $$x$$ from first principles Calculate the sum of areas of the three squares #### ↓ ↓ Find the equation of the curve formed by a cable suspended between two points at the same height Solve the equation for real values of $$x$$ Solve the equation for $$x\epsilon\mathbb{R}$$ #### ↓ ↓ Determine the angle $$x$$ Calculate the following limit Calculate the following limit #### ↓ ↓ Calculate the integral Challenging problem Prove that #### ↓ ↓ Prove that $$e$$ is an irrational number Find the derivative of $$y$$ with respect to $$x$$ Find the limit of width and height ratio #### ↓ ↓ How Tall Is The Table ? Why 0.9999999...=1 Solve the equation for $$x \in \mathbb{R}$$ #### ↓ ↓ Is the walk possible? Calculate the following Is $$\pi$$ an irrational number ? #### ↓ ↓ How far apart are the poles ? Solve for $$x \in \mathbb{R}$$ What values of $$x$$ satisfy this inequality #### ↓ ↓ Prove that the function $$f(x)=\frac{x^{3}+2 x^{2}+3 x+4}{x}$$ has a curvilinear asymptote $$y=x^{2}+2 x+3$$ Why does the number $$98$$ disappear when writing the decimal expansion of $$\frac{1}{9801}$$ ? Only one in 1000 can solve this math problem #### ↓ ↓ Error to avoid that leads to: What's the problem ? #### ↓ ↓ Determine the square's side $$x$$ Find out what is a discriminant of a quadratic equation. Calculate the rectangle's area #### ↓ ↓ Determine the square's side $$x$$ Wonderful math fact: 12542 x 11 = 137962 #### ↓ ↓ Find the volume of the interior of the kiln Calculate the sum What is the new distance between the two circles ? #### ↓ ↓ Calculate the area of the Squid Game diagram blue part Calculate the limit Find the infinite sum #### ↓ ↓ Calculate the integral Prove that Can we set up this tent ? #### ↓ ↓ Find the value of $$h$$ Find the length of the black segment Prove that pi is less than 22/7 #### ↓ ↓ Find the value of $$x$$ Amazing ! Prove that #### ↓ ↓ What is the weight of all animals ? Determine the length $$x$$ of the blue segment How many triangles does the figure contain ? #### ↓ ↓ if we draw an infinite number of circles packed in a square using the method shown below, will the sum of circles areas approach the square's area? What is the value of the following infinite product? Which object weighs the same as the four squares? #### ↓ ↓ What is $$(-1)^{\pi}$$ equal to? Can you solve it? Can you solve it? #### ↓ ↓ Great Math Problem Calculate the integral $$\int_{0}^{1}(-1)^{x} d x$$ Find the general term of the sequence #### ↓ ↓ Find the radius of the blue circles Determine the area of the green square Find the area of the square #### ↓ ↓ Calculate the integral Calculate the integral Calculate #### ↓ ↓ What is the radius of the smallest circle ? Find the Cartesian equation of the surface Calculate the integral #### ↓ ↓ Calculate the limit Can you solve this ? Solve the quintic equation for real $$x$$ #### ↓ ↓ How many students study no language ? Probability of seeing a car in 10 minutes What is the number of where the car stands ? What is the value of $$x$$ ? Prove that Is it possible to solve for $$x$$ so that $$ln(x)$$, $$ln(2x)$$, and $$ln(3x)$$ form a right triangle? How many times will circle A revolve around itself in total ? Find the distance $$BC$$ Solve for a^6, a : positive number Solve the quadratic equation by Completing the Square Prove that The area of a circle by slicing Determine the length and width of the rectangular region of the house a+b=20, find the maximum value of a²b Why 1+2+3+4+... is not equal to -1/12 Elon Musk's interview question Proof of Thales' Theorem How to check if two line segments intersect? What is the minimum number of operations required to reach 100? Fundamental Theorem of Calculus

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https://cstheory.stackexchange.com/questions/2698/computational-power-of-neural-networks

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# Computational Power of Neural Networks? Let's say we have a single-layer feed forward neural network with k inputs and one output. It calculates a function from $\lbrace 0,1\rbrace ^{n}\rightarrow\lbrace 0,1\rbrace$, it's fairly easy to see that this has at least the same computational power as $AC^0$. Just for fun, we'll call the set of functions computable by a single layer neural network "$Neural$". It seems, however, that it might have more computational power than $AC^0$ alone. So ... is $AC^0 \subseteq Neural$ or is $Neural = AC^0$? Also has this kind of complexity class been studied before? • A note on terminology -- important information is how many hidden layers there are. Zero hidden layer neural network with one output is just a linear threshold function, and is often (confusingly) called one layer or two layer neural network/perceptron, depending on whether inputs/outputs are considered layers. Also, in AI literature, neural networks are typically defined in terms of sigmoid functions which means that input/outputs are real valued. One hidden layer networks are known to be universal approximators in a sense that any continuous function can be approximated arbitrarily close Nov 4, 2010 at 18:33 There are some references I could find: General-purpose computation with neural networks: A survey of complexity theoretic results, 2003 and Counting hierarchies: polynomial time and constant depth circuits, 1993. It appears that neural networks are considered threshold circuits; i.e. those circuits using MAJORITY gates. In (2) it is the case that a depth $d$ neural network has complexity $TC^0_d$ (here's a link to link to complexity zoo entry about $TC^0$). Since $TC^0$ containts $ACC^0$, which is $AC^0$ with arbitrary MOD gates, then $AC^0 \subset TC^0$. Also, it is mentioned in the zoo that such circuits with depth 3 are strictly more powerful than those of depth 2. In On the computational power of sigmoid versus Boolean threshold circuits,1991 it is mentioned that for a constant depth $d$, Boolean and real-valued threshold circuits (with polynomially bounded weights) are essentially the same. • Interesting, thanks. This is what I was looking for. More interesting still is that $TC^0$ has a complete problems ... hmm ... Nov 4, 2010 at 5:34

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https://kateathome.com/fry/what-is-the-limiting-reactant-in-vinegar-and-baking-soda.html

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# What is the limiting reactant in vinegar and baking soda? Contents One underlying assumption is that the baking soda is the only limiting reactant. In other words, there is essentially an unlimited supply of acetic acid in the vinegar bottle, and the reaction output is only dictated by the amount of baking soda you add – every mole added results in a mole of carbon dioxide produced. ## Which is the limiting reagent in baking soda and vinegar? That a reaction occurs shows that the baking soda was the limiting reactant. Add vinegar to the last flask that originally had 10.0 g of baking soda. That a reaction occurs shows that the vinegar was the limiting reactant. ## What is the limiting reagent between acetic acid and sodium bicarbonate? Indeed, sodium bciarbonate will act as a limiting reagent, determining the number of moles of acetic acid that react. ## How do you determine the limiting reactant? Calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity. Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation. ## What is the chemical equation of baking soda and vinegar? Vinegar or Acetic Acid has the chemical formula CH3COOH . Baking soda is a base also known as Sodium Bicarbonate and has the chemical formula ‎NaHCO3 . During this reaction the products are sodium acetate ( C2H3NaO2 ). Sodium acetate is made of 1 sodium ion, 2 carbon atoms, 3 hydrogen atoms, and 2 oxygen atoms. ## What reaction is vinegar and baking soda? When vinegar and baking soda are first mixed together, hydrogen ions in the vinegar react with the sodium and bicarbonate ions in the baking soda. The result of this initial reaction is two new chemicals: carbonic acid and sodium acetate. The second reaction is a decomposition reaction. ## Is vinegar a reactant or product? Acetic acid mixed with water is vinegar. Usually vinegar is a solution of about 5% acetic acid and 95% water. When a reactant is in solution, the water is usually not listed as a reactant. Which atoms make up a molecule of acetic acid (vinegar)? ## What is the limiting reactant and excess reactant? The limiting reagent in a chemical reaction is the reactant that will be consumed completely. … Therefor it limits the reaction from continuing. Excess Reagent. The excess reagent is the reactant that could keep reacting if the other had not been consumed. ## Which is the limiting reagent? The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed. ## What is limiting reagent explain with an example? The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A+4B→3C+4D, when 5 moles of A react with 6 moles of B, then. THIS IS INTERESTING:  How long does Paxo stuffing take to cook? ## What is the chemical formula of vinegar? In flasks 1 and 2, a small amount of Mg is used and therefore the metal is the limiting reagent. In flask 3, the reagents are added in a stoichiometric ratio. In flask 4, excess Mg is added and HCl becomes the limiting reagent. ## What affects the rate of the baking soda and vinegar reaction? Therefore, the reaction is quicker due to the increase in collisions. The colder vinegar should not produce more carbon dioxide. The vinegar and bicarbonate soda reaction is endothermic*, meaning that the reaction requires heat to form products. Cold vinegar is a disadvantage. Categories Fry

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## What is Time Study? This was originally proposed by Frederick Taylor and was later modified to include a performance rating (PR) adjustment. The time study is considered to be one of the most-widely used means of work-measurement. Basically, by using time study, an analyst will be taking a small sample of a single worker's activity and using it to derive a standard for tasks of that nature. The steps of time study are as follows: Step-1: First select the job to be studied. Breakdown the work content of the job into smallest possible elements. Then inform the worker and define the best method. Step-2: Observe the time for appropriate number of cycles (such as 25 to 50). Step-3: Determine the average cycle time (CT) CT=(∑Times)/(No.of cycles) Step-4: Determine- the normal time (NT). NT = CT (PR) Where, PR is the performance rating Step-5: Determine the standard time using the following formula. ST = NT (AF) Where AF =1/(1-% Allowance) AF being the allowance factor In step-2, n can be calculated using the following formula. Where AF =(Z2 [n1 ∑x2-(∑x)2]/h2 (∑x)2 Where, n1 = Preliminary sample size x = recorded stopwatch times h = half the precision interval in % (for example, if the precision interval is ±5%, then h = 0.05) z = the standard normal statistic for the desired confidence level. ## Email Based, Online Homework Assignment Help in Time Study Transtutors has a team of highly skilled and certified tutors who can clear your doubt regarding Time Study. You can submit your school, college or university level homework or assignment to us and we will make sure that you get the answers you need which are timely and also cost effective Also you can directly interact with our tutors for a one to one session and get answers to all your Production and operation management. ## Related Topics All Management Topics More Q&A

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Community Profile # ilanqa89 22 total contributions since 2015 View details... Contributions in View by Solved Remove any row in which a NaN appears Given the matrix A, return B in which all the rows that have one or more <http://www.mathworks.com/help/techdoc/ref/nan.html NaN... 10달 ago Solved Circle area using pi Given a circle's radius, compute the circle's area. Use the built-in mathematical constant pi. 4년 이하 ago Solved Check if sorted Check if sorted. Example: Input x = [1 2 0] Output y is 0 4년 이하 ago Solved Weighted average Given two lists of numbers, determine the weighted average. Example [1 2 3] and [10 15 20] should result in 33.333... 4년 이하 ago Solved Create times-tables At one time or another, we all had to memorize boring times tables. 5 times 5 is 25. 5 times 6 is 30. 12 times 12 is way more th... 4년 이하 ago Solved Swap the first and last columns Flip the outermost columns of matrix A, so that the first column becomes the last and the last column becomes the first. All oth... 4년 이하 ago Solved Column Removal Remove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; and ... 4년 이하 ago Solved Select every other element of a vector Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s... 4년 이하 ago Solved Determine if input is odd Given the input n, return true if n is odd or false if n is even. 4년 이하 ago Solved How to find the position of an element in a vector without using the find function Write a function posX=findPosition(x,y) where x is a vector and y is the number that you are searching for. Examples: fin... 4년 이하 ago Solved Lightning strike distance: Writing a function Write a function named LightningDistance that outputs "distance" given input "seconds". Seconds is the time from seeing lightnin... 4년 이하 ago Solved Sum all integers from 1 to 2^n Given the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10. 4년 이하 ago Solved Fibonacci sequence Calculate the nth Fibonacci number. Given n, return f where f = fib(n) and f(1) = 1, f(2) = 1, f(3) = 2, ... Examples: Inpu... 4년 이하 ago Solved Determine whether a vector is monotonically increasing Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f... 4년 이하 ago Solved Make a checkerboard matrix Given an integer n, make an n-by-n matrix made up of alternating ones and zeros as shown below. The a(1,1) should be 1. Example... 4년 이하 ago Solved Find all elements less than 0 or greater than 10 and replace them with NaN Given an input vector x, find all elements of x less than 0 or greater than 10 and replace them with NaN. Example: Input ... 4년 이하 ago Solved Triangle Numbers Triangle numbers are the sums of successive integers. So 6 is a triangle number because 6 = 1 + 2 + 3 which can be displa... 4년 이하 ago Solved Reverse the vector Reverse the vector elements. Example: Input x = [1,2,3,4,5,6,7,8,9] Output y = [9,8,7,6,5,4,3,2,1] 4년 이하 ago Solved Times 2 - START HERE Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:... 4년 이하 ago Solved Given a and b, return the sum a+b in c. 4년 이하 ago Solved Find the sum of all the numbers of the input vector Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ... 4년 이하 ago Solved Project Euler: Problem 10, Sum of Primes The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below the input, N. Thank you <http:/... 4년 이하 ago

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Instructions Question Stem Consider the lines $$L_1$$ and $$L_2$$ defined by $$L_1 : x \sqrt{2} + y - 1 = 0$$ and $$L_2 : x \sqrt{2} - y + 1 = 0$$ For a fixed constant $$\lambda$$, let C be the locus of a point P such that the product of the distance of P from $$L_1$$ and the distance of P from $$L_2$$ is $$\lambda^2$$. The line $$y = 2x + 1$$ meets C at two points R and S, where the distance between R and S is $$\sqrt{270}$$. Let the perpendicular bisector of RS meet C at two distinct points $$R′$$ and $$S′$$. Let D be the square of the distance between $$R′$$ and $$S′$$. Question 9

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+0 # Pre Calc 0 109 1 +414 A fruit stand sells two varieties of strawberries: standard and deluxe. A box of standard strawberries sells for \$6, and a box of deluxe strawberries sells for \$10. In one day the stand sold 110 boxes of strawberries for a total of \$820. How many boxes of each type were sold? standard strawberries   ___  boxes deluxe strawberries _____    boxes Dec 9, 2019 #1 +109064 +1 Similar to the last problem.....we have this equation (N)(6)   +  ( 110 - N)(10)   = 820 6N + 1100 - 10N  =  820 -4N + 1100  = 820           subtract  1100 from both  sides -4N  = -280      divide both sides by  -4 N  = 70  =  number of boxes of standards 110 - N   = 110  - 70  =  40  = number of boxes of deluxes Dec 9, 2019

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https://fr.slideserve.com/nubia/how-do-we-start-this-proof-assume-a-n-is-a-subgroup-of-s-n-assume-o-s-n-n-nonempty

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How do we start this proof? Assume A n is a subgroup of S n .  Assume o(S n ) = n! Nonempty: # How do we start this proof? Assume A n is a subgroup of S n .  Assume o(S n ) = n! Nonempty: Télécharger la présentation ## How do we start this proof? Assume A n is a subgroup of S n .  Assume o(S n ) = n! Nonempty: - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. How do we start this proof? • Assume An is a subgroup of Sn. •  • Assume o(Sn) = n! • Nonempty: 2. What is the next step in the proof? • Closed: • Show An is a subgroup of Sn. • (c) Assume o(Sn) = n! • (d) Nonempty: 3. What is the next step in the proof? • Let a, b  H (b) Let f, g  An • (c) Let f, g  Sn (d) Let a*b  H • (e) Let f  g  An (f) Let f  g  Sn 4. What is the next step in the proof? • Identity: • Associative: • (c) Inverses: • (d) Nonempty: 5. What is the next step in the proof? • Let a  H (b) Let f  An • (c) Let f  Sn (d) Let a-1 H • (e) Let f -1  An (f) Let f -1  Sn 6. How many elements does S5 have? • 5 (b) 10 (c) 20 (d) 60 • (e) 120 (f) 200 (g) 500 (h) 546 7. How many elements does A5 have? • 5 (b) 10 (c) 20 (d) 60 • (e) 120 (f) 200 (g) 500 (h) 546

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# Theory Central_Limit_Theorem theory Central_Limit_Theorem imports Levy ```(* Title: HOL/Probability/Central_Limit_Theorem.thy Authors: Jeremy Avigad (CMU), Luke Serafin (CMU) *) section ‹The Central Limit Theorem› theory Central_Limit_Theorem imports Levy begin theorem (in prob_space) central_limit_theorem_zero_mean: fixes X :: "nat ⇒ 'a ⇒ real" and μ :: "real measure" and σ :: real and S :: "nat ⇒ 'a ⇒ real" assumes X_indep: "indep_vars (λi. borel) X UNIV" and X_integrable: "⋀n. integrable M (X n)" and X_mean_0: "⋀n. expectation (X n) = 0" and σ_pos: "σ > 0" and X_square_integrable: "⋀n. integrable M (λx. (X n x)⇧2)" and X_variance: "⋀n. variance (X n) = σ⇧2" and X_distrib: "⋀n. distr M borel (X n) = μ" defines "S n ≡ λx. ∑i<n. X i x" shows "weak_conv_m (λn. distr M borel (λx. S n x / sqrt (n * σ⇧2))) std_normal_distribution" proof - let ?S' = "λn x. S n x / sqrt (real n * σ⇧2)" define φ where "φ n = char (distr M borel (?S' n))" for n define ψ where "ψ n t = char μ (t / sqrt (σ⇧2 * n))" for n t have X_rv [simp, measurable]: "⋀n. random_variable borel (X n)" using X_indep unfolding indep_vars_def2 by simp interpret μ: real_distribution μ by (subst X_distrib [symmetric, of 0], rule real_distribution_distr, simp) (* these are equivalent to the hypotheses on X, given X_distr *) have μ_integrable [simp]: "integrable μ (λx. x)" and μ_mean_integrable [simp]: "μ.expectation (λx. x) = 0" and μ_square_integrable [simp]: "integrable μ (λx. x^2)" and μ_variance [simp]: "μ.expectation (λx. x^2) = σ⇧2" using assms by (simp_all add: X_distrib [symmetric, of 0] integrable_distr_eq integral_distr) have main: "∀⇩F n in sequentially. cmod (φ n t - (1 + (-(t^2) / 2) / n)^n) ≤ t⇧2 / (6 * σ⇧2) * (LINT x|μ. min (6 * x⇧2) (¦t / sqrt (σ⇧2 * n)¦ * ¦x¦ ^ 3))" for t proof (rule eventually_sequentiallyI) fix n :: nat assume "n ≥ nat (ceiling (t^2 / 4))" hence n: "n ≥ t^2 / 4" by (subst nat_ceiling_le_eq [symmetric]) let ?t = "t / sqrt (σ⇧2 * n)" define ψ' where "ψ' n i = char (distr M borel (λx. X i x / sqrt (σ⇧2 * n)))" for n i have *: "⋀n i t. ψ' n i t = ψ n t" unfolding ψ_def ψ'_def char_def by (subst X_distrib [symmetric]) (auto simp: integral_distr) have "φ n t = char (distr M borel (λx. ∑i<n. X i x / sqrt (σ⇧2 * real n))) t" by (auto simp: φ_def S_def sum_divide_distrib ac_simps) also have "… = (∏ i < n. ψ' n i t)" unfolding ψ'_def apply (rule char_distr_sum) apply (rule indep_vars_compose2[where X=X]) apply (rule indep_vars_subset) apply (rule X_indep) apply auto done also have "… = (ψ n t)^n" by (auto simp add: * prod_constant) finally have φ_eq: "φ n t =(ψ n t)^n" . have "norm (ψ n t - (1 - ?t^2 * σ⇧2 / 2)) ≤ ?t⇧2 / 6 * (LINT x|μ. min (6 * x⇧2) (¦?t¦ * ¦x¦ ^ 3))" unfolding ψ_def by (rule μ.char_approx3, auto) also have "?t^2 * σ⇧2 = t^2 / n" using σ_pos by (simp add: power_divide) also have "t^2 / n / 2 = (t^2 / 2) / n" by simp finally have **: "norm (ψ n t - (1 + (-(t^2) / 2) / n)) ≤ ?t⇧2 / 6 * (LINT x|μ. min (6 * x⇧2) (¦?t¦ * ¦x¦ ^ 3))" by simp have "norm (φ n t - (complex_of_real (1 + (-(t^2) / 2) / n))^n) ≤ n * norm (ψ n t - (complex_of_real (1 + (-(t^2) / 2) / n)))" using n by (auto intro!: norm_power_diff μ.cmod_char_le_1 abs_leI simp del: of_real_diff simp: of_real_diff[symmetric] divide_le_eq φ_eq ψ_def) also have "… ≤ n * (?t⇧2 / 6 * (LINT x|μ. min (6 * x⇧2) (¦?t¦ * ¦x¦ ^ 3)))" by (rule mult_left_mono [OF **], simp) also have "… = (t⇧2 / (6 * σ⇧2) * (LINT x|μ. min (6 * x⇧2) (¦?t¦ * ¦x¦ ^ 3)))" using σ_pos by (simp add: field_simps min_absorb2) finally show "norm (φ n t - (1 + (-(t^2) / 2) / n)^n) ≤ (t⇧2 / (6 * σ⇧2) * (LINT x|μ. min (6 * x⇧2) (¦?t¦ * ¦x¦ ^ 3)))" by simp qed show ?thesis proof (rule levy_continuity) fix t let ?t = "λn. t / sqrt (σ⇧2 * n)" have "⋀x. (λn. min (6 * x⇧2) (¦t¦ * ¦x¦ ^ 3 / ¦sqrt (σ⇧2 * real n)¦)) ⇢ 0" using σ_pos by (auto simp: real_sqrt_mult min_absorb2 intro!: tendsto_min[THEN tendsto_eq_rhs] sqrt_at_top[THEN filterlim_compose] filterlim_tendsto_pos_mult_at_top filterlim_at_top_imp_at_infinity tendsto_divide_0 filterlim_real_sequentially) then have "(λn. LINT x|μ. min (6 * x⇧2) (¦?t n¦ * ¦x¦ ^ 3)) ⇢ (LINT x|μ. 0)" by (intro integral_dominated_convergence [where w = "λx. 6 * x^2"]) auto then have *: "(λn. t⇧2 / (6 * σ⇧2) * (LINT x|μ. min (6 * x⇧2) (¦?t n¦ * ¦x¦ ^ 3))) ⇢ 0" by (simp only: integral_zero tendsto_mult_right_zero) have "(λn. complex_of_real ((1 + (-(t^2) / 2) / n)^n)) ⇢ complex_of_real (exp (-(t^2) / 2))" by (rule isCont_tendsto_compose [OF _ tendsto_exp_limit_sequentially]) auto then have "(λn. φ n t) ⇢ complex_of_real (exp (-(t^2) / 2))" by (rule Lim_transform) (rule Lim_null_comparison [OF main *]) then show "(λn. char (distr M borel (?S' n)) t) ⇢ char std_normal_distribution t" qed (auto intro!: real_dist_normal_dist simp: S_def) qed theorem (in prob_space) central_limit_theorem: fixes X :: "nat ⇒ 'a ⇒ real" and μ :: "real measure" and σ :: real and S :: "nat ⇒ 'a ⇒ real" assumes X_indep: "indep_vars (λi. borel) X UNIV" and X_integrable: "⋀n. integrable M (X n)" and X_mean: "⋀n. expectation (X n) = m" and σ_pos: "σ > 0" and X_square_integrable: "⋀n. integrable M (λx. (X n x)⇧2)" and X_variance: "⋀n. variance (X n) = σ⇧2" and X_distrib: "⋀n. distr M borel (X n) = μ" defines "X' i x ≡ X i x - m" shows "weak_conv_m (λn. distr M borel (λx. (∑i<n. X' i x) / sqrt (n*σ⇧2))) std_normal_distribution" proof (intro central_limit_theorem_zero_mean) show "indep_vars (λi. borel) X' UNIV" unfolding X'_def[abs_def] using X_indep by (rule indep_vars_compose2) auto show "integrable M (X' n)" "expectation (X' n) = 0" for n using X_integrable X_mean by (auto simp: X'_def[abs_def] prob_space) show "σ > 0" "integrable M (λx. (X' n x)⇧2)" "variance (X' n) = σ⇧2" for n using ‹0 < σ› X_integrable X_mean X_square_integrable X_variance unfolding X'_def by (auto simp: prob_space power2_diff) show "distr M borel (X' n) = distr μ borel (λx. x - m)" for n unfolding X_distrib[of n, symmetric] using X_integrable by (subst distr_distr) (auto simp: X'_def[abs_def] comp_def) qed end ```

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Civil Engineering Mechanical Engineering Chemical Engineering Networking Database Questions Computer Science Basic Electronics Digital Electronics Electronic Devices Circuit Simulation Electrical Enigneering Engineering Mechanics Technical Drawing # Placement Papers - Caritor @ : Home > Placement Papers > Caritor > View Paper CARITOR PAPER - 2004 Rated : +0 , -0 1.a cube object 3" * 3" * 3" is painted with green in all the outer surfaces. If the cube is cut into cubes of 1"*1"*1", how many 1" cubes will have at least one surface painted. a. 8     b.26      c.27       d. none ans.b 2. Single table tennis tournament is held at IT solutions in which 32 players participated. If a single playe eliminated as soon as the player loses a match. How many matches are required to determine the winner. a. 32      b.    16         c. 31        d. 15 ans.c 3. There are 200 employees in a company. An external vender is chosen to serve coffee twice a day.100coffee cups were offered by the company , but as an incentive to have the cups in fact at the end of the day, the company offered 30 paise for every cup remained safely and charged 90 paise for every broken cup. At the end  of  the day ,  the vender received RS.24 . how many cups did the vender break. a. 20 b.5 c.10 d.14 ans.c 4. A box contains 16 balls of 4 different colors green blue yellow &red each. if you were to close your eyes and pick them at random, how many marbles must you take out to be sure that there at least two of one colour  among  the marbles picked out. A. 4 b. 5 c. 6 d. 14 ans.b 5. If 8 tyres were used on a bus (6 tyres ) which has traveled 16000 km , how many km did each tyre sustain. if  all the tyres were used equally in sustaining this distance . a. 2000 b.16000 c.12000 d.10000  ans.c 6. A company purchased 3 computer tables in 1995. as the company wanted to renovate the office , sold those tables at RS.2400 each making a profit of 20% of one, no profit on second table and 20%loss on third table. What is the company get in this transaction. A. no loss no profit b.RS.200 loss c.RS.800profit d. RS.400 loss ans.b 7. A farmer owns a square field of side with a pole in one of the corners to which he tied his cow with a rope whose    length is about is 10 m . what is the area available for the cow to grace .assume pi=3. A.150 sq.m          b.125sq,m         c.75sq.m         d. not enough data.      ans.c 8.The average of x & y is 12.if z=9 what is the          average of x ,y, z a.11 b.6.5 c.5 d. not enough data ans. a 9.In a certain shop note books that normally sell for 59 cents each or on sale at 2 for 99cents.how can be saved by purchasing 10 of these note books at the sale price. a.\$0.85      b.\$1.0         c.\$0.95         d.\$1.15      ans. c 10.The cost in \$ of manufacturing x fridges is 9000+400x . the amount received when selling these x fridges is 500x  \$, what is the least no of fridges that must be manufactured & sold so that the amount received is at least   equal to the manufacturing cost. a. 10 b.18 c.15 d.90 ans. d 11 The sides of the right triangular field containing the right angle are x &x+10. its area is 5500sq.m.the equation to determine is a. x(x+10)=5500        b. x(x+10)=2750       c. x(x+10)=11000     d. x(x+20)=5500 ans.c 12.The length and breadth of a rectangular plot are in the ratio of 7:5. if the length is reduced by 5 m& breadth is increased by 2 m then the area is reduced by 65 sq.m. the length and breadth of the rectangular plot are a.25,35           b.21,15          c.35,25          d.49,35 ans.c 13. 6 men earn as much as 8 women ,two women earn as much as 3 boys&4 boys earn as much as 5 girls. if a girl earns RS.50 a day then the earning of the man would be a.115          b.125         c.135       d.150 ans.b 14. a & b can separately do a piece of work in 10 & 15 days respectively. They work together for sometimes and b stops. If a completes the rest of work in 5 days, then b has worked for a.5    b.4      c.3         d.2 (days). Ans.c 15. A Farmer owns a square land of 15 m each side with a pole in one of the corners to which he tied his cow with a  rope whose length is about 10 m. What is the area available for the cow to graze. (Assume pi = 3) a)  125          sq. m b)  150          sq.m c)             Not enough data d)    75 sq.          m Ans.75 16.Five friends Lokesh, Manoj, Neeraj, Raveesh and Rohit have agreed to work together on a part-time job offered by a local restaurant. The restaurant works five days a week and this group has the following schedule  when they can work: 1. Neeraj and Raveesh can work on Monday, Tuesday and Wednesday 2. Raveesh and Rohit can work on Monday, Wednesday and Thursday 3. Rohit and Lokesh can work on Monday, Friday and Thursday 4. Lokesh and Manoj can work on Friday, Tuesday and Thursday 5. Neeraj and Manoj can work on Friday, Tuesday and Wednesday Which one of the five friends cannot work on Thursdays? 17.A Fraction has the denominator greater than its numerator by 4. But if you add 10 to the denominator, the value of the fraction would then become 1/8. What is the fraction. 1/5,3/7,1/3,2/6 18. Concentrations of three wines A, B and C are 10, 20 and 30 percent respectively. They are mixed in the ratio 2 : 3 : x resulting in a 23% concentration solution. Find x. Ans.4 19.  In the figure below, marked triangles are equilateral congruent triangles. If the area of the shaded region is A, Find the remaining area 1.A(3+4√2) 2.A(3+4√2) 3.6A 4.A(3+2√3)            Ans:A 20 .            Four towers in a commercial complex are connected by a network of escalators in the manner shown below. Determine the number of ways you can go from one tower to the other and back, without using the same escalator more than once and not using any other means of transportation? Ans:10 A partly true or follows logically. B partly untrue or opposite follows logically            . C            can't say anything The big economic difference between nuclear and fossil fueled power stations is that the nuclear reactors are more expensive to build and decommission, but cheaper to run. So disputes over relatively of the two systems revolved not just around the prizes of coal and uranium today and tomorrow, but also around the way in which the future income should be compared with income. 7. the main difference between nuclear and fossil          fueled is an economic          one. A B C 8. the price of coal is not relevant to discussions to          about the efficiency of nuclear          reactors. A B C 9. if nuclear reactors were cheaper to build and decommission than fossil fueled power stations, they could definitely have economic advantage. A B  At any given moment are being bombarded by physical and psychological stimuli computing for one attention. Although our eyes are capable of handling more than 5 millions of data per sec, our brains are capable of interpreting only about 500 bits per sec. with similar disparities between other senses and brain it is easy to see that select visual, auditory or tactile stimuli that we wish to compute at any specific time. 10.physical stimuli usually win the competition for          our attention. A B C 11.the capacity of human brain is sufficient to interpret nearly all the stimuli the senses can register under optimum condition. A B C 12.eyes are able to hope with greater input of          information than ears. A B C 15. A farmer owns a square field of side with a pole in one of the corners to which he tied his cow with a rope whose length is about is 10 m . what is the area available for the cow to grace .assume pi=3. A.150 sq.m b.125sq,m c.75sq.m d. not          enough data. ans.c 16  The average of x & y is 12.if z=9 what          is the average of x ,y, z a.11 b.6.5          c.5 d. not enough data ans.a 17.In a certain shop note books that normally sell for 59 cents each or on sale at 2 for 99cents.how can be saved  by purchasing 10 of these note books at the sale price. a.\$0.85 b.\$1.0          c.\$0.95 d.\$1.15 ans. c 18.The cost in \$ of manufacturing x fridges is 9000+400x . the amount received when selling these x fridges is 500x   \$ , what is the least no of fridges that must be manu-factured & sold so that the amount received is at least equal  to the manufacturing cost. a. 10 b.18 c.15 d.90 ans. d 19.The sides of the right triangular field containing the right angle are x &x+10. its area is 5500sq.m.the equation to determine is a. x(x+10)=5500       b. x(x+10)=2750     c. x(x+10)=11000   d. x(x+20)=5500 ans.c 20.The length and breadth of a rectangular plot are in the ratio of 7:5. if the length is reduced by 5 m& breadth is  increased by 2 m then the area is reduced by 65 sq.m. the length and breadth of the rectangular plot are a.25,35         b.21,15         c.35,25        d.49,35 ans.c 21.6 men earn as much as 8 women ,two women earn as much as 3 boys&4 boys earn as much as 5 girls. if a girl earns RS.50 a day then the earning of the man would be a.115         b.125         c.135        d.150 ans.b 22.a & b can separately do a piece of work in 10 & 15 days respectively. They work together for sometimes and b stops. If a completes the rest of work in 5 days ,then b has worked for a.5       b.4        c.3       d.2 (days). Ans.c 23.The question using the data from the table the table had the details of population, birth per 1000 populations, deaths per 1000 population, percentage of population etc, for different countries. 1. which country had          the highest no. of people aged 60 or          over Ans. A 2. how many like          births occurred in 1985 in Spain and Italy .Ans          C 3. what was the net effect on the UK population of like birth and death rates in 1985. ans B Next was a quest   from the data from the graph Graph related to production in 1000 of month Ans: 1.D 2.D 3.C Then there was a question relative to the diagrams following logical diagrams From Edward Thorpe Ans: 30. D 31. B 32. D 33. A 34. B 35. D Numerical Ability The cost, in dollars, of manufacturing x refrigerators is 9,000 + 400x. The amount received when selling these x refrigerators is 500x dollars. What is the least number of refrigerators that must be manufactured and sold so that the amount received is at least equal to the manufacturing cost? 10 18 90 50 A Fraction has the denominator greater than its numerator by 4. But if you add 10 to the denominator, the A - If you think the statement is patently true or follows logically given the information or opinions contained in the passage Select B - If the statement is patently untrue or the opposite follows logically, given the information or opinions contained in the passage Select C - If you cannot say whether the statement is true or untrue or follows logically without further information The big economic difference between nuclear and fossil-fueled power stations is that the nuclear reactors are more expensive to build and decommission, but cheaper to run. So dispute over the relative efficiency of the two systems revolve not just around the prices of coal and uranium today and tomorrow, but also around the way in which the future income value of the fraction would then become 1/8. What is the fraction. 1/5 3/7 1/3 2/6 Verbal Ability Select           should be compared with current income. If nuclear reactors were cheaper to build and decommission than fossil fuelled power stations, they would definitely have the economic advantage. In this test, you are given two words outside the brackets. You have to fill in the brackets with a word which has a similar meaning (maybe in a different context) as that of the words on the right and left hand side of the brackets. FINAL ( ) ULTIMATE end Last Finish Quantitative Ability A takes 4 days to finish a job and B takes 5 days. If both of them work together on the same job, what proportion of the work is done by A? 5/9 4/9 6/9 7/9 Like this?   +0   -0

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# Does FOL extended with least-fixed points satisfy the Compactness Theorem? I am aware that first-order logics (FOL) satisfies the compactness theorem. That is, if a FOL theory is insatisfiable, a finite subset of the axioms of such theory is insatisfiable too. Is it the case that FOL extended with least-fixed point (LFP) satisfies the compactness theorem too? • Chapter 4 of these lecture notes indicates that the answer is negative. Commented Sep 28, 2020 at 22:51 Least fixed point operators can let us define, for example, the standard part of a model of $$\mathsf{PA}$$. The standard part of a model $$\mathfrak{A}=(A;+^\mathfrak{A},\times^\mathfrak{A},0^\mathfrak{A},1^\mathfrak{A})$$ of $$\mathsf{PA}$$ is just the smallest subset of $$A$$ containing $$0^\mathfrak{A}$$ and closed under $$n\mapsto n+^\mathfrak{A}1^\mathfrak{A}$$. But this is exactly a least fixed point definition: the standard part of $$\mathfrak{A}$$ is the least fixed point of the operator $$\Phi(P): x=0\vee\exists y(P(y)\wedge y+1=x).$$ This in turn lets us pin down the standard model of $$\mathsf{PA}$$ up to isomorphism by the theory $$\mathsf{PA}$$ + "Every element is standard." But this contradicts compactness: no compact logic can ever pin down an infinite structure up to isomorphism (this is really just the observation that the upwards Lowenheim-Skolem theorem is just a compactness corollary). • @441Juggler Sorry, I just saw this. This is a standard trick for applying compactness - expand the language. Consider the language of $\mathsf{PA}$ augmented by a new constant symbol $c$, and let $T$ be the theory gotten by adding to $\mathsf{PA}$ each sentence of the form $c>\underline{n}$ for $n\in\mathbb{N}$, where $\underline{n}$ is the numeral corresponding to $n$ (so e.g. $\underline{3}$ is the term $(1+1)+1$). Then $T$ is finitely consistent, hence satisfiable, but if we let $M$ be a model of $T$ then $M$ (or rather, $M$'s reduct to the language of arithmetic) must be nonstandard. Commented Nov 12, 2020 at 16:14

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# L beam - Strange result I have the following system (at P and Q there is a ball joint; the length of L beam is l; the PB and PA forces are applied at l/2 and are known; Fz is not zero and is known): The system has 0 degrees of freedom because each ball joint removes three degrees of freedom (6-2*3=0). I replaced each ball joint with three unknown forces. The problem is that the second cardinal equation along x gives: $$-F_z \, l = 0$$ How can I solve the problem? Thank you very much for your time. • For planar force system you can write 3 equilibrium equations and only 3 reactions can be found this way. Your example is statically indeterminate. One of your bearings has to have only one reaction component, the other can have two for your problem to become determinate. May 23, 2018 at 18:35 • Hello @Katarina, my system is in the space and thus has 6 degrees of freedom. I wrote 6 equations, 3 for translation equilibrium and 3 for rotation equilibrium. The only equation which gives me problem is the rotational equation along x. Maybe do I miss somithing? Thank you. May 23, 2018 at 18:40 • I see, if this is 3D problem, you are missing PAz and PBz. May 23, 2018 at 18:49 • And you have - Fz x l+PAz x l/2 =0, or whichever direction of PAz you choose. May 23, 2018 at 18:53 • If points P and Q are ball joints and you have a nonzero force in the Z direction, what is preventing the model from rotating about the X axis through points P and Q? May 23, 2018 at 20:12 The system is still statically indeterminate, but not due to lack of DOF. You have a singularity. Having correct DOFs only assures you have a chance of being statically determinate, but it does not guarantee this condition. In this case, the singularity is because the beam is allowed to rotate about the x axis unless $F_z$ is an unknown reaction. As such, in this case, $F_zl = 0$ means $F_z = 0$. Here are your other equations, in no particular order: \begin{alignat}{4} \sum& F_x &&= A + PB_x + D + PA_x &&= 0 \\ \sum& F_y &&= B + PB_y + E + PA_y &&= 0 \\ \sum& F_z &&= C + G + (F_z = 0) &&= 0 \\ \sum& M_{y@Q} &&= Cl &&= 0 \\ \sum& M_{x@Q} &&= F_zl &&= 0 \\ \sum& M_{z@Q} &&= -Bl + PB_yl/2 - PA_xl/2 &&= 0 \end{alignat} We can immediately see $C = F_z = 0$. Because of $\sum F_z$, $G = 0$. We now have four unknowns and three equations. We can solve for $B$ immediately using: $$\sum M_{z@Q} = -Bl + PB_yl/2 - PA_xl/2 = 0$$ Rearranging, we can use this result for $B$ to solve for $E$ using: $$\sum F_y = B + PB_y + E + PA_y = 0$$ But the final equation is singular: $$\sum F_x = A + PB_x + D + PA_x = 0$$ This cannot be resolved using statics, but require static indeterminate methods. • Hello @Mark, your answer is excellent. Thank you so much. May 23, 2018 at 20:26

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Algebra Tutorials! Home Rational Expressions Graphs of Rational Functions Solve Two-Step Equations Multiply, Dividing; Exponents; Square Roots; and Solving Equations LinearEquations Solving a Quadratic Equation Systems of Linear Equations Introduction Equations and Inequalities Solving 2nd Degree Equations Review Solving Quadratic Equations System of Equations Solving Equations & Inequalities Linear Equations Functions Zeros, and Applications Rational Expressions and Functions Linear equations in two variables Lesson Plan for Comparing and Ordering Rational Numbers LinearEquations Solving Equations Radicals and Rational Exponents Solving Linear Equations Systems of Linear Equations Solving Exponential and Logarithmic Equations Solving Systems of Linear Equations DISTANCE,CIRCLES,AND QUADRATIC EQUATIONS Solving Quadratic Equations Quadratic and Rational Inequalit Applications of Systems of Linear Equations in Two Variables Systems of Linear Equations Test Description for RATIONAL EX Exponential and Logarithmic Equations Systems of Linear Equations: Cramer's Rule Introduction to Systems of Linear Equations Literal Equations & Formula Equations and Inequalities with Absolute Value Rational Expressions SOLVING LINEAR AND QUADRATIC EQUATIONS Steepest Descent for Solving Linear Equations The Quadratic Equation Linear equations in two variables Try the Free Math Solver or Scroll down to Resources! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: Our users: I love it. It is much easier to move around and the colors are easier on the eyes. H.M., Texas Algebrator is truly an educational software. My students feel at ease while using it. Its like having an expert sit next to you. T.P., New York Your program so far has been great. I purchased this software for my 13 yr. old daughter who is having some problems in math. We have a tutor coming over to the house and between your software and him she got her first "A" in a very hard chapter test. 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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 8-0. Introduction Introduction to Further Applications of Trigonometry class="introduction" class="key-equations" title="Key Equations"class="key-concepts" title="Key Concepts"class="review-exercises" title="Review Exercises"class="practice-test" title="Practice Test"class="try"class="section-exercises" <figure class="splash" id="Figure_08_00_001" style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 1; word-spacing: 0px; -webkit-text-stroke-width: 0px;"> <figcaption>General Sherman, the world’s largest living tree. (credit: Mike Baird, Flickr)</figcaption> </figure> The world’s largest tree by volume, named General Sherman, stands 274.9 feet tall and resides in Northern California.1 Just how do scientists know its true height? A common way to measure the height involves determining the angle of elevation, which is formed by the tree and the ground at a point some distance away from the base of the tree. This method is much more practical than climbing the tree and dropping a very long tape measure. In this chapter, we will explore applications of trigonometry that will enable us to solve many different kinds of problems, including finding the height of a tree. We extend topics we introduced in Trigonometric Functions and investigate applications more deeply and meaningfully. ## Footnotes 1. 1 Source: National Park Service. "The General Sherman Tree." http://www.nps.gov/seki/naturescience/sherman.htm. Accessed April 25, 2014.

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Tagged Questions 39 views Find the power series for a definite integral I am a bit unsure when integration is used together with summation. Here is my question: Find power series for $\int_0^{1} \frac{\sin x}{x}dx$ in the form $\sum_{k=1}^{\infty} a_kx^k$ Here is what I ... Express the indefinite integral $\int\sin x^2~dx$ as a power series What does this mean? I never saw this in my class/notes so I don't understand the conversion from integral to power series. Also if the integral were defined from $0$ to $1$, what new steps do I add? ... After some algebra, Wolfram online integrator gave me the following: \tag{1} \int (1-a-t)^{N-2}\ \sqrt{2t-t^2}\ \text{d} t = c\ \cdot t^{3/2}\ \operatorname{F}_1 \left( \frac{3}{2}; -\frac{1}{2}, ...

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# Tagged Questions 90 views ### When does finite presentability of the associated graded Lie algebra of a group imply the group is finitely presented? Let $G$ be a finitely generated group; let $L(G)$ denote the graded Lie algebra (over $\mathbb{Q}$) associated to the lower central series of $G$. I would like to know conditions for when the finite ... 267 views ### Lie algebraic Grassmannian Assume that $L$ is a Lie algebra structure on $\mathbb{R}^{n}$, and $1<k<n$ is given. We define $Gr(k,n)_{L}$, the space of all $k$ dimensional Lie subalgebra of $(\mathbb{R}^{n}, L)$. For ... 191 views ### Finding the 2nd homotopy group $\pi_2(G^\mathbb{C}/P)$ Let $G$ be a compact connected and simply connected Lie group and $G^\mathbb{C}$ be the complexification of Lie group (with is diffeomorphic with $G^\mathbb{C}\cong T^*G$) then I am looking for ... 254 views ### Lie algebras vs. graph complexes A ribbon graph is a graph in which every vertex has valence at least three and is equipped with a cyclic ordering of its adjacent half edges. The ribbon graph complex $\mathcal{G}_*$ is the chain ... 121 views ### About the Lie algebra of polyvector fields I would like to know if someone already did some computations of the group of Lie algebra automorphisms of the algebra of polyvector fields on $\mathbb{R}^n$ equiped with the Schouten bracket (or ... 118 views ### Linear independence in (graded) Lie algebras I asked a mixed-up version of this question earlier. The Lie algebras I have in mind are the homotopy Lie algebras of wedges of finitely many spheres (in dimensions greater than $1$). Thus each ... 100 views ### Maurer-Cartan elements of the extension of an $L_{\infty}$-algebra Let $g$ be a nilpotent $L_{\infty}$-algebra. For every commutative differential graded algebra $A$, one can form the extension $g\otimes A$ and endow it with a nilpotent $L_{\infty}$-algebra ... 735 views ### HIgher Homotopy Groups and Representation Theory Let $G$ be a compact Lie group, and $g$ its associated Lie algebra. In what ways do the higher homotopy groups $\pi_{n}(G)$ with $n>1$ appear in the representation theory of $G$? As an example, ... 233 views ### Stabilizers for Nilpotent Adjoint Orbits of Semisimple Groups Let $G$ be a connected, simply-connected, complex, semisimple Lie group with Lie algebra $\frak{g}$. Suppose that $X\in\frak{g}$ is a nilpotent element (ie. that $ad_X:\frak{g}\rightarrow\frak{g}$ is ... 340 views ### Proof for which primes H*G has torsion In 1960 Borel proved a beautiful result: Theorem. Let G be a simple, simply connected Lie group. Suppose that p is a prime that does not divide any of the coefficients of the highest root (expressed ... 1k views ### Weight lattice and the first fundamental group Let $G$ be a compact connected Lie group and let $T$ be a maximal torus of $G$, with Lie algebras $\frak{g}$ and $\frak{t}$ respectively. Then, $\frak{t}$ can be considered as a Cartan subalgebra of ... 719 views ### Torsion for Lie algebras and Lie groups This question is about the relationship (rather, whether there is or ought to be a relationship) between torsion for the cohomology of certain Lie algebras over the integers, and torsion for ... 292 views 2k views ### Cohomology of Lie groups and Lie algebras The length of this question has got a little bit out of hand. I apologize. Basically, this is a question about the relationship between the cohomology of Lie groups and Lie algebras, and maybe ...

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# Parallel methods and bounds of evaluating polynomials. • 42 Pages • 3.58 MB • English by Dept. of Computer Science, University of Illinois at Urbana-Champaign , Urbana Polynomials -- Data proces Classifications The Physical Object Series University of Illinois at Urbana-Champaign. Dept. of Computer Science. Report no. 437 LC Classifications QA76 .I4 no. 437, QA161 .I4 no. 437 Pagination v, 42 p. Open Library OL5396347M LC Control Number 72637327 An illustration of an open book. Books. An illustration of two cells of a film strip. Video. An illustration of an audio speaker. Audio. An illustration of a " floppy disk. Parallel methods and bounds of evaluating polynomials Item Preview remove-circle Share or Embed This : Numerical Methods for Roots of Polynomials - Part I (along with volume 2 covers most of the traditional methods for polynomial root-finding such as Newton’s, as well as numerous variations on them invented in the last few decades. Perhaps more importantly it covers recent developments such as Vincent’s method, simultaneous iterations, and matrix : \$ ReportNo.1*37 PARALLELMETHODSANDBOUNDS OFEVALUATINGPOLYNOMIALS by KiyoshiMaruyama March, DepartmentofComputerScience UniversityofIllinoisatUrbana. Perhaps more importantly it covers recent developments such as Vincent's method, simultaneous iterations, and matrix methods. There is an extensive chapter on evaluation of polynomials, including parallel methods and errors. There are pointers to robust and efficient programs. Horner's method is a fast, code-efficient method for multiplication and division of binary numbers on a microcontroller with no hardware of the binary numbers to be multiplied is represented as a trivial polynomial, where (using the above notation) =,x (or x to some power) is repeatedly factored out. In this binary numeral system (base 2), =, so powers of 2 are. PARALLEL POLYNOMIAL EVALUATION On the other hand, if the boundedness of the parallelism does not interfere with the computation, that is t ~. Numerical analysis - Numerical analysis - Approximation theory: This category includes the approximation of functions with simpler or more tractable functions and methods based on using such approximations. When evaluating a function f(x) with x a real or complex number, it must be kept in mind that a computer or calculator can only do a finite number of operations. Abstract. We consider the problem of fast parallel evaluation of multivariate polynomials over a field F. We define “maximal-degree” (max deg) of a multivariate polynomial f as max i i=1, first lower bound result states that if a circut G evaluates a multivariate polynomial f, where its nodes are capable of performing (+,*), then the depth(G) is not less than log 2 [max deg (f)]. The book introduces new techniques which imply rigorous lower bounds on the complexity of some number theoretic and cryptographic problems. These methods and techniques are based on bounds of character sums and numbers of solutions of some polynomial equations over finite fields and. We exhibit a new method for showing lower bounds for the time complexity of polynomial evaluation procedures. Time, denoted by L, is measured in terms of nonscalar arithmetic operations. Download Numerical Methods For Roots Of Polynomials Book For Free in PDF, EPUB. In order to read online Numerical Methods For Roots Of Polynomials textbook, you need to create a FREE account. Read as many books as you like (Personal use) and Join Over Happy Readers. We cannot guarantee that every book is in the library. We present a new method to obtain lower bounds for the time complexity of polynomial evaluation procedures. Time, denoted by L, is measured in terms of nonscalar arithmetic contrast with known methods for proving lower complexity bounds, our method is purely combinatorial and does not require powerful tools from algebraic or diophantine geometry. This paper presents a number of bounds on the parallel processor evaluation of arithmetic expressions. Several previous papers show that if the evaluation of an expression using a serial computer requires t operations, by using a number of processors in parallel, the expression may be evaluated in time proportional to \$\log _2 t\$. Since \$\log _2 t\$ is an obvious lower bound, it is of. () Polynomial optimization methods for determining lower bounds on decentralized assignability. 54th Annual Allerton Conference on Communication, Control, and Computing (Allerton), method was used by V. ### Description Parallel methods and bounds of evaluating polynomials. EPUB Strassen and S. Winograd for proving various other fundamental lower bounds in algebraic computing. b) I have accelerated polynomial evaluation by using preconditioning. The work is surveyed in my paper in Russian Math. Surveys,and in the book by D.E. Knuth, The. Book Review Polynomial and Matrix Computations Volume 1: Fundamental Algorithms by D. Bini and V. Pan publisher: Birkh/iuser, Boston, pages, ISBN Reviewed by: I. Emiris and A. Galligo INRIA Sophia-Antipolis This book presents algorithms for manipulating polynomials and matrices that are efficient in terms of asymptotic computational complexity; it includes sequential. In this paper, we give a new, simple, and efficient method for evaluating the pth derivative of the Jacobi polynomial of degree n. The Jacobi polynomial is written in terms of the Bernstein basis. In terms of com-plexity, our method induces a recursion tree of almost optimal size O(n log(nτ)), where n denotes the degree of the polynomial and τ the bitsize of its coefficients. The latter bound constitutes an im-provement by a factor of τ upon all existing subdivision methods for the task of. to obtain the characteristic polynomial in O(m. "log(m)) operations in K, which is in O(m!). Similarly, we consider a nondecreasing function d 7!M(d) and an algorithm which multiplies two polynomials in K[x] of degree at most d using at most M(d) operations in K; our algorithms rely on this subroutine for polynomial multiplication. HYAFIL, L., AN}) KUNG, H.T. Bounds on the speed-ups of parallel evaluation of recurrences. Second USA-Japan Computer Conference Proceedings,Google Scholar. overall layout of the book, e.g., in the restriction to classes of methods which are actually found in software packages, as well as in the treatment of individ- ual issues, like the predictor-corrector mode of linear multistep methods. RLaB implements horner's scheme for polynomial evaluation in its built-in function polyval. What is important is that RLaB stores the polynomials as row vectors starting from the highest power just as matlab and octave do. This said, solution to the problem is >> a = [6, -4, 7, ] 6 -4 7. Abstract. We exhibit a new method for showing lower bounds for the time complexity of polynomial evaluation procedures. Time, denoted by L, is measured in terms of nonscalar arithmetic time complexity function considered in this paper is L contrast with known methods for proving lower complexity bounds, our method is purely combinatorial and does not require. Figure shows this strategy graphically. One minor detail concerns degree-bounds. The product of two polynomials of degree-bound n is a polynomial of degree-bound 2n. Before evaluating the input polynomials A and B, therefore, we first double their degree-bounds to 2n. A Taylor polynomial approximates the value of a function, and in many cases, it’s helpful to measure the accuracy of an approximation. This information is provided by the Taylor remainder term: f(x) = Tn(x) + Rn(x) Notice that the addition of the remainder term Rn(x) turns the approximation into an equation. Here’s the formula for [ ]. The parallel evaluation of arithmetic expressions without division. IEEE Trans. Comput. C (May ), Google Scholar; 6 HOBBS, L. (Ed.) Parallel processor systems, technologies and applications. Spartan Books, New York, Google Scholar; 7 KNUTH, D.E. An empirical study of FORTRAN programs. Software i (April ), properties as the standard QR method can be challenging. The companion matrix for a polynomial p(x) expressed in the monomial basis can be decomposed as the sum of a unitary matrix and a rank one correction; hence, fast QR solvers for unitary-plus-rank-one structured matrices provide quadratic methods for computing roots of polynomials. The eigenvalue approach can be easily combined with other methods for the evaluation of the roots of polynomials (see the book [8] and the cited earlier papers). Those methods can be used in order to choose the initial approximation i. * for the shifted power method, to test the. Algebra is a branch of math in which letters and symbols are used to represent numbers and quantities in formulas and equations. ### Details Parallel methods and bounds of evaluating polynomials. FB2 The assemblage of printable algebra worksheets encompasses topics like translating phrases, evaluating and simplifying algebraic expressions, solving equations, graphing linear and quadratic equations, comprehending linear and quadratic functions, inequalities. @article{osti_, title = {A method for deriving lower bounds for the complexity of monotone arithmetic circuits computing real polynomials}, author = {Gashkov, Sergey B and Sergeev, Igor' S}, abstractNote = {This work suggests a method for deriving lower bounds for the complexity of polynomials with positive real coefficients implemented by circuits of functional elements over the. In probability theory, the Chernoff bound, named after Herman Chernoff but due to Herman Rubin, gives exponentially decreasing bounds on tail distributions of sums of independent random variables. It is a sharper bound than the known first- or second-moment-based tail bounds such as Markov's inequality or Chebyshev's inequality, which only yield power-law bounds on tail decay.Find an irreducible polynomial with a random search on each parallel kernel: Search for a Mersenne prime starting at a given prime exponent: Try different methods for .Calculus Precalculus: Mathematics for Calculus (Standalone Book) Upper and Lower Bounds Show that the given values for a and b are lower and upper bounds for the real zeros of the polynomial. P (x) = 3 x 4 − 17 x 3 + 24 x 2 − 9 x + 1; a = 0, b = 6.

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133 Reputation 17 years, 34 days Sometimes replacing a... Sometimes replacing a variable with another expression can help to simplify the expression. If I replace x in " f(x) for  x>=1 " using x=u+1, I get f(u+1), u>=0. When I then take d/dx f(x), and then set m=3 for example, simplify(eval(diff(f(x),x),m=3)); 3/8*(1-4*x+6*x^2-4*x^3+x^4)/x/(1+x)^4 Looking at this output I can't really see that it's positive for x>=1. But this is easy to see if I do x = u+1, where x>=1 means u>=0: simplify(eval(diff(f(u+1),u),m=3)); 3/8*u^4/(u+1)/(2+u)^4 Here it is easy to see that for m=3, d/du f(u+1)>=0 for u>=0 and so d/dx f(x)>=0 for x>=1. For m=4: simplify(eval(diff(f(u+1),u),m=4)); 1/4*(u^2-8*u-8)*u^2/(u+1)/(2+u)^4 so the derivative of f with m=4 is positive for x> 5+2*sqrt(6). Perhaps it helps to perform... Perhaps it helps to perform d/du f(u+1), u>=0. With this trick, I see that df/dx>0 for m=1,2,3. truncate numbers... "trunc" does it. trunc(-2.01);                  -2 trunc(2.01);                  2 Compare "floor": floor(2.01);                      2 floor(-2.01);                   -3 This looks like homework, so... This looks like homework, so you should better explain what you did so far and where you got stuck. Just some hint for maple: See ?limit - and don't forget the argument of "cos" and use some brackets around the denominator. It depends... In general: Try it with solve, e.g.: solve([r*phi=1,r*(1-phi^2)=0],[r,phi]); ` [[r = 1, phi = 1], [r = -1, phi = -1]]` If solve doesn't give a solution: this could have different reasons, depending on the equations. Maple can also give numeric solutions, see ?fsolve. procedure... Hi, First of all: LinearAlgebra (at least in Maple 10) has no command rowdim but RowDimension. A:= assigns a value to the input variable A (same for B) which doesn't make sense and causes an error "invalid left hand side in assignment". With A and B of type Matrix you don't need to convert them to arrays. The procedure should perform a matrix product, but your procedure only assigns a value for C[h,j] (the bottom right corner of the matrix C). mult := proc (A, B) local h, i, j: global C: h := LinearAlgebra[RowDimension](A); i := LinearAlgebra[ColumnDimension](A); j := LinearAlgebra[ColumnDimension](B); C:=Matrix(...) end proc: Some hint for 1... Instead of sum, add is better in some cases. But for the Fibonacci numbers you need a loop like ```for k from ... to ... do ... end``` arcsin... arcsin(W) does not evaluate to real values because W>1. arcsin(W/2) can be plotted. 2) The VariationalCalculus... 2) The VariationalCalculus package (maple 8 and 10 have it, so maple 9.5 should have it too) ```L:=m/2*(diff(x(t),t)^2+diff(y(t),t)^2)-k/2*x(t)^2-k2/2*y(t)^2; EulerLagrange(L,t,[x(t),y(t)]); / 2 \ / 2 \ |d | |d | {-k x(t) - m |--- x(t)|, -k2 y(t) - m |--- y(t)|, | 2 | | 2 | \dt / \dt / //d \2 /d \2\ 2 2 /d \2 /d \2 1/2 m ||-- x(t)| + |-- y(t)| | - 1/2 k x(t) - 1/2 k2 y(t) - |-- x(t)| m - |-- y(t)| m = K[1]} \\dt / \dt / / \dt / \dt /``` I don't know if it's allowed to use this package for homework though. linear eq. system... This is a system of 16 linear equations with 16 unknown variables. Solving with LinearAlgebra works better for those: ```with(LinearAlgebra): eqs:=[seq](eqn||i,i=1..16): (Mat,Vec):=GenerateMatrix(eqs,[a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p]); LinearSolve(Mat,Vec); Error, (in LinearAlgebra:-LA_Main:-LinearSolve) inconsistent system Rank(Mat); 15``` Connected points... In Maple 10 connected points with different colours can be produced with "display" using another "pointplot" (but without different coloured line segments): ```k := [[1, 5], [2, 4], [3, 7], [4, 8]];# points must be typed that way in Maple10 display(pointplot(k,color=[red,blue,magenta,pink]), pointplot(k,connect=true));``` solve nasty expr... With e. g. `assume(alpha::positive):` you can tell Maple that your parameters have to be positive. In this case, Maple apparently can't give symbolic solutions even with these assumptions. If you provide numeric values for alpha, kappa etc, you can plot it and find numeric solutions with fsolve (plotting helps to give the initial point). Example (without thinking of whether the parameters are realistic) expr:=eval(eq,{alpha=1,kappa[B]=7,kappa=3,mu=4,mu[B]=1}); plot(eq1,eta=0..1); ```fsolve(expr); 0. fsolve(expr,eta=0.9); 0.9487214131 ``` I don't know why... I don't know why but ```int(sqrt(t^5+6*t)*(5*t^4+6),t=0..u) 2/3*u*(u^4+6)*(u*(u^4+6))^(1/2) eval(int(sqrt(t^5+6*t)*(5*t^4+6),t=0..u),u=1);``` does evaluate to 14/3*7^(1/2) directly. Without Array... This solution doesn't use numpoints and arrays: (pds:-value(u,t=10))(7) returns a list [x = 7., t = 10., u = .34935] so I wrote  procedures for u and v: Ufunc:=l->eval(u,(pds:-value(u,t=10))(l)): Vfunc:=l->eval(v,(pds:-value(v,t=10))(l)): The parametric plot is now with(plots): plot([Ufunc,Vfunc,-10..10]); Thanks... evalc is good but the RealDomain solution has the disadvantage that it is set globally so everytime I must type :-sqrt(-3) instead of sqrt(-3), and use RealDomain in simplify(long real mess*exp(I*k*z)) end; was quite slow. For an expression like (long trig(theta) term)* exp(ikz) which I had just now I prefer Joe Riel's solution because it leaves the exp term as it is. 1 2 Page 1 of 2 

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## forums.naturalpoint.com Discussion and Support for the OptiTrack, SmartNav and TrackIR brands by NaturalPoint ## yaw-pitch-roll and quaternion inconsistent Hi. I am using Tracking Tool 2.3.3 and I find that the yaw-pitch-roll and quaternion are inconsistent. According to this post: https://forums.naturalpoint.com/viewtopic.php?f=56&t=7834&hilit=orientation, the last response on the first page by NaturalPoint-Brent at 4:23pm 2013 Wed Mar27, yaw is about y axis, pitch about z axis, and roll about x axis. The rotation matrix is computed by R = Rx*Ry*Rz. but when I compare the rotation matrix computed using quaternions, call it Rq, I find that Rq != R. The computation of rotation matrix from quaternion is from page 5 of paper https://www.atmos.colostate.edu/ECE555/reading/article_8.pdf, or page 496 of book "Principle of Robot Motion, Theory, Algorithms, and Implementation" by Howie Choset, Kevin Lynch et al. According to this post: https://forums.naturalpoint.com/viewtopic.php?f=56&t=8381, second response by beckdo at 4:01 2010 Tue Jul 06. The yaw-pitch-roll is computed by the quaternions. I believe the formula is right, because it is consistent with page 26 of this paper: https://www.astro.rug.nl/software/kapteyn/_downloads/attitude.pdf. But using this formula, the resulting yaw-pitch-roll is different from what is exported by TrackingTool. I used a single frame of the exported csv file. and I used Matlab to obtain the above result. Below is the code: Code: Select all `qx = -0.42941201;qy = 0.10853183;qz = -0.37245581;qw = 0.81553835;% Angle theta is in degrees% yaw is about y axisyaw = 35.40805435;  % pitch is about z axispitch = 30.9502964;% roll is about x axisroll = -65.63815308;%% Euler anglestheta_y = yaw * pi / 180theta_z = pitch * pi / 180theta_x = roll * pi / 180Ry = [  cos(theta_y)  0             sin(theta_y);        0             1             0           ;       -sin(theta_y)  0             cos(theta_y)];Rz = [  cos(theta_z) -sin(theta_z)  0           ;        sin(theta_z)  cos(theta_z)  0           ;        0             0             1           ];Rx = [  1             0             1           ;        0             cos(theta_x) -sin(theta_x);        0             sin(theta_x)  cos(theta_x)];     Rtheta =   Rx*Rz*RyRtheta_2 = Rx*Ry*RzRtheta_3 = Ry*Rz*Rx Rtheta_4 = Ry*Rx*RzRtheta_5 = Rz*Ry*RxRtheta_6 = Rz*Rx*Ry%% Quaternionsq0 = qw;q1 = qx; q2 = qy;q3 = qz;Rq11 = 2*(q0^2 + q1^2) - 1;Rq12 = 2*(q1*q2 - q0*q3);Rq13 = 2*(q1*q3 + q0*q2);Rq21 = 2*(q1*q2 + q0*q3);Rq22 = 2*(q0^2 + q2^2) - 1;Rq23 = 2*(q2*q3 - q0*q1);Rq31 = 2*(q1*q3 - q0*q2);Rq32 = 2*(q2*q3 + q0*q1);Rq33 = 2*(q0^2 + q3^2) - 1;Rq = [Rq11 Rq12 Rq13;      Rq21 Rq22 Rq23;      Rq31 Rq32 Rq33]  %% quaternion to Euler-angleq2y = atan2(-Rq13, Rq11)q2z = asin(Rq12)q2x = atan2(-Rq32, Rq22)` The resulting R=Rx*Ry*Rz is Code: Select all `Rtheta =    0.1196   -0.5143    1.3119   -0.3549    0.3538    0.8654   -0.6208   -0.7813    0.0648` and Rq from quaternion is Code: Select all `Rq =    0.6990    0.5143    0.4969   -0.7007    0.3538    0.6196    0.1429   -0.7813    0.6077` . As can be seen Rtheta != Rq. Moreover, no sequence of rotation results in the same rotation matrix as Rq: Code: Select all `Rtheta_2 =    0.2021   -0.1212    1.3944   -0.2405    0.6252    0.7425   -0.6735   -0.6583    0.3362Rtheta_3 =    0.6990   -0.7007    0.5561    0.5143    0.3538    1.2955   -0.4969   -0.6196    0.1108Rtheta_4 =    0.4275   -0.8718    1.0540    0.2121    0.3538    0.9110   -0.8787   -0.3388   -0.2432Rtheta_5 =    0.6990   -0.6648    0.4355    0.4192    0.0823    1.3233   -0.5794   -0.7425   -0.2432Rtheta_6 =    0.4735   -0.2121    0.8140   -0.3315    0.3538    1.3539   -0.2390   -0.9110    0.3362` The yaw-pitch-roll in radian are computed as Code: Select all `theta_y =    0.6180theta_z =    0.5402theta_x =   -1.1456` but the converted yaw-pitch-roll from quaternion returns: Code: Select all `q2y =   -0.6180q2z =    0.5402q2x =    1.1456` . I think the above suggests either quaternion or yaw-pitch-roll exported by TrackingTool is incorrect. Could Natural Point enlighten me how to handle this please? I would really appreciate it! I attach the Matlab code. Thank you! puffin Posts: 3 Joined: Fri Oct 16, 2015 7:54 am ## Re: yaw-pitch-roll and quaternion inconsistent Hello puffin, Thank you for using OptiTrack and thank you for reaching out to us regarding your questions. The software you are using, TrackingTools, is the early version of our software and is quite old. It has errors that have since been corrected, and you may be experiencing something related to this. Would it be possible for you to upgrade to the latest working version of Motive? Overall, it would be a good idea to upgrade to the current version of our software, Motive 1.8 Final. This would require the purchase of a license renewal, but would provide you with the benefits of the latest software including improved usability and data quality, as well as fixes for any problems that existed in the early software. You would also have access to all new releases within the year of your license renewal, so you would continue to benefit from the ongoing improvements being made to the software. Best regards, Steven -- Steven Andrews OptiTrack | Customer Support Engineer steven.andrews NaturalPoint Employee Posts: 434 Joined: Mon Jan 19, 2015 11:52 am ## Re: yaw-pitch-roll and quaternion inconsistent Hi Steven, Thank you for your suggestion. I will consider that. Still, I would like to know if I can trust either the quaternion or the yaw-pitch-roll, or neither. Do you have the answer to that? Thanks! puffin Posts: 3 Joined: Fri Oct 16, 2015 7:54 am ## Re: yaw-pitch-roll and quaternion inconsistent Update: Take away: The quaternion is correct. Minor finding: The yaw-pitch-roll is incorrect if yaw is about y axis, pitch is about z-axis, and roll is about x-axis, and the rotation is obtained by R = Rx*Rz*Ry. I verified the correctness of the quaternion by writing a simple animation in Matlab. puffin Posts: 3 Joined: Fri Oct 16, 2015 7:54 am ## Re: yaw-pitch-roll and quaternion inconsistent Hi puffin, I'm glad to hear our suggestion of creating a fresh Rigid Body and comparing the 0 rotations to your calculated rotations worked out. Thank you very much for sharing the results. Cheers, Steven -- Steven Andrews OptiTrack | Customer Support Engineer steven.andrews NaturalPoint Employee Posts: 434 Joined: Mon Jan 19, 2015 11:52 am

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# Slope of Perpendicular Lines – Definition, Formula, Examples, FAQs Home » Math Vocabulary » Slope of Perpendicular Lines – Definition, Formula, Examples, FAQs ## What Is the Slope of Perpendicular Lines? The slopes of two perpendicular lines are negative reciprocals of each other. In simple words, the product of the slopes of two perpendicular lines equals -1. Two lines are said to be perpendicular if they intersect each other at a right angle (90). ## Formula for Finding the Slope of Perpendicular Lines If the slopes of the two perpendicular lines are m1 and m2, then the relationship between the slopes is given by the formula $m_{1} \times m_{2} = -1$ ## Slope of Perpendicular Lines: Negative Reciprocals The slopes of perpendicular lines are negative reciprocals of each other. Thus, the slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line. A reciprocal is defined as the multiplicative inverse of a given number. The multiplicative inverse (or reciprocal) of a non-zero number “a” is written as $\frac{1}{a}$. To obtain the negative reciprocal of a particular value, just add a minus sign to its reciprocal. So, the negative reciprocal of “a” can be written as $\frac{-1}{a}$. ## How to Find the Slope of a Line Perpendicular to the Given Line If the slope of one line is known, the slope of the line perpendicular to it is calculated as the negative reciprocal of the first line. The general form of the equation of a line is given as $ax + by + c = 0$ Convert this equation into the slope-intercept form of the equation of a line. $y = \frac{-ax}{b} \;-\; \frac{c}{b}$ So, we get the slope for the line would be $m_{1} = -\;\frac{a}{b}$ The general formula for the slope of perpendicular lines is $m_{1} . m_{2} = -1$ $\Rightarrow -\;\frac{a}{b} . m_{2} = -\;1$ $\Rightarrow m_{2} = \frac{b}{a}$ Therefore, the slope of the perpendicular line would be $\frac{b}{a}$. Example: Find the slope of a line perpendicular to the line $y = -\;2x + 1$. Given line: $y = -\;2x + 1$. Slope of the given line $= m = -\;2$ Slopes of perpendicular lines are negative reciprocals of each other. Thus, slope of the line perpendicular to the given line is $-\;\frac{1}{-\;2} = \frac{1}{2}$. ## Derivation of the Formula for Finding the Slope of Perpendicular Lines The angle between the two lines with slopes m1 and m2 is obtained using the formula tan $\theta = \frac{m_{1}-m_{2}}{1 + m_{1} m_{2}}$ If two lines are perpendicular, then $\theta = 90^{\circ}$. tan $90^{\circ} = \frac{m_{1}-m_{2}}{1 + m_{1}m_{2}}$ Note that tan $90^{\circ}$ is not defined. $\infty = \frac{m_{1}-m_{2}}{1 + m_{1}m_{2}}$ Thus, $1 + m_{1} m_{2} = 0$ $m_{1} m_{2} = -1$ ## What Is the Slope of Parallel Lines and Perpendicular Lines? • If two lines are parallel, they have the same slope. In other words, the slopes of two parallel lines are equal. If lines l and m are parallel, such that the slope of line l is m1 and the slope of line m is m2, then $m_{1} = m_{2}$. • On the other hand, the slopes of two perpendicular lines are negative reciprocals of each other. If lines l and m are perpendicular, such that the slope of line l is m1 and the slope of line m is m2, then $m_{1} m_{2} = -\;1$ or $m_{1} = \frac{-1}{m_{2}}$ ## Facts about Slope of Perpendicular Lines • Perpendicular lines form right angles at the point they intersect. • If two lines are parallel, their slopes are equal. • If the slope of a line is m, the slope of the line perpendicular to it is -1m. • A vertical line line is always perpendicular to a horizontal line. ## Conclusion In this article, we learned how to find the slope of perpendicular lines, the associated formula, and its derivation. Let’s solve a few examples and practice MCQs on these concepts for better comprehension. ## Solved Examples on Slope of Perpendicular Lines Example 1: What will be the slope of the line  perpendicular to the line $6x – 2y = 4$? Solution: Given line: $6x – 2y – 4 = 0$ $6x – 2y = 4$ $\Rightarrow -\; 2y = 4\;-\;6x$ $\Rightarrow y = 3x\;-\;2$ Slope of the first line $= m_{1} = 3$ Slope of the perpendicular line $= \frac{-1}{m_{1}}$ $\Rightarrow m_{2} = -\;\frac{1}{3}$ Therefore, the slope of the perpendicular line would be $-\;\frac{1}{3}$. Example 2: What will be the equation of a line passing through the point (5, 2) and with the slope of the perpendicular line equal to -3? Solution: The slope of the perpendicular line is $m_{1} = \;-\;3$. From the formula of the slope of the perpendicular lines $m_{1} .m_{2} = -1$ $\Rightarrow -\;3 \times m_{2} = -\;1$ $m_{2} = \frac{1}{3}$ The line passes through the point $(x_{1},\; y_{1}) = (5,\; 2)$ and has slope $m = \frac{1}{3}$. To find the equation of the required line, we will use the slope-point form. $(y\;-\;y_{1}) = m(x\;-\;x_{1})$ $\Rightarrow (y\;-\;2) = \frac{1}{3} (x \;-\;5)$ $\Rightarrow 3(y\;-\;2) = 1(x\;-\;5)$ $\Rightarrow 3y\;-\;6 = x\;-\;5$ $\Rightarrow x\;-\;3y + 6\;-\;5 = 0$ $\Rightarrow x\;-\;3y +1 = 0$ Therefore, the equation of the line is $x\;-\;3y + 1 = 0$. Example 3: Find the slope of a line perpendicular to the line passing through points (1, 2) and (3, 4). Solution: Slope of a line passing through the points $(x_{1},\;y_{1})$ and $(x_{2},\;y_{2})$ is given by Slope $= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ Here, $(x_{1},\;y_{1}) = (1,\; 2)$ and $(x_{2},\;y_{2}) = (3,\; 4)$ Slope $= \frac{4 – 2}{3 – 1}$ $m = \frac{2}{2} = 1$ Slope of the required line $= \frac{-1}{m} = -\;1$ ## Practice Problems on Slope of Perpendicular Lines 1 ### What is the slope of the line perpendicular to the line whose slope is $\frac{6}{7}$? $\frac{6}{7}$ $-\;\frac{6}{7}$ $\frac{7}{6}$ $-\;\frac{7}{6}$ CorrectIncorrect Correct answer is: $-\;\frac{7}{6}$ $m_{1}.m_{2} = -\;1$ $\Rightarrow \frac{6}{7} \times m_{2} = -\;1$ $\Rightarrow m_{2} = -\;\frac{7}{6}$ 2 ### Find the slope of the line perpendicular to $3x \;-\;4y + 5 = 0$. $\frac{3}{4}$ $\frac{-4}{3}$ $\frac{4}{3}$ $\frac{-3}{4}$ CorrectIncorrect Correct answer is: $\frac{-4}{3}$ $3x \;-\;4y + 5 = 0 \;m_{1} = -\;\frac{a}{b} \Rightarrow m_{1} = -\;\frac{3}{-4} = \frac{3}{4}$. $\Rightarrow m_{1} m_{2} = -\;1$ $\Rightarrow \frac{3}{4} \times m_{2} = -\;1$ $\Rightarrow m_{2} = -\;\frac{4}{3}$ 3 ### Slopes of perpendicular lines are reciprocals multiplicative inverses additive inverses negative reciprocals CorrectIncorrect Correct answer is: negative reciprocals Slopes of perpendicular lines are negative reciprocals of each other. 4 ### If the slopes of two perpendicular lines are m and n, then m + n = 0 mn = 1 mn = -1 m - n = 0 CorrectIncorrect Correct answer is: mn = -1 If the slopes of two perpendicular lines are m and n, then $mn = -\;1$. ## Frequently Asked Questions about Slope of Perpendicular Lines Perpendicular lines do not have the same slope. The slopes of the perpendicular line are negative reciprocals of each other. The slope of parallel lines is equal to each other. Any lines that are parallel to each other have the same slope but different y-intercepts. If the product of slopes is -1, the two lines are perpendicular. The slopes of two perpendicular lines are negative reciprocals of each other. If the slope of a line is m, the slope of the line perpendicular to it is the negative reciprocal of m, given by $\frac{-1}{m}$.

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Types of Data Structure A data structure is basically used to collect and organize the data on which we are going to perform the operations. For example, we all know what an array is; it is one of the most-used data structures. We use arrays to collect similar data types and organize them in a linear structure. Similar to object-oriented programming, we use a class to store different kinds of data types and functions (methods). In programming, we require different types of data structures in order to make an effective program. In this post, we are going to discuss them. Let's begin! Types of Data Structures Data structures are broadly of two types: 1. Primitive 2. Non-primitive Primitive Data Structures Primitive data structures are further classified into the following types: 1. Integer 2. (Double) Floating-point Numbers 3. Character 4. Pointer What are Primitive Data Structures? These types of data structures are the basic data structures, and most of these are built-in in many high-level programming languages. Primitive data structures can be defined as data types present in the programming languages. The main property of a primitive data structure is they can not be further divided, they are elementary. Basic Types of Primitive Data Structures: • Integer: The integer holds all the integer values, basically all the positive and negative numbers without decimals. e.g. 11 • Floating Points Numbers: Floating point numbers also cover all the real numbers, positive fractions, as well as negative with the decimal point. E.g. = 11.0 • Character: It holds every character, may it be any number, alphabet, or any special symbol, but it would be represented inside the double or single inverted comma. eg = ’\$’ • Pointers: Pointers hold (point to) the memory address of the variables, pointers, etc. Non-Primitive Data Structures We can classify all non-primitive data structures into the following categories: 1. Linear Data Structures 1. Array 2. List 3. Stack 4. Queue 2. Non-Linear Data Structures 1. Tree 2. Graph 3. File Data Structures What are Non-Primitive Data Structures? Non-primitive data structures are the complex data structures we use in programming. Mostly, all the non-primitive data structures are user-defined data structures, though many languages provide built-in support for these data structures, and thus, they are considered user-defined data structures. They are derived from primitive data structures. Types of Non-Primitive Data Structures: As mentioned above, there are 3 types of non-primitive data structures: 1. Linear Data Structures In the linear data structure, elements are stored in a sequential manner. Myth-buster: their name doesn't say that they store elements in a linear or contiguous memory location. Types of Linear Data Structures The linear data structure is further divided into 4 categories: • Array: An array is a homogeneous collection of elements. In simple words, an array can store only similar data types at once. An array stores all the elements in a linear sequence and in a contiguous memory location. Due to storing elements in contiguous memory locations, the operation, such as retrieving data elements, is very easy. There are many disadvantages of an array, such as it can only store similar data types at once and is not that memory efficient when it comes to an arbitrary number of elements. • List: The functionality of a list is similar to an array. It also stores elements in a linear sequence. List, however, uses dynamic memory allocation to store its elements at different memory locations. Though elements are stored at different memory locations, all are arranged in a sequence and linked with one another. • Stack: A stack is similar to a list but follows the LIFO principle to store and retrieve elements. LIFO stands for Last In First Out, which means the element stored last in the stack would be retrieved first. To perform the LIFO principle, the stack uses two operations: push and pop. push() is used to insert an element in the stack, while pop() is used to retrieve one or many. • Queues: The queue data structure is just the opposite of the stack since it uses the FIFO principle. FIFO stands for First In, First Out. In a queue, the element stored first in the structure would be retrieved first. 2. Non-Linear Data Structures In a non-linear data structure, the storage of elements doesn't follow a sequential order. Types of Non-Linear Data Structures • Graph: Used for network representation. A graph data structure basically uses two components: vertices and edges. In the graph, edges are used to connect vertices. • Tree: The tree data structure uses a hierarchical form of structure to represent its elements. One of the famous tree data structures is the binary tree. A tree uses a node-like structure to make a hierarchical form, where each node represents a value. The uppermost node of the tree is known as the root node, and the bottom node is known as the leaf node. Tree data structures are the most complex and efficient data structures we use in programming. As such, we use these to solve many real-time problems. 3. File Data Structures A file is a collection of records. Using a file, we can store data in the .txt format. Many programming languages come with a built-in file-handling structure. We can perform read, write, and append operations on files so we can add and retrieve data from them. The file is the easiest method of creating and retrieving data, but it is not productive when it comes to handling a huge amount of data. Conclusion Data structures are the most important part of a programming language and, during tech interviews, the favorite topic of the interviewer. That's probably because the ability to choose the right data structure is one of the vital skills a programmer should possess. In a big project or furthering development, especially dealing with huge amounts of data, developers who have complete knowledge of all basic data structures are able to make the best decisions. We hope that the aforementioned information helps you gain clarity of the topic. Good luck!

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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Please make a donation to keep the OEIS running. We are now in our 56th year. In the past year we added 10000 new sequences and reached almost 9000 citations (which often say "discovered thanks to the OEIS"). Other ways to donate Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A097250 Smallest m such that A097249(m) = n; from n=1 onwards, twice the primorials, 2*A002110(n). 6 1, 4, 12, 60, 420, 4620, 60060, 1021020, 19399380, 446185740, 12939386460, 401120980260, 14841476269620, 608500527054420, 26165522663340060, 1229779565176982820, 65178316954380089460, 3845520700308425278140, 234576762718813941966540, 15716643102160534111758180 (list; graph; refs; listen; history; text; internal format) OFFSET 0,2 COMMENTS A097249(a(n))=n and A097249(m)=1. - G. C. Greubel, Apr 23 2017 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 FORMULA a(n) = if n=0 then 1 else 2*A002110(n). MATHEMATICA Join[{1}, 2 Denominator[Accumulate[1/Prime[Range[20]]]]] (* Vincenzo Librandi, Mar 25 2017 *) Join[{1}, 2*FoldList[Times, 1, Prime[Range[50]]]] (* G. C. Greubel, Apr 23 2017 *) CROSSREFS Cf. A002110, A088860, A097249. From a(1)=4 onwards, row 3 of A276945. Sequence in context: A276707 A083484 A088860 * A188328 A291772 A222645 Adjacent sequences:  A097247 A097248 A097249 * A097251 A097252 A097253 KEYWORD nonn AUTHOR Reinhard Zumkeller, Aug 03 2004 EXTENSIONS Name amended by Antti Karttunen, Sep 24 2016 a(18)-a(19) from Vincenzo Librandi, Mar 25 2017 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified November 27 05:27 EST 2020. Contains 338678 sequences. (Running on oeis4.)

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Games Problems Go Pro! ## Ask Professor Puzzler Do you have a question you would like to ask Professor Puzzler? Click here to ask your question! Shawna from Michigan asks, "The upper case sigma symbol is used for summation, but I've also seen the upper case pi symbol used in a similar way. What does that mean?" Hi Shawna. The answer is pretty short and straight-forward. The pi symbol, when used that way, means multiplying a set of numbers (instead of adding them). If you're wondering why pi is used, the reason is simple; it's the same reason we use sigma for sums. S for sigma, S for sum. P for Pi, P for product. Here's an example; take a look at this expression: This means that we multiply together all the numbers we get by plugging in x = 3, x = 4, x = 5, and x = 6. The numbers are: 3 - 2 = 1 4 - 2 = 2 5 - 2 = 3 6 - 2 = 4 So the product is 1 x 2 x 3 x 4 = 24 Of course, that was a fairly simple problem, and they can get a lot more complicated (interesting) than that. But that will be sufficient to answer your question, I think! Professor Puzzler # Blogs on This Site Reviews and book lists - books we love! The site administrator fields questions from visitors. Like us on Facebook to get updates about new resources

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# Mathematical Induction ## Prove that ∑(-1/2)^j = [2^(n+1) + (-1)^n]/3×2^n whenever n is a nonnegative integer In this case, we will use Mathematical Induction. PRINCIPLE OF MATHEMATICAL INDUCTION: “To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: BASIS STEP: We verify that P (1) is true. INDUCTIVE STEP: We show that the conditional statement P (k) → P […] ## a) Find a formula for 1/(1×2) + 1/(2×3) + 1/n(n+1) by examining the values of this expression for small values of n. b)Prove the formula you conjectured in part (a) Let’s solve this exercise! a) Find a formula by examining the values of this expression for small values of n n=0 it is undefined. n=1: 1/(1×2) = 1/2 n=2: 1/(1×2) + 1/(2×3) = 1/2 + 1/6 = 4/6=2/3 n=3: 1/2 + 1/6 +1/12 = 3/4 n=4: 3/4 +1/20 = 4/5 1/(1×2) + 1/(2×3) + 1/n(n+1) ## Prove that 2−2·7+2·7^2 −···+2(−7)^n =(1− (−7)^(n+1))/4 whenever n is a nonnegative integer To make this proof, we will use the principle of mathematical induction. PRINCIPLE OF MATHEMATICAL INDUCTION: “To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: BASIS STEP: We verify that P (1) is true. INDUCTIVE STEP: We show that the conditional statement ## Let P(n) be the statement that 1^3 +2^3 +···+n^3 = (n(n + 1)/2)^2 for the positive integer n PRINCIPLE OF MATHEMATICAL INDUCTION: “To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: BASIS STEP: We verify that P (1) is true. INDUCTIVE STEP: We show that the conditional statement P (k) → P (k + 1) is true for all positive ## a) Find a formula for 1/2 + 1/4 + 1/8 + … + 1/2^n by examining the values of this expression for small values of n. b) Prove the formula you conjectured in part (a). We will use the mathematical induction to answer b). So, let’s start with the definition of the principle of mathematical induction. PRINCIPLE OF MATHEMATICAL INDUCTION: “To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: BASIS STEP: We verify that P (1) is ## Prove that 3+3 · 5+3 · 5^2+···+3 · 5^n=3(5^(n+1)−1)/4 whenever n is a nonnegative integer PRINCIPLE OF MATHEMATICAL INDUCTION: “To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: BASIS STEP: We verify that P (1) is true. INDUCTIVE STEP: We show that the conditional statement P (k) → P (k + 1) is true for all positive ## Let P(n) be the statement that 1^2 +2^2 +···+n^2 = n(n + 1)(2n + 1)/6 for the positive integer n. PRINCIPLE OF MATHEMATICAL INDUCTION: “To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: BASIS STEP: We verify that P (1) is true. INDUCTIVE STEP: We show that the conditional statement P (k) → P (k + 1) is true for all positive ## There are infinitely many stations on a train route. Suppose that the train stops at the first station and suppose that if the train stops at a station, then it stops at the next station. Show that the train stops at all stations Let’s revisit the principle of mathematical induction. PRINCIPLE OF MATHEMATICAL INDUCTION: “To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: BASIS STEP: We verify that P (1) is true. INDUCTIVE STEP: We show that the conditional statement P (k) → P (k ## Prove that 1^2−2^2+3^2−···+(−1)^(n−1)n^2=(−1)^(n−1) n(n + 1)/2 whenever n is a positive integer We should always start with the definitions when solving exercises. So, let’s see the principle of mathematical induction. PRINCIPLE OF MATHEMATICAL INDUCTION To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: The definition above is from the textbook Discrete Mathematics and its ## a) Find a formula for the sum of the first n even positive integers. Prove the formula that you conjectured in part (a) After finding the formula in a), we will need to use induction to prove that it holds for the first n even positive numbers. So, let’s refresh the principle of mathematical induction. PRINCIPLE OF MATHEMATICAL INDUCTION To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we

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Bankers Discount Sample Questions And Answers #### Bankers Discount Sample Questions And Answers • Posted by  Pandit 29 Jan, 2012 Ex. 1. A bill for Rs. 6000 is drawn on July 14 at 5 months. It is discounted on 5th October at 10%. Find the banker's discount, true discount, banker's gain and the money that the holder of the bill receives. Sol. Face value of the bill = Rs. 6000. Date on which the bill was drawn = July 14 at 5 months. Nominally due date = December 14. Legally due date = December 17. Date on which the bill was discounted = October 5. Unexpired time : Oct. Nov. Dec. 26 + 30 + 17 = 73 days =1/ 5Years B.D. = S.I. on Rs. 6000 for 1/5 year = Rs. (6000 x 10 x1/5 x1/100)= Rs. 120. T.D. = Rs.[(6000 x 10 x1/5)/(100+(10*1/5))] =Rs.(12000/102)=Rs. 117.64. B.G. = (B.D.) - (T.D.) = Rs. (120 - 117.64) = Rs. 2.36. Money received by the holder of the bill = Rs. (6000 - 120) = Rs. 5880. Ex. 2. If the true discount on a certain sum due 6 months hence at 15% is Rs. 120, what is the banker's discount on the same sum for the same time and at the same rate? Sol. B.G. = S.I. on T.D. = Rs.(120 x 15 x 1/2 x 1/100) = Rs. 9. (B.D.) - (T.D.) = Rs. 9. B.D. = Rs. (120 + 9) = Rs. 129. Ex. 3. The banker's discount on Rs. 1800 at 12% per annum is equal to the true discount on Rs. 1872 for the same time at the same rate. Find the time. Sol. S.I. on Rs. 1800 = T.D. on Rs. 1872. P.W. of Rs. 1872 is Rs. 1800. Rs. 72 is S.I. on Rs. 1800 at 12%. Time =[(100 x 72)/ (12x1800)]year 1/3year = 4 months. Ex. 4. The banker's discount and the true discount on a sum of money due 8 months hence are Rs. 120 and Rs. 110 respectively. Find the sum and the rate percent. Sol. Sum =[( B.D.*T.D.)/(B.D.-T.D.)] = Rs.[(120x110)/(120-110)] = Rs. 1320. Since B.D. is S.I. on sum due, so S.I. on Rs. 1320 for 8 months is Rs. 120. Rate =[(100 x120)/( 1320 x 2/3)% = 13 7/11%. Ex. 5. The present worth of a bill due sometime hence is Rs. 1100 and the true discount on the bill is Rs. 110. Find the banker's discount and the banker's gain. Sol. T.D. =√(P.W.*B.G) B.G. =(T.D.)2/ P.W. = Rs.[(110x110)/ 1100] = Rs. 11. B.D.= (T.D. + B.G.) = Rs. (110 + 11) = Rs. 121. Ex. 6. The banker's discount on Rs. 1650 due a certain time hence is Rs. 165. Find the true discount and the banker's gain. Sol. Sum = [(B.D.xT.D.)/ (B.D.-T.D.)] = [(B.D.xT.D.)/B.G.] T.D./B.G. = Sum/ B.D. =1650/165 =10/1 Thus, if B.G. is Re 1, T.D. = Rs. 10. If B.D.is Rs. ll, T.D.=Rs. 10. If B.D. is Rs. 165, T.D. = Rs. [(10/11)xl65] =Rs.150 And, B.G. = Rs. (165 - 150) = Rs, 15. Ex. 7. What rate percent does a man get for his money when in discounting a bill due 10 months hence, he deducts 10% of the amount of the bill? Solution: Let amount of the bill = Rs.100 Money deducted =Rs.10 Money received by the holder of the bill = Rs.100-10 = Rs.90 SI on Rs.90 for 10 months = Rs.10 Rate =[(100*10)/(90*10/12)%=13 1/3%

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Google Classroom GeoGebra Classroom # Cópia de Exploring Projectile Motion Concepts In this simulation you can look at the paths of four projectiles, all shot over level ground. The projectiles are launched at four different angles, 20 degrees, 30 degrees, 45 degrees and 60 degrees. You can choose whether you want them all to have the same horizontal range, the same maximum height, or the same initial velocity. Their initial velocities will be automatically adjusted based on your choice. For each scenario (same range / same max. height / same initial speed) you should be able to answer questions such as: Which has the greatest initial horizontal velocity? Which has the greatest initial vertical velocity? Which has the greatest initial speed? Which has the least initial horizontal velocity? Which has the least initial vertical velocity? Which has the least initial speed? Which has the greatest time in the air? Which has the least time in the air?

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statistics posted by on . 14) A population of N = 20 scores has a mean of µ = 15. One score in the population is changed from X = 8 to X =28. What is the value for the new population mean? • statistics - , (20(15) +28 -8)/20 (300+28-8)/20 320/20 = 16

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### A web app to explore the possibilities I talk about this in my maths teaching blog: see the "How Many Tunes" post at www.thewessens.net/blog I used a very restricted musical space - just 2 bars, 12 notes (white notes from A below middle C to E an octave above), some constraints for musicality and restrictions on rhythm, and wrote a web app to explore (and listen to) the possibilities. It is at my website "The Mathenæum" at www.thewessens.net/maths and is called "Computer Composer". The blog post links to it as well. It's a lot of fun, and quite incredible how many tunes are possible. Here's the conclusion from the blog: Adding these together, we have a total of 22 059 rhythms (from 3 notes to 16 notes), leading to more than 50 000 000 000 000 000 tunes (that’s 16 zeroes — 50 quadrillion!). Including harmonies, we end up with a final number of 2 300 000 000 000 000 000 songs (2.3 quintillion). And remember, these are not just random arrangements, but we have used rules to greatly increase the chance of a resulting tune sounding musical (matching harmony, starting on C etc). That’s a lot of tunes! A very lot. So many that it’s actually quite difficult to imagine, but this might help. - If we were to only keep one in every million songs (assuming the others are too similar or not musically interesting) we would still have more than 2 trillion songs in total. - If we started composing in this way the very instant the universe began, writing one song every second right up until now, we would still be less than 20% through all the possibilities! - If we printed out all the songs in a line, each song taking up 15cm, the length of paper would go from Earth to Proxima Centauri (the nearest star after the Sun — about 4.23 light years away) and back more than 4 times. That’s about 36 light years. Ken

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### Asymmetrical Information - Megan McArdle 12.21.1210:01 AM ET # Whatever Happened to the "B" Battery? AA, C, D . . . where's the B? If you have even a hint of curiosity in your soul, you cannot help but wonder why there are AA, C and D batteries, but no "B".  Well, it turns out that there were B Batteries, which were used for things like radios.  Vintage radio enthusiasts can find instructions for making your own B Batteries here.  An antique radio site explains what they were used for: It will help to understand a little about the difference between the A and B power supplies. The A supply provides low-voltage DC to heat the filaments inside the radio tubes. It can be as low as 1.5 volts. The B supply provides higher-voltage DC for the "plate" circuits of the radio. The B supply can be 22.5, 45, 67.5, or 90 volts. Why the difference in voltage between A and B? The answer has to do with the way that tubes work. When you connect the A battery, the filament of the tube is heated to release negatively charged electrons. When the B battery is connected, it puts a positive charge on the plate of the tube. Electrons travel through the partial vacuum inside the tube, flowing from the filament to the positively charged plate. Many tubes also have small structures, known as grids, between the filament and the plate. The grid regulates the number of electrons that strike the plate. Thus, every radio tube must be supplied with two different voltages (A and B), and most will need three (A, B, and C). The A voltage heats the filament to release electrons. The B voltage gives the plate a positive charge to attract electrons from the filament. All early radios apparently used batteries, often getting recharged by the car.  Then technology changed: Early battery sets had several drawbacks. A dead battery could leave you radio-less in the middle of a crucial broadcast. Lead acid cells could leak acid, which might drip out of the radio cabinet onto your lovely Persian rug. Worst of all, if you accidentally reversed the A and B battery connectors, you could fry your radio's precious tubes. Recognizing these problems, radio makers, many of which also made and sold tubes, sought to develop battery-less radio sets. Perhaps more radio tubes could do part of the job of expensive disposable batteries. Radio tubes offer two important features. A tube can act as an amplifier, taking a tiny voltage, such as an incoming radio wave, and increasing it sufficiently to be heard through a headphone or speaker. A tube also can act as a diode, which changes alternating current (AC) into a series of half-cycle pulses that approximate the direct current (DC) that flows from a battery. Alternating current became increasingly available in homes during the 1920s, and radio engineers soon developed new radio tubes, called rectifiers, which could convert AC to DC. Soon, the stores were filled with battery-less radio sets using rectifier tubes. Everyone loved these new radios, except the battery manufacturers. Eventually, of course, the transistors entirely eliminated the need for the heavy voltage B batteries provided.  The B battery seems to have fallen out of the ANSI standards, which started being formulated in the late 1920s, as did the A battery.  So now you know why there's a letter gap on the drugstore shelf. The act of creating standards is always going to create these sort of little artifacts; once people start using them widely, you can't revise away the gap when it turns out that no one is using that battery size any more.  It's only been a century since we started seriously standardizing, and already our naming regimes have a lot of these "missing" items.  Presumably, as the centuries march on, they will be littered with these hole-shaped relics of earlier technologies.

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# Statics: Moment about an axis ## Homework Statement Two 20 lb forces are applied to a ratchet wrench down at A and up at B to loosen bolt β. Find the sum of the moments of the two forces about the axis (x) of the bolt. ## Homework Equations What are the moments at point A, B, and Q? How do I find the moment about the x axis and not at the bolt β? ## The Attempt at a Solution Not sure how to set up the problem because of the different pivots Q and B which are angled. First step: What torque is transmitted down the 8" wrench extension? Have you studied torques as vectors? Yes I believe I have, but we call them moments? Could I just find the moment on the 8" wrench extension and project it on the x axis by multiplying it by cos(45) to find the answer? Yes, I believe that's right. This is my attempt at the problem, I feel like it is wrong because of my Force vector. #### Attachments • 20150907_205826.jpg 16.7 KB · Views: 358 Unfortunately, I can open, but can't read your attachment. What is the moment in the wrench extension?

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Tangrams Can you make five differently sized squares from the tangram pieces? Polydron This activity investigates how you might make squares and pentominoes from Polydron. Baked Bean Cans Is there a best way to stack cans? What do different supermarkets do? How high can you safely stack the cans? Three Squares Three Squares What is the greatest number of squares you can make by overlapping three squares of the same size? You could try this interactivity. Click on the edge of any square to drag it. Full screen version If you can see this message Flash may not be working in your browser Please see http://nrich.maths.org/techhelp/#flash to enable it. Why do this problem? This problem is fantastic for reinforcing the properties of a square. Its exploratory nature will engage children and encourage them to experiment with mathematical ideas. Possible approach You could begin by using the interactivity to arrange just two squares in different ways and asking children to count the number of squares made in each case. It would be helpful if leaners were invited to draw round each square they could see on the interactive whiteboard so that the squares were made easily visible. There might be some debate about which are squares and this gives the group the opportunity to remind each other of a square's properties. Once they are familiar with the idea, introduce the main problem and suggest they work in pairs. If it is not possible for everyone to be at a computer to use the interactivity, then you could print squares on three different OHTs for children to manipulate themselves. It would also be useful to have squared paper available for jottings, rough working and recording. In the plenary, you could use the interactivity to share solutions. It would also be worth talking about how children went about the problem. Did they record as they went along? If so, what and why? You may find that some learners drew an arrangement so that they could count the squares more easily by marking in colour. Others might have recorded an arrangement as a reminder of the largest number of squares they had found so far. Key questions How many squares can you make by overlapping two large squares? How do you know that is a square? Can you move the large squares so that you create more squares? Possible extension Some children could try using four squares in the same way, or they could use equilateral triangles instead. Possible support Learners could start by looking at two squares and using this interactivity will help .

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# urgent!!!!! acceleration on a string- Isaac Physics Question Watch #1 Hey guys urgent help needed please anyone help me with these isaac physics please thank you. A particle Q of mass 2.0kg is resting on a frictionless surface. It is attached to one end of a piece of elastic string, with the other end of the elastic being attached to a fixed point. Q is then pulled back and released from rest with the elastic extended 0.50m beyond its natural length. After the particle has been accelerated, the string becomes slack. Rather than being allowed to continue in some kind of oscillatory motion, the string is cut when it first becomes slack. The particle is travelling at a speed of 4.0ms −1 when this occurs. Assuming no losses to friction or other dissipative forces: part A: Calculate the energy stored in the elastic before Q was released from rest. part B: What is the maximum tension in the elastic during the process? part c: Q then makes an elastic collision with a stationary body P and rebounds back along the same track at 3.0ms −1 . Calculate the mass of P. part D: Calculate the velocity of P after the collision. part E: If Q had instead been accelerated from rest by a small motor giving a power of 40W, how long would it have taken to reach a speed of 4.0ms −1? 0 1 year ago #2 (Original post by future-doctor <3) Hey guys urgent help needed please anyone help me with these isaac physics please thank you. A particle Q of mass 2.0kg is resting on a frictionless surface. It is attached to one end of a piece of elastic string, with the other end of the elastic being attached to a fixed point. Q is then pulled back and released from rest with the elastic extended 0.50m beyond its natural length. After the particle has been accelerated, the string becomes slack. Rather than being allowed to continue in some kind of oscillatory motion, the string is cut when it first becomes slack. The particle is travelling at a speed of 4.0ms −1 when this occurs. Assuming no losses to friction or other dissipative forces: part A: Calculate the energy stored in the elastic before Q was released from rest. part B: What is the maximum tension in the elastic during the process? part c: Q then makes an elastic collision with a stationary body P and rebounds back along the same track at 3.0ms −1 . Calculate the mass of P. part D: Calculate the velocity of P after the collision. part E: If Q had instead been accelerated from rest by a small motor giving a power of 40W, how long would it have taken to reach a speed of 4.0ms −1? A: We can assume conservation of energy, so kinetic energy of Q is energy stored in elastic. B:using F=kx and E=1/2 k x^2 you should be able to use the energy and max length to calculate the max tension. C: Conservation of momentum. D:More Conservation of momentum E: Use kinetic energy formula and Power * time = energy 0 1 year ago #3 (Original post by tande33) A: We can assume conservation of energy, so kinetic energy of Q is energy stored in elastic. .... In physics, we don't assume conservation of energy. Even friction is present, energy is still conserved and the law of conservation of energy still holds. 0 1 year ago #4 (Original post by Eimmanuel) In physics, we don't assume conservation of energy. Even friction is present, energy is still conserved and the law of conservation of energy still holds. Your statement isn't consistent - you say that we don't "assume" it, then say that it's true. The question says: (Original post by future-doctor <3) Assuming no losses to friction or other dissipative forces So, I'm not seeing your point. 0 #5 (Original post by tande33) A: We can assume conservation of energy, so kinetic energy of Q is energy stored in elastic. B:using F=kx and E=1/2 k x^2 you should be able to use the energy and max length to calculate the max tension. C: Conservation of momentum. D:More Conservation of momentum E: Use kinetic energy formula and Power * time = energy i am still confuse how do i find the energy stored from elastic because i only know for the elastic extended 0 #6 0 1 year ago #7 (Original post by RogerOxon) Your statement isn't consistent - you say that we don't "assume" it, then say that it's true. The question says: So, I'm not seeing your point. If it is true, there is no point of assuming it is correct. The word "assume" has the implication that it is not right. I have not encountered a well-written physics text that assumes the conservation of energy. The question is the question and what I am quoting is what I am quoting. If you want to "over-interpret" it, I cannot help. 0 1 year ago #8 (Original post by future-doctor <3) Hey guys urgent help needed please anyone help me with these isaac physics please thank you. A particle Q of mass 2.0kg is resting on a frictionless surface. It is attached to one end of a piece of elastic string, with the other end of the elastic being attached to a fixed point. Q is then pulled back and released from rest with the elastic extended 0.50m beyond its natural length. After the particle has been accelerated, the string becomes slack. Rather than being allowed to continue in some kind of oscillatory motion, the string is cut when it first becomes slack. The particle is travelling at a speed of 4.0ms −1 when this occurs. Assuming no losses to friction or other dissipative forces: part A: Calculate the energy stored in the elastic before Q was released from rest. part B: What is the maximum tension in the elastic during the process? part c: Q then makes an elastic collision with a stationary body P and rebounds back along the same track at 3.0ms −1 . Calculate the mass of P. part D: Calculate the velocity of P after the collision. part E: If Q had instead been accelerated from rest by a small motor giving a power of 40W, how long would it have taken to reach a speed of 4.0ms −1? Hello, where did you find this in Isaac Physics? 0 1 year ago #9 (Original post by future-doctor <3) Please post your thinking and what you have done in regard to the question. I believe in isaac physics website there are hints and have you read the hints. If yes, explain what is confusing you in hints. 0 1 year ago #10 (Original post by Eimmanuel) If it is true, there is no point of assuming it is correct. The word "assume" has the implication that it is not right. I have not encountered a well-written physics text that assumes the conservation of energy. The question is the question and what I am quoting is what I am quoting. If you want to "over-interpret" it, I cannot help. Sorry, but none of that makes sense. Let's move on though. Last edited by RogerOxon; 1 year ago 0 1 year ago #11 (Original post by future-doctor <3) (Original post by future-doctor <3) A particle Q of mass 2.0kg is resting on a frictionless surface. It is attached to one end of a piece of elastic string, with the other end of the elastic being attached to a fixed point. Q is then pulled back and released from rest with the elastic extended 0.50m beyond its natural length. After the particle has been accelerated, the string becomes slack. Rather than being allowed to continue in some kind of oscillatory motion, the string is cut when it first becomes slack. The particle is travelling at a speed of 4.0ms −1 when this occurs. Assuming no losses to friction or other dissipative forces: part A: Calculate the energy stored in the elastic before Q was released from rest. part B: What is the maximum tension in the elastic during the process? part c: Q then makes an elastic collision with a stationary body P and rebounds back along the same track at 3.0ms −1 . Calculate the mass of P. part D: Calculate the velocity of P after the collision. part E: If Q had instead been accelerated from rest by a small motor giving a power of 40W, how long would it have taken to reach a speed of 4.0ms −1? A: This is wholly transferred to KE; B: The tension is linear with extension, so calculate the average force, and, from that, find the maximum; C: Look at the definition of an elastic collision. D: As above; E: You know the mass and change in KE, so can calculate the time for the motor to provide that. 1 1 year ago #12 (Original post by Eimmanuel) In physics, we don't assume conservation of energy. Even friction is present, energy is still conserved and the law of conservation of energy still holds. What I meant by that is that all the elastic energy becomes kinetic energy, that's all, but thanks for pointing my error in terminology! 0 X new posts Back to top Latest My Feed ### Oops, nobody has postedin the last few hours. Why not re-start the conversation? see more ### See more of what you like onThe Student Room You can personalise what you see on TSR. Tell us a little about yourself to get started. ### Poll Join the discussion #### Feeling behind at school/college? What is the best thing your teachers could to help you catch up? Extra compulsory independent learning activities (eg, homework tasks) (3) 3.75% Run extra compulsory lessons or workshops (11) 13.75% Focus on making the normal lesson time with them as high quality as possible (14) 17.5% Focus on making the normal learning resources as high quality/accessible as possible (9) 11.25% Provide extra optional activities, lessons and/or workshops (29) 36.25% Assess students, decide who needs extra support and focus on these students (14) 17.5%

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# How many sample do we need for normality of t-ratio? [duplicate] I am currently learning about confidence intervals for the population mean. Assume we do not know the variance of the population. Let $$\bar{x}$$ be the sample mean, $$s$$ be the sample variance and $$n$$ the sample size. I learnt the following: • Use $$\bar{x}\pm t_{\alpha,n-1}\frac{s}{\sqrt{n}}$$ if the data is normally distributed, where $$t_{\alpha,n-1}$$ is the $$\alpha$$ t-score from the $$T$$ distribution with $$n-1$$ degrees of freedom. • If the data is not normally distributed, then with a large enough sample we can use the z-interval $$\bar{x}\pm z_{\alpha}\frac{s}{\sqrt{n}}$$ where $$z_{\alpha}$$ is a suitable z-score. The reason is that the T-ratio $$T=\bar{x}-\mu/(s/\sqrt{n})$$ becomes approximately normal with a large enough sample. The lecture notes mention that we can consider $$n=30$$ a large enough sample size but without providing any explanation. So my question is simple: Why $$n=30$$? • It is generally not true that $n=30$ is sufficient. – Tim Commented Apr 5, 2023 at 19:19 • If $\alpha=0.05$ you get $z_\alpha=1.96$ in your two-tailed example, which people sometimes round to $2$. The equivalent $t_{\alpha,30}\approx 2.04$ which also rounds to $2$ Commented Apr 5, 2023 at 19:34

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# [R] How to get the normal direction to a plane? Douglas Bates bates at stat.wisc.edu Sat Jul 3 13:35:59 CEST 2004 ```Fred wrote: > Dear All > > Maybe the following is a stupid question. > Assume I have 3 coordinate points (not limited to be in 2D or 3D space) > a, b, c. > It is known that these 3 points will define a plane. > The problem is how to get the normal direction that is orthogonal to > this plane. > > Is there an easy way to calculate it using the values of a, b, and c? In the general problem of k vectors of length n, create the n by k-1 matrix of differences v_i - v_k for i = 1,...,k-1, take its QR decomposition, and form Q. The last n + 1 - k columns of Q are orthogonal to the affine subspace that contains the k original points. ```

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# Illustrative Mathematics Grade 7, Unit 1, Lesson 13: Draw It to Scale Learning Targets: • I can create a scale drawing of my classroom. • When given requirements on drawing size, I can choose an appropriate scale to represent an actual object. Related Pages Illustrative Math #### Lesson 13: Draw It to Scale Let’s draw a floor plan. Illustrative Math Unit 7.1, Lesson 13 (printable worksheets) #### Lesson 13.1 Which Measurements Matter? Which measurements would you need in order to draw a scale floor plan of your classroom? List which parts of the classroom you would measure and include in the drawing. Be as specific as possible. #### Lesson 13.2 Creating a Floor Plan (Part 1) 1. On a blank sheet of paper, make a rough sketch of a floor plan of the classroom. Include parts of the room that the class has decided to include or that you would like to include. Accuracy is not important for this rough sketch, but be careful not to omit important features like a door. 2. Trade sketches with a partner and check each other’s work. Specifically, check if any parts are missing or incorrectly placed. Return their work and revise your sketch as needed. 3. Discuss with your group a plan for measuring. Work to reach an agreement on: • Which classroom features must be measured and which are optional. • The units to be used. • How to record and organize the measurements (on the sketch, in a list, in a table, etc.). • How to share the measuring and recording work (or the role each group member will play). 1. Gather your tools, take your measurements, and record them as planned. Be sure to double-check your measurements. 2. Make your own copy of all the measurements that your group has gathered, if you haven’t already done so. You will need them for the next activity. #### Lesson 13.3 Creating a Floor Plan (Part 2) Your teacher will give you several paper options for your scale floor plan. 1. Determine an appropriate scale for your drawing based on your measurements and your paper choice. Your floor plan should fit on the paper and not end up too small. 2. Use the scale and the measurements your group has taken to draw a scale floor plan of the classroom. Make sure to: • Show the scale of your drawing. • Label the key parts of your drawing (the walls, main openings, etc.) with their actual measurements. • Show your thinking and organize it so it can be followed by others. #### Are you ready for more? 1. If the flooring material in your classroom is to be replaced with 10-inch by 10-inch tiles, how many tiles would it take to cover the entire room? Use your scale drawing to approximate the number of tiles needed. 2. How would using 20-inch by 20-inch tiles (instead of 10-inch by 10-inch tiles) change the number of tiles needed? Explain your reasoning. #### Lesson 13.4 Creating a Floor Plan (Part 3) 1. Trade floor plans with another student who used the same paper size as you. Discuss your observations and thinking. 2. Trade floor plans with another student who used a different paper size than you. Discuss your observations and thinking. 3. Based on your discussions, record ideas for how your floor plan could be improved. The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

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## 5 Traditional Methods of Primary Data Collection In recent years, we have seen a sudden increase in commercial applications in machine learning and AI. Machine learning models heavily rely on data. Therefore, data scientists have invented new methods of data collection. Some of these data collection methods did not exist before the rise of the digital revolution. ... read more ## Aggregating Data using Bar Charts And Histograms Bar chart and histogram chart both use vertical bars and both are used to aggregate data. The main difference between them is that the bar chart is defined over 2D data - one dimension applies to x-axis and the other to y-axis. On the other hand, histogram is only apply ... read more ## 4 Important Scales of Measurement in Statistics Assigning of numbers to observations, and sometime scaling them as well, is known as measurement. Scale of measurement also defines what type of random variable we are dealing with. In the following, we will briefly discuss four important scales of measurements: 1. Nominal Scale Nominal scale allows us classify observations into ... read more ## 4 Steps to Construct Index Numbers Index number is a statistical measure of average change in a variable or a group of variables with respect to time or space. The variable may be the enrolment of staff in an organization, the cost of salaries for staff, prices of a particular commodity or a group of commodities, ... read more ## Visualizing Data using Pie Chart A circular statistical image divided into sections to illustrate the numerical ratio is known as a pie chart. In a pie chart, the arc length of each segment is similar to the ratio of the quantity that it shows. In statistics, pie charts are used to visualize data, specifically the ... read more

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sameer Posted in : Java Beginners August 19, 2009 at 12:32 PM Hi Friend, We are providing you a code that will prompt the user to enter the latitude and longitude coordinates of the source and destination and accordingly return the distance between them. import java.util.*; public class CalculateDistance{ public static void main(String args[]){ System.out.println("Enter the Longitude and Latitude of Source location:"); Scanner input=new Scanner(System.in); double lat1=input.nextDouble(); double long1=input.nextDouble(); System.out.println("Enter the Longitude and Latitude of destination location:"); double lat2=input.nextDouble(); double long2=input.nextDouble(); CalculateDistance cal=new CalculateDistance(); double dist=cal.distance( lat1, -long1 ,lat2 , -long2, 'K'); System.out.println(dist + " Kilometers\n"); } public static double distance(double lat1, double lon1, double lat2, double lon2, char unit) { double theta = lon1 - lon2; dist = Math.acos(dist); dist = dist * 60 * 1.1515; if (unit == 'K') { dist = dist * 1.609344; } return (dist); } private static double degreeToRadian(double deg) { return (deg * Math.PI / 180.0); } return (rad * 180.0 / Math.PI); } } Thanks Road distance using Java - Java Beginners Road distance using Java  Dear All i am calculating the straight line distance between two latlong points using the simple mathematical formula but i need the Real Road Driving Distance // protected double distance(double pX Delhi to Udaipur by road distance Delhi to Udaipur by road distance  delhi to udaipur by road distance Distance in meters - Java Beginners Distance in meters  Write a program that asks the user to enter a distance in meters. The program will then present the following menu of selections... the program The program will convert the distance to kilometers, inches Distance conversion - Java Beginners Distance conversion  Write an application that reads a distance... the distance to kilometres, feet and inches, according to the user?s selection...() for the user selection. ? Write a method to read a distance in metres from Another distance conversion - Java Beginners Another distance conversion  Write an application that reads a distance in metres. The program will then convert the distance to kilometres, feet...[]){ System.out.print("Enter distance in meters : "); BufferedReader br = new BufferedReader(new GUI and how to convert a distance - Java Beginners application that can be is used to convert a distance unit in miles into its...=new JTextField(20); t5=new JTextField(20); l1=new JLabel("Enter distance...){ String value=t1.getText(); double distance=Double.parseDouble(value Edit Distance Edit Distance  I want java programming ask from user input two string and the program find the edit distance between two strings and find table and optimal solution using GUI class distance class distance  class distance-integer data member km1,km2,km3,m1,m2,m3. run()-to input the two distance and add them in the third. display()-to display the total distance. display the constructor if required SMART VEHICLE ROAD SIGN IDENTIFICATION  JAVA code for "SMART VEHICLE ROAD SIGN IDENTIFICATION Delhi to Udaipur by road  Please guide me an easy and the shortest road route from Delhi to Udaipur. Thanks List of Version of com.mapzen>on-the-road dependency List of Version of com.mapzen>on-the-road dependency How to make a damaged road       If you want to convert a fresh road to damaged mode come... a picture: First we have take a fresh road picture as I have taken here Hill stations near Delhi distance Hill stations near Delhi distance   Hi, Can any one tell the distances of the Hill stations near Delhi? Thanks using eval in Java using eval in Java  using eval in Java

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Here are three “little” problems that my friend John (from UK) gave me yesterday. $(1)$ How many 5-digit numbers can be constructed using only $1, 2$ and $3$ such that each of those three digits is used at least once? $(2)$ Find all pairs of positive integers $(m,n)$ such that $1 + 5\cdot 2^m = n^2$. $(3)$ Find all positive integers $a, b$ such that $a^b = b^a$.

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Hello below is my code for a program that reads in integers rom keyboard input and creates two polynomials from that input and then does some maths functions on them. I've got the addition working, having trouble with the multiplication though. Any help would be appreciated. thanks `````` import java.util.*; import java.io.File; import java.io.FileInputStream; import java.io.FileNotFoundException; import java.text.*; import java.io.IOException; import java.util.Iterator; import java.util.Scanner; public class Polynomial { private int TOLERANCE = 0; /****************** the Monomial (inner) class ********************************/ private class Monomial { private DecimalFormat precision = new DecimalFormat("#.####"); private int deg; // degree of polynomial private int coeff; // coefficients private Monomial next; // Pointer to next term public Monomial(int coeff, int deg, Monomial next) { this.coeff = coeff; //coefficient this.deg = deg; // Degree this.next = next; // Pointer } public String toString() { String form = precision.format(Math.abs(coeff)); if(deg == 0) return form ; else if(deg == 1) return form + "x"; else return form +"x^" + deg; } public boolean equals(Monomial mono) { return coeff == mono.coeff && deg == mono.deg; } } //********************************************************************************************* public Polynomial() { } /*************************************************************************** * Adds a new term into the polynomial, assuming that the polynomial * is sorted in order from smallest to largest exponent. **************************************************************************/ public void addTerm(int coeff, int deg) { if( Math.abs(coeff) < TOLERANCE ) return; { return; } Monomial prev = null; while( cur != null && deg > cur.deg) { prev = cur; cur = cur.next; } if( cur == null || deg != cur.deg ) prev.next = new Monomial(coeff, deg, cur); else { cur.coeff += coeff; if( Math.abs(cur.coeff) < TOLERANCE ) if(prev != null) prev.next = cur.next; else } } public String toString() { StringBuffer sb = new StringBuffer(); for(Monomial tmp = head; tmp != null; tmp = tmp.next) if(tmp.coeff < 0 ) sb.append(" - " + tmp.toString()); else sb.append(" + " + tmp.toString()); return sb.toString(); } //********************************************************************************************* /********************************************************************************************* * Return the degree of this polynomial **********************************************************************************************/ //*********************************************************************************************/ /********************************************************************************************* * Multiplies Polynomial 1 to Polynomial 2 * The method does not change the original polynomial. **********************************************************************************************/ //*********************************************************************************************/ /********************************************************************************************* * Adds Polynomial 1 to Polynomial 2 * The method does not change the original polynomial. **********************************************************************************************/ { Polynomial res = clone(); for(Monomial tmp = poly.head; tmp != null; tmp = tmp.next) return res; } public Polynomial clone() { Polynomial res = new Polynomial(); for(Monomial tmp = head; tmp != null; tmp = tmp.next) return res; } public boolean equals(Polynomial poly) { while(tmp1 != null && tmp2 != null) { if( !tmp1.equals(tmp2) ) return false; tmp1 = tmp1.next; tmp2 = tmp2.next; } return true; } //*********************************************************************************************/ /********************************************************************************************* * Multiplies by a Constant * The method does not change the original polynomial. **********************************************************************************************/ public Polynomial multiply(int num) { Polynomial res = clone(); for(Monomial tmp = res.head; tmp != null; tmp = tmp.next) tmp.coeff *= num; return res; } //*********************************************************************************************/ /********************************************************************************************* * Returns a new polynomial that is the derivative of this polynomial. **********************************************************************************************/ public Polynomial diff() { Polynomial res = new Polynomial(); for(Monomial tmp = head; tmp != null; tmp = tmp.next) { if(tmp.deg != 0) res.addTerm(tmp.coeff * tmp.deg, tmp.deg - 1 ); } return res; } //*********************************************************************************************/ /********************************************************************************************* * Returns a new polynomial that is the result of Monomial times polynomial. ** The method does not change the original polynomial. **********************************************************************************************/ /* public Polynomial times(Polynomial poly) { { Polynomial result = new Polynomial(); for(Monomial tmp2 = poly.head; tmp2 != null; tmp2 = tmp2.next) for(Monomial tmp = head; tmp != null; tmp = tmp.next) while(tmp != null && tmp2 != null) { result.addTerm(tmp.coeff * tmp2.coeff, tmp.deg + tmp2.deg); tmp = tmp.next; tmp2 = tmp2.next; } return result; } } */ //*********************************************************************************************/ /********************************************************************************************* * Driver Program to Test Polynomial Class **********************************************************************************************/ public static void main(String[] args) { //---------------- Code to chose between keyboard input or file input------------------------ System.out.println("If you would like you enter the polynomial via a text file 'Polyinput.txt' PRESS 1. "); System.out.println("If you would like you enter the polynomial via the keyboard, PRESS ANY OTHER INTEGER... "); Scanner inputquestion=new Scanner(System.in); System.out.print("Enter Option: "); // Prompts for selection choice int option=inputquestion.nextInt(); // Stores input in option if (option == 1) // if loop to jump into read from text file { String content = new String(); String name = new String(); File file = new File("polyinput.txt"); try { Scanner sc = new Scanner(new FileInputStream(file)); while (sc.hasNext()) { name = sc.next(); content = sc.nextLine(); } sc.close(); } catch (FileNotFoundException fnfe) { fnfe.printStackTrace(); } catch (Exception e) { e.printStackTrace(); System.out.println("\nProgram terminated Safely..."); } Iterator<String> i = list.iterator(); while (i.hasNext()) { System.out.println(name + i.next() + "\n"); } } //------------ End if to enter input text file code--------------------------------------- else //--------------start of else for keyboard input-------------------------------------- { //------- Code to Prompt for Polynomial 1 ----------------------------------------------------------------- Polynomial polynomial1 = new Polynomial(); System.out.println( ); System.out.println( ); System.out.println("Enter first Polynomial"); System.out.println("(Enter Zero (0) as Exponent value to move to second Polynomial"); System.out.println( ); //Insert do while loop exponent not equal 0 int ExponentTest = 1; do { Scanner inputCoefficent=new Scanner(System.in); System.out.print("Enter Coefficient: "); // Prompts for coefficent int Coefficient=inputCoefficent.nextInt(); // Stores coefficent in coefficent Scanner inputExponent=new Scanner(System.in); System.out.print("Enter Exponent: "); // Prompts for Exponent int Exponent=inputExponent.nextInt(); // Stores Exponent in Exponent if (Exponent < 1) { ExponentTest = 0; } } while (ExponentTest > 0); // ------ end do while loop --------------- System.out.println( ); System.out.println( " Entered Polynomial is: " + polynomial1 ); //---------------------------------------------------------------------------------------------------------- //----------- Code to Prompt for Polynomial 2 ----------------------------------------------------------------- Polynomial polynomial2 = new Polynomial(); System.out.println( ); System.out.println( ); System.out.println("Enter second Polynomial"); System.out.println("(Enter Zero (0) as Exponent value to complete Polynomial"); System.out.println( ); //Insert do while loop exponent not equal 0 int ExponentTest2 = 1; do { Scanner inputCoefficent=new Scanner(System.in); System.out.print("Enter Coefficient: "); // Prompts for coefficent int Coefficient=inputCoefficent.nextInt(); // Stores coefficent in coefficent Scanner inputExponent=new Scanner(System.in); System.out.print("Enter Exponent: "); // Prompts for Exponent int Exponent=inputExponent.nextInt(); // Stores Exponent in Exponent if (Exponent < 1) { ExponentTest2 = 0; } } while (ExponentTest2 > 0); // ------ end do while loop --------------- System.out.println( ); System.out.println( " Entered Polynomial is: " + polynomial2 ); //---------------------------------------------------------------------------------------------------------- //------- Code to Prompt for Monomial (ax^k) to multiply both Polynomials by--------------------------------- Polynomial monomial1 = new Polynomial(); System.out.println( ); System.out.println( ); System.out.println("Enter details for the Monomial equation"); Scanner inputMonomial=new Scanner(System.in); System.out.print("Enter Coefficent: "); //Prompts for Coefficient int num1=inputMonomial.nextInt(); // Stores Coefficient in num1 System.out.print("Enter Exponent: "); // Prompts for Exponent int num2=inputMonomial.nextInt(); //Stores Exponenent in num2 monomial1.addTerm(num1, num2); // Creates the Monomial equation System.out.println(); System.out.println( "Monomial: " ); System.out.println( monomial1 ); System.out.println( ); //---------------------------------------------------------------------------------------------------- //------- Code to Prompt for constant to multiply by and then reads in the constant and stores in Constant1 System.out.println("Enter integer to be used as the constant"); Scanner inputConstant=new Scanner(System.in); System.out.print("Enter Constant: "); // Prompts for constant int Constant1=inputConstant.nextInt(); // Stores Constant in Constant1 System.out.println( "Constant used to multiply polynomials: " ); System.out.println( Constant1 ); System.out.println( ); //---------------------------------------------------------------------------------------------------------- //------------------------ Returns display for operations onto the screen----------------------------------- System.out.println( ); System.out.println( "Degree of Polynomial 1 ("+ polynomial1 + "): " /*+ polynomial1.degree()*/); System.out.println( ); System.out.println( "Degree of Polynomial 2 ("+ polynomial2 + "): " ); System.out.println( ); System.out.println( "Multiply Polynomial 1 ("+ polynomial1 + ") by Polynomial 2 ("+ polynomial2 + "): " ); System.out.println( ); System.out.println( "Multiply Polynomial 1 ("+ polynomial1 + ") by Monomial ("+ monomial1 + "): " ); System.out.println( monomial1 + " * " + polynomial1 + " = " + MultiplyMonomialResult1 ); System.out.println( ); System.out.println( ); System.out.println( "Multiply Polynomial 2 ("+ polynomial2 + ") by Monomial ("+ monomial1 + "): " ); System.out.println( monomial1 + " * " + polynomial2 + " = " + MultiplyMonomialResult2 ); System.out.println( ); System.out.println( ); System.out.println( "Polynomial 1 (" + polynomial1 + ") plus Polynomial 2 (" + polynomial2 + "): " ); System.out.println( polynomial1 ); System.out.println( polynomial2 ); System.out.println( "=" ); System.out.println( ); System.out.println( "multiply Polynomial 1 ("+ polynomial1 + ") by constant entered into console (ie " + Constant1 + "): " ); Polynomial MultiplyConstant1 = polynomial1.multiply(Constant1); System.out.println( Constant1 + " * " + polynomial1 + " = " + MultiplyConstant1 ); System.out.println( ); System.out.println( "multiply Polynomial 2 ("+ polynomial2 + ") by constant entered into console (ie " + Constant1 + "): " ); Polynomial MultiplyConstant2 = polynomial2.multiply(Constant1); System.out.println( Constant1 + " * " + polynomial2 + " = " + MultiplyConstant2 ); System.out.println( ); System.out.println( "Derivative of Polynomial 1: " ); Polynomial diffresult1 = polynomial1.diff(); System.out.println( "Polynomial 1: " + polynomial1 ); System.out.println( "Derivative = " + diffresult1 ); System.out.println( ); System.out.println( "Derivative of Polynomial 2" ); Polynomial diffresult2 = polynomial2.diff(); System.out.println( "Polynomial 2: " + polynomial2 ); System.out.println( "Derivative = " + diffresult2 ); System.out.println( ); //-------------------------------End Screen output of operations for keyboard entry------------------------- } // End of else for keyboard input.------------------------------------------------------------------------ } } `````` ## All 7 Replies You should post only relevant part where you have trouble. it's gonna be hard to tell which part of the code shows where the multiplication has gone awry could you post the current output and then post what it should look like Cheers. I have been working on this since i posted it and yes realise i prob shouldnt have posted the whole massive chunk at once. problem i have now is that it is only multiplying polynomial 1 by the first term of polynomial 2 so 2x + 2 * 5x + 5 = 10x2 + 10x instead of 10x2 + 20x + 10 and here is the code that does the multiplication: `````` public Polynomial multiply(Polynomial poly) { Polynomial res = clone(); for(Monomial tmp = res.head; tmp != null; tmp = tmp.next) double num = 0.5; for(Monomial tmp = res.head; tmp != null; tmp = tmp.next) tmp.coeff *= num; return res; } `````` commented: Good work! +6 Solved most of issues, going to recondense single issues alot easier to rea d commented: More good work! +0 hey Bradoz, can u post the full solution including multiplication? thank you!!!! if any one want the full polynomial addition,multiplication,sub code than contact me Let me just add the usual health warning to the previous post. anand.khatik is a first-time poster; we have no information, good or bad, about his skills or credibility. Nobody has checked the email address he gives. Anyone copying someone else's code and submitting it as their own homework deserves the automatic fail they will probably get. Be a part of the DaniWeb community We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, learning, and sharing knowledge.

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# ball mill media load calculation formula • ### Ball Mill Design/Power Calculation The basic parameters used in ball mill design (power calculations) rod mill or any tumbling mill sizing are material to be ground characteristics Bond Work Index bulk density specific density desired mill tonnage capacity DTPH operating solids or pulp density feed size as F80 and maximum chunk size product size as P80 and maximum and finally the type of circuit open/closed Get Price • ### Quick and Easy Black Powder Ball Mill — Skylighter Inc. To load the mill jar optimally fill it half full of a dense media such as 1/2-inch lead balls. 50-caliber muzzle loading bullets will work but they are soft and will both wear down over time and leave lead dust in whatever it is you are milling. It s better to antimony-hardened 1/2-inch diameter lead balls made specifically for pyrotechnic Get Price • ### Page 1 Ball Milling Theoryfreeshell Ball Milling Theory Page 2 Optimized Jar Loading Figure 3 An overcharged mill. drum. Undercharging your mill in this manner will increase the milling times Figure 4 Undercharged media with overcharged load. relative to a properly charged mill. One key to efficient milling is a properly charged milling jar. "Charge" refers to Get Price • ### Ball Mill Grinding Media Calculation Ball Mill Effects Of Grinding Media Shapes On Ball Mill. Media shape and mill power the objective of this dissertation was to investigate how media shape affects grinding ball size distribution inside an industrial mill was analysed in terms of shapes and sizes load behaviour mill power and breakage as affected by media shapes were studied in a pilot laboratory mill an inductive proximity probe Get Price • ### (PDF) Effect of circulating load and classification The ball mill is the most common ore grinding technology today and probably more than 50 of the total world energy consumption for ore grinding is consumed in ball mills. Get Price • ### how to estimate the wear rate for Ball millPage 1 of 10 Dec 08 2009 · Re how to estimate the wear rate for Ball mill. Most practical way to by measuring empty height and calculating the volumetric filling of grinding media.Alternative way to make track of mill main drive kW consumption reduction and add makeup charge according to the extent of fall in power consumption. Get Price • ### Ball Mill Grinding Media Calculation Ball Mill Effects Of Grinding Media Shapes On Ball Mill. Media shape and mill power the objective of this dissertation was to investigate how media shape affects grinding ball size distribution inside an industrial mill was analysed in terms of shapes and sizes load behaviour mill power and breakage as affected by media shapes were studied in a pilot laboratory mill an inductive proximity probe Get Price • ### Grinding in Ball Mills Modeling and Process Control the milling process takes place during rotation as a result of the transfer of kinetic energy of the moving grinding media into the grinding product. The design of a ball mill can vary significantly depending on the size the equipment used to load the starting material (feeders) and Get Price • ### Circulating loadSlideShare Nov 18 2014 · Slide 2 Content Introduction Circulating load calculate Most factories the purpose of control Effective factors for circulating load Test design Conclusion References 3. Slide 3 circulating load ratio As the ratio of the amount of solids going through the ball mill divided by the amount of solids going through the circuit. Get Price • ### Iron Ore Ball Mill Grinding Media Calculation Formula-ball Iron Ore Ball Mill Grinding Media Calculation Formula. Iron ore ball mill grinding media calculation formulairon ore ball mill grinding media calculation formulaIron ore ball mill grinding media calculation formula aug 29 2016 grinding balls design calculation price of ball mill and stone crusher in india effects of grinding grinding theory pdf ball load calculation get price and support Get Price • ### TECHNICAL NOTES 8 GRINDING R. P. King the mill is used primarily to lift the load (medium and charge). Additional power is required to keep the mill rotating. 8.1.3 Power drawn by ball semi-autogenous and autogenous mills A simplified picture of the mill load is shown in Figure 8.3 Ad this can be used to establish the essential features of a model for mill Get Price • ### ball mill media charge calculation pdf grinding volume calculation in a ball mill . May 19 2014 Calculations for mill motor power mill speed and media charge the Ball Mill Instruction Manual (PDF)BICO Inc. Get Price • ### AMIT 135 Lesson 7 Ball Mills CircuitsMining Mill Number size and mass of each ball size depends on mill load and whether or not the media is being added as the initial charge. For the initial chargin of a mill Coghill and DeVaney (1937) defined the ball size as a function of the top size of the feed i.e. Get Price • ### ball mill media load calculation formula ball mill media load calculation formula. TECHNICAL NOTES 8 GRINDING R. P. KingMineral Technologies particles of mediathe rods balls or pebbles. This motion can be A simplified picture of the mill load is shown in Figure In equation 8.14 D is the diameter inside the mill liners and Le is the effective length of the mill Get Price • ### How To Calculate For Ball Mill Media Charge Ball Mill Grinding Media Charge CalculationHenan Mining . How To Calculate Charge In Ball Mill Ball milling Digitalfire A ball mill is simply a container that is filled with pebbles either of porcelain or stones eg Flint into which a charge powder or slurry is put and that is to determine the amount of balls needed distribution of ball sizes and other operating Ball mill media charge calculation Get Price • ### Jyoti Ceramic Industries Pvt. Ltd. Grinding media quantity = 3307 kgs To calculate the motor power required for a cylindrical type ball mill the following formula can be applied W = 0.04116 x D3 x L x n x (0.6d 0.4d1) where W = Required motor power in HP D = Internal dia of the mill in mtr L = Internal length of the mill in mtr d = Specific gravity of grinding media Get Price • ### Circulating loadSlideShare Nov 18 2014 · Slide 2 Content Introduction Circulating load calculate Most factories the purpose of control Effective factors for circulating load Test design Conclusion References 3. Slide 3 circulating load ratio As the ratio of the amount of solids going through the ball mill divided by the amount of solids going through the circuit. Get Price • ### Ball charges calculatorsthecementgrindingoffice - Ball top size (bond formula) calculation of the top size grinding media (balls or cylpebs) -Modification of the Ball Charge This calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill efficiency Get Price • ### how to estimate the wear rate for Ball millPage 1 of 10 Dec 08 2009 · Re how to estimate the wear rate for Ball mill. Most practical way to by measuring empty height and calculating the volumetric filling of grinding media.Alternative way to make track of mill main drive kW consumption reduction and add makeup charge according to the extent of fall in power consumption. Get Price • ### Estimating Ball ConsumptionMolycop Sep 17 2018 · For an improved understanding the wear rate constant is directly related to the grinding media consumption rate expressed in grams of steel per kWh drawn by the mill using the simple expression where dr represents the diameter of the new make-up balls periodically charged to the mill. Get Price • ### grinding ball mill load calculation formula html ball mill grinding media calculation formula. media shape affects grinding Ball size distribution inside an industrial mill was analysed in terms of shapes and siz Load behaviour mill power and breakage as affected by media shapes were studied in a pilot laboratory mill An inductive proximity probe . Get Price • ### Calculate Top Ball Size of Grinding MediaEquation Method An online calculator lets you calculate Top Ball Size of Grinding Media for your mill. Use this Equation Method to properly grind your ore. Get Price • ### TECHNICAL NOTES 8 GRINDING R. P. King the mill is used primarily to lift the load (medium and charge). Additional power is required to keep the mill rotating. 8.1.3 Power drawn by ball semi-autogenous and autogenous mills A simplified picture of the mill load is shown in Figure 8.3 Ad this can be used to establish the essential features of a model for mill Get Price • ### Grinding in Ball Mills Modeling and Process Control the milling process takes place during rotation as a result of the transfer of kinetic energy of the moving grinding media into the grinding product. The design of a ball mill can vary significantly depending on the size the equipment used to load the starting material (feeders) and Get Price • ### Iron Ore Ball Mill Grinding Media Calculation Formula-ball Iron Ore Ball Mill Grinding Media Calculation Formula. Iron ore ball mill grinding media calculation formulairon ore ball mill grinding media calculation formulaIron ore ball mill grinding media calculation formula aug 29 2016 grinding balls design calculation price of ball mill and stone crusher in india effects of grinding grinding theory pdf ball load calculation get price and support Get Price • ### Grinding in Ball Mills Modeling and Process Control the milling process takes place during rotation as a result of the transfer of kinetic energy of the moving grinding media into the grinding product. The design of a ball mill can vary significantly depending on the size the equipment used to load the starting material (feeders) and Get Price • ### Grinding Ball Mill Load Calculation Formula Grinding ball mill load calculation formula ball mill circulating load 911 metallurgist apr 29 2018 calculating a grinding circuits circulating loads based on screen analysis of its pulp samples collected around the ball mill or rod mill and a new approach to the calculation of work index and the potential ured sizes x1 and x2 used to Get Price • ### Jyoti Ceramic Industries Pvt. Ltd. Calculations of media charge To calculate media charge for batch mill. M = 0.000929 x D2 x L where M = Grinding media charge in kgs D = Internal Dia of the Mill in cms. after lining L = Internal length of the mill in cms. after lining. To calculate grinding media charge for continuous type ball mill M Get Price • ### A Method to Determine the Ball Filling in Miduk Copper load percentage calculation that belongs to mill diameter (𝐷𝑚) and distance between top point in the mill and load surface (H) 5-7 . This equation is profitable for determining of whole mill load volume which is included stony and metallic balls. Mill load ( ) = 113.7127.3 H Get Price • ### Ball charges calculatorsthecementgrindingoffice - Ball top size (bond formula) calculation of the top size grinding media (balls or cylpebs) -Modification of the Ball Charge This calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill efficiency Get Price • ### Optimization of mill performance by using to the volumetric mill filling which influences grinding media wear rates throughput power draw and product grind size from the circuit. Each of these performance parameters peaks at different filling values. In order to contin-uously optimize mill operation it is vital to obtain regular measurements of the ball load and pulp position. Get Price • ### ball mill media load calculation formula ball mill media load calculation formula. TECHNICAL NOTES 8 GRINDING R. P. KingMineral Technologies particles of mediathe rods balls or pebbles. This motion can be A simplified picture of the mill load is shown in Figure In equation 8.14 D is the diameter inside the mill liners and Le is the effective length of the mill Get Price • ### THE OPTIMAL BALL DIAMETER IN A MILL The ball impact energy on grain is proportional to the ball diameter to the third power 3 E K 1 d b. (3) The coefficient of proportionality K 1 directly depends on the mill diameter ball mill loading milling rate and the type of grinding (wet/dry). None of the characteristics of the material being ground have any influence on K 1. Get Price • ### Ball Mill Recirculating Load Calculation Pdf Recirculating Load In Ball Milling Formula. Recirculating load in ball milling formula ball mill recirculating load calculation pdf primary wet autogenous mill wam and a series of ball mills wam is is a ratio of the mill drive power to the milling circuit output without regard to circulating load at the constant milling barratt presented Get Price • ### Estimating Ball ConsumptionMolycop Sep 17 2018 · For an improved understanding the wear rate constant is directly related to the grinding media consumption rate expressed in grams of steel per kWh drawn by the mill using the simple expression where dr represents the diameter of the new make-up balls periodically charged to the mill. Get Price • ### REGARDING OF GRINDING MEDIA IN THE MILLPage 1 of 2 Mar 11 2014 · re REGARDING OF GRINDING MEDIA IN THE MILL. Dear you may used for just reference point . 113 AND 126 ARE COME FOR CALCULATION FROM GRINDING CHART VS GRINDING VOLUME IN BALL MILL . AS A CONSTANT FOR CHART. GKS . Reply Get Price • ### Ball Nose Finishing Mills Speed Feed CalculatorDAPRA The calculator will automatically provide the necessary speed and feed in the green fields. For assistance setting up your milling program contact a Dapra applications specialist or call (800) . Click here to download a chart of recommended Ball Nose cutting speeds and feeds (PDF). Get Price

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# balloonplot: Plot a graphical matrix where each cell contains a dot whose... In gplots: Various R Programming Tools for Plotting Data ## Description Plot a graphical matrix where each cell contains a dot whose size reflects the relative magnitude of the corresponding component. ## Usage ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32``` ```balloonplot(x, ...) ## S3 method for class 'table' balloonplot(x, xlab, ylab, zlab, show.zeros=FALSE,show.margins=TRUE,...) ## Default S3 method: balloonplot(x,y,z, xlab, ylab, zlab=deparse(substitute(z)), dotsize=2/max(strwidth(19),strheight(19)), dotchar=19, dotcolor="skyblue", text.size=1, text.color=par("fg"), main, label=TRUE, label.digits=2, label.size=1, label.color=par("fg"), scale.method=c("volume","diameter"), scale.range=c("absolute","relative"), colsrt=par("srt"), rowsrt=par("srt"), colmar=1, rowmar=2, show.zeros=FALSE, show.margins=TRUE, cum.margins=TRUE, sorted=TRUE, label.lines=TRUE, fun=function(x)sum(x,na.rm=T), hide.duplicates=TRUE, ... ) ``` ## Arguments `x` A table object, or either a vector or a list of several categorical vectors containing grouping variables for the first (x) margin of the plotted matrix. `y` Vector or list of vectors for grouping variables for the second (y) dimension of the plotted matrix. `z` Vector of values for the size of the dots in the plotted matrix. `xlab` Text label for the x dimension. This will be displayed on the x axis and in the plot title. `ylab` Text label for the y dimension. This will be displayed on the y axis and in the plot title. `zlab` Text label for the dot size. This will be included in the plot title. `dotsize` Maximum dot size. You may need to adjust this value for different plot devices and layouts. `dotchar` Plotting symbol or character used for dots. See the help page for the points function for symbol codes. `dotcolor` Scalar or vector specifying the color(s) of the dots in the plot. `text.size, text.color` Character size and color for row and column headers `main` Plot title text. `label` Boolean flag indicating whether the actual value of the elements should be shown on the plot. `label.digits` Number of digits used in formatting value labels. `label.size, label.color` Character size and color for value labels. `scale.method` Method of scaling the sizes of the dot, either "volume" or "diameter". See below. `scale.range` Method for scaling original data to compute circle diameter. `scale.range="absolute"` scales the data relative to 0 (i.e, maps [0,max(z)] –> [0,1]), while `scale.range="relative"` scales the data relative to min(z) (i.e. maps [min(z), max(z)] –> [0,1]). `rowsrt, colsrt` Angle of rotation for row and column labels. `rowmar, colmar` Space allocated for row and column labels. Each unit is the width/height of one cell in the table. `show.zeros` boolean. If `FALSE`, entries containing zero will be left blank in the plotted matrix. If `TRUE`, zeros will be displayed. `show.margins` boolean. If `TRUE`, row and column sums are printed in the bottom and right margins, respectively. `cum.margins` boolean. If `TRUE`, marginal fractions are graphically presented in grey behind the row/column label area. `sorted` boolean. If `TRUE`, the rows will be arranged in sorted order by using the levels of the first y factor, then the second y factor, etc. The same process is used for the columns, based on the x factors `label.lines` boolean. If `TRUE`, borders will be drawn for row and column level headers. `hide.duplicates` boolean. If `TRUE`, column and row headers will omit duplicates within row/column to reduce clutter. Defaults to `TRUE`. `fun` function to be used to combine data elements with the same levels of the grouping variables `x` and `y`. Defaults to `sum` `...` Additional arguments passed to `balloonplot.default` or `plot`, as appropriate. ## Details This function plots a visual matrix. In each `x`,`y` cell a dot is plotted which reflects the relative size of the corresponding value of `z`. When `scale.method="volume"` the volume of the dot is proportional to the relative size of `z`. When `scale.method="diameter"`, the diameter of the dot is proportional to the the relative size of `z`. The "volume" method is default because the "diameter" method visually exaggerates differences. ## Value Nothing of interest. ## Note `z` is expected to be non-negative. The function will still operate correctly if there are negative values of `z`, but the corresponding dots will have 0 size and a warning will be generated. ## Author(s) Gregory R. Warnes [email protected] ## References Function inspired by question posed on R-help by Ramon Alonso-Allende [email protected]. `plot.table` ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53``` ```# Create an Example Data Frame Containing Car x Color data carnames <- c("bmw","renault","mercedes","seat") carcolors <- c("red","white","silver","green") datavals <- round(rnorm(16, mean=100, sd=60),1) data <- data.frame(Car=rep(carnames,4), Color=rep(carcolors, c(4,4,4,4) ), Value=datavals ) # show the data data # generate balloon plot with default scaling balloonplot( data\$Car, data\$Color, data\$Value) # show margin label rotation & space expansion, using some long labels levels(data\$Car) <- c("BMW: High End, German","Renault: Medium End, French", "Mercedes: High End, German", "Seat: Imaginary, Unknown Producer") # generate balloon plot with default scaling balloonplot( data\$Car, data\$Color, data\$Value, colmar=3, colsrt=90) # Create an example using table xnames <- sample( letters[1:3], 50, replace=2) ynames <- sample( 1:5, 50, replace=2) tab <- table(xnames, ynames) balloonplot(tab) # Example of multiple classification variabls using the Titanic data library(datasets) data(Titanic) dframe <- as.data.frame(Titanic) # convert to 1 entry per row format attach(dframe) balloonplot(x=Class, y=list(Survived, Age, Sex), z=Freq, sort=TRUE) # colorize: surviors lightblue, non-survivors: grey Colors <- Titanic Colors[,,,"Yes"] <- "skyblue" Colors[,,,"No"] <- "grey" colors <- as.character(as.data.frame(Colors)\$Freq) balloonplot(x=list(Age,Sex), y=list(Class=Class, Survived=gdata::reorder.factor(Survived,new.order=c(2,1)) ), z=Freq, zlab="Number of Passengers", sort=TRUE, dotcol = colors, show.zeros=TRUE, show.margins=TRUE) ```

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# ENTROPY: QUANTUM GRAVITY: COSMOLOGY CONSTANT ## Tienzen (Jeh-Tween) Gong Section one: Quantum gravity Gravity is all about mass. But, masses are carried by fermions, the quantum particles. So, any gravity theory which is not based on particle theory is not correct, seehttps://tienzengong.wordpress.com/2016/03/16/nothingness-vs-nothing-there-the-quantum-gravity/ . Most of mainstream ‘quantum gravity’ theories try to quantize the GR (General Relativity) spacetime sheet in one way of another (discrete space or time, fixed coupling constants, etc.), such as those theories below. Loop Quantum Gravity: space itself is discrete Asymptotically Safe Gravity: pick a high-energy fixed point for the coupling constant Causal Dynamical Triangulations: It’s similar to LQG in that space itself is discrete, but time must be discrete as well! On the other hand, the M-string quantum gravity does describe the spacetime sheet with particles (the M-string). But, the M-string itself does not have a ‘mass-rising’ mechanism of its own but borrows the {Higgs Nonsense} which is a total bullcrap, seehttps://www.linkedin.com/pulse/before-lhc-run-2-begins-enough-jeh-tween-gong ; that is, M-string quantum gravity is just a hallow hype. Gravity is all about moving the ‘entire universe’ from {[HERE (now), NOW] to [HERE (next), Next]}, see http://prebabel.blogspot.com/2013/11/why-does-dark-energy-make-universe.html . This accelerates the expansion of universe and gives rise to the ‘uncertainty principle’. Yet, this ‘arrow of time’ is the result of {nothing to something transformation} process, the answer for {why is there something rather than nothing?} The {Nothingness} is defined as {timelessness and immutability}. The timelessness is expressed with 4-real/ghost time dimensions. That is, at every t, it is in fact timelessness in essence. And, this is expressed with an equation ‘zero’. Delta S = (i^n1, i^n2, i^3) C delta T, {C, light speed; S, space; T, time; n1, n2, n3 take the value (0, 1, 2, 3)} … equation ‘zero’ With this equation, 64 subspaces are created; 48 are particles (fermions), and 16 are spacetime (dark energy), see http://prebabel.blogspot.com/2012/04/48-exact-number-for-number-of.html and https://medium.com/@Tienzen/here-is-the-correct-answer-5d1a392f700#.a0jmn237w . The moving of this 4-real/ghost time dimensions forms a {garden hose} as below. This {garden hose} universe has eleven (11) dimensions: One, the external space dimensions (3) Two, the internal (of hose) space dimensions (3) Three, the real/ghost time dimensions (4) Four, the ‘nothingness’ which surrounds the entire {garden hose) is (1) So, the total = 3 + 3 + 4 + 1 = 11 Although there are spatial dimensions, the definition for the dimensions here is a ‘linguistic’ definition, the ‘codes’ which are needed to describe a system (such as universe), see http://prebabel.blogspot.com/2012/04/origin-of-spatial-dimensions-and.html . That is, 11 codes are enough to DESCRIBE this universe: 4-spacetime dimension and 7 color-codes (Red, Yellow, Blue, White, G1, G2, G3). It is the same for the {life universe}: (A, G, T, C, M (male), F (female), K (kids)). With the {framework} being set up, the next is to set up the measuring rulers {ħ (Planck constant), C (light speed)}. Then these rulers must be locked up: e (electric charge) = F (ħ, C), the first lock. A second lock is need: Alpha = F (e, ħ, C) With these two locks and the fixed framework, this universe is set FREE for its evolution, see http://prebabel.blogspot.com/2012/04/arbitrariness-and-final-unification-in.html . Of course, the litmus test for the above is that the nature constants and parameters must be derived from the above scheme. Cabibbo and Weinberg angles, see http://prebabel.blogspot.com/2011/10/theoretical-calculation-of-cabibbo-and.html Alpha calculation, see http://prebabel.blogspot.com/2012/04/alpha-fine-structure-constant-mystery.html . For Planck CMB data, see https://tienzengong.wordpress.com/2015/04/22/dark-energydark-mass-the-silent-truth/ Entropy was initially defined as a macro-thermo-phenomena, as lack of order or predictability; gradual decline into disorder with the following equation, with the dimension of energy divided by temperature, which has a unit of joules per kelvin (J K−1). In the modern microscopic interpretation of entropy in statistical mechanics, entropy is the amount of additional information needed to specify the exact physical state of a system, given its thermodynamic specification. Understanding the role of thermodynamic entropy in various processes requires an understanding of how and why that information changes as the system evolves from its initial to its final condition. Again, it is often said that entropy is an expression of the disorder, or randomness of a system, or of our lack of information about it. This statistical entropy is expressed as Boltzmann’s entropy equation (S = k log W), and the ‘S’ is an irreducible essence of any large ‘system’. Then, the concept of entropy is spreading into many (almost all) systems, especially the ‘information’ system. Again, it is all about the inherent tendency of any system towards the dissipation of useful energy or information. So, entropy is thus far a phenomenological parameter without a deep connections to the other fundamental laws of physics (SM particles theory and gravity). Yet, very recently, there is a ‘quantum gravity’ which is based on entropy: the  entropic gravity (EG), which views that gravity was not a fundamental force, but rather emerged as a phenomenon linked to entropy. Unfortunately, those EG physicists still view the entropy with the above phenomenological sense, not knowing the true ‘essence’ of entropy. Thus, there is no chance for EG to success. For constructing any system (a universe or else), it needs three departments. D-one, the big framework: the 11 dimensions, the nothing to something transformation. D-two, the locked measuring rulers. D-three, a bookkeeper. Entropy is the result of the bookkeeping. This bookkeeping is all about the ‘action count’. State one, a particle is at location (1), no action State two, that particle moves to location (2), action one (1) State three, that particle moves back to location (1), action (2) State one and state three have the identical ‘physical’ states, but the action count for them is different. And, the count is an arrow, always in the increase, a true arrow. In a big system (especially a thermo-system), the measurement of the action count (entropy) is described with the Boltzmann’s entropy equation. So, the arrow of time and the arrow of entropy are totally different arrows. But, but, but, when a Pepsi can sits on my desk without any movement in the past 10 years, is there any action count for it? The answer is big YES in this G-string quantum gravity, as it is in fact moving in ‘time’, and every movement is an action. So, many people is confusing about that the two arrows are the same arrow. No, they are completely different arrows. One, arrow of ‘time’: an emergent of the ‘timelessness’. Two, arrow of ‘entropy’: the ever increasing the action count. In the example above (from state one to state three), it gained nothing but still paid with higher action count. Phenomenologically, this is viewed as ‘dissipation’, and thus the entropy has not much to do with the fundamental laws of physics. On the other hand, the TOTAL action count of this universe defines the STATUS of this universe. So, this {TOTAL COUNT} must be the most important number in the BOOK of this universe. How can we calculate this {TOTAL COUNT}? In the G-string quantum gravity, the universe moves {from [here (now), now] to [here (next), next]} with quantum action {ħ}, and each action is one {quantum information}, the quantum-bit. Then, the largest unit of this quantum information is {ħ C} in one unit of time. Finally, the NUMBER of quantum ACTIONs per time-dimension in a unit of time is {1/ (ħ C)}. So, the {TOATL quantum action COUNT} of this universe (TC) = {1/ (ħ C)} ^4 x T; T, life time of universe; there are 4-time-dimensions. So, TC = {1/ (ħ C) ^4} x T = 4.34/9.714 x 10^103 x 10^17 = 0.446 x 10 ^ 120 Planck’s constant   ħ = 1.0545718 × 10-34 m2 kg / s The speed of light = 299 792 458 m / s = 2.99 x 10 ^ 8 m/s So, (ħ C) ^ 4 = {(1.05 x 2.99) x (10 ^ (-34 + 8))} ^4 = 97.14 x 10 ^ (-104) = 9.714 x 10 (-103) T (life time of universe, 13.799±0.021 billion years) = 4.34 x 10 ^17 s (second) This universe thus far should have a {TOATL quantum action COUNT} in an order of 10^120. Section three: Cosmological Constant Cosmology constant (Λ) is the value of the energy density of the vacuum of space. A positive vacuum energy density resulting from a cosmological constant implies a negative pressure, and vice versa. In terms of Planck units, and as a natural dimensionless value, the cosmological constant, λ, is on the order of 10^−120 or 10^−122. Thus far, the Cosmology constant (Λ) is a phenomenological parameter without a theatrical meaning, although it was ad hoc defined in the GR equation. Again, it cannot be derived with any mainstream quantum gravity theories. M-string theory cannot even determine its sign (being positive or negative). In this G-string quantum gravity, Cosmology constant (Λ) is defined as the ‘SHARE’ per quantum action. So, Cosmology constant (Λ) = {1/ (TOATL quantum action COUNT)} = {1/ [1/ (ħ C) ^4 x T]} = {1/0.446 x 10 ^ 120} = 2.242 x 10^-120 Section four: Conclusion Now, quantum gravity, entropy and Cosmology constant (Λ) are linked together. One, Quantum gravity: moving this universe from {[here (now), now] to [here (next), next]} with a quantum action ħ. Two, the total quantum action count is manifested as entropy. Three, the Cosmology constant (Λ) is the SHARE per quantum action in this quantum gravity.

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+0 # I need help on maths +1 152 2 The ratio of the number of marbles received by John and Peter was 4:7 respectively. The ratio of the number of marbles received by Peter and Sam was 9:5. John gave 1/12 of his marbles to Sam. Peter gave 1/9 of his marbles to Sam. Eventually, Sam (a)    Find the ratio of the number of John's marbles to Sam's marbles at first. (b)    Find the total number of marbles received by the 3 boys. Nov 4, 2021 #1 +1 Ratios can be scaled up or down by multiplication or division. So for example the ratio 4 : 7 is the same as 8 : 14 (having multiplied by 2), or 12 : 21 ( having multiplied by 3) and so on. With problems like this, where two ratios are involved, the thing to do is to arrange for two of the components, (in different ratios) to be equal to each other. So, multiply the first ratio by 9 and the second by 7 to arrive at J : P = 36 : 63, and P : S = 63 : 35. The two can then be shunted together so J : P : S = 36 : 63 : 35. The actual number of marbles for each person will then be (for some positive integer k), 36k for John, 63k for Peter and 35k for Sam. John then gives 3k and Peter gives 7k to Sam. That means that Sam now has 35k + 3k + 7k = 45k marbles and that has to equal 135, so k = 3, etc. . Nov 5, 2021 #2 +121 +1 Ratio of marbles of Peter by Sam, P : S = 9 : 5 Now,  J : P : S = 36 : 63 : 35 (a)    Ratio of John by Sam : J : S = 36 : 35 (b)    Let the number of marbles with John = 36x Peter = 63x Sam = 35x When John Gabe 1/12 of his marble to Sam and Peter gave 1/9 of his marbles to Sam. Then marbles with Sam = 35x + (36x * 1/12) + (63 * 1/9) = 35x + 3x + 7x = 45x According to Question 45x = 135 => x = 135/45 = 3 Total marbles + (36x + 63x + 35x) = 134x = 134 * 3 = 402 Nov 6, 2021

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Free Algebra Tutorials! Try the Free Math Solver or Scroll down to Tutorials! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: who invented algebra? Related topics: polonomial factorer | algebra 9th grade practice problems | algebra 6th grade inequalities | factoring online | math 6th grade calculator free online | multiply and divide rational expressions | College Algebra Clep Free Practice Test | terms cubed Author Message abusetemailatdrics Registered: 24.03.2006 From: Northern Illinois Posted: Friday 29th of Dec 15:24 I think God would have been in a really bad mood decided to come up with something called algebra to trouble us! I’ve spent days working on this algebra problem which relates to who invented algebra? and I still can’t solve it. I’m particularly having problems with side-angle-side similarity, least common denominator and 3x3 system of equations. Can anyone show me the way on how to go about solving such problems? I’ve tried various means that I could think of, but none helped. I need some urgent help now. Anybody? Vofj Timidrov Registered: 06.07.2001 From: Bulgaria Posted: Sunday 31st of Dec 11:58 I have no clue why God made math , but you will be delighted to know that someone also came up with Algebrator! Yes, Algebrator is a software that can help you crack math problems which you never thought you would be able to. Not only does it provide a solution the problem, but it also gives a detailed description of how it got to that solution. All the Best! Mibxrus Registered: 19.10.2002

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# XVA for Bermudan Swaptions Welcome back. Sorry for the hiatus which was longer than planned. Anyway and finally here is the (presumably) last episode in the series of posts on proxy pricing of exotics for XVA simulations. In my last posts I wrote a bit on the general design principles I came up with and how I used it to implement the FX TaRF. Today I will describe how to do similar things for bermudan swaptions using the same design principles. Some general thoughts beforehand though. Reflections on what we are doing here at all. What we have is some model $M$ in which we want to compute some XVA or potential exposure figure $\Xi$ for our exotic deal $I$. Possibly and probably together with a whole portfolio of other deals which interact in a non linear way, i.e. $\Xi( \sum_i I_i ) \neq \sum_i \Xi( I_i)$ To do this we need NPVs $\nu$ of each $I_i$ at some time $t$ conditional on some model state $m(t)$ in our model $M$: $\nu = \nu(I_i,t,m(t))$ $M$ might be operated in a real world measure (e.g. for exposure measurement) or a risk neutral measure (e.g. for CVA calculation), dependent on the nature of the figure to compute. The model state $m(t)$ implies a market data scenario like certain discount and forwarding term structures and swaption or cap volatility surfaces and the like. Ideally the NPV $\nu$ is computed in the same pricing model we use for our regular PL computation or trading activities, calibrated to the market data scenario created by $M$ in the way we would do it for usual pricing. That means in general we have an “outer” model $M$ and an “inner” model $N_i$ for each instrument $I_i$. Even if the instruments all allow for the same model for the inner evaluation, they will probably be different in their calibrated parameters: For example if we have a portfolio composed of single currency swaps, callable swaps and swaptions we could choose the Hull White model to evaluate the callable swaps and swaptions and price the swaps directly on the yield curves. But each callable swap / swaption would in general require a different set of calibration instruments, so we would have a set of local, inconsistent models for each individual swaption valuation. We could try to use a global model in which all instruments can be priced consistently and maybe use this model as the outer model for XVA calculation at the same time, provided that the XVA figure $\Xi$ allows for risk neutral valuation at all. This could be a Libor market model which is one of the most flexible models in terms of global calibration. The global calibration alone would be a delicate task however. If more than one currency is involved and maybe even other asset classes the task gets even more demanding. And what do you do if you want a real world exposure figure? This approach seems quite heavy, not impossible, but heavy. So it seems reasonable to stick with the concept of separate outer and inner models in general. The question is then how to efficiently compute the inner single NPVs $\nu(I_i,t,m(t))$ conditional on market scenarios generated by the outer model. Full pricing is not an option even for relatively simple inner models like Hull White one factor models. Another idea could be to compute some reference NPVs in time direction $t$ assuming some arbitrary market data scenarios (e.g. simply using today’s market data rolled forward), together with a “rich” set of sensitivities which allow to estimate NPVs under arbitrary market scenarios by Taylor approximation and interpolating in time direction (or use a theta sensitivity). Maybe automatic differentiation comes into play here. The idea I followed in the previous posts is different, more simple, and goes like this. During an initial single deal pricing performed by a monte carlo simulation we collect path data and build a regression model of conditional NPVs on the model state, pricing time and possibly more ancillary variables (like the accumulated amount in the case of the FX TaRF) just as it is done in the Longstaff Schwartz method for early exercise pricing (or almost like this, see below). Then, when asked for an NPV at a pricing time $t$ under a given market scenario $m(t)$ (and possibly more conditions like the already accumulated amount of a target structure) we imply the model state belonging to the market scenario and use our regression model to produce the desired NPV very efficiently. The limitation of this method is that the given market data can not be replicated in full beauty in general. In case of the FX TaRF we could replicate any given FX spot rate, but not its volatility which is fixed and implied by the original pricing model for example. In case of bermudan swaptions in the Hull White model we will see in this post that we can replicate one given reference rate, i.e. the general level of the yield curve, but not the shape of the curve, the spread to other yield curves involved and in particular not the volatility structure, which are all fixed again and implied by the model used for initial pricing. Actually what can be replicated is exactly corresponding to what is explicitly modelled as a stochastic quantity in the model: Since for the TaRF we used a Black model, possibly with local volatility but not with a stochastic factor for the volatility, the volatility is fixed. The FX spot on the other hand as the explicitly modelled quantity can be replicated to be any desired value. In case of the Hull White model we can imply the level of the yield curve, since this is modelled through the one factor in the model, but we can not match a given shape, since the model does not have the flexibility change the shape of the curve (the shape does change in fact, but in a fixed manner, not driven by stochastic factors in the model). The same with the volatility structure, it is implied by the initial pricing model again. If one accepts these limitations the method offers a very fast and handy way to approximate the desired scenario NPVs however. Let’s come to an example how this works for bermudan swaptions. I’d like to explain this going through an example I wrote. We start like this // Original evaluation date and rate level Real rateLevelOrig = 0.02; Date refDateOrig(12, January, 2015); Settings::instance().evaluationDate() = refDateOrig; // the yield term structure for the original pricing // this must _not_ be floating, see the warning in Handle<YieldTermStructure> ytsOrig(boost::make_shared<FlatForward>( refDateOrig, rateLevelOrig, Actual365Fixed())); The original pricing date is 12-Jan-2015 and the yield term structure on this date is at $2\%$ flat. We set QuantLib’s original pricing date to this date and construct the yield term structure. One important point is that the yield term structure has a fixed reference date, i.e. it does not change when the evaluation date changes. The reason is that this yield term structure will be linked to the initial pricing model which in turn will be used in the proxy pricing engine later on, which finally relies on the fact that the model does not change when shifting the evaluation date for scenario pricing. It is like having a fixed frame, as seen from the outer model described above, within which we do local pricings at different times and under different market scenarios, but relying on the frame surronding it to not change. Likewise the rates describing the yield term structure should not change, nor should the initial pricing model be recalibrated or change its parameters during the scenario pricings. The pricing model will be a Hull White model // the gsr model in T-forward measure, T=50 chosen arbitrary here // the first model uses the fixed yts, used for the mc pricing // generating the proxy boost::shared_ptr<Gsr> gsrFixed = boost::make_shared<Gsr>( ytsOrig, stepDates, sigmas, reversion, 50.0); The monte carlo pricing engine will be set up like this boost::shared_ptr<PricingEngine> mcEngine = MakeMcGaussian1dNonstandardSwaptionEngine<>(gsrFixed) .withSteps(1) // the gsr model allows for large steps .withSamples(10000) .withSeed(42) .withCalibrationSamples(10000) .withProxy(true); We do not need a fine time grid since the stochastic process belonging to the GSR model allows for exact evolution over arbitrary large time intervals. We use $10000$ steps to calibrate the Longstaff-Schwartz regression models and also for the final pricing. The seed is $42$, why not. The last attribute says that we want not only an usual pricing of the swaption, but also generate a proxy information object which can be used for scenario pricing later on. Since the generation consumes some additional time, it is optional to do it. Now we can do a good old pricing on our instrument swaption2 (I skipped how this was created). swaption2->setPricingEngine(mcEngine); Real npvOrigMc = swaption2->NPV(); Real errorOrigMc = swaption2->errorEstimate(); What I also did not show is that I created a reference integral engine to verify the prices of the mc engine and later on of the proxy engine also. The reference engine relies on a floating yield term structure, both with respect to the reference date and the rate level. The same holds for the proxy engine, so we can move forward in time, change the rate level and compare the pricings in the two engines. The result of the inital pricing is Bermudan swaption proxy pricing example Pricing results on the original reference date (January 12th, 2015): Integral engine npv = 0.0467257 (timing: 97723mus) MC engine npv = 0.0459135 error estimate 0.000809767 (timing: 5.55552e+06mus) The reference NPV from the integral engine is 467bp, slightly higher than the monte carlo price 459bp. This is perfectly expected since the regression model based approximate exercise decisions are sub-optimal, so the option NPV will be underestimated systematically. Together will the error estimate (one standard deviation) of 8bp this all looks quite ok. One word about the computation times. The monte carlo simulation (10000 paths both for calibration and pricing) takes 5.5 seconds. This is somewhat in the expected region and I am completely relying on the standard monte carlo framework of the library here. The integral engine on the other hand takes 97 milli seconds, which is quite slow. This can easily be brought down to 5 milliseconds just by using less integration points and covered standard deviations (here I used 129 points and 7 standard deviations), without loosing too much accuracy. This seems quite ok, since neither the GSR model nor the integral engine are much optimized for speed currently. To create the proxy engine we can say boost::shared_ptr<PricingEngine> proxyEngine = boost::make_shared<ProxyNonstandardSwaptionEngine>( swaption2->proxy(), rateLevelRef, maturityRef, 64, 7.0, true); where we just pass the proxy result from the pricing above together with quotes representing the rate level of the scenario pricing and the maturity to which the rate level belongs. Remember that we can only prescribe the level, but not the shape of the yield curve in the scenario pricing. So I allowed to chose a maturity, e.g. 10 years, and prescribe the (continuously compounded) zero yield w.r.t. this maturity. The proxy engine will then imply the Hull White’s model state such that this rate level is matched for the given maturity. The shape of the yield curve is implied by the initial model though and can not be changed. I am repeating myself, ain’t I ? Do I ? Repeat ? God. The next two parameters refer to the way we do the scnenario pricing between two of the original structure’s exexercise dates: We determine the next available exercise date (if any) and build a grid on the exercise date covering $n$ standard deviations (here $5.0$) of the state variable around its mean, all this conditional on the pricing time and the model state (which is implied by the rate level as explained above). We then compute the NPV on each of these grid points using the regression model from the initial monte carlo pricing. Finally we interpolate between the points using cubic splines and integrate the resulting function against the state variable’s density (which is possibly in closed form by the way). Let’s see what we get under some scenarios for the evaluation date and rates. One scenario is simply created by e.g. saying Settings::instance().evaluationDate() = Date(12, June, 2015); rateLevelRefQuote->setValue(0.02); maturityRefQuote->setValue(10.5); which moves the evaluation date half a year ahead to 12-Jun-2015 and requires the 10.5y zero rate (which corresponds to the maturity of the swaption as the reference point) to be 2%. For the scenarios in the example we get Pricing results on June 12th, 2015, reference rate 0.02 with maturity 10.5 Integral engine npv = 0.0444243 Proxy engine npv = 0.0430997 (timing: 127mus) Pricing results on June 11th, 2019, reference rate 0.025 with maturity 6.5 Integral engine npv = 0.0372685 Proxy engine npv = 0.0361514 (timing: 170mus) Pricing results on June 11th, 2019, reference rate 0.02 with maturity 6.5 Integral engine npv = 0.0223442 Proxy engine npv = 0.0223565 (timing: 159mus) Pricing results on June 11th, 2019, reference rate 0.015 with maturity 6.5 Integral engine npv = 0.0128876 Proxy engine npv = 0.0141377 (timing: 169mus) Pricing results on January 11th, 2020, reference rate 0.02 with maturity 6 Integral engine npv = 0.0193142 Proxy engine npv = 0.0194446 (timing: 201mus) Pricing results on January 11th, 2021, reference rate 0.02 with maturity 5 Integral engine npv = 0.0145542 Proxy engine npv = 0.0137726 (timing: 208mus) Pricing results on June 11th, 2024, reference rate 0.02 with maturity 1.5 Integral engine npv = 0.00222622 Proxy engine npv = 0.00292998 (timing: 282mus) which looks not too bad. The proxy engine’s computation time is also not bad, around 100-300 microseconds. Again, this can be made faster by using less integration points, but already seems competitive enough for XVA simulations. The proxy regression model deserves some comments. It is not just the Longstaff-Schwartz regression model, since this only looks at states where the exercise value is positive. This is essential for a good fit of the regression function (which I chose to be simply a quadratic function), since left from the exercise point the option value (which equals the continuation value in this area) flattens and would destroy the global fit. What I did until now is just calibrate two separate models, one in the region corresponding to positive exercise values and another in the complementary region. This is similar to the regression model for the FX TaRF, where I also used two quadratic polynomials, if you remember (if not, you can read about it here or more detailled in the paper). This solution is not perfect and can be improved probably. The following picture shows the regression models from the example. The different exercise dates are distinguished by color (0 denotes the first exercise date, in dark color, 1 the second, and so on until 9 which denotes the 10th exercise date, in bright color). What you can see is the discontinuity at the cutoff point between exercise and non-exercise area. It is exactly this jump that one should work on I guess. But not now. In technical terms I relied on Klaus’ implementation of the Longstaff Schwartz algorithm in longstaffschwarzpathpricer.hpp. This needed some amendments to allow for non-american call rights which it was designed for I believe. Apart of this I mainly inserted a hook in the pricer virtual void post_processing(const Size i, const std::vector<StateType> &state, const std::vector<Real> &price, const std::vector<Real> &exercise) {} with an empty default implementation. This function is called during the calibration phase, passing the collected data to a potential consumer, which can do useful things with it like building a regression model for proxy pricing like in our use case here. To do this I derived from Klaus’ class as follows class LongstaffSchwartzProxyPathPricer : public LongstaffSchwartzPathPricer<typename SingleVariate<>::path_type> { public: LongstaffSchwartzProxyPathPricer( const TimeGrid &times, const boost::shared_ptr<EarlyExercisePathPricer<PathType> > &pricer, const boost::shared_ptr<YieldTermStructure> &termStructure); const std::vector<boost::function1<Real, StateType> > basisSystem() const { return v_; } const std::vector<Array> coefficientsItm() const { return coeffItm_; } const std::vector<Array> coefficientsOtm() const { return coeffOtm_; } const StateType cutoff() const { return cutoff_; } which offers inspectors for the two regression models coefficients, the cutoff point separating them and the underlying function system. The implementation of the hook function is simple void LongstaffSchwartzProxyPathPricer::post_processing( const Size i, const std::vector<StateType> &state, const std::vector<Real> &price, const std::vector<Real> &exercise) { std::vector<StateType> x_itm, x_otm; std::vector<Real> y_itm, y_otm; cutoff_ = -QL_MAX_REAL; for (Size j = 0; j < state.size(); ++j) { if (exercise[j] > 0.0) { x_itm.push_back(state[j]); y_itm.push_back(price[j]); } else { x_otm.push_back(state[j]); y_otm.push_back(price[j]); if(state[j]>cutoff_) cutoff_ = state[j]; } } if (v_.size() <= x_itm.size()) { coeffItm_[i - 1] = GeneralLinearLeastSquares(x_itm, y_itm, v_).coefficients(); } else { // see longstaffschwartzpricer.hpp coeffItm_[i - 1] = Array(v_.size(), 0.0); } if (v_.size() <= x_otm.size()) { coeffOtm_[i - 1] = GeneralLinearLeastSquares(x_otm, y_otm, v_).coefficients(); } else { // see longstaffschwartzpricer.hpp coeffOtm_[i - 1] = Array(v_.size(), 0.0); } } and does nothing more than setting up the two data sets on which we estimate the regression models. The coefficients are read in the monte carlo engine and serve as the essential part in the proxy object which can then be passed to the proxy engine. And we are done. You can find the full example code here. # Adjoint Greeks IV – Exotics Today I feature a sequel to the adjoint greeks drama (see my prior posts on this). Before I start I would like to point you to a new and excellent blog authored by my colleague Matthias https://ipythonquant.wordpress.com/. You will want to follow his posts, I am certain about that. I am still on my way to convert the essential parts of the library to a template version. Since this is boring work I created a small friend who helps me. No I did not go mad. It is a semi-intelligent emacs-lisp script that does some regular expression based search-and-replacements which speeds up the conversion a lot. Emacs is so cool. Today I am going to calculate some derivatives of a bermudan swaption in the Gsr / Hull White model. This is the first post-vanilla application and admittedly I am glad that it works at last. One point I have to make today is that AD is slow. Or to put it differently, doubles can be tremendously fast. Lookit here: void multiply(T *a, T *b, T *s) { for (int i = 0; i < N; ++i) { for (int k = 0; k < N; ++k) { for (int j = 0; j < N; ++j) { s[i * N + j] += a[i * N + k] * b[k * N + j]; } } } } This code multiplies two matrices a and b and stores the result in s. It is the same algorithm as implemented in QuantLib for the Matrix class. When feeding two 1000 x 1000 double matrices the code runs 1.3s on my laptop if compiled with gcc 4.8.2 using -O1. With -O3 I get 950ms. With clang 3.7.0 (fresh from the trunk) I get 960ms (-O1) and 630ms (-O3). With T=CppAD::AD on the other hand the running time goes up to at least (-O3) 10.9s (gcc) and 14.3s (clang). Code optimization seems to be a delicate business. These timings refer to the situation where we use the AD type without taping, i.e. only as a wrapper for the double in it. This seems to indicate that at least for specific procedures it is not advisable to globally replace doubles by their active counterpart if one is not really interested in taping their operations and calculating the derivatives. Performance may break down dramatically. What is the reason behind the huge difference ? I am not really the right person to analyze this in great detail. The only thing I spotted when looking into the generated assembler code is that with double there are SIMD (single instruction multiple data) instructions for adding and multiplying in the nested loops (like addpd and mulpd, it’s a long time since I programmed in assembler and it was on a 6510 so I am not really able to read a modern x86 assembler file). With CppAD::AD there doesn’t seem to be such instructions around. So part of the perfomance loss may be due to the inability of streaming AD calculations. For sure this is not the only point here. The second point to make today is that AD has pitfalls that may in the end lead to wrong results, if one uses the AD framework blindly. Let’s come back to our specific example. The underlying source code can be found here if you are interested or want to run it by yourself. It is a bermudan swaption, ten years with yearly exercise dates. The model for pricing will be the Gsr or Hull White model. We just want to compute the bucket vegas of the bermudan, i.e. the change in its NPV when the implied market volatility of the canonical european swaptions used for the model calibration is increased by one percent. The rate level is set to $3\%$ and the strike is out of the money at $5\%$. The volatilities are produced by SABR parameters $\alpha=3\%$, $\beta=60\%$, $\nu=12\%$, $\rho=30\%$. All of this is arbitrary, unrealistic and only for explanatory purposes … The naive AD way of doing this would be to declare the input implied volatilities as our quantities of interest CppAD::Independent(inputVolAD); and then go through the whole model calibration gsrAD->calibrateVolatilitiesIterative( and pricing yAD[0] = swaptionAD->NPV(); to get the derivative d bermudan / d impliedVol CppAD::ADFun<Real> f(inputVolAD, yAD); vega = f.Reverse(1, w); When I first wrote about AD I found it extremely attractive and magical that you could compute your sensitivities like this. That you can actually differentiate a zero search (like in the case of yield curve bootstrapping) or an optimization (like the Levenberg-Marquardt algorithm which is used here). However there are dark sides to this simplicity, too. Performance is one thing. The calibration step and the pricing including the gradient calculation takes AD model calibration = 1.32s We can also do it differently, namely calibrate the model in an ordinary way (using ordinary doubles), then compute the sensitivity of the bermudan to the model’s sigma and additionally compute the sensitivity of the calibration instruments to the model’s sigma. Putting everything together yields the bermudan’s bucketed vega again. I will demonstrate how below. First I report the computation time for this approach: model calibration = 0.40s This leaves us with a performance gain of around 15 percent (7.32s vs 8.43s). This is not really dramatic, still significant. And there is another good reason to separate the calibration from the greek calculation which I will come to below. Note also, that the pricing which takes 5.95s with AD (including derivatives) is much faster without AD where it only consumes 0.073s. This is a factor of 80 which is much worse than the theoretical factor of 4 to 5 mentioned in earlier posts (remember, we saw 4.5 for plain vanilla interest rate swap npv and delta computation). This is again due to optimization issues obviously. The background here is that the swaption pricing engine uses cubic spline interpolation and closed-form integration of the resulting cubic polynominals against the normal density for the roll back. Again a lot of elementary calculations, not separated by OO-things that would hinder the compiler from low level optimizations. You would surely need quite a number of sensitivities to still get a performance gain. But lets follow the path to compute the bucket vegas further down. First I print the bucket vegas coming from the naive AD way above – direct differentiation by the input implied volatilities with the operation sequence going through the whole model calibration and the pricing. vega #0: 0.0153% vega #1: 0.0238% vega #2: 0.0263% vega #3: 0.0207% vega #4: 0.0209% vega #5: 0.0185% vega #6: 0.0140% vega #7: 0.0124% vega #8: 0.0087% This looks plausible. We started the calibration from a constant sigma function at $1\%$. This is actually far away from the target value (which is around $0.40\%$), so the optimizer can walk around before he settles at the minimum. But we could have started with a sigma very close to the optimal solution. What would happen then ? With the target value as an initial guess (which is unlikely to have in reality, sure) we get vega #0: 0.0000% vega #1: 0.0238% vega #2: 0.0263% vega #3: 0.0207% vega #4: 0.0209% vega #5: 0.0448% vega #6: 0.0000% vega #7: 0.0000% vega #8: 0.0000% Some vegas are zero now. vega #5 is even completely different from the value before. This is a no go, because in a productive application you wouldn’t notice if some deals are contributing a zero sensitivity or a false value. What is happening here is that the function we differentiate depends on more input variables than only the primary variables of interest (the implied vol), like the initial guess for the optimization. These might alter the derivative drastically even if the function value (which is the model’s sigma function on an intermediate level, or ultimately the bermudan’s npv) stays the same. For example if the initial guess is so good that the optimizers tolerance is already satisfied with it, the output will stay the same, no matter if the input is pertubed by an infinitesimal shift $dx$. Here pertubed does not really mean bumped and $dx$ really is infinitesimal small. It is only a concept to get a better intuition of what is going on during the process of automatic differentiation. Another example is the bisection method for zero searching. This method will always yield zero AD derivatives in the sense if $x$ is an input (e.g. a vector of swap quotes) and $y$ is the output (e.g. a vector of zero rates) linked by a relation $f(x,y) = 0$ the iff $x$ is pertubed by $dx$ then the checks in the bisection algorithm will yield exactly the same results for $x+dx$ as for $x$. Therefore the computed $y$ will be exactly the same, thus the derivative zero. I don’t know if this explanation is good, but this is how I picture it for myself. It seems to be like this: If it feels too magical what you do with AD, you better don’t do it. What is the way out ? We just avoid all optimization and zero searches, if possible. And it is for interest rate deltas and vegas: Just calibrate your curve in the usual way, then apply AD to compute sensitivities to zero rates (which then does not involve a zero search any more). If you want market rate deltas, compute the Jacobian matrix of the market instruments in the curve by the zero rates as well, invert it and multiply it with the zero delta vector. You do not even have to invert it, but just only have to solve one linear equation system. We make this procedure explicit with our example above. The first step would be to compute d bermudan NPV / d model’s sigma. This is done on the already calibrated model. Technically we separate the calibration step (done with the usual doubles) and the AD step (done on a copy of the model feeded with the model parameters from the first model, but not itself calibrated again). This is what we get for d bermudan / d sigma vega #0: 1.6991 vega #1: 1.2209 vega #2: 0.8478 vega #3: 0.5636 vega #4: 0.4005 vega #5: 0.2645 vega #6: 0.1573 vega #7: 0.0885 vega #8: 0.0347 Next we compute the Jacobian d helperNpv / d sigma 297.4% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 179.7% 183.3% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 126.9% 129.4% 132.2% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 90.8% 92.6% 94.6% 96.7% 0.0% 0.0% 0.0% 0.0% 0.0% 64.9% 66.2% 67.6% 69.1% 71.2% 0.0% 0.0% 0.0% 0.0% 47.5% 48.4% 49.5% 50.6% 52.1% 53.3% 0.0% 0.0% 0.0% 31.9% 32.6% 33.3% 34.0% 35.0% 35.9% 36.3% 0.0% 0.0% 19.2% 19.6% 20.0% 20.4% 21.0% 21.5% 21.8% 22.3% 0.0% 8.7% 8.9% 9.0% 9.2% 9.5% 9.8% 9.9% 10.1% 10.5% helperVega = f.Jacobian(sigmas); This is a lower triangular matrix, because the ith calibration helper depends only on the sigma function up to its expiry time. The inverse of this matrix is also interesting, although we wouldn’t need it in its full beauty for our vega calculation, representing d sigma / d helperNpv 33.6% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% -33.0% 54.6% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% -53.4% 75.6% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% -74.0% 103.5% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% -100.4% 140.4% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% -137.2% 187.5% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% -185.1% 275.3% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% -269.1% 448.1% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% -431.6% 953.1% This says how the different sigma steps go up when the helper belonging to the step goes up (these are the positive diagonal elements) and how the next sigma step would need to go down when the same helper as before goes up, but the next helper stays the same. The contra-movement is roughly by the same amount. Next we need d bermudan / d sigma. AD delivers vega #1: 1.6991 vega #2: 1.2209 vega #3: 0.8478 vega #4: 0.5636 vega #5: 0.4005 vega #6: 0.2645 vega #7: 0.1573 vega #8: 0.0885 vega #9: 0.0347 And the last ingredient would be the calibration helpers’ vegas, which can be computed analytically (this is the npv change when shifting the input market volatility by one percent up), d helperNpv / d marketVol vega #1: 0.0904% vega #2: 0.1117% vega #2: 0.1175% vega #3: 0.1143% vega #4: 0.1047% vega #5: 0.0903% vega #6: 0.0720% vega #7: 0.0504% vega #8: 0.0263% Now multiplying everything together gives d bermudan / d impliedVol = d bermudan / d sigma x d sigma / d helperNpv x d helperNpv / d impliedVol which is vega #0: 0.0153% vega #1: 0.0238% vega #2: 0.0263% vega #3: 0.0207% vega #4: 0.0209% vega #5: 0.0185% vega #6: 0.0140% vega #7: 0.0124% vega #8: 0.0087% same as above. Partly because I just copy-pasted it here. But it actually comes out of the code also, just try it.

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Upcoming SlideShare × # Integers powerpoint 36,534 views Published on An introduction to integers - what they are, and how to add them Published in: Education, Technology • Full Name Comment goes here. Are you sure you want to Yes No • thanks I found this really helpful Are you sure you want to  Yes  No ### Integers powerpoint 1. 1. Mrs. Haataja 6th Grade Math 2. 2.  Are numbers  Can be positive or negative  Do NOT have decimals  Do NOT have fractions 3. 3. INTEGERS NON- INTEGERS  5  10059  -33  -478  29  4.5  -3.33  2/3  -6 ½  -83.9 4. 4. When you count, you say 1, 2, 3, 4, 5… You are listing INTEGERS  To count in the other direction, it is -1, -2, -3, -4, - 5… 5. 5. POSITIVE TEMPERATURE READING ALL NUMBERS IN THIS PHOTO ARE POSITIVE AVERAGE PER GAME VALUE IS NEGATIVE (THIS IS AN OLD IPOD!) 6. 6. POSITIVE NEGATIVE  When the temperature is above zero  When you are above ground  When you earn money  In hockey, when your team scores a goal while you are on the ice  When the temperature is below zero  When you are below ground  When you spend money  In hockey, when the other team scores a goal while you are on the ice 7. 7.  Positive integers increase in increments of one as you move to the right  Negative integers mirror this moving to the left of zero, however, even though they appear to be getting bigger, they are actually getting smaller  -5 is one number smaller than -4, for example 8. 8. You already know how to add positive integers…you just haven’t called them integers before! 9. 9.  5 + 3 = 8  You just added the integers positive 5 and positive 3 and found out the answer was positive 8! 10. 10.  Adding negative integers is just like adding positive integers…only difference…the answer will have a negative sign  -6 +(-5) = -11  All you do is ignore the negative signs and add the numbers, then put the negative sign back in your answer  THIS ONLY WORKS IF BOTH NUMBERS ARE NEGATIVE!! 11. 11.  On the previous slide we saw a diagram of the problem 6 + (-2)  To solve this, they started at zero and moved to the right 6 places – to represent the positive 6  Next, they moved left 2 places – this represented the -2  They landed on the number 4 (positive 4) which is the answer 12. 12.  Using a number line is a great way to practice when you are first learning to add numbers with different signs – but there is another way  We can use the absolute value of a number and subtraction to solve addition problems involving integers of different signs 13. 13.  Absolute value is the distance between a number and zero  It does not matter which direction you are moving to get back to zero – all that is important is how many steps it takes to get there 14. 14.  It is written as vertical lines on either side of a number  |12|, is 12  |-23|, is 23  Absolute values are ALWAYS positive 15. 15.  Here’s how with the help of absolute value…  1) find the absolute value of each number being added  2) subtract the smaller absolute value from the larger absolute value  3) the sign of the number with the larger absolute value is the sign of the answer 16. 16.  21 + 29 = 50  -21 + (-29) = -50  29 + (-21) = 8 |29| = 29 |-21| = 21 29 – 21 = 8 29 is +, so answer is +  21 + (-29) = -8 |-29| = 29 |21| = 21 29 – 21 = 8 29 is -, so answer is - 17. 17. All photos from flickr.com, all have a creative commons license  Slide 2: Photographer: Ciccio Pizzettaro Title: Question Mark Taken: 1/23/2010  Slide 3: Photographer: Kenyee Title: Floating Integers Taken: 8/22/2008  Slide 6 Photographer Rob Friesel found_drama Title: Is that a…positive integer? Taken: 1/25/2013  Slide 6: Photographer: Frank Wales Title: I should have quit while I was ahead Taken: 11/19/2007 18. 18.  Slide 8: Photographer: Ethan Hein Title: Even Integers Subgroup Taken: 4/11/2008  Slide 13: Photographer gfinder Title: Goalfinder Math integer-numberline 1 Taken: 3/1/2012  Slide 16 Photographer: gfinder Title: Goalfinder Math Absolute-value-of-an- integer Taken: 3/1/2012  Slide 20 Photographer danmachold Title: clock Taken: 2/3/2011

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Motor Calculations, Part 2: Feeders What is the correct way to size motor feeders and related overcurrent protection? Mike Holt for EC&M Magazine In Part 1, we looked at how to size motor overloads. We also looked at how to size circuit short-circuit and ground-fault protection for motor branch circuits. The key lesson learned there is motor overload protection requires separate calculations from short-circuit and ground-fault protection. Understanding this fact clears up a common source of confusion and a common point of error. But, another source of confusion arises when it comes to sizing short-circuit and ground-fault protection for a feeder that supplies more than one motor. Let's look again at branch calculations, then resolve the feeder issues so your calculations will always be correct. Branch-circuit conductors and protection devices Per 430.6(A), branch-circuit conductors to a single motor must have an ampacity of not less than 125% of the motor FLC as listed in Tables 430.147 through 430.150. To illustrate this, let's size the branch-circuit conductors (THHN) and short-circuit ground-fault protection device for a 3-hp, 115V, 1Ø motor. The motor nameplate FLC is 31A, and we're using dual-element fuses for short-circuit and ground-fault protection. See Figure 7-16. • The FLC current per Table 430.148 is 34A. • 34A x 125% = 43A. • Per Table 310.16 (60°C terminals), the conductor must be a 6 AWG THHN rated 55A. You size the branch-circuit short-circuit and ground-fault protection devices by using multiplication factors based on the type of motor and the type of protection device per the motor FLC listed in Table 430.52. When the protection device values determined from Table 430.52 do not correspond with the standard rating of overcurrent protection devices listed in 240.6(A), you must use the next higher overcurrent protection device. To illustrate this, let's use the same motor as in the previous example. • Per [240.6(A)], multiply 34A x 175% • You need a 60A dual-element fuse. To explore this example further, see Example No. D8 in Annex D. Once you've sized the motor overloads, branch circuit conductors, and branch circuit protective devices, you are ready to move on to the next step. Motor feeder conductor calculations From 430.24, we see conductors that supply several motors must have an ampacity that is not less than: • 125% of the highest-rated motor FLC [430.17], plus • the sum of the FLCs of the other motors (on the same phase), as determined by [430.6(A)], plus • the ampacity required to supply the other loads on that feeder. Let's look at Figure 7-12 and solve the following problem: For what ampacity must you size the feeder conductor, if it is supplying the following two motors? One is 71/2-hp, 230V (40A), 1Ø; the other is 5-hp, 230V (28A), 1Ø. The terminals are rated for 75ºC. (a) 50A (b) 60A (c) 70A (d) 80A How did we arrive at this answer? Let's walk through the solution. • The largest motor is 40A. • 40A x 1.25 + 28A = 78A. • Rounding up to the next ten gives us 80A. What size conductor would give us this ampacity? (a) 2 AWG (b) 4 AWG (c) 6 AWG (d) 8 AWG Let's see how to select the correct answer. By looking at Table 310.16, we find a 6 AWG at 75ºC provides 65A of ampacity, so it is too small. However, a 4 AWG conductor provides 85A of ampacity, which will accommodate the 80A we need. Therefore, you need to size this feeder conductor at 4 AWG. Next, we have to answer another question. What size overcurrent protection device (OCPD) must we provide for a given feeder? Using a slightly more complex example, try sizing the feeder conductor (THHN) and protection device (inverse-time breakers 75°C terminal rating) for the following motors (Figure 7-17): • Three 1-hp, 120V, 1Ø motors • Three 5-hp, 208V, 1Ø motors • One wound-rotor 15-hp, 208V, 3Ø motor Did you get the right answers? Let's walk through this and see. Our references are 240.6(A), 430.52(C)(1), Tables 430.148, and Table 430.52. We'll start by determining the ampacities required for each size of motor, then walk through each step until we arrive at the correct OCPD size. • 1-hp: FLC is 16A. 16A x 250% = 40A • 5-hp: FLC is 30.8A. 30.8A x 250% = 77A. Next size up = 80A. • 15-hp: FLC is 46.2A. 46.2A x 150% (wound-rotor) = 69A: Next size up = 70A. Now, let's look at the feeder conductor. Conductors that supply several motors must have an ampacity of not less than 125% of the highest-rated motor FLC [430.17], plus the sum of the other motor FLCs [430.6(A)]. See Figure 7-18. Continuing on with this example, add up all the ampacities, multiplying the highest rated motor by 125%. Thus, (46.2A x 1.25) + 30.8A + 30.8A + 16A = 136A. From Table 310.16, we see we need 1/0 AWG THHN because at 150A it is the smallest conductor that accommodates the 136A of ampacity we're working with. When sizing the feeder conductor, be sure to include only the motors that are on the same phase. For that reason, we used only four motors in this calculation. You must provide the feeder with a protective device having a rating or setting not greater than the largest rating or setting of the branch-circuit short-circuit and ground-fault protective device (plus the sum of the full-load currents of the other motors of the group) [430.62(A)]. Remember, motor feeder conductors must have protection against the overcurrent resulting from short circuits and ground-faults but not those resulting from motor overload. Let's illustrate this with a sample motor feeder protection calculation. What size feeder protection (inverse-time breaker) do you need for the following two motors? They are a 5-hp, 230V, 1Ø motor and a 3-hp, 230V, 1Ø motor. Refer to Figure 7-13. (a) 30A breaker (b) 40A breaker (c) 50A breaker (d) 80A breaker Let's walk through the solution. 1. Get the motor FLC from Table 430.148. • A 5-hp motor FLC is 28A. • A 3-hp motor FLC is 17A. 2. Size the branch-circuit protection per the requirements of 430.52(C)(1), Table 430.52, and 240.6(A)] • 5-hp: 28A x 2.5 = 70A. • 3-hp: 17A x 2.5 = 42.5A. The next size up is 45A. 3. Size the feeder conductor per 430.24(A). • The largest motor is 28A. • (28A x 1.25) + 17A = 52A. • Table 310.16 shows 6 AWG rated 55A at 60ºC as the smallest conductor with sufficient ampacity. 4. Size the feeder protection per 430.62. • It must not be greater than the 70A protection of the branch circuit plus the 17A of the other motor (which is the total of all loads on that feeder). • 70A + 17A = 87A. • The next size down is 80A, so that is the breaker you choose. Here is where many people get confused and say something must be wrong. How can you be safe if you are selecting the next size down instead of the next size up? Remember, you have already accounted for all the loads and the NEC requires that you not exceed the protection of the branch circuit. Again, keep in mind that you are not calculating for motor overload protection, here. Motor calculations are different from other calculations. With motor feeders, you are calculating for protection from short circuits and ground faults, only-not overload. That seemed like a lot of steps, but by proceeding methodically you will arrive at the correct answer. See Example D8 in Annex D of the NEC, for more detail. Also, see Figure 7-19. When sizing the feeder protection, be sure to include only the motors that are on the same phase. For that reason, we used only four motors in this calculation. Putting it all together Motor calculations get confusing when you forget there's a division of responsibility in the protective devices. To get your motor calculations right, you must separately calculate the motor overload protection (typically located near the motor), the branch circuit protection (from short circuits and ground faults), and the feeder circuit protection (from short circuits and ground faults). Remember that overload protection is at the motor, only. Any time you find yourself confused, just refer to NEC Figure 430.1. It allows you to clearly see the division of responsibility between different forms of protection in motor circuits. Example D8 illustrates this with actual numbers. Keeping this division of responsibility in mind will allow you to make correct motor calculations every time.

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# Pulse Counting Logic for Water Meter Hi All I’m trying to implement pulse counting on my water meter and I’m struggling to get daily feed totals and also I want to measure how much water I use for showers but for the logic that I’ve attached below, once someone uses the shower, it will add the previous water use for that day. Dave - Sorry I have no solutions just wanted you to know you are not alone! I’m struggling with the same water challenges. My input is not a pulse but a running total. I’ve only been able to do these two types or graphs: EDIT: In your Water Meter (Node 8) I seem to remember the Accumulator should appear first. I think this had to do with power outages and dropping the count. I am trying to find a reference. I am wrong. Also, what is the “x .0001” in line 4 calculate? Is each pulse a Liter? or a 1000th of a Liter? Yeh with the pulse input it will be just a count of the usage, the x0.0001?? My water metre pulses every litre, so the above value converts it into cubic meters, I.E. 2000 litres used is 2m3. TBH the only real use of pulse is for gas and water as they are measured by volume but I do agree that I’m struggling to find the right processors in EmonCMS. Regards Dave Surely 1000 litres per cu.m, not 10,000? Yeh it is 0.001 on my logic, there’s an extra zero that’s crept in The error was mine. The image in the first post is 0.001. Dave - There is a well written guide written by Steve @smitt1979 in the old forum that may help → Water Meter Guide Updated v2.0 The issue mentioned in that linked thread, at reboot, when using a WhAccumulator will probably be due to emoncms not knowing what the last raw pulsecount was since it is not persisted, as far as i know it is only held in RAM using redis. Rather than use a WhAccumulator, try using a “Total pulse count to pulse increment” (Tpctpi) followed by a regular accumulator. The Tcptci will record each raw pulsecount,That way at reboot it can calculate the correct pulse increment regardless of whether it is a quick reboot or a prolonged outage as the increment will always be based on the value used to calculate the last increment, This is not data that can be used for anything else except debugging so some may argue it’s a waste of diskspace, but I say it’s a small price to pay for reliable data with a built-in debugging tool. @Dave The main issue with your your processing (reboot issues aside) is the feeding of one “ratchet type” processor into another. Both the Tcptci and WhAccumulator will only process increases in the input values as expected a lower value will not, A WhAccumulator just discards the first lower value and bases the next calc on that new lower value where as a Tpctpi will still process the first lower value and also base the next calc on it. since a pulsecount will fluctuate up and down there will be a significant amount of data ignored if these are daisy chained… for example passing values 2,4,2,4,2,4 into a “WhAccumulator” process one after the other would give you 2+2+0+2+0+2 = 8, passing the same the totals into a Tcptci would give you 2+2+2+2+2+2 = 12, should be 18 if they are individual increments rather than an incrementing total. On the first “shower” processlist changing #21 to a straight Accumulator should work ok, including the kWh to kWh/d which is effectively a simple “daily total” accumulator, the second “water meter” would translate to • Total pulse count to pulse increment “WaterMeterRawPulse” (much more useful than a “log to feed”) • Max daily value “WaterMeterMaxDailyLitres” (assuming one pulse is one litre) • Accumulator “WaterMeterTotalLitres” (replaces the first WhAccumulator) • x 0.001 (scale to cubic meters) • Accumulator “WaterMeterTotalM3” (replaces the second WhAccumulator) • Max daily value “WaterMeterMaxDailyM3” This only applies to self hosted emoncms as unfortunately the Tpctpi process is not available if using emoncms.org so you have to use the WhAccumulator, but as the server is “rebooted” less it should be less of an issue. The ignored first lower values could still be an issue though. Thanks Paul I’ve adjusted the logic but I seem to be having the WaterMeterCost feed counting up without any pulses? Max daily value must go before the accumulator as the accumulator returns an ever increasing total rather than the increase, so for that same reason we can’t daisy chain accumulators for different scales (sorry didn’t spot this as I haven’t tried it) swapping out the second and subsequent accumulators to log to feeds will get all the accumulating totals working correctly as it will blindly scale the value returned by the preceding process and record it. eg • Total pulse count to pulse increment “WaterMeterRawPulse” • Max daily value “WaterMeterMaxDailyLitres” • Accumulator “WaterMeterTotalLitres” • x 0.001 (scale to cubic meters) • Log to feed “WaterMeterTotalM3” (replaces the second accumulator) • x 3.13 (price per M3?) • Log to feed “WaterMeterTotalCost” but that will remove the ability to record the daily max’s, so you need to find another way of doing it eg • Total pulse count to pulse increment “WaterMeterRawPulse” • Max daily value “WaterMeterMaxDailyLitres” • Accumulator “WaterMeterTotalLitres” • x 0.001 (scale to cubic meters) • Total pulse count to pulse increment “WaterMeterM3Raw” • Max daily value “WaterMeterMaxDailyM3” • Accumulator “WaterMeterTotalM3” • x 3.13 (price per M3?) • Total pulse count to pulse increment “WaterMeterCostRaw” • Max daily value “WaterMeterMaxDailyCost” • Accumulator “WaterMeterTotalCost” or perhaps to avoid 2 more semi-redundant Tpctpi feeds you could subtract the last accumulated feed value before logging the max daily and accumulating the increment. • Total pulse count to pulse increment “WaterMeterRawPulse” • Max daily value “WaterMeterMaxDailyLitres” • Accumulator “WaterMeterTotalLitres” • x 0.001 (scale to cubic meters) • feed “WaterMeterTotalM3” (to give you an increase in M3) • Max daily value “WaterMeterMaxDailyM3” • Accumulator “WaterMeterTotalM3” • x 3.13 (price per M3?) • feed “WaterMeterTotalCost” (to give you an increase in cost) • Max daily value “WaterMeterMaxDailyCost” • Accumulator “WaterMeterTotalCost” Thanks Paul I’ve updated the logic as attached, I’ll have a play about with it tonight to make sure that it works as it should do. I’m still getting issues with bouncing from the meter, I’ve put a resistor and a capacitor in to try and cure it but no joy. There is a sketch about that Robert Wall produced to provide a software debounce but it doesn’t work on my EmonTX v3.4? You mean this one? It was for the V3.2, In the V3.4, the pulse input has moved from DIO2 to DIO3 and changed from IRQ0 to IRQ1, and you’ll need to change RFu_JeeLib back to the standard JeeLib, and if you’re using the RFM69CW you’ll need to add the line for that too. I think that’s all that needs to be changed, but I could well have missed something. Hi Robert Yes, that is the one, I’ve been fiddling with it this evening and managed to make the emontx run, but I’ve been unsuccessful in getting it to transmit or set the pulse pin to No.3. If you’ve fiddled, I can’t help because I don’t know what you have changed. As the sketch “runs”, my first guess is you have JeeLib correctly changed, it should transmit if you’ve got the right radio set (12B or 69CW). Looking again at the sketch, it might be easier to transplant the pulse bits from that sketch into a modern V3.4 sketch, (but take out any existing interrupt-driven pulse counting first) because a lot of other things have changed since that was written. Basically, the bits you need are: 1. The piece at the beginning of setup( ) that sets up the Timer 1, down to and including “interrupts();” 2. The piece towards the bottom of loop( ) that accumulates the count of pulses, again down to and including “interrupts();” 3. The interrupt service routine immediately after the end of loop( ) function. Of course, you’ll still need to change the input pin in the ISR to whichever one your switch is connected to, which is probably 3. Thanks Robert I think from my experience last night its a little over my head, TBH I couldn’t even work out where I needed to change the IRQ pin. I would be happy thou to slip you some cash for your time if your willing to update the code and of cause share with the wider community. Oh dear. I’m desperately trying to get back onto sorting out the problems with the CM Library. If nothing else goes belly-up on me, I’ll try to do something over the next few days, but no promises. Keep an eye on this thread, because I customised a sketch for someone a few weeks ago, and they’ve not acknowledged it, so that could have been a complete waste of time, and the thought of that rather puts one off offering again. Here you are. It seems to work for me: emonTxV3_4_DiscreteSamplingPolledPulse.ino.zip (7.2 KB) Thankyou very much @Robert.Wall I’ve just uploaded the sketch and it appears to be working. I’ll drag the freezer out of the way to get to the water meter tomorrow to confirm that its resolved the bouncing issue. I owe you a beer! Dave Hi Robert I’ve been playing about with it this morning but the pulse is not counting up. I currently have a pulse of one, Ive disconnected the meter and shorted the Gnd and Pin3 manually to make sure that pulses under 300ms arnt causing a problem but with no avail. It won’t work on battery power and requires the ac adapter (you did read the comments?). The minimum on and off times are 400 ms, if either is shorter, it won’t be counted. But after that, the input can stay high or low indefinitely. The underlying assumption is that switch bounce will last less than 100 ms. If the maximum time that contact bounce can last is known, the clock rate of Timer1 can be changed (line 164, e.g. 3124 for 20 Hz / 50 ms) and that will alter the times proportionately (i.e. the minimum pulse and minimum space are both always 4 × the maximum time allowed for contact bounce). Hi Robert OK i see, not being an expert on reed switches, can I assume that the bounce time will always be significantly shorter than a pulse from the water meter, baring in mind that my water pressure here could potentially deliver 1 litre a second so I can also assume that the pulse time is approx 100ms? The setup I have here is 4x CTs, 1x Pulse, AC & USB power. At the moment I cant seem to register any pulses, I can physically short out ground and pin 3 for one second but still nothing registers.

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GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 16 Feb 2019, 22:43 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History ## Events & Promotions ###### Events & Promotions in February PrevNext SuMoTuWeThFrSa 272829303112 3456789 10111213141516 17181920212223 242526272812 Open Detailed Calendar • ### Free GMAT Algebra Webinar February 17, 2019 February 17, 2019 07:00 AM PST 09:00 AM PST Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT. • ### Free GMAT Strategy Webinar February 16, 2019 February 16, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. # By what percent is 25 greater than 15? Author Message TAGS: ### Hide Tags CR & LSAT Forum Moderator Status: He came. He saw. He conquered. -- Studying for the LSAT -- Corruptus in Extremis Joined: 31 Jul 2017 Posts: 473 Location: United States Concentration: Finance, Economics By what percent is 25 greater than 15?  [#permalink] ### Show Tags 23 Oct 2017, 13:45 2 00:00 Difficulty: 5% (low) Question Stats: 90% (00:44) correct 10% (01:08) wrong based on 70 sessions ### HideShow timer Statistics By what percent is 25 greater than 15? (A) 10% (B) 25% (C) 33 1/3% (D) 60% (E) 66 2/3% _________________ D-Day: November 18th, 2017 Need a laugh and a break? Go here: https://gmatclub.com/forum/mental-break-funny-videos-270269.html Need a CR tutor? PM me! Intern Joined: 02 Feb 2016 Posts: 34 Location: United States GMAT 1: 710 Q49 V38 GPA: 3.5 Re: By what percent is 25 greater than 15?  [#permalink] ### Show Tags 23 Oct 2017, 14:02 (25-15)/15*100 = 66 2/3% Intern Joined: 15 May 2017 Posts: 33 GMAT 1: 770 Q51 V45 By what percent is 25 greater than 15?  [#permalink] ### Show Tags 23 Oct 2017, 14:03 By what percent is 25 greater than 15? (A) 10% (B) 25% (C) 33 1/3% (D) 60% (E) 66 2/3% Use the following English to algebra translation to write out the requisite equation: What: x Percent: ÷ 100 Is: = Greater Than: - Therefore: x/100 = (25 - 15)/15 If necessary, cross multiply to solve. Ideally, memorize common fraction to decimal conversions. 10/15 can be reduced to 2/3 and 2/3 = 66.67%. _________________ M.S. Journalism, Northwestern University '09 Director of Online Tutoring, MyGuru http://www.myguruedge.com Find out more about online GMAT tutoring with MyGuru http://www.myguruedge.com/gmat-prep/online-gmat-tutoring Senior Manager Joined: 07 Jul 2012 Posts: 376 Location: India Concentration: Finance, Accounting GPA: 3.5 Re: By what percent is 25 greater than 15?  [#permalink] ### Show Tags 23 Oct 2017, 18:08 $$\frac{(25-15)}{15}$$*100= 66$$\frac{2}{3}$$% _________________ Kindly hit kudos if my post helps! Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 4921 Location: United States (CA) Re: By what percent is 25 greater than 15?  [#permalink] ### Show Tags 21 Jan 2019, 17:46 By what percent is 25 greater than 15? (A) 10% (B) 25% (C) 33 1/3% (D) 60% (E) 66 2/3% We can create the following expression: (25 - 15)/15 x 100 = 10/15 x 100 = 2/3 x 100 = 66 ⅔% Alternate Solution: We see that 25 is 10 more than 15. Thus, 25 is 10/15 greater than 15. We see that 10/15 = 2/3 ≈ 0.67, or 67% _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Re: By what percent is 25 greater than 15?   [#permalink] 21 Jan 2019, 17:46 Display posts from previous: Sort by

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rar - 7 months ago 55 R Question # Calculating number of three consecutive values above a threshold (in raster stack) I need to compute number of three consecutive days when value of each pixel in a raster stack ( `x` ) is above a given threshold (defined by another raster `y` ). I tried using `rle` for the purpose with `calc` as follows after stacking `x` and `y` together into new raster `a` : ``````library(raster) fn<-function(a) with(rle(a), sum(lengths>=3 & values>a[[nlayers(a)]])) calc(b,fn) `````` However, I am getting the error: Error in .calcTest(x[1:5], fun, na.rm, forcefun, forceapply) : cannot use this function Reproducible sample: ``````x1 <- raster(nrows=10, ncols=10) x2=x3=x4=x5=x6=x1 x1[]= runif(ncell(x1)) x2[]= runif(ncell(x1)) x3[]= runif(ncell(x1)) x4[]= runif(ncell(x1)) x5[]= runif(ncell(x1)) x6[]= runif(ncell(x1)) x=stack(x1,x2,x3,x4,x5,x6) y=x1 y[]= runif(ncell(x1)) a<-stack(x,y) `````` You could try this: `````` x1 <- raster(nrows=10, ncols=10) x2=x3=x4=x5=x6=x1 x1[]= runif(ncell(x1)) x2[]= runif(ncell(x1)) x3[]= runif(ncell(x1)) x4[]= runif(ncell(x1)) x5[]= runif(ncell(x1)) x6[]= runif(ncell(x1)) x=stack(x1,x2,x3,x4,x5,x6)*4 y=x1 y[]= runif(ncell(x1))*2 a<-stack(x,y) library(raster) fn<-function(x) { seq <- rle(as.numeric(x)>as.numeric(x)[[nlayers(a)]]) n = length(seq\$lengths > 3 & seq\$values == TRUE) return(n) } calc(a,fn) class : RasterLayer dimensions : 10, 10, 100 (nrow, ncol, ncell) resolution : 36, 18 (x, y) extent : -180, 180, -90, 90 (xmin, xmax, ymin, ymax) coord. ref. : +proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0 data source : in memory names : layer values : 1, 7 (min, max) `````` (note that I modified the example dataset to get some good sequences) HTH !

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## Calculus with Applications (10th Edition) $y(1.4)=0.584$ We are given $\frac{dy}{dx}=1+\frac{y}{x}$ so that $g(x,y)=1+\frac{y}{x}$ Since $x=1, y=0$ $g(x_{0},y_{0})=1$ and $y_{1}=y_{0}+g(x_{0},y_{0})h=0+1\times0.1=0.1$ Now $x_{1}=1.1, y_{1}=0.1$ and $g(x_{1},y_{1})=1.09$ Then $y_{2}=0.1+(1.09)\times0.1=0.209$ $y_{3}=0.209+(1.174)\times0.1=0.326$ $y_{4}=0.326+(1.251)\times0.1=0.451$ $y_{5}=0.451+(1.322)\times0.1=0.584$ $y(1.4)= y_{5}=0.584$

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how to set nan to areas outsid a polygon on the image Asked by zawaiter on 31 Jan 2012 Latest activity Commented on by samiy goller on 9 Sep 2012 i have the vertices and facec of a figure i want to set all area outsid this boudaries to nan and save the image matrix. image matrix contain the polygon defined by vertices and faces and nan outside it.plz some help Products Answer by Kevin Moerman on 31 Jan 2012 Then its more complicated. When you say vertices and faces do you mean a tesselation? e.g. a Delaunay tesselation? Of the head volume or just the outer surface? For volume (e.g. tetrahedrons) tesselations you could use: t = tsearchn(X,TES,XI); which returns NaN for points outside of the tesselation (actually outside of the convex hull in case of a Delaunay tesselation). I've tried this for some non-Delaunay derived tesselations but I'm not sure if that is allowed/100% correct so you should check. (you could also look at using [k,d]= dsearchn(X,TES,XI) which allows you to check for distance to your vertices for each point and then formulate a treshold). If you only have vertices and faces for the boundary of the head (e.g. surface triangulation) then you could add a known inner point to which all triangles can form a valid tetrahedron and then create your own tesselation. But this may not work for all boundary triangles (I imagine it gets complicated at the ears). Alternatively if you work with image/voxel coordinates only and you have the indices for the voxels in the head. Lets say these are IND and M is your 3D image matrix then you could create your image matrix containing nan's as follows: M_nan=nan(size(M)); %Matrix filled with nan's the size of M M_nan(IND)=M(IND); %Replace entries at IND with M(IND) Or if L is a logic array describing the head entries then simply M(~L)=nan; Kevin Answer by Image Analyst on 31 Jan 2012 If you know the coordinates for each slice, then you can use poly2mask() to get a binary image of your "in polygon" area. Then say ```sliceImage(~binaryMask) = nan; ``` Do that for every slice. 1 Comment samiy goller on 9 Sep 2012 hi dear, if i want separate spine from back,in MRI image. what are we do? Answer by Kevin Moerman on 31 Jan 2012 If its a 2D image you can use the inpolygon command. Your polygon (which needs to form the boundary of your patch data) is defined by all vertices that have edges that are only part of 1 face. Answer by zawaiter on 31 Jan 2012 thanks kevin for care actually i have 3d array contain a scanned MRI data (head).some data belong to the head and some belong to the film surrounding the head, i want to set all data not belong to the head (outside the head vertices)to nan.so it is a 3d image.can u explain more. Answer by zawaiter on 31 Jan 2012 kevin thanks again,i will need sometime to read and understand ur answer,alittle complicated for me, but i just want tell you for now that i have got the vertices and faces simply by isosurface(D,70). Answer by zawaiter on 2 Feb 2012 thank you very much kevin and image analyst your comments was of great help to me.

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The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A254825 Number of length n 1..(6+1) arrays with every leading partial sum divisible by 2, 3 or 5 1 %I #4 Feb 08 2015 10:27:01 %S 5,26,121,547,2644,13504,69858,361448,1827707,8901990,42510428, %T 205423322,1020648167,5181556417,26415622622,133002076952, %U 656895200707,3202519957155,15628783471819,77223050176456,386855148392695 %N Number of length n 1..(6+1) arrays with every leading partial sum divisible by 2, 3 or 5 %C Column 6 of A254827 %H R. H. Hardin, <a href="/A254825/b254825.txt">Table of n, a(n) for n = 1..210</a> %F Empirical: a(n) = 171*a(n-5) +3374*a(n-6) +23214*a(n-7) +87203*a(n-8) +214299*a(n-9) +377802*a(n-10) +503302*a(n-11) +522736*a(n-12) +432416*a(n-13) +289894*a(n-14) +159820*a(n-15) +73076*a(n-16) +27617*a(n-17) +8448*a(n-18) +2006*a(n-19) +345*a(n-20) +38*a(n-21) +2*a(n-22) %e Some solutions for n=4 %e ..4....4....5....4....5....6....2....3....4....4....4....2....3....5....3....6 %e ..2....4....4....5....4....2....3....2....5....4....4....1....6....1....6....4 %e ..6....1....5....5....5....2....7....5....3....4....6....3....1....6....3....6 %e ..2....3....4....7....7....4....4....6....4....3....7....4....5....4....4....5 %K nonn %O 1,1 %A _R. H. Hardin_, Feb 08 2015 Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified June 2 20:13 EDT 2023. Contains 363100 sequences. (Running on oeis4.)

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## Download Question Paper of "Data Structure 2" , Question Paper of BCA 2nd Semester, Subject Code : BC - 204, Paper ID B0208, Paper 4 • Monday, June 27, 2016 Roll No.................... Total No. of Questions : 07 Paper ID [B0208] (Please fill this Paper ID in OMR, Sheet) BCA (Sem. – 2nd) DATA STRUCTURES (BC - 204) Time : 03 Hours Instruction to Candidates: 1) Section - A is Compulsory. 2) Attempt any Four questions from Section - B. Section - A Q1) a)     What is a post order traversal? b)    What is the best and average case of binary search? c)     What is a binary tree? d)    How a binary tree can be represented as alary structure? e)     What is an complexity of an algorithm? f)      What is a big O notation? g)     What are the Pointers of queue? h)    How a doubly link list is represented? i)       What is a Binary Search Tree? j)       What is the basic principle of bubble sort. Section - B Q2)    Write algorithm for : (a) Binary search. (b) Merge sort. Q3)    Construct the binary tree for the following expression (2x - 3z + 5) (1* -, + 8). Q4)    Suppose a binary tree T is in the memory. Write a recursive algorithm or C program which find the number of nodes in T. Q5)    What are the various operation possible on a doubly link list. Explain with the algorithm and graphically. Q6)    What are the various operations possible on stacks. Q7)    What is the post fix and prefix representation of the following expression (A * (b + c) ) + (b/d) * a* z.

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Matthew L. Wright Assistant Professor, St. Olaf College # Probability Theory ## Math 262 ⋅ Fall 2020 Top Today Bottom Friday August 21 Introduction; what is probability? Do the following before next class: Monday August 24 Sample spaces and events; axioms of probability Do the following before next class: • Review the solutions to the problems from class. • Complete Homework 1 (due Wednesday). • Read §1.3 in the textbook, and watch the accompanying video Counting Methods. Answer the three questions embedded in the video before class on Wednesday. Wednesday August 26 Counting methods; permutations and combinations Homework 1 due at 5pm Bonus video: John Urschel-NFL Math Whiz Do the following before next class: Friday August 28 More counting methods Do the following before next class: • Review the solutions to the problems from class. • Finish Homework 2 (due Monday). • Watch the video Conditional Probability and answer the questions embedded in the video before coming to class on Monday. Also read the examples in §1.4 in the textbook. Monday August 31 Conditional probability; Bayes' Theorem Homework 2 due at 5pm Do the following before next class: • Review the solutions to the problems from class. • Begin Homework 3 (due Friday). • Watch the video Independence and answer the questions embedded in the video before coming to class on Wednesday. Also read §1.5 in the textbook. Wednesday September 2 Independence Do the following before next class: • Review the solutions to the problems from class. • Begin Homework 3 (due Friday). • Watch the video Simulation of Random Events and answer the questions embedded in the video before coming to class on Friday. Also read §1.6 in the textbook. • On Friday, we meet in RNS 410. Wear your face mask, and bring your computer with R or Mathematica. Friday September 4 Simulation of random events Meet in RNS 410 — masks required! Homework 3 due at 5pm Do the following before next class: • Please complete the Week 3 Survey. Answers are voluntary and anonymous. • Watch the video Discrete Random Variables and answer the questions embedded in the video before coming to class on Monday. Also read §2.1 and §2.2 in the textbook. • Begin Homework 4 (due Wednesday). Monday September 7 Random variables; discrete distributions Do the following before next class: • Please complete the Week 3 Survey, if you haven't done so already. Answers are voluntary and anonymous. • Watch the video Expected Value and Standard Deviation and answer the questions embedded in the video before coming to class on Wednesday. Also read §2.3 in the textbook. • Finish Homework 4 (due Wednesday). Wednesday September 9 Expected value, variance, and standard deviation Homework 4 due at 5pm Bonus video: Yitang Zhang: An Unlikely Math Star Rises Do the following before next class: • Review the solutions from the problems in class, especially those involving Chebyshev's Inequality. • Watch the video The Binomial Distribution and answer the questions embedded in the video before coming to class on Friday. Also read §2.4 in the textbook. • Begin Homework 5 (due Monday). Friday September 11 Binomial distribution Do the following before next class: • Review the solutions from the problems in class. • Watch the video The Poisson Distribution and answer the questions embedded in the video before coming to class on Monday. Also read §2.5 in the textbook. • Finish Homework 5 (due Monday). • Take a look at Homework 6 (due Wednesday). Monday September 14 Poisson distribution Homework 5 due at 5pm Do the following before next class: Wednesday September 16 Review day Homework 6 due at 5pm Bonus video: Moon Duchin: "Political Geometry" Work on Exam 1. Complete it before class on Monday. You may either hand in your exam at class on Monday, or submit your work electronically using this Moodle assignment link. Friday September 18 Exam 1: take-home, covering sections 1.1 through 2.4 in the textbook (no class meeting today) Do the following before next class: Monday September 21 Hypergeometric distribution Do the following before next class: • Review the solutions to the problems from class on Monday. (See notes and solutions above.) • Watch the video The Negative Binomial Distribution and answer the questions embedded in the video. Also read §2.6.2 in the textbook. • Begin Homework 7 (due Friday). Wednesday September 23 Negative binomial and geometric distributions Bonus: video How Not to Be Wrong: The Power of Mathematical Thinking - with Jordan Ellenberg and article The Psychology of Statistics by Jordan Ellenberg Do the following before next class: Friday September 25 Moment-generating functions Homework 7 due at 5pm Do the following before next class: • Video and reading to be announced. Monday September 28 Moment-generating functions Do the following before next class: • Video and reading to be announced. Wednesday September 30 Simulation of discrete random variables Homework 8 due at 5pm Do the following before next class: • Video and reading to be announced. Friday October 2 Continuous random variables; uniform distribution Do the following before next class: • Video and reading to be announced. Monday October 5 Expected values of continuous distributions Homework 9 due at 5pm Do the following before next class: • Video and reading to be announced. Wednesday October 7 Normal distribution Do the following before next class: • Video and reading to be announced. Friday October 9 Exponential distribution Homework 10 due at 5pm Do the following before next class: • Video and reading to be announced. Monday October 12 Gamma distribution Do the following before next class: • to be announced Wednesday October 14 Review day Homework 11 due at 5pm Do the following before next class: • to be announced Friday October 16 Exam 2 Do the following before next class: • to be announced Monday October 19 Transformations of continuous random variables Do the following before next class: • to be announced Wednesday October 21 Joint distributions Do the following before next class: • to be announced Friday October 23 Expected values, covariance, and correlation Do the following before next class: • To be announced Monday October 26 Linear combinations of random variables Do the following before next class: • to be announced Wednesday October 28 Properties of linear combinations Do the following before next class: • to be announced Friday October 30 Conditional distributions and expectation Do the following before next class: • to be announced Monday November 2 Limit theorems Do the following before next class: • to be announced Wednesday November 4 Limit theorems Do the following before next class: • to be announced Friday November 6 Transformations: distribution function method Do the following before next class: • to be announced Monday November 9 Bivariate tranformation theorem Do the following before next class: • to be announced Wednesday November 11 Order statistics Do the following before next class: • to be announced Friday November 13 Catch up Do the following before next class: • to be announced Monday November 16 Review Final exam information will be posted. Thursday November 19 Final exam 2:00 – 4:00pm

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0 # What is 8-9000? Wiki User 2011-03-24 13:33:22 9008 Alberta Schmitt Lvl 10 2021-02-27 00:07:26 Study guides 20 cards ## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials ➡️ See all cards 3.74 1020 Reviews Wiki User 2011-03-24 13:33:22 8 - 9000 = -8,992

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# Thread: show that R square is zero 1. ## show that R square is zero A two-variable regression with no intercept is given as: $Y_i=B_2X_i+u_i$ Show that the R-squared ( $R^2$) for this model is equal to zero. I think I should show that the covariance of X and Y are zero? 2. ## Re: show that R square is zero Hey wopashui. There are quite a few formulas for R^2 and covariance would indeed be one way to show this. 3. ## Re: show that R square is zero Originally Posted by chiro Hey wopashui. There are quite a few formulas for R^2 and covariance would indeed be one way to show this. I can't end up getting zero by using the covariance, can you show me some steps of approaching this? 4. ## Re: show that R square is zero What information do you have about the model and its constraints? Econ 4706? 6. ## Re: show that R square is zero No I mean the actual course material you are covering (specific topics, concepts, stuff like that). I have no idea what that course is.

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# Time Conversion m #### tjc154 ##### Active Member I would like to write a formula that converts time from HH:MM to minutes format in a different cell. Ex: 3:50 should convert to 210. Thanks, Tom ### Excel Facts How to create a cell-sized chart? Tiny charts, called Sparklines, were added to Excel 2010. Look for Sparklines on the Insert tab. #### Scott Huish ##### MrExcel MVP =A1*1440 Format as General although I think that sample should result in 230, not 210 #### tjc154 ##### Active Member 210 would be the correct value- 60 x 3.5. How did you come up with 1440 for this formula? Thanks, Tom #### Scott Huish ##### MrExcel MVP 3:50 is 3 hours 50 minutes 3 hours is 180 minutes + 50 minutes = 230 There are 1440 minutes in a day. Replies 6 Views 282 Replies 2 Views 137 Replies 3 Views 191 Replies 6 Views 447 Replies 4 Views 221 ### Forum statistics 1,190,736 Messages 5,982,654 Members 439,789 Latest member Justhavinaltlfun2002 ### We've detected that you are using an adblocker. We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com. ### Which adblocker are you using? 1)Click on the icon in the browser’s toolbar. 2)Click on the icon in the browser’s toolbar. 2)Click on the "Pause on this site" option. Go back 1)Click on the icon in the browser’s toolbar. 2)Click on the toggle to disable it for "mrexcel.com". Go back ### Disable uBlock Origin Follow these easy steps to disable uBlock Origin 1)Click on the icon in the browser’s toolbar. 2)Click on the "Power" button. 3)Click on the "Refresh" button. Go back ### Disable uBlock Follow these easy steps to disable uBlock 1)Click on the icon in the browser’s toolbar. 2)Click on the "Power" button. 3)Click on the "Refresh" button. Go back

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## Probability problem involving multiple coronavirus tests in the same household Mark Tuttle writes: Here is a potential homework problem for your students. The following is a true story. Mid-December, we have a household with five people. My wife and myself, and three who arrived from elsewhere. Subsequently, various diverse symptoms ensue – nothing too serious, but everyone is concerned, obviously. Video conference for all five of us with an Urgent Care doc who interviews each of regarding identity, onset, severity, etc. of symptoms. This leads to him scheduling virus tests for all of us. Doc says he’s seen too many cases where patients with symptoms postpone tests because they think it’s not serious, then it both gets serious and others are infected. No surprise there. We all agree to self-quarantine if anyone gets a positive test. (Maybe we should have anyway, but that’s another story.) The symptoms resolve themselves, and all five tests are negative. In email follow-up from the doc (good for him) he observes that since all five tests are negative they probably do not include a false negative. I recount this story – and the doc’s observation re reduced likelihood of false negative – to various smart folks, but no statistics/probability nerds, with fascinating results. It shows how hard it is for normal folks to deal with these concepts. One friend who heard the story consulted his quantitative expert who confirmed the doc’s observation. I think it’s a wonderful example because we probably all had the same “bug”, but, apparently, not THE virus. Thus, if we all had THE virus, it does make it more difficult to get five (false) negatives. Again, the implication from the doc was that he had had to deal with false negatives, generally – apparently, they were frequent enough to be a concern (ignoring the technical challenge of confirming a false negative in real practice). The test was some kind of PCR thing; I could get the exact name if that ever mattered. So, the homework problem is to assign probabilities to the various inputs in a way that justifies the doc’s observation that false negatives were unlikely (or not), and unlikely vs. the false negative rate for a single test on a single random person with symptoms. 1. TPassin says: On the face of it, the bare statement that with five negatives, probably none of them were false seems silly. But in context, it makes sense. If one of the family really had had covid, then probably some or all of the others would have gotten it from them. In that case, the chance that all of the covid ones would all have had a false negative test would not be high. • I suppose it depends on the transmission dynamics. For example if you were exposed yesterday, and tested everyone today… it’s too early to have enough virus to detect. If one person was exposed 2 days ago, it’s too early for that person to have transmitted to others and have them be definitely detectable, so if that one initial person gets a false negative the others would be expected to be negative too. Now, if everyone was exposed 4 days ago… then it’s unlikely they’d all be false negative, yes. If one person was exposed 6 days ago and the others were in close contact, then yes it’s also unlikely for them all to be false negative. The viral dynamics are such that from initial exposure to peak viral concentration is something like 3 days. so it’s something like 6 days from initial exposure of one person to detectability for others in the family. 2. Emily says: There’s information I don’t have about how COVID tests work and what causes them to be falsely negative. Basically, are they randomly falsely negative, or are they non-random in ways that could be correlated within a household? Like, are there particular strains that are likely to cause false negatives? If that were the case, then the likelihood of a bunch of false negatives among people who’ve infected each other is the same as the chance of one false negative. (Or, and this doesn’t apply here, but are there people who are because of some genetic thing more likely to test falsely negative, in which case that could also be correlated within a household.) If those aren’t the case than this seems correct to me. Conditional on n people having COVID, the chances that none of them will test positive for it goes down as n goes up, if these are independent events – it’s a (1-p)^n problem. I am assuming they all have the same thing, but that seems likely since they all have something and they live together. • Thomas Lumley says: For the Covid PCR test – The false negative assay rate is very low: ie, if you get a negative test, it’s safe to assume there was little or no viral SARS-Cov-2 RNA on the swab – The false negative rate due to not picking up virus on the swab is appreciable. It’s probably reasonable to model it as roughly independent between tests, but it varies strongly over the course of the infection. Early in the symptomatic period and just before it, the false negative rate gets as low as 20%, but it’s much higher immediately after infection. • Daniel says: I am not so sure whether it is reasonable to model it as random. A swap as you said depends on the person who receives the test and the one taking the sample. Someone can make the same mistake over and over again and therefore cause false negatives. On the other hand it is possible that due to the kind of infection it differs where it sits. Also there was a publication a while ago investigating testing procedures in different centres in the same city (Muenchhoff et al.) seeing that the outcome was correlated to the location. They argue that simply due to the kits which were used this could cause missing of infections. Considering that they were probably tested together one could argue that they were exposed to the same bias at least to some degree. The main issue as you pointed out though is the time point of infection in relation to the time of when the sample was taken. • There are at least two causes for little or no virus on a swab: high cycle threshold (Ct) value or bad swab technique. Check out figures 1 and 2 of this paper, which back up what you (Thomas Lumley) are saying https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7427302/ Figure 1 plots Ct value vs. days since onset of symptoms (not since infection). Figure 2 plots sensitivity of PCR tests as a function of Ct. They’re plotting days since onset of symptoms because there’s not enough viral load right after infection to test positive. These were from April/May 2020 and aren’t based on huge amounts of data. I’m looking for something more recent that had enough info for me to estimate sensitivity as a function of days since infection (not just symptom onset). 3. Alain says: For this sort of matter, you should probably also consider the timing of the events (possible exposure, symptoms onset, testing). If you test too early for covid-19, i.e. before the viral load is high enough, the test may come back negative, but I’m not sure if it officially qualifies as a false negative, i.e. if this aspect is already included in the advertised sensibility of the test. 4. Matt Skaggs says: “the homework problem is to assign probabilities to the various inputs in a way that justifies the doc’s observation” I just don’t get this. Why make up numbers that no two people will interpret the same? This is a logic problem. 1. The circumstantial evidence – everyone got similar symptoms at approximately the same time – makes it unlikely that a single member of the household got coronavirus while the other members of the household became sick from a different virus. 2. The false positive rate has to be less <10% for a test to even be used. Combine 1. and 2. and the likelihood of there being a single family member with coronavirus getting a false negative is vanishingly small. How would putting numbers on this make it any better? • JP says: A few points: 1. “Exercise to the reader” is just to engage with the problem I think, not to converge on a canonical model. That said, sharing approaches to problem modeling is how consensus is built. 2. False positive rate false negative rate. False positive rate is tied to specificity, false negative rate is tied to sensitivity IIRC. You can tune a test to be more likely to report false positives or false negatives – ideally we minimize both, but usually there is a priority and these two don’t move completely independently. Finally, you pivot from logic to probability in your last sentence, and hand-wave away with “vanishingly small.” I would rather have a number on that ;-) • Matt Skaggs says: “Finally, you pivot from logic to probability in your last sentence, and hand-wave away with “vanishingly small.” I would rather have a number on that ;-)” Well, let’s parse this. One of three things happened: 2. One or more, but not all had coronavirus. The provided fact that everyone had similar symptoms supports 1. and 3. and refutes 2. The provided fact that everyone tested negative supports 3. and refutes 1. and 2. So only 3. is supported by the known facts, unless there was a false negative. Now we have to deal with bad test results. For 1. to be true, all five tests had to be botched. Not false negatives, per se, but botched tests. Five out of five properly performed tests being false negatives is not worth considering. For 2. to be true, two things have to be true. One is that the test of the coronavirus-infected person(s) had to be a false negative, and the second is that the other family members had to be sick at the same time with a different disease. So all the doctor claimed was that it was more likely that no one had coronavirus than that 1. all the tests were botched, or 2. there was both a false negative AND two different diseases in the household at the same time. Works for me. Now if you want to embellish the logic with false positive rates or whatever, it does no harm. But it doesn’t add much either. This is a great example to use when explaining to folks how to diagram causality! • Joshua says: >Not false negatives, per se, but botched tests. Five out of five properly performed tests being false negatives is not worth considering. Not that I’m arguing the basic logic, but as people have mentioned there could have been a shared site-collection problem for all the tests, or a contamination problem at the lab where all the tests were evaluated in close sequence. How do you determine those possibilities aren’t worth considering? • Matt Skaggs says: “How do you determine those possibilities aren’t worth considering?” It is not that they are not worth considering. Everything is worth considering. Not only would the tests have been botched, but they would have been botched in a way that produced valid-looking results, without the testing entity having any clue. And botched tests do not mean that testees had coronavirus. The scenario is very unlikely, that’s all. Or a comet could have hit the lab. That was always the running joke when I was building these diagrams (retired now). • Dzhaughn says: Most answers above are not noting that this group of 5 is sharing viruses. (They all got sorta sick at the same time; viruses are shared frequently within households; COVID is thought to be eminently sharable.) I think we’d want to estimate p(n) = P( n housemates get COVID | 1 gets COVID ), for n = 1 to 5. If p(5) = 1, then what happens in the story is as good as 5 negative tests. p(3) = 0 means the pandemic is a probably a hoax, right? So we’re somewhere in between. I guess I am modeling false negatives as noise. • Dzhaughn says: (This wasn’t meant to reply to Matt Skaggs in particular, sorry! But his remark is suggestive of this. • Specificity (accuracy on negative cases) for PCR tests is very good—well over 99% by most estimates. Sensitivity (accuracy on positive cases) is very bad—well below 80% by most estimates, even at peak viral load, given the variation in Ct values for infected individuals. It’s the sensitivity (false negatives) that matters here, not the specificity (false positives). Check out the plots of the raw data in the first two figures of this paper: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7427302/ Before figure 1, the paper says in the first week after symptom onset, (geometric) mean Ct is 28.2. That gives a roughly 60% test sensitivity according to figure 2. But you can also see that the individual variation in Ct value is huge from figure 1, even during the highest viral load around time of symptom onset. 5. Now this is an intriguing exercise b/c it does require an explanation of assumptions being made by the physician and family members. In my observation only a handful of experts have the knowledge to undertake this exercise. They have knowledge of the biology of the virus as well as the kinetics of PCR and Antigen testing. So in some sense, I wonder of what use it is for statisticians to assign probabilities without the input of this handful. What do we mean by ‘bug’? And if the PCR was used, did the physician get the CT values alongside? The case presents an overarching question that I have had. If you have a coronavirus other than SARS2, will the PCR test be able to specify accurately that it is SARS2? To date no one has answered this question. Period. Now if it is a question that is silly or stupid. Just tell me. Frankly, from early on, I came to the conclusion that the PCR was the wrong SCREENING tool: it is typically referred to as a Diagnostic. Given that the PCR is being used equally as a screening tool, the PCR test has been identified as potentially resulting in false positives b/c the PCR test is so sensitive. It can yield a false positive for up to 2 or 3 months. 6. BenK says: Great question! The main issue – are PCR tests just finicky and fail randomly, or do they fail for good reasons? The main answer is – they fail for good reasons. Were the swabs taken by the same technician? That could lead to a coordinated failure of the tests. Were they run through the same lab, using the same PCR primers, etc? Some tests were much worse than others… If all the infections were ‘in phase’ from a single source – since many false negatives come from people with low titers – then having tested all of them is unhelpful. However, if the first case was from outside and the rest travelled around the home, then the five tests might be informative if any of them are positive. Further, a question of variants – particularly if negative results were being called because of the ‘drop out’ antigen – then having 5 tests is also potentially not helpful. So these are several ways the simple multiplication of false negative rates could be undermined by the real world. • Great answers! It’s all about correlation, including viral load, swab, and lab tech! We only get (1 – p)^n decay in false negatives with n tests of sensitivity p if the tests are independent. If they were 100% correlated, n tests would still have a (1-p) chance of a false negative. 7. Joshua says: > So, the homework problem is to assign probabilities to the various inputs in a way that justifies the doc’s observation that false negatives were unlikely (or not), and unlikely vs. the false negative rate for a single test on a single random person with symptoms. I guess this is a statistics course? From my point of view, what’s interesting is how much difficulty many people have understanding the statistical implications of multiple tests. I’ve come across many people who think that if there’s an X probability of inaccuracy, there’s no change to that probability with multiple tests. Something like: “Why take the tests multiple times if it’s only accurate half the time? It’s just not as good as the PCR test which is more accurate.” I guess it’s about conditional probability. Like when people say “What’s the point of wearing a mask if it doesn’t guarantee to keep me from getting infected/infect other people?” It’s very hard to get across the message that a marginal benefit in an individual risk event compounds across a population. • Ben R says: “It’s very hard to get across the message that a marginal benefit in an individual risk event compounds across a population.” Does anyone know of an explainer (video, article, simulation tool?) for this concept? Does it mean that if each individual has a 5% lower chance of getting infected, there is a greater than 5% reduction in overall cases? • BoseQC35 says: Why Masks Work BETTER Than You’d Think • Joshua says: Interesting video. I was actually thinking of a different issue. Say one person (unknowingly infectious) decides not to wear a mask to the supermarket. Say the risk reduction from a mask is quite small, but it so happens that not wearing a mask resulted in one other person at the market getting infected who wouldn’t have gotten infected had that person worn a mask. That now infected person then goes home and infects one other person in their pod, and that now-infected person interacts with two other pods and infects one person in each of those pods. And so on. A somewhat different dynamic than what is discussed in the video. 8. Jonathan (another one) says: Ah, but here’s a different story. In January, my wife and I (there are no others in the household) came down with the same flu-like and respiratory symptoms on the same day. We got tested three days later. My test came back negative. But we got no result on my wife for another three days, at which point her test came back positive. Very mild symptoms continued for another week, during which time we both quarantined. I have assumed that there is a much higher chance that I had a false negative than that my wife had a false positive. Given that we had virtually identical symptoms, there is really a negligible chance that one result wasn’t incorrect; given the fact that the symptoms were pretty classic mild Covid symptoms, I assume the false report was mine. Does anyone disagree? • That sounds right. • Dzhaughn says: It would be relevant to know the rate of COVID in your area that that time, and then to estimate how many people you encountered. When someone tells you “Yeah, it was probably a false negative,” reply “Oh, that’s great, I can assume I’m immune now!” and see if their expression or story changes. Or if they slap you. • Jonathan (another one) says: January was of course a spike period here in NY, and I left out the info that we ate at a restaurant three nights before symptoms started that, a night from which (apparently) several others had confirmed positives. While that greatly raise the chances of a false negative I left it out of my description because I wanted to see people’s take on the bare facts. And yeah, I absolutely assumed I was immune, which greatly relieved psychological stress while waiting for the vaccine line to diminish. 9. Peter Gerdes says: The problem here is less probability interpretation and more one of language use. Assume C_i is the i-th person in the household has Covid. T_i is the i-th person tests positive. For algebraic simplicity I’ll assume only 2 family members from here on but the point clearly generalizes. What the doctor really means is that P(C_0|~T_0) > P(C_0|~T_0&~T_1). What a pedantic translation would interpret him as saying is that P(~T_0|C_0 & ~T_1) <P(~T_0|C_0). This later claim is going to be false under simplified toy model style assumptions since it would require how my Covid test turns out to be affected by something besides whether I'm infected. OTOH the indented claim is clearly true on reaeonable toy model assumptions since for you not to have Covid you must have both had a false negative and either not infected your partner or they just also have had a false negative and the effect grows with number of family members. • Peter Gerdes says: So I’d say it’s not an indictment of our probability intuitions but the naturalness of the way we formally talk about probability. 10. Sean P Mackinnon says: Huh, interesting homework problem. On one hand, you could note that the familywise error rate increases with the number of tests. So the chance that at least one is a false negative naturally increases with the number of tests. So if you take that perspective, the doctors statement seems wrong. But… There are important background assumptions. 3 people traveled, and all five people in the household develop symptoms. So if you assume that they all have the SAME disease (a reasonable assumption, given the circumstances) it is indeed much more unlikely to have five false negatives if everyone really did have COVID. Good example of how context and underlying assumptions are important for a student. Where can you find the best CBD products? CBD gummies made with vegan ingredients and CBD oils that are lab tested and 100% organic? Click here.

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https://glenallsopxeidosnapoli.com/mathpower-7-answer-key.html

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# Mathpower 7 Answer Key Free Similar PDF's 3211 dl's @ 1714 KB/s 1885 dl's @ 4285 KB/s 4594 dl's @ 2889 KB/s Guide to the Grades 3–8 Testing Program. Strand and Performance Indicator Map with Answer Key. Grade 7, Book 1. Question. Type. Points. Strand. Content. samplesg.pdf MathWorks 11 Workbook - Pacific Educational Press - University of 7. 63 m. 21 m. 3.0. 19.5 ft. 78 ft. 0.25. 3. 7. 15 or approx. 0.47. 4. 5.5 ft. 5. 50 cm. 6. 20.5 m. 7. 143. 1200 or approx. 0.12. Answer Key. MathWorks 11 Workbook. 1. MW11-WB-AK.pdf Math Makes Sense 7 - SD43 Teacher Sites Pearson Math Makes Sense 7 prior to publication. ... Grade 7 Reviewers .... 342. Answers. 346. Illustrated Glossary. 376. Index. 382. Acknowledgments. 385 ... mms7.pdf Grade 7 Mathematics: Support Document for Teachers - Manitoba Number (N). BLM 7.N.1.1: Math Language Crossword Puzzle (with Answer Key). BLM 7.N.1.2: Divisibility Questions. BLM 7.N.1.3: Applying Divisibility Rules. full_doc.pdf Answers. 6. a) The length is twice the width. b) c) d) m; m. 7. a) b) c) d) It is a function because it passes the vertical line test. e) f) It is not a function because (5 , ... 08-034_15_AFSB_AnsKey_612-683.pdf 5th Grade Math Power Standards at a Glance - Perrysburg Schools Ask and answer questions about key details in a text read aloud or ... 5th Grade Math POWER Standards at a Glance ... NBT.7. Add, subtract, multiply and divide to hundredths using concrete models and traditional methods. 3. 3. 3. 3. NBT.5. Math Power Standards - 5th Grade.pdf progrAm - Mathematical Association of America Aug 2, 2013 ... multiple correct answers to the same complex question. Competitive ..... Page 7 ... and “Math Power: How to Help Your Child Love Math Even If. 2013mathfest_program.pdf Grade 7 Mathematics Curriculum Guide - Department of Education MATHEMATICS GRADE 7 CURRICULUM GUIDE 2013 ..... Mental mathematics enables students to determine answers without paper and pencil. It improves ... Saxon Math Course 3 Ebook - Chisago Lakes School District ... Achieve Inc. 1 2 3 4 5 6 7 8 9 048 12 11 10 09 08 07 06 ..... Table of Contents xi . Lessons 61–70, Investigation 7. Section 7. Math Focus: ..... Answers will vary. Student eBook Course 3.pdf MATHEMATICAL POWER FOR ALL STUDENTS K-12 answers to real-world problems) ..... statistics). All FCAT math items at Grade 7 are tested in multiple choice (MC) or gridded response (GR) format pp - Test Item  ... fullpowr.pdf Math Focus 7 Text Book - Holy Spirit Roman Catholic Separate b) Use your answer in part (a) to create a divisibility rule for 20. Use a four-digit number ... 7. Number Relationships. 1.2. Apply divisibility rules to determine if 3 or 9 is a factor of a whole number. ...... How have your answers and explanations  ... Math_Focus_7.pdf mathematics - mstworkbooks.co.za + 3 154 7 235 5 331 9 456 2 572 3 638 4 764 2 885 4 921. 228. 3 382 .... Try to give answers that you trust to these questions, without doing any calculations. Maths1_Gr8_LB.pdf

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https://mathematica.stackexchange.com/questions/242162/making-propertyvalue-of-vertexlabels-respect-the-order-of-vertexlist-graphs

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# Making PropertyValue of VertexLabels respect the order of VertexList (Graphs) Consider the following graph: Clear[a, b, c] graph = Graph[{b -> a, a -> c}, VertexLabels -> {b -> 2, a -> 1, c -> 3}] However PropertyValue[graph, VertexLabels] produces {a -> 1, c -> 3, b -> 2} Which is not consistent with the order of VertexList[graph] which produces {b, a, c} I use the following to extract the vertex labels from a graph: vertexLabels[g_] := #[[2]] & /@ PropertyValue[g, VertexLabels] This does not produce the label list in an order that is consistent with VertexList. The result is: {1, 3, 2} Rather than {2, 1, 3} Which would follow the order of vertices in VertexList[g] Is there a way to address this? And is there a reason why PropertyValue does not follow the VertexList order consistently? I tried Clear[a, b, c] graph = Graph[{b -> a, a -> c}, VertexLabels -> {b -> 2, a -> 1, c -> 3}] I assumed that by adjusting the VertexLabels, so their assignment follows the vertex order of VertexList (first b, then a, then c), the result of PropertyValue would be match the order of VertexList. This is not the case. The code produces: {a -> 1, c -> 3, b -> 2} as before. I could adapt the code to reorganise the output to match the order of VertexList. Is there another way to ensure this? • PropertyValue[{graph, #}, VertexLabels] & /@ VertexList[graph1]? – kglr Commented Mar 20, 2021 at 16:56 • Great. Odd that PropertyValue picks a different order. Commented Mar 20, 2021 at 17:17 • Weird indeed. You get {a -> 1, c -> 3, b -> 2} regardless of the order you specify the weights PropertyValue[Graph[{b -> a, a -> c}, VertexLabels -> #], VertexLabels] & /@ Permutations[{a -> 1, b -> 2, c -> 3}]. This is so even if you specify the vertex list in the first argument of Graph: PropertyValue[Graph[{a, b, c}, {b -> a, a -> c}, VertexLabels -> #], VertexLabels] & /@ Permutations[{a -> 1, b -> 2, c -> 3}] – kglr Commented Mar 20, 2021 at 17:39 • ... also regardless of the order the edges are given: PropertyValue[Graph[Reverse@{b -> a, a -> c}, VertexLabels -> #], VertexLabels] & /@ Permutations[{a -> 1, b -> 2, c -> 3}] – kglr Commented Mar 20, 2021 at 17:41 A work-around: First a simple fix to get what you need: graph = Graph[{b -> a, a -> c}, VertexLabels -> {b -> 2, a -> 1, c -> 3}]; PropertyValue[{graph, #}, VertexLabels] & /@ VertexList[graph] {2, 1, 3} VertexList[graph] {b, a, c} If you need list of rules: Thread[VertexList[graph] -> (PropertyValue[{graph, #}, VertexLabels] & /@ VertexList[graph])] {b -> 2, a -> 1, c -> 3} We get the same result (1) regardless of the order vertex labels are given, (2) regardless of the order edges are input, (3) regardless of whether a vertex list is specified in the first argument of Graph and (4) regardless of whether we use PropertyValue or AnnotationValue: PropertyValue[Graph[{b -> a, a -> c}, VertexLabels -> #], VertexLabels] & /@ Permutations[{a -> 1, b -> 2, c -> 3}] PropertyValue[Graph[Reverse@{b -> a, a -> c}, VertexLabels -> #], VertexLabels] & /@ Permutations[{a -> 1, b -> 2, c -> 3}] PropertyValue[Graph[{a, b, c}, {b -> a, a -> c}, VertexLabels -> #], VertexLabels] & /@ Permutations[{a -> 1, b -> 2, c -> 3}] (and same lines with PropertyValue replaced with AnnotationValue) all give {{a -> 1, c -> 3, b -> 2}, {a -> 1, c -> 3, b -> 2}, {a -> 1, c -> 3, b -> 2}, {a -> 1, c -> 3, b -> 2}, {a -> 1, c -> 3, b -> 2}, {a -> 1, c -> 3, b -> 2}} Note: This also happens with VertexShapeFunction: PropertyValue[Graph[{b -> a, a -> c}, VertexShapeFunction -> {b -> "Square", a -> "Star", c -> "Triangle"}], VertexShapeFunction] {b -> "Square", c -> "Triangle", a -> "Star"} We get the same result for all the variations above. However, it does not happen with other properties like VertexShape or VertexStyle. What is going on? The source of the mysterious ordering {a, c, b} seems to be the ordering of vertices in sorted edge list of the input graph: VertexList[Sort[EdgeList[Graph[{b -> a, a -> c}, VertexLabels -> {a -> 1, b -> 2, c -> 3}]] ]] {a, c, b} We get {a, c, b} for all the combinations explored below: VertexList[Sort[EdgeList[Graph[{b -> a, a -> c}, VertexLabels -> #]] ]] & /@ Permutations[{a -> 1, b -> 2, c -> 3}] VertexList[Sort[EdgeList[Graph[Reverse@{b -> a, a -> c}, VertexLabels -> #]] ]] & /@ Permutations[{a -> 1, b -> 2, c -> 3}] VertexList[Sort[EdgeList[Graph[{a, b, c}, {b -> a, a -> c}, VertexLabels -> #]] ]] & /@ Permutations[{a -> 1, b -> 2, c -> 3}] VertexList[Sort[EdgeList[Graph[#, {b -> a, a -> c}, VertexLabels -> {a -> 1, b -> 2, c -> 3}]] ]] & /@ Permutations[{a, b, c}] • + thanks for the list of rules comment. Would love to know what goes on internally. It is as if PropertyValue/AnnotationValue performs a sort internally, fixing matters in a specific order. None that looks familiar though. Commented Mar 20, 2021 at 18:28

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https://layton.fandom.com/wiki/Puzzle:Brother_and_Sister

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## FANDOM 2,504 Pages Brother and Sister is a puzzle in Professor Layton and the Curious Village. ## Puzzle A boy and his big sister are sitting around the kitchen table chatting. "You know, Sis, if I took away two years from my age and gave them to you, you'd be twice my age, huh!" "Well, why don't you just give me one more on top of that? Then I'll be three times your age." So just how old is each sibling? ## Hints Click a Tab to reveal the Hint. Let's see if we can't pare this puzzle down a bit. When you take two years away from the brother's age and add them to the big sister's, she becomes twice his age. Additionally, when you take three years away from the brother and give them to the sister, she becomes three times older than he is. You could try and solve this with an algebraic equation, but that's no way to tackle a puzzle! Try to reason your way through this one. Move two years from the brother's age, and the difference in age becomes four years. Move three years, and the difference widens to six years. Four years makes the sister twice as old as the boy. Six years makes her three times as old. The brother and sister were born in the same year. ## Solution ### Incorrect For each year the brother gives to his sister, his age decreases by one. When he loses two years, the sister becomes twice his age. When he loses three years, his sister becomes three times his age. If you're feeling stumped, try graphing the information you have out on paper. ### Correct That's right! US Version The conditions in the puzzle only work out if both the brother and sister are currently six. The two siblings must have been born within a year of each other. UK Version The conditions in the puzzle only work out if both the brother and sister are currently six years old. The two siblings are actually twins! A big thanks to http://professorlaytonwalkthrough.blogspot.com Community content is available under CC-BY-SA unless otherwise noted.

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https://survivalblog.com/2021/04/20/survivalblog-readers-editors-snippets-6/

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# SurvivalBlog Readers’ & Editors’ Snippets This weekly column is a collection of short snippets: practical self-sufficiency items, how-tos, lessons learned, tips and tricks, and news items — both from readers and from SurvivalBlog’s editors. We may select some long e-mails for posting as separate letters. First, and foremost, here is a very useful piece from my friend Commander Zero’s excellent Notes From The Bunker blog: Canned beef back at Costco. o  o  o Americans will likely have to navigate a maze of vaccine “passports” o  o  o “Tunnel Rabbit’s informative article on solar power is helpful for someone just starting and trying to keep some lights on with minimal cash outlay.  I’ve lived off-grid for 23 years and have over 4KW of panels, many on a tracker, and a large Iron Edison battery bank.  Also have 1095 watts of solar on my RV using linear actuators to raise the panels.  {Available from W.W. Grainger}  I’m writing to explain something most solar installers don’t consider.  There is a difference between AC and DC voltage.  AC starts at zero and follows a sine wave to maximum and then returns to zero and follows a sine wave to maximum in the opposite direction and returns to zero.  You can measure the area under the curve and play with the numbers, but the bottom line means you need larger [gauge] wire for DC which stays at one voltage and amperage.  The wire size ratings from Underwriters Lab are for fire prevention on AC circuits, not efficiency.  They say #14 will carry 15 amps,  #12  20 amps,  #10  30 amps,  #6  60 amps.  For a continuous load, such as lights, decrease #12 from 20 amps to 16 amps.  For efficient DC, such as from solar panels to the controller, triple the wire size.  Instead of #12 for 20 amps, use #6.  A favorite trick of solar panel installers on RVs is to use small wire on the DC circuits so they can sell more panels to get enough power.  A small voltage drop from the panel to the solar controller will make a large drop in the output because batteries need a voltage above their rated voltage to charge them. Like many things in life, it comes down to money.  If you are starting with a small system and think you will add more panels later, install larger wire now.  If you are using a large enough system to wire the solar panels in series for 24 volts or 48 volts, you will save on wire because raising the voltage decreases the amperage.  Consider this when buying a solar controller.  Most people are impressed with free energy from solar and like no electric bill every month.  If the solar controller, wire, and inverter are sized properly it is easy to add a couple solar panels and more batteries later.  Prior planning prevents poor performance.” o  o  o o  o  o SaraSue wrote: “I have finally come to my decision to move out of Idaho and to Tennessee, after much gnashing of teeth.  I love Idaho.  The impetus – my children and grandchildren are relocating from California to Tennessee, and I cannot be that far away from them.  I am their only involved grandparent.  It is a really quick flight from Idaho to California.  Conversely, flying across country is horrible and I have no idea if vaccine passports will be required for air travel in the future, so I want to make sure I can get to them.  I am getting my home in Idaho ready to put on the market.  I know the demand is extremely high.  My realtor, who sold me this home, is just waiting for that phone call to come over with the drone and camera, then they will list it.  For SurvivalBlog readers, my home is in an HOA golf resort and not quite an acre, although still considered “rural”, so it’s decidedly not a sustainable property.  I stand to make a huge profit, which will help me purchase something more sustainable (and rural) in Tennessee – a much longer growing season!  Real estate is crazy in TN now too and after having lived through the housing market crash after buying a home in Nevada at height of market, I will buy at the low end of the market for sure, just in case.  After having lived through want and plenty, I’m very comfortable with minimalist living as long as I have land for growing food. I don’t need a lot and I don’t mind fixing things up.  The exact location is not decided yet because it’s dependent upon my grandchildren’s father’s job situation. ‘ We are praying ‘like crazy’ over this move.  I covet your prayers.” o  o  o Reader C.B mentioned this troubling yet unsurprising news: Democrats to propose legislation expanding the Supreme Court o  o  o SurvivalBlog reader R.K. in southwestern Ohio wrote: “Just thought you would appreciate anecdotal news about the reselling of ammunition.  I spoke to a friend of mine today who is a manager at a nationwide outdoor sporting goods store.  He told me that they have daily lines of 30-100 people who wait for the store to open and purchase ammunition and reloading supplies.  He said that one of the people he sees every morning that always purchases and complains the whole time about prices, is also that individual he saw on a recent drive home from work with tables set up in his front yard selling boxes of various calibers for a 216% mark-up.  And apparently, he does not have any issues with his sales.” o  o  o Blog reader John M. e-mailed me to mention:

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https://brainmass.com/physics/work/work-done-constant-force-lifting-carrying-pumpkin-field-9276

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Explore BrainMass # Work done by a constant force: Lifting and carrying a pumpkin from a field Not what you're looking for? Search our solutions OR ask your own Custom question. This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! I think I have both answers, but I really need (b) explained. You go to a pumpkin patch to select your Halloween pumpkin. You lift the 3.2kg pumpkin to a height of 1.2 m, then carry it 50.0 m on level ground to the checkout stand. (a) Calculate the work you do on the pumpkin as you lift it from the ground, and (b) How much work do you do on the pumpkin as you carry it from the field. © BrainMass Inc. brainmass.com December 24, 2021, 4:49 pm ad1c9bdddf https://brainmass.com/physics/work/work-done-constant-force-lifting-carrying-pumpkin-field-9276 #### Solution Preview a.) When you lift the pumpkin from the ground, you will work against the gravity, because gravitational force is acting ... #### Solution Summary The expert analyzes lifting and carrying a pumpkin from a field. The solution shows all the calculations and the answers. \$2.49

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https://www.usingenglish.com/forum/threads/106534-I-m-confused

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1. ## I'm confused I'm a little confused about the following sentence. Current liabilities exceeded current assets at January 31, 2009, by \$6.4 billion, a decrease of \$4.0 billion from January 31, 2008, largely due to a reduction in commercial paper outstanding at January 31, 2009. Does that mean current liabilities decreased by \$4.0 billion from January 31, 2008 or the difference between current liabilities and current assets decreased by \$4.0 billion to \$6.4 billion or is it just a bad sentence that could be interpreted in both ways? 2. ## Re: I'm confused Originally Posted by kendara I'm a little confused about the following sentence. Current liabilities exceeded current assets at January 31, 2009, by \$6.4 billion, a decrease of \$4.0 billion from January 31, 2008, largely due to a reduction in commercial paper outstanding at January 31, 2009. Does that mean current liabilities decreased by \$4.0 billion from January 31, 2008 or the difference between current liabilities and current assets decreased by \$4.0 billion to \$6.4 billion or is it just a bad sentence that could be interpreted in both ways? This means that as at January 31 2008 liabilities were exceeding the current assets by \$10.4 billion. As at January 31 2009 liabilities are exceeding the current assets by \$6.4 billion. This gives you a decrease in debt (in liabilities) of \$4.0 billion since the previous year (since Jan 31 2008). Both underlined seem to be correct. 1st sentence tells you more about the decrease in liabilities, 2nd sentence tells gives you more info (is more focused) on the decrease between the current assets and liabilities. You can interpret this in many ways. It all depends which information is more important for the business. Decrease in liabilities or the gap between assets and liabilities. I hope this helps. Please note, I am not an english theacher. 3. ## Re: I'm confused Hi ! If I understand this correctly. I would say that "the difference between current liabilities and current assets decreased by \$4.0 billion to \$6.4 billion" is correct. 4. ## Re: I'm confused I managed to find the acutual financial data. Wal-Mart Stores, Inc. - Walmart Second Quarter Earnings Exceed Consensus Estimates Based on that, the second sentence "the difference between current liabilities and current assets decreased by \$4.0 billion to \$6.4 billion" seems to be correct (not the first one). Thank you very much. 5. ## Re: I'm confused Originally Posted by kendara I managed to find the acutual financial data. Wal-Mart Stores, Inc. - Walmart Second Quarter Earnings Exceed Consensus Estimates Based on that, the second sentence "the difference between current liabilities and current assets decreased by \$4.0 billion to \$6.4 billion" seems to be correct (not the first one). Thank you very much. No,this indicates a net decrease in current liabilities by \$4 billions as on 31st Jan’09 (as on both the dates current liability was more than current assents)as explained below. As on 31st January’08: current labializes – current assets = X (Say) As on 31st january’09: current liabilities – current assets = \$6 = X-\$4(less/decrease) So X = \$10 billion which means there is a net decrease in current liabilities by \$4 billions and this is a good sign #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •

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CC-MAIN-2016-44

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https://oeis.org/A206898

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The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A206898 Number of nX6 0..3 arrays avoiding the pattern z-2 z-1 z in any row, column or nw-to-se diagonal 1 3636, 13220496, 35974254384, 100388008195387, 277547203982336971, 768965608329861602588, 2129766555218823044199240, 5899351779197139995930378687, 16340661622490256210423617956636 (list; graph; refs; listen; history; text; internal format) OFFSET 1,1 COMMENTS Column 6 of A206900 LINKS R. H. Hardin, Table of n, a(n) for n = 1..54 EXAMPLE Some solutions for n=4 ..3..1..3..0..1..0....3..0..2..0..2..2....0..0..3..0..1..1....2..1..2..0..0..1 ..1..3..2..3..1..3....3..0..0..0..3..2....2..0..2..2..1..0....2..0..0..1..3..0 ..2..2..1..1..3..3....0..2..3..2..3..3....3..2..2..0..2..3....2..0..3..0..2..0 ..0..1..0..3..3..1....2..3..0..0..1..3....3..0..1..3..0..2....3..1..1..2..0..2 CROSSREFS Sequence in context: A232418 A206844 A206836 * A207288 A239992 A253498 Adjacent sequences: A206895 A206896 A206897 * A206899 A206900 A206901 KEYWORD nonn AUTHOR R. H. Hardin Feb 13 2012 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified April 20 17:12 EDT 2024. Contains 371845 sequences. (Running on oeis4.)

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# Econ posted by . How do you calculate the Herfindahl- Hirschmann Index ? • Econ - ## Similar Questions 1. ### health can anyone find me a chart or something that tells me the average weight that a girl a certain height should be? 2. ### economics what is the price index for 2007 when \$675.00 is total expenditure for that year and \$600.00 for 2006. These total expenditures comes from a fixed market basket of goods. How do you calculate price index? 3. ### econ If the average level of nominal income in a nation is \$21,000 and the price level index is 154, the average real income is about:- a)\$12,546 b)\$13,636 c)\$15,299 d)\$17,823 4. ### economics in healthcare 6. Calculate Herfindahl Index and Concentration Ratios for these markets. Drug Companies H.I. = ? 5. ### econ Microsoft has been a giant in the software industry. Can we define Microsoft as a monopoly? 6. ### Algebra The body mass index I can be used to determine an individuals risk for heart disease. An index less than 25 indicates low risk. the formula for body mass index is I=700w/H^. W=weight and H=height, Jeromine is 67 inches tall. What weights …

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× ### Let's log you in. or Don't have a StudySoup account? Create one here! × or 19 0 22 # Class Note for CMPSCI 601 at UMass(22) Marketplace > University of Massachusetts > Class Note for CMPSCI 601 at UMass 22 No professor available These notes were just uploaded, and will be ready to view shortly. Either way, we'll remind you when they're ready :) Get a free preview of these Notes, just enter your email below. × Unlock Preview COURSE PROF. No professor available TYPE Class Notes PAGES 22 WORDS KARMA 25 ? ## Popular in Department This 22 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Massachusetts taught by a professor in Fall. Since its upload, it has received 19 views. × ## Reviews for Class Note for CMPSCI 601 at UMass(22) × × ### What is Karma? #### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more! Date Created: 02/06/15 CMPSCI 601 Recall From Last Time Lecture 21 To prove A is NPcomplete 0 Prove A E NP 0 Prove B 3 A where B is known to be NPcomplete The following problems are NPComplete 0 SAT CookLevin Theorem 0 3SAT 0 3COLOR 0 CLIQUE 0 Subset Sum 0 Knapsack decision version Knapsack Given n objects object 01 02 on weight 1121 mg mm 20 value 111 v2 on W max weight I can carry in my knapsack Optimization Problem chooseS g 1n to maximize 2 vi z39es such that Z 211239 3 W z39es Decision Problem Given 11717 W V can I get total value 2 V while total weight is 3 W Proposition 211 Khapsack is NPComplete Proof Let I 2 m1 mm T be an instance of Subset Sum Problem ENS g 1nZSmi 2 T i6 LetfI 2 m1 mmml mnTTgt be aninstance ofKnapsack Claim I E Subset Sum lt2 f I E Knapsack Fact 212 Even though Khapsack is NPComplete there is an e icieht dynamic programming algorithm that can closely approximate the maximum possible V CMPSCI 601 Approximability Lecture 21 Fact NPcomplete decision problems are all equiva lent Belief NPcomplete problems require exponential time in the worst case Fact Dif culty of NP Approximation problems varies Widely De nition 213 A is an NPoptz mz zatl on problem iff 0 For each instance as n 2 19 g 2pm is the set of feasible solutions We can test in P Whether 8 E F 0 Each 8 E has a cost 08 6 Z The cost 08 is computable in F For minimization problems OPTa 2 Sang 08 For maximization problems OPTa 2 833 08 A De nition 214 Let M be a polynomialtime algorithm st on any instance as M is an eapproximation algorithm iff for all as 10M OPTW maxOPTarcMar S 6 4 Oltelt1 e 2 01 is an excellent approximation Minimization problems at most times optimal Maximization prob lems at least 99 percent of optimal e 2 Minimization problems no more than twice opti mal Maximization problems at least half optimal e 2 99 not a very good approximation Minimization problems at most 100 times more than optimal Maxi mization problems at least one percent of optimal Four Classes of NP Optimization Problems INAPPROX E n0 PTIME eapprox alg if P g NP E 36162 O lt 61 lt 62 lt exists PTIME eggapprox alg n0 PTIME elapprox alg if P g NP PTAS E V6 gt OeXists PTIME eapprox alg FPTAS E V6 gt OeXists uniform eapprox alg running in time p01yn FPTAS stands for Fully PolynomialTime Approxima tion Scheme Clique TSP INAPPROX exists P appmx algfar n0 8 lt 1 MAX SAT ATSP VerteXCover APPROX same but nat all 8lt I PT As ETSP all 8lt I Kn apsac poly in n 18 CMPSCI 601 Vertex Cover Lecture 21 Input an undirected graph G 2 V Output a minimum size 0 C V such that C touches every edge Greedy nodes of high degree About log n times opti mal in the example above There are about n log n total nodes The n fat ones are a vertex cover but the greedy algorithm takes most of the others rst Better Find a Maximal Matching l C 2 Z 2 while E 5 d0 3 pick um E E 4 C 2 C U um 5 delete u 11 from G The edges picked are a maximal matching a disjoint set of edges to which we can t add another disjoint edge If there are m edges in this matching we ve used 2m nodes in C but any algorithm would have to use at least m C 3 2opt G e 2 1 Best known approx ratio 2 10 A Hamilton circuit for an undirected graph G is a cycle that starts and ends at some vertex v and visits every other vertex exactly once HC 2 G G has a Hamilton Circuit Fact 215 HC is NPComplete Nicest proof is in Sipser TSP 2 G 2 V EwL Ghas aHC ofweight g L G 2 V E n 1V1716thla V EL11101321 n 1 ifuv E E w uv 2 h 72106 otherwise Observation 216 For any undirected graph G G 6 HC lt2 MG 6 TSP Corollary 217 If TSP has a polynomialtime e approximation algorithm for any 6 lt 1 then P NP Thus TSP E INAPPROX ll G VE n 1V let 80 2 V 9019136 2 70106 1000000 1mm 6 E we uv 2 1000001 otherw1se Let FairTSP be the subset of TSP st no edge weight is more than 00001 percent more than any other edge weight Observation 218 For any undirected graph G G E HC 42gt 8G E TSP 42gt 89 E Fair TSP Observation 219 FairTSP is NPcomplete as a decision problem and as an Optimization problem it has a polynom ial time 10396appr0ximati0n algorithm 12 i Vk J ATSP TSP where dz k g dij dj 1 Claim 2110 Minimum Spanning Tree is a lower bound for ATSP CMST g CATSP Proof Visualize optimal tour Delete one edge and we have a spanning tree A 13 Theorem 2111 cMST g CATSP g 2cMST Proof The multigraph 2MST made by taking two copies of each edge in the tree is connected and all its nodes have even degree Thus it has an Euler s tour providing an e 2 approxi mation algorithm A 14 Aside A multigraph G 2 V E is a graph except that E may be a multiset ie there can be more than one edge between a certain pair of vertices An Euler s Tour of an undirected multigraph G is a tour that starts and ends at the same vertex and traverses each edge exactly once Fact G has an Euler s tour iff G is connected and each vertex of G has even degree If G has an Euler s tour then such a tour can be computed in linear time 15 Christo des Algorithm 1976 In the MST only worry about the odd degree nodes There are an even number of vertices of odd degree In polynomial time we can nd a minimum weight per fect matching M on the odddegree nodes MST UM is an Eulerian graph MST g TSP M g grsn Thus we get a tour at most 15 times optimal e 2 g 16 Euclidean TSP ETSP Euclidean distance in plane Wm 31 56 y 50 50 y 31 ETSP has a PolynomialTime Approximation Scheme PTAS Arora 1997 17 CMPSCI 601 Interactive Proofs Lecture 21 Il X readonly input nite 039 random bits control H Proof work tape MerlinArthur games MA Babai Decision problem D input string 3 Merlin Prover chooses the polynomiallength string H that Maximizes the chances of convincing Arthur that a is an element of D Arthur Veri er computes the Average value of his possible computations on H as Arthur is a polynomial time probabilistic Turing machine 18 De nition 2112 We say that A accepts D iff the follow ing conditions hold 1 If a E D there exists a proof H39 such that A accepts for every random string a Pro A az a 2 Accept 2 1 2 If a 4 D for every proof H A rejects for most of the random strings a Pro AH a 2 Accept lt i 19 Any decision problem D E NP has a deterministic polynomial time veri er satisfying De nition By adding randomness to the veri er we can greatly re strict its computational power and the number of bits of H that it needs to look at while still enabling it to accept all of NP We say that a veri er A is r02 qnrestricted iff for all inputs of size n and all proofs H A uses at most Orn random bits and examines at most Oqn bits of its proof H Let PCPrn be the set of boolean queries that are accepted by r01 qnrestricted veri ers Fact 2113 PCP Theorem NP 2 PCPlogn 1 The proof of this theorem is pretty messy certainly more than we can deal with here But we can look at the appli cations of the PCP Theorem to approximation problems 20 MAXBSAT given a 3CNF formula nd a truth as signment that maximizes the number of true clauses x1 V932 V927 x1 V324 V927 aT1aTgaT4 xQVaTgvm aT2933a5 mvmvm KVTQVasg Vacjvm Proposition 2114 MAXBSAT has a polynomialtime e 2 approximation algorithm Proof Be greedy set each variable in turn to the better value A You can do better a random assignment gets 78 of the clauses Open for Years Assuming NP 7 P is there some 6 O lt 6 lt 1 st MAXBSAT has no PTIME eapproximation algorithm 21 Theorem 2115 The PCP theorem NP 2 PCPlog n 1 is equivalent to the fact that If P NP then Forsome e1gt e gt O MAXBSAT has no polynomialtime eapproximation algorithm Fact 2116 MAXBSAT has a PTIME approximation algorithm with e 2 i and no better ratio can be achieved unless P 2 NP References 0 Approximation Algorithms for NP Hara Problems Dorit Hochbaum 6d PWS 1997 0 Juraj Hromkovic Algorithmics for Hard Problems Springer 2001 0 Sanjeev Arora The Approximability ofNPhard Prob lems STOC 98 wwwcsprincetoneduNarora 22 × × ### BOOM! 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# Math Snap ## Let $f(x)=\frac{5}{2} x^{2}-e^{x}$. Find the value of $x$ for which $f^{\prime \prime}(x)$ equals zero. Select one: a. 0 b. $e^{5}$ c. $\ln 5$ d. $5 e$ #### STEP 1 Assumptions 1. The function given is $f(x) = \frac{5}{2} x^2 - e^x$. 2. We need to find the value of $x$ for which the second derivative $f''(x)$ equals zero. #### STEP 2 First, we need to find the first derivative $f'(x)$ of the function $f(x)$. Using the power rule and the exponential rule for differentiation: $f'(x) = \frac{d}{dx} \left( \frac{5}{2} x^2 \right) - \frac{d}{dx} \left( e^x \right)$ #### STEP 3 Differentiate each term separately. $\frac{d}{dx} \left( \frac{5}{2} x^2 \right) = \frac{5}{2} \cdot 2x = 5x$ $\frac{d}{dx} \left( e^x \right) = e^x$ #### STEP 4 Combine the results to get the first derivative $f'(x)$. $f'(x) = 5x - e^x$ #### STEP 5 Next, we need to find the second derivative $f''(x)$ by differentiating $f'(x)$. $f''(x) = \frac{d}{dx} \left( 5x - e^x \right)$ #### STEP 6 Differentiate each term separately. $\frac{d}{dx} \left( 5x \right) = 5$ $\frac{d}{dx} \left( e^x \right) = e^x$ #### STEP 7 Combine the results to get the second derivative $f''(x)$. $f''(x) = 5 - e^x$ #### STEP 8 Set the second derivative $f''(x)$ equal to zero to find the value of $x$. $5 - e^x = 0$ #### STEP 9 Solve for $x$. $e^x = 5$ #### STEP 10 Take the natural logarithm of both sides to solve for $x$. $x = \ln 5$ ##### SOLUTION Thus, the value of $x$ for which $f''(x)$ equals zero is $\ln 5$. The correct answer is: c. $\ln 5$

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