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9 Pages

perm_03

Course: CMU 301, Fall 2009
School: Iowa State
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Word Count: 602

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Iowa State - CMU - 301
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Iowa State - CMU - 115
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Iowa State - CMU - 801
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Iowa State - CMU - 484
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Iowa State - CMU - 301
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Iowa State - CMU - 106
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Iowa State - CMU - 801
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Iowa State - CMU - 257
q p 6 o) P n4$#1 %X58 % 5mR 5W % #20X( @) P Q F V 3 E & V ( ) 3 1 1 ) & G p#g%dE j00%9E Heqd ) 9Iiq%@98 i4 $hR R f 1 ) ( p U 1 t Gh 8 6 ) V R f g%dE 20X( A9E HeqHd ) %Iiw8 A@98 TgutQ 1 ) p v 1 t Gh R V h &#$hR V l I k 4 G I) $V p & G V
Iowa State - CMU - 484
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Iowa State - RU - 477
Summary of Random VariablesDe nition. A random variable is a real-valued function on a sample space. De nition. The cumulative distribution function or c.d.f of the random variable X is de nedMATH 477 Section E17 8 98for all real numbers b, ,1
Iowa State - RU - 152
MATH 152 Section C.17 7 97Review Sheet 2Answers are not guaranteed, but if you nd an incorrect answer, please let me know. 1. Determine whether or not the sequence fa g converges and nd its limit if it does converge where a = n sinn. Answer: Div
Iowa State - CMU - 301
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Iowa State - CMU - 801
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Iowa State - RU - 477
Homework 3 Solutionsp. 177 4. False. Suppose X = 0 with probability 1, so1MATH 477 Section E17 8 98y=FX(x)x 0Then FX 0 = 1 but lim, FX x = 0 and lim+ FX x = 1; indeed, the limit fails to exist.x!0 x!0p. 178 9. = IE 1 X , = 1 IE X , =
Iowa State - RU - 152
MATH 152 Section C.16 3 97Homework 1 Sections 6.5 & 7.1SOLUTIONSx6.5 12. Find the numbers b such that the average value of f x = 2 + 6x , 3x2 on the interval0; b is equal to 3.Solution.So, rst we nd favg in terms of b. 1 Z b f x dx favg =
Iowa State - CMU - 106
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Iowa State - CMU - 301
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Iowa State - RU - 477
Summary of Continuous Random Variables, Part I 5.1, 5.2, 5.3, 5.5De nition. X is a continuous random variable if there exists a nonnegative function f , de nedMATH 477 Section E17 9 98for all real x 2 ,1; 1 having the property that for any set
Iowa State - CMU - 801
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Iowa State - RU - 477
Final Exam SOLUTIONS Take-Home Portion10 pts. 1. A game show host has a contestant on stage and o ers her three doors to choose. Behind one of those doors is a fabulous prize. Behind each of the other two is a copy of A First Course in Probability,
Iowa State - RU - 373
Bisection method: 0 1.5 1 1.25 2 1.375 3 1.4375 4 1.46875 5 1.45312 6 1.44531 7 1.44922 8 1.44727 9 1.44824 10 1.44775 11 1.44751 12 1.44739 13 1.44745 14 1.44742 15 1.4474 16 1.44741 17 1.44741 18 1.44741 19 1.44741 Convergence. Newton's method: 0 1
Iowa State - RU - 373
{VERSION 2 3 "SUN SPARC SOLARIS" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Output" 2 20 " 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {PSTY
Iowa State - RU - 373
{VERSION 2 3 "SUN SPARC SOLARIS" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Output" 2 20 " 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {PSTY
Iowa State - CMU - 257
maximize 36r+42p-16d-30fsubject tormaxdem) r<=100000rmindel) r>=50000pmaxdem) p<=20000pmindel) p>=5000d_inv) d<=40000f_inv) f<=60000rmax_v_p) 2dr-8fr<=0pmax_v_p) 2dp-8fp<=0rmin_o_r) -6dr+5fr>=0pmin_o_r) -6dp+5fp>=0d_gas) d-dr-dp=0f_gas)
Iowa State - CMU - 257
max 20x1-150x2-40x3st2) 4x1-2x2+6x3=83) 2x1-3x2+2x3<104) -3x1-5x2+4x3<4endfree x3 LP OPTIMUM FOUND AT STEP 0 OBJECTIVE FUNCTION VALUE 1) 460.0000 VARIABLE VALUE REDUCED COST X1 11
Iowa State - CMU - 257
maximize 2L+Dsubject to malt) 2L+3D<=75hops) 3L+D<=60yeast) 2L+1.66666667D<=50end LP OPTIMUM FOUND AT STEP 0 OBJECTIVE FUNCTION VALUE 1) 43.22581 VARIABLE VALUE REDUCED COST L 16.
Iowa State - CMU - 257
min 1.5y1+0.7y2subject to 2) 50y1+20y2>=6003) 30y1+20y2>=4004) 50y1+10y2>=350 LP OPTIMUM FOUND AT STEP 0 OBJECTIVE FUNCTION VALUE 1) 18.50000 VARIABLE VALUE REDUCED COST Y1 10.00000
Iowa State - CMU - 257
!p. 138 #4max 4x1+3x2-x3st2) x1+x2+x3=1003) 2x1+2x2+x3<2004) 4x2+3x3>505) -5x1+2x2>40endfree x3 LP OPTIMUM FOUND AT STEP 4 OBJECTIVE FUNCTION VALUE 1) 322.8571 VARIABLE VALUE REDUCED COST
Iowa State - CMU - 257
maximize 2L+Dsubject to malt) 2L+3D<=75hops) 3L+D<=60yeast) 2L+1.66666667D<=50end LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE 1) 43.33333 VARIABLE VALUE REDUCED COST L 16.
Iowa State - RU - 477
Final Exam SOLUTIONS In-Class Portion10 pts. 1. Mathematicians, physicists and engineers frequent a bar. The probability that an engineer will leave a tip is 90. The probability that a physicist will leave a tip is 80. The probability that a mathema
Iowa State - RU - 477
Final Exam Take-Home PortionThe Rules:1. The answers to this part of the exam will be due on Wednesday, July 29th when you enter class. No answers will be accepted for any reason after 8:30. NO EXCEPTIONS. You may submit answers sooner, of course.
Iowa State - MATH - 314
Math 314Sketch of Solutions to Midterm 21. Complete the table by finding all nonnegative integers m and n for which the graphs below have the following properties: bipartite n3 m, n 0 n2 n-connected n=3 m n 0; or (m, n) = (0, 1) n0 =n n=3 n =
Iowa State - MATH - 301
Math 301 Part A.Homework 2Due 10/04/041. Let a and b be nonzero integers. Prove that d = gcd(a, b) if and only if a = a/d and b = b/d are relatively prime integers. 2. (a) Use induction to prove that 02 + 12 + 22 + + n2 = n 0. (b) Use induc
Iowa State - MATH - 301
Math 301Homework 1Due 09/13/04Part A. 1. In each part, prove that the two statements are not logically equivalent. To do that, find counterexamples P and Q which make one of the statements true and the other false. Does one of the statements in
Iowa State - MATH - 304
Math 304 1. Proof by induction.Sketch of Solutions to Homework 2Base. The statement is true for n = 1 since 13 = 1 = 12 (1 + 1)2 /4. Step. Assume the statement is true for some integers n 1. Then 13 + 23 + + n3 = 13 + 23 + + n3 + (n + 1)3
Iowa State - MATH - 304
Math 304 1. Solution 1. We have Sketch of Solutions to Homework 41 -1 1 -1 2 (-4x)n = 2 (-4)n xn , = (1 - 4x)- 2 = n n 1 - 4x n=0 n=0so we need to prove that1 -2 (-4)n = n2n . nBut -1 2 n = = = so we are done. Solution 2. We know that
Iowa State - MATH - 314
Math 314Graph TheorySpring 2003Sketch of Solutions to Assignment 11. (a) not graphic (sum of degrees is odd). (b) graphic. (c) graphic. (d) graphic. 2. Let G be any graph on p vertices, and let and be the maximum and minimum degree of G, res
Iowa State - MATH - 304
Math 304Sketch of Solutions to Homework 11. (a) (Problem 8.1.1.) Use the Pigeonhole Principle. i. There are 26 possible initials (pigeonholes). To make sure at least 2 people have the same first initial, we need 26 + 1 = 27 people. ii. There are
Iowa State - MATH - 314
Math 314 1. The answers are as follows: cubic never m=n=3 n=4 acyclic never m = 0, 1 or n = 0, 1 n = 0, 1, 2Sketch of Solutions to Midterm 1Cn Km,n KnEulerian n3 even m, n 2; or m + n = 1 odd nHamiltonian n3 m = n 2; or m + n = 1 n = 1 or n
Iowa State - MATH - 301
Math 301Homework 4Due 12/06/04For denition of a conjugacy class and notation cl(a), see Problem 3 on p. 89 of the textbook. Recall that elements are called conjugate if they belong to the same conjugacy class. Equivalently, for x, y G, y is co
Iowa State - MATH - 314
Math 314Homework 3Due 3/27/031. Is the tennis graph of Homework 2 bipartite? Prove your answer. If no, what is the smallest number of edges you need to remove to obtain a bipartite graph? Prove your answer. 2. Determine the chromatic numbers o
Iowa State - MATH - 304
Math 304nSketch of Solutions to Midterm 22n n1. Let cn = k=0 2k 2(nk) . The sum on the right is exactly the convolution of the sequence { k nk with itself, i.e. {cn } = { 2n } { 2n }, so n n }cn xn =n=0 n=02n n x n n=02n n x n=
Iowa State - MATH - 314
Math 314Sketch of Solutions to Homework 31. The tennis graph is not bipartite because it has odd cycles, for example, the top and bottom C7 (the long thin boxes). Since these two cycles are disjoint, we need to delete at least 1 edge from each of
Iowa State - MATH - 314
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Iowa State - MATH - 304
Math 304Final ExamDue 12/17/2003This exam is due Wednesday, December 17, in my office, Carver 456, by 5pm. You may consult the texts for this course, the handouts, the homework solutions, your notes taken in lecture and your homework. Do not us
Iowa State - MATH - 314
Math 314Graph TheorySpring 20031Decomposition of GraphsDefinition 1.1 A graph G is decomposable into H1 , H2 , . . . , Hk if G has subgraphs H1 , H2 , . . . , Hk such that 1. each edge of G belongs to one of the Hi 's for some i = 1, 2, . .
Iowa State - MATH - 301
Math 301Midterm 1Due 10/22/04This exam is due Friday, October 22, in class. You may consult the text for this course, your notes taken in lecture, your homework, and sketches of solutions of homework problems. Do not use other books or papers o
Iowa State - MATH - 552
Math 552Homework 1Due 2/7/20051. Use the formal power series identity etx ety = et(x+y) in C[x, y][t] to derive the binomial formula n n k nk n (x + y) = x y . k k=0 2. (a) Consider walks in the (x, y)-plane with unit steps (1, 0) east, (1, 0)
Iowa State - MATH - 317
UNIV ID q1, 8/29 q2, 9/5 q3, 9/12 q4, 9/19 q5, 9/26 m1, 10/3 q6, 10/10 q7, 10/17 q8, 10/24 q9,10/31 q10,11/7 m2,11/14 q11,12/5 q12,12/12 q total f,12/15 total grade Total 10 10 10 10 10 80 10 10 10 10 10 80 10 10 100 150 100 letter 3494 10 10 70 8 7
Iowa State - MATH - 304
Math 304Midterm 1 (Take-home)Due 10/7/2003This exam is due Tuesday, October 7, in class. You may consult the text for this course, your notes taken in lecture and your homework. Do not use any other books or papers or materials from a library o
Iowa State - STAT - 643
STAT 643, Spring 2009, Homework 04Due: Thur. Feb 12, 20091. The -algebra generated by a r.v. Y : (, F) (R, B) is (Y ) = {Y -1 (B) = { : Y () B} : B B}. Verify that {Y -1 (B), B B} is a -algebra, and it is the smallest -algebra such that Y is m
Iowa State - STAT - 643
STAT 643, Spring 2009, Homework 03Due: Thur. Feb 5, 20091. Suppose 1 , 2 , are i.i.d. r.v.s such that E {znj , j n} are constants s.t. as n i= 0 and E2 i= 2 (0, ). Suppose2 max1jn znj 0, as n . n 2 j=1 znjLet Sn =n j=1 znj j
Iowa State - NR - 86223
4-H Materials Order Form 2008-09 Program Year (Each member is limited to $5 worth of materials, additional materials may be purchased.) (Place an X in front of the materials you'd like to order.) Name Total AmountAerospace 4H-81 Rockets Away! ($6
Iowa State - ADE - 43628
Elementary Engineering: Bridges and StructuresTarget Audience: Grades 3-5Description: Introduce your students to the exciting career of engineering as they develop abilities of technological design. Challenge your students to design a structure to
Iowa State - CS - 518
ComS 418/518 : HW1 Solutions Question 1. (10 points) Exercise 1.3, page 15. Answer 1 We have n segments, so there are 2n vertices. The algorithm is as follows: 1. 2. 3. 4. Sort the 2n points lexicographically, first by x, then by y coordinate. Remove
Iowa State - CS - 342
Com S 342 meeting -*- Outline -*-* Call by Reference (3.8, pp. 107-114) Note: see below for conventions used to draw pictures.* Informal overview* pictures and basic meaning- CALL BY REFERENCElet t = 5 u = 6 v = 7 w =
Iowa State - CS - 342
1Com S 342 Fall 2007Name:Exam 1 on Language Design and Scheme BasicsThis test has 8 questions and pages numbered 1 through 6.Principles of Programming LanguagesRemindersThis test is closed book and notes. If you need more space, use the b
Iowa State - CS - 342
Com S 342 - Principles of Programming Languages HOMEWORK 1: SCHEME BASICSDue: problem 1, 3-5 at beginning of class on Tuesday, September 4, 2007.In this homework, you will learn some of the basics of Scheme andfu
Iowa State - CS - 342
Com S 342 - Principles of Programming Languages HOMEWORK 0: GETTING STARTEDDue: problems 1-2, 3, 8, 10-12 at beginning of class August 28, 2007.In this homework, you will get around a bit on the Com S departmentmachines, send us vital
Iowa State - CS - 342
Com S 342 - Principles of Programming Languages HOMEWORK 8: ENVIRONMENT PASSING INTERPRETERS IIDue: Problems 2,3,5,6 at the beginning of class, Thursday, November 8, 2007. Problems 8,9 at the beginning of class, Tuesday
Iowa State - CS - 342
Topics for Exam 2 in Com S 342This exam covers topics from homework 2-4 upto and including the lectures on tail recursion.REMINDERSFor this test, you can use one (1) page (8.5 by 11 inches, one (1)side, no less than 9pt font)