Math in Society Contents Problem Solving Extension: Taxes 30 Voting Theory 35 Weighted Voting 59 Apportionment 75 Fair Division 93 Graph Theory 117 Scheduling 155 Growth Models 173 Finance 197 Statistics 227 Describing Data 247 Probability 279 Sets 319 Historical Counting Systems 333 Fractals 367 Cryptography 387 Solutions to Selected Exercises 407 David Lippman David Lippman David Lippman Mike Kenyon, David Lippman David Lippman David Lippman David Lippman David Lippman David Lippman David Lippman, Jeff Eldridge, onlinestatbook.com David Lippman, Jeff Eldridge, onlinestatbook.com David Lippman, Jeff Eldridge, onlinestatbook.com David Lippman Lawrence Morales, David Lippman David Lippman David Lippman, Melonie Rasmussen Edition 2.4 David Lippman Pierce College Ft Steilacoom Copyright © 2013 David Lippman This book was edited by David Lippman, Pierce College Ft Steilacoom Development of this book was supported, in part, by the Transition Math Project and the Open Course Library Project Statistics, Describing Data, and Probability contain portions derived from works by: Jeff Eldridge, Edmonds Community College (used under CC-BY-SA license) www.onlinestatbook.com (used under public domain declaration) Apportionment is largely based on work by: Mike Kenyon, Green River Community College (used under CC-BY-SA license) Historical Counting Systems derived from work by: Lawrence Morales, Seattle Central Community College (used under CC-BY-SA license) Cryptography contains portions taken from Precalculus: An investigation of functions by: David Lippman and Melonie Rasmussen (used under CC-BY-SA license) Front cover photo: Jan Tik, http://www.flickr.com/photos/jantik/, CC-BY 2.0 This text is licensed under a Creative Commons Attribution-Share Alike 3.0 United States License To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/3.0/us/ or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California, 94105, USA You are 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terms of this work The best way to this is with a link to this web page: http://creativecommons.org/licenses/by-sa/3.0/us/ About the Author/Editor David Lippman received his master’s degree in mathematics from Western Washington University and has been teaching at Pierce College since Fall 2000 David has been a long time advocate of open learning, open materials, and basically any idea that will reduce the cost of education for students It started by supporting the college’s calculator rental program, and running a book loan scholarship program Eventually the frustration with the escalating costs of commercial text books and the online homework systems that charged for access led to action First, David developed IMathAS, open source online math homework software that runs WAMAP.org and MyOpenMath.com Through this platform, he became an integral part of a vibrant sharing and learning community of teachers from around Washington State that support and contribute to WAMAP These pioneering efforts, supported by dozens of other dedicated faculty and financial support from the Transition Math Project, have led to a system used by thousands of students every quarter, saving hundreds of thousands of dollars over comparable commercial offerings David continued further and wrote the first edition of this textbook, Math in Society, after being frustrated by students having to pay $100+ for a textbook for a terminal course Together with Melonie Rasmussen, he co-authored PreCalculus: An Investigation of Functions in 2010 Acknowledgements David would like to thank the following for their generous support and feedback • Jeff Eldridge, Lawrence Morales, and Mike Kenyon, who were kind enough to license me use of their works • The community of WAMAP users and developers for creating some of the homework content used in the online homework sets • Pierce College students in David’s online Math 107 classes for helping correct typos and identifying portions of the text that needed improving, along with other users of the text • The Open Course Library Project for providing the support needed to produce a full course package for this book Preface The traditional high school and college mathematics sequence leading from algebra up through calculus could leave one with the impression that mathematics is all about algebraic manipulations This book is an exploration of the wide world of mathematics, of which algebra is only one small piece The topics were chosen because they provide glimpses into other ways of thinking mathematically, and because they have interesting applications to everyday life Together, they highlight algorithmic, graphical, algebraic, statistical, and analytic approaches to solving problems This book is available online for free, in both Word and PDF format You are free to change the wording, add materials and sections or take them away I welcome feedback, comments and suggestions for future development If you add a section, chapter or problems, I would love to hear from you and possibly add your materials so everyone can benefit New in This Edition Edition has been heavily revised to introduce a new layout that emphasizes core concepts and definitions, and examples Based on experience using the first edition for three years as the primarily learning materials in a fully online course, concepts that were causing students confusion were clarified, and additional examples were added New “Try it Now” problems were introduced, which give students the opportunity to test out their understanding in a zero-stakes format Edition 2.0 also added four new chapters Edition 2.1 was a typo and clarification update on the first 14 chapters, and added additional new chapters No page or exercise numbers changed on the first 14 chapters Edition 2.2 was a typo revision A couple new exploration exercises were added Edition 2.3 and 2.4 were typo revisions Supplements The Washington Open Course Library (OCL) project helped fund the creation of a full course package for this book, which contains the following features: • • • • • Suggested syllabus for a fully online course Possible syllabi for an on-campus course Online homework for most chapters (algorithmically generated, free response) Online quizzes for most chapters (algorithmically generated, free response) Written assignments and discussion forum assignments for most chapters The course shell was built for the IMathAS online homework platform, and is available for Washington State faculty at www.wamap.org and mirrored for others at www.myopenmath.com The course shell was designed to follow Quality Matters (QM) guidelines, but has not yet been formally reviewed Problem Solving Problem Solving In previous math courses, you’ve no doubt run into the infamous “word problems.” Unfortunately, these problems rarely resemble the type of problems we actually encounter in everyday life In math books, you usually are told exactly which formula or procedure to use, and are given exactly the information you need to answer the question In real life, problem solving requires identifying an appropriate formula or procedure, and determining what information you will need (and won’t need) to answer the question In this chapter, we will review several basic but powerful algebraic ideas: percents, rates, and proportions We will then focus on the problem solving process, and explore how to use these ideas to solve problems where we don’t have perfect information Percents In the 2004 vice-presidential debates, Edwards's claimed that US forces have suffered "90% of the coalition casualties" in Iraq Cheney disputed this, saying that in fact Iraqi security forces and coalition allies "have taken almost 50 percent" of the casualties Who is correct? How can we make sense of these numbers? Percent literally means “per 100,” or “parts per hundred.” When we write 40%, this is 40 equivalent to the fraction or the decimal 0.40 Notice that 80 out of 200 and 10 out of 100 80 10 40 25 are also 40%, since = = 200 25 100 Example 243 people out of 400 state that they like dogs What percent is this? 243 60.75 This is 60.75% = 0.6075 = 400 100 Notice that the percent can be found from the equivalent decimal by moving the decimal point two places to the right Example Write each as a percent: a) a) 1 = 0.25 = 25% 4 b) 0.02 c) 2.35 b) 0.02 = 2% http://www.factcheck.org/cheney_edwards_mangle_facts.html © David Lippman c) 2.35 = 235% Creative Commons BY-SA Percents If we have a part that is some percent of a whole, then part , or equivalently, percent = = part percent ⋅ whole whole To the calculations, we write the percent as a decimal Example The sales tax in a town is 9.4% How much tax will you pay on a $140 purchase? Here, $140 is the whole, and we want to find 9.4% of $140 We start by writing the percent as a decimal by moving the decimal point two places to the left (which is equivalent to dividing by 100) We can then compute: = tax 0.094 = (140 ) $13.16 in tax Example In the news, you hear “tuition is expected to increase by 7% next year.” If tuition this year was $1200 per quarter, what will it be next year? The tuition next year will be the current tuition plus an additional 7%, so it will be 107% of this year’s tuition: $1200(1.07) = $1284 Alternatively, we could have first calculated 7% of $1200: $1200(0.07) = $84 Notice this is not the expected tuition for next year (we could only wish) Instead, this is the expected increase, so to calculate the expected tuition, we’ll need to add this change to the previous year’s tuition: $1200 + $84 = $1284 Try it Now A TV originally priced at $799 is on sale for 30% off There is then a 9.2% sales tax Find the price after including the discount and sales tax Example The value of a car dropped from $7400 to $6800 over the last year What percent decrease is this? To compute the percent change, we first need to find the dollar value change: $6800-$7400 = -$600 Often we will take the absolute value of this amount, which is called the absolute change: −600 = 600 Problem Solving Since we are computing the decrease relative to the starting value, we compute this percent out of $7400: 600 = 0.081 = 8.1% decrease This is called a relative change 7400 Absolute and Relative Change Given two quantities, Absolute change = ending quantity − starting quantity Relative change: absolute change starting quantity Absolute change has the same units as the original quantity Relative change gives a percent change The starting quantity is called the base of the percent change The base of a percent is very important For example, while Nixon was president, it was argued that marijuana was a “gateway” drug, claiming that 80% of marijuana smokers went on to use harder drugs like cocaine The problem is, this isn’t true The true claim is that 80% of harder drug users first smoked marijuana The difference is one of base: 80% of marijuana smokers using hard drugs, vs 80% of hard drug users having smoked marijuana These numbers are not equivalent As it turns out, only one in 2,400 marijuana users actually go on to use harder drugs Example There are about 75 QFC supermarkets in the U.S Albertsons has about 215 stores Compare the size of the two companies When we make comparisons, we must ask first whether an absolute or relative comparison The absolute difference is 215 – 75 = 140 From this, we could say “Albertsons has 140 more stores than QFC.” However, if you wrote this in an article or paper, that number does not mean much The relative difference may be more meaningful There are two different relative changes we could calculate, depending on which store we use as the base: 140 = 1.867 75 This tells us Albertsons is 186.7% larger than QFC 140 Using Albertsons as the base, = 0.651 215 This tells us QFC is 65.1% smaller than Albertsons Using QFC as the base, http://tvtropes.org/pmwiki/pmwiki.php/Main/LiesDamnedLiesAndStatistics Notice both of these are showing percent differences We could also calculate the size of 215 Albertsons relative to QFC: = 2.867 , which tells us Albertsons is 2.867 times the size 75 75 of QFC Likewise, we could calculate the size of QFC relative to Albertsons: = 0.349 , 215 which tells us that QFC is 34.9% of the size of Albertsons Example Suppose a stock drops in value by 60% one week, then increases in value the next week by 75% Is the value higher or lower than where it started? To answer this question, suppose the value started at $100 After one week, the value dropped by 60%: $100 - $100(0.60) = $100 - $60 = $40 In the next week, notice that base of the percent has changed to the new value, $40 Computing the 75% increase: $40 + $40(0.75) = $40 + $30 = $70 In the end, the stock is still $30 lower, or $30 = 30% lower, valued than it started $100 Try it Now The U.S federal debt at the end of 2001 was $5.77 trillion, and grew to $6.20 trillion by the end of 2002 At the end of 2005 it was $7.91 trillion, and grew to $8.45 trillion by the end of 2006 Calculate the absolute and relative increase for 2001-2002 and 2005-2006 Which year saw a larger increase in federal debt? Example A Seattle Times article on high school graduation rates reported “The number of schools graduating 60 percent or fewer students in four years – sometimes referred to as “dropout factories” – decreased by 17 during that time period The number of kids attending schools with such low graduation rates was cut in half.” a) Is the “decrease by 17” number a useful comparison? b) Considering the last sentence, can we conclude that the number of “dropout factories” was originally 34? http://www.whitehouse.gov/sites/default/files/omb/budget/fy2013/assets/hist07z1.xls Problem Solving a) This number is hard to evaluate, since we have no basis for judging whether this is a larger or small change If the number of “dropout factories” dropped from 20 to 3, that’d be a very significant change, but if the number dropped from 217 to 200, that’d be less of an improvement b) The last sentence provides relative change which helps put the first sentence in perspective We can estimate that the number of “dropout factories” was probably previously around 34 However, it’s possible that students simply moved schools rather than the school improving, so that estimate might not be fully accurate Example In the 2004 vice-presidential debates, Edwards's claimed that US forces have suffered "90% of the coalition casualties" in Iraq Cheney disputed this, saying that in fact Iraqi security forces and coalition allies "have taken almost 50 percent" of the casualties Who is correct? Without more information, it is hard for us to judge who is correct, but we can easily conclude that these two percents are talking about different things, so one does not necessarily contradict the other Edward’s claim was a percent with coalition forces as the base of the percent, while Cheney’s claim was a percent with both coalition and Iraqi security forces as the base of the percent It turns out both statistics are in fact fairly accurate Try it Now In the 2012 presidential elections, one candidate argued that “the president’s plan will cut $716 billion from Medicare, leading to fewer services for seniors,” while the other candidate rebuts that “our plan does not cut current spending and actually expands benefits for seniors, while implementing cost saving measures.” Are these claims in conflict, in agreement, or not comparable because they’re talking about different things? We’ll wrap up our review of percents with a couple cautions First, when talking about a change of quantities that are already measured in percents, we have to be careful in how we describe the change Example 10 A politician’s support increases from 40% of voters to 50% of voters Describe the change 10% Notice that since the We could describe this using an absolute change: 50% − 40% = original quantities were percents, this change also has the units of percent In this case, it is best to describe this as an increase of 10 percentage points 10% = 0.25 = 25% increase This is the 40% relative change, and we’d say the politician’s support has increased by 25% In contrast, we could compute the percent change: Lastly, a caution against averaging percents Example 11 A basketball player scores on 40% of 2-point field goal attempts, and on 30% of 3-point of field goal attempts Find the player’s overall field goal percentage It is very tempting to average these values, and claim the overall average is 35%, but this is likely not correct, since most players make many more 2-point attempts than 3-point attempts We don’t actually have enough information to answer the question Suppose the player attempted 200 2-point field goals and 100 3-point field goals Then they made 200(0.40) = 80 2-point shots and 100(0.30) = 30 3-point shots Overall, they made 110 shots 110 out of 300, for a = 0.367 = 36.7% overall field goal percentage 300 Proportions and Rates If you wanted to power the city of Seattle using wind power, how many windmills would you need to install? Questions like these can be answered using rates and proportions Rates A rate is the ratio (fraction) of two quantities A unit rate is a rate with a denominator of one Example 12 Your car can drive 300 miles on a tank of 15 gallons Express this as a rate 300 miles 20 miles Expressed as a rate, We can divide to find a unit rate: , which we could 15 gallons 1 gallon miles also write as 20 , or just 20 miles per gallon gallon Proportion Equation A proportion equation is an equation showing the equivalence of two rates or ratios Example 13 Solve the proportion x = for the unknown value x This proportion is asking us to find a fraction with denominator that is equivalent to the fraction We can solve this by multiplying both sides of the equation by 6, giving x = ⋅ = 10 414 P1 T4 P2 10 19 21 T2 T3 P2 T5 T7 T6 T9 T10 T8 T1 13 18 26 Priority List: T4, T3, T7, T2, T6, T5, T1 11 P1 T3 P2 22 T4 T1 T2 T6 T5 32 T7 17 12 30 11 Priority List: T5, T1, T3, T10, T2, T8, T4, T6, T7, T9 12 14 P1 T1 P2 T4 T3 T7 22 T6 T2 17 29 T10 T8 T5 T9 13 22 13 Priority List: C, D, E, F, B, G, A P1 C P2 B 16 10 A D E F G 21 13 15 a T1 (3) [25] T4 (12) [22] T2 (9) [24] T6 (8) [8] T7(10) [10] T3 (11) [21] T5 (5) [15] b Critical path: T1, T4, T7 Minimum completion time: 25 c Critical path priority list: T1, T2, T4, T3, T5, T7, T6 P1 P2 T1 20 15 T4 T7 T5 T3 T2 30 T6 20 28 Solutions to Selected Exercises 415 17 a T1 (8) [24] T2 (6) [22] T9 (2) [2] T5 (9) [16] T6 (3) [15] T8 (5) [12] T10 (7) [7] T3 (7) [21] T4 (4) [4] T7 (2) [14] b Critical path: T1, T5, T10 Minimum completion time: 24 c Critical path priority list: T1, T2, T3, T5, T6, T7, T8, T10, T4, T9 19 Critical path priority list: B, A, D, E, C, F, G 15 20 P1 P2 B E D A C F G 10 19 Growth Models a P0 = 20 Pn = Pn-1 + b Pn = 20 + 5n a P1 = P0 + 15 = 40+15 = 55 P2 = 55+15 = 70 b Pn = 40 + 15n c P10 = 40 + 15(10) = 190 thousand dollars d 40 + 15n = 100 when n = years Grew 64 in weeks: per week a Pn = + 8n b 187 = + 8n n = 23 weeks a P0 = 200 (thousand), Pn = (1+.09) Pn-1 where n is years after 2000 b Pn = 200(1.09)n c P16 = 200(1.09)16 = 794.061 (thousand) = 794,061 d 200(1.09)n = 400 n = log(2)/log(1.09) = 8.043 In 2008 Let n=0 be 1983 Pn = 1700(2.9)n 2005 is n=22 P22 = 1700(2.9)22 = 25,304,914,552,324 people Clearly not realistic, but mathematically accurate 416 11 If n is in hours, better to start with the explicit form P0 = 300 P4 = 500 = 300(1+r)4 500/300 = (1+r)4 1+r = 1.136 r = 0.136 a P0 = 300 Pn = (1.136)Pn-1 b Pn = 300(1.136)n c P24 = 300(1.136)24 = 6400 bacteria d 300(1.136)n = 900 n = log(3)/log(1.136) = about 8.62 hours 13 a P0 = 100 Pn = Pn-1 + 0.70 (1 – Pn-1 / 2000) Pn-1 b P1 = 100 + 0.70(1 – 100/2000)(100) = 166.5 c P2 = 166.5 + 0.70(1 – 166.5/2000)(166.5) = 273.3 15 To find the growth rate, suppose n=0 was 1968 Then P0 would be 1.60 and P8 = 2.30 = 1.60(1+r)8, r = 0.0464 Since we want n=0 to correspond to 1960, then we don’t know P0, but P8 would 1.60 = P0(1.0464)8 P0 = 1.113 a Pn = 1.113(1.0464)n b P0= $1.113, or about $1.11 c 1996 would be n=36 P36 = 1.113(1.0464)36 = $5.697 Actual is slightly lower 17 The population in the town was 4000 in 2005, and is growing by 4% per year Finance A = 200 + 05(200) = $210 I=200 t = 13/52 (13 weeks out of 52 in a year) P0 = 9800 200 = 9800(r)(13/52) r = 0.0816 = 8.16% annual rate P10 = 300(1 + 05 / 1)10 (1) = $488.67 a P20 = 2000(1 + 03 / 12) 20 (12 ) = $3641.51 in 20 years b 3641.51 – 2000 = $1641.51 in interest P8 = P0 (1 + 06 / 12) 8(12 ) = 6000 P0 = $3717.14 would be needed ( ) 200 (1 + 0.03 / 12) 30 (12 ) − = $116,547.38 0.03 / 12 b 200(12)(30) = $72,000 c $116,547.40 - $72,000 = $44,547.38 of interest 11 a P30 = ( ) d (1 + 0.06 / 12) 30 (12 ) − d = $796.40 each month 0.06 / 12 b $796.40(12)(30) = $286,704 c $800,000 - $286,704 = $513,296 in interest 13 a P30 = 800,000 = Solutions to Selected Exercises 417 ( ) 30000 − (1 + 0.08 / 1) −25(1) = $320,253.29 0.08 / b 30000(25) = $750,000 c $750,000 - $320,253.29 = $429,756.71 15 a P0 = 17 P0 = 500,000 = ( d − (1 + 0.06 / 12) −20 (12 ) 0.06 / 12 ( ) d = $3582.16 each month ) 700 − (1 + 0.05 / 12) −30 (12 ) = a $130,397.13 loan 0.05 / 12 b 700(12)(30) = $252,000 c $252,200 - $130,397.13 = $121,602.87 in interest 19 a P0 = ( ) d − (1 + 0.02 / 12) −48 21 P0 = 25,000 = = $542.38 a month 0.02 / 12 23 a Down payment of 10% is $20,000, leaving $180,000 as the loan amount d − (1 + 0.05 / 12) −30 (12 ) d = $966.28 a month b P0 = 180,000 = 0.05 / 12 d − (1 + 0.06 / 12) −30 (12 ) c P0 = 180,000 = d = $1079.19 a month 0.06 / 12 ( ) ( ) 25 First we find the monthly payments: d − (1 + 0.03 / 12) −5(12 ) d = $431.25 P0 = 24,000 = 0.03 / 12 431.25 − (1 + 0.03 / 12) −2 (12 ) Remaining balance: P0 = = $10,033.45 0.03 / 12 ( ) ( 27 6000(1 + 0.04 / 12) 12 N = 10000 (1.00333) = 1.667 12 N log((1.00333) ) = log(1.667 ) 12 N log(1.00333) = log(1.667 ) log(1.667 ) = about 12.8 years N= 12 log(1.00333) 12 N ) 418 ( ) 60 − (1 + 0.14 / 12) −12 N 0.14 / 12 3000(0.14 / 12) = 60 − (1.0117) −12 N 3000(0.14 / 12) = 0.5833 = − (1.0117) −12 N 60 0.5833 − = −(1.0117) −12 N − (0.5833 − 1) = (1.0117) −12 N log(0.4167) = log (1.0117) −12 N log(0.4167) = −12 N log(1.0117 ) log(0.4167) = about 6.3 years N= − 12 log(1.0117 ) 29 3000 = ( ) ( ) ( ) 50 (1 + 0.08 / 12) 5(12 ) − = $3673.84 31 First years: P5 = 0.08 / 12 Next 25 years: 3673.84(1 + 08 / 12) 25(12 ) = $26,966.65 ( ) 10000 − (1 + 0.08 / 4) −10 ( ) = $273,554.79 needed at 0.08 / d (1 + 0.08 / 4)15( ) − retirement To end up with that amount of money, 273,554.70 = 0.08 / He’ll need to contribute d = $2398.52 a quarter 33 Working backwards, P0 = ( Statistics a Population is the current representatives in the state’s congress b 106 c the 28 representatives surveyed d 14 out of 28 = ½ = 50% e We might expect 50% of the 106 representatives = 53 representatives This suffers from leading question bias This question would likely suffer from a perceived lack of anonymity This suffers from leading question bias Quantitative 11 Observational study 13 Stratified sample ) Solutions to Selected Exercises 419 15 a Group 1, receiving the vaccine b Group is acting as a control group They are not receiving the treatment (new vaccine) c The study is at least blind We are not provided enough information to determine if it is double-blind d This is a controlled experiment 17 a Census b Observational study Describing Data a Different tables are possible Score 30 40 50 60 70 80 90 100 Frequency b This is technically a bar graph, not a histogram: 30 40 50 c 100 60 70 80 90 30 40 50 90 60 80 70 100 420 a 5+3+4+2+1 = 15 b 5/15 = 0.3333 = 33.33% Bar is at 25% 25% of 20 = students earned an A a (7.25+8.25+9.00+8.00+7.25+7.50+8.00+7.00)/8 = $7.781 b In order, 7.50 and 8.00 are in middle positions Median = $7.75 c 0.25*8 = Q1 is average of 2nd and 3rd data values: $7.375 0.75*8 = Q3 is average of 6th and 7th data values: $8.125 5-number summary: $7.00, $7.375, $7.75, $8.125, $9.00 a (5*0 + 3*1 + 4*2 + 2*3 + 1*5)/15 = 1.4667 b Median is 8th data value: child c 0.25*15 = 3.75 Q1 is 4th data value: children 0.75*15 = 11.25 Q3 is 12th data value: children 5-number summary: 0, 0, 1, 2, d 11 Kendra makes $90,000 Kelsey makes $40,000 Kendra makes $50,000 more Probability b, 13 13 a 150 = 44.8% 335 26 65 = 11 13 1 – 11 = 12 12 17 1 ⋅ = 6 36 21 17 16 17 17 ⋅ = ⋅ = 49 48 49 147 = 52 13 15 1 – 19 25 40 = 65 65 3 ⋅ = = 6 36 12 Solutions to Selected Exercises 421 23 a 4 16 ⋅ = = 52 52 2704 169 48 192 12 ⋅ = = 52 52 2704 169 48 48 2304 144 c ⋅ = = 52 52 2704 169 13 13 169 d ⋅ = = 52 52 2704 16 48 39 1872 117 e = ⋅ = 52 52 2704 169 b 25 4 16 ⋅ = 52 51 2652 11 14 154 ⋅ = 25 24 600 14 11 154 b ⋅ = 25 24 600 11 10 110 c ⋅ = 25 24 600 14 13 182 d ⋅ = 25 24 600 e no males = two females Same as part d 27 a 29 P(F and A) = 10 65 31 P(red or odd) = of 14 33 P(F or B) = 10 + − = Or red and odd-numbered blue marbles is 10 out 14 14 14 14 18 + + 10 + 12 44 26 22 44 + − = Or P(F or B) = = 65 65 65 65 65 65 35 P(King of Hearts or Queen) = 37 a P(even | red) = 5 + = 52 52 52 b P(even | red) = 39 P(Heads on second | Tails on first) = They are independent events 422 41 P(speak French | female) = 14 43 Out of 4,000 people, 10 would have the disease Out of those 10, would test positive, while would falsely test negative Out of the 3990 uninfected people, 399 would falsely test positive, while 3591 would test negative 9 a P(virus | positive) = = 2.2% = + 399 408 3591 3591 b P(no virus | negative) = = 99.97% = 3591 + 3592 45 Out of 100,000 people, 300 would have the disease Of those, 18 would falsely test negative, while 282 would test positive Of the 99,700 without the disease, 3,988 would falsely test positive and the other 95,712 would test negative 282 282 = 6.6% P(disease | positive) = = 282 + 3988 4270 47 Out of 100,000 women, 800 would have breast cancer Out of those, 80 would falsely test negative, while 720 would test positive Of the 99,200 without cancer, 6,944 would falsely test positive 720 720 P(cancer | positive) = = 9.4% = 720 + 6944 7664 49 ⋅ ⋅ ⋅ = 96 outfits 51 a · · = 64 b · · = 24 53 26 · 26 · 26 · 10 · 10 · 10 = 17,576,000 55 4P4 or · · · = 24 possible orders 57 Order matters 7P4 = 840 possible teams 59 Order matters 12P5 = 95,040 possible themes 61 Order does not matter 12C4 = 495 63 50C6 = 15,890,700 65 27C11 · 16 = 208,606,320 67 There is only way to arrange CD's in alphabetical order The probability that the CD's are in alphabetical order is one divided by the total number of ways to arrange CD's Since alphabetical order is only one of all the possible orderings you can either use permutations, or simply use 5! P(alphabetical) = 1/5! = 1/(5 P 5) = 120 Solutions to Selected Exercises 423 69 There are 48C6 total tickets To match of the 6, a player would need to choose of ⋅ 42 252 those 6, 6C5, and one of the 42 non-winning numbers, 42C1 = 12271512 12271512 71 All possible hands is 52C5 Hands will all hearts is 13C5 1287 2598960 28 73 $3 + $2 + ( −$1) =−$ = -$0.19 37 37 37 37 75 There are 23C6 = 100,947 possible tickets 100946 Expected value = $29,999 + ( −$1) = -$0.70 100947 100947 77 $48 ( 0.993) + (−$ 302)(0.007) = $45.55 Sets {m, i, s, p} One possibility is: Multiples of between and 10 Yes True True 11 False 13 A ⋃ B = {1, 2, 3, 4, 5} 15 A ⋂ C = {4} 17 Ac = {6, 7, 8, 9, 10} 19 Dc ⋂ E = {t, s} 23 (F ⋂ E)c ⋂ D = {b, c, k} 21 (D ⋂ E) ⋃ F = {k, b, a, t, h} F 25 F E D 29 One possible answer: (A ⋂ B) ⋃ (B ⋂ C) 27 E D 424 31 (A ⋂ Bc) ⋃ C 33 39 n(A ⋂ B ⋂ Cc) = 41 n(G ⋃ H) = 45 37 n(A ⋂ C) = 35 43 136 use Redbox M 6 LotR 45 SW a) had seen exactly one b) had only seen SW Historical Counting Partial answer: Jars: singles, @ x2, @ x6, @ x12 3+6+12+12 = 33 113 3022 53 1100100 11 332 13 111100010 15 7,1,10 base 12 = 1030 base 10 17 6,4,2 base 12 = 914 base 10 19 175 base 10 = 1,2,7 base 12 = 21 10000 base 10 = 5,9,5,4 base 12 = 23 135 = 6,15 base 20 = Solutions to Selected Exercises 425 25 360 = 18,0 base 20 = 27 10500 = 1,6,5,0 base 20 29 1,2,12 base 20 = 452 base 10 31 3,0,3 base 20 = 1203 base 10 33 35 Fractals Step Step Step Step 426 Step Step Four copies of the Koch curve are needed to create a curve scaled by log(4 ) D= ≈ 1.262 log(3) 11 Eight copies of the shape are needed to make a copy scaled by D = log(8) ≈ 1.893 log(3) imaginary –2 + 3i 2+i real –3i 13 15 a) – i 17 a) + 12i b) – 4i b) 10 − 2i imaginary c) 14 + 2i 19 (2 + 3i )(1 − i ) = + i It appears that multiplying by – i both scaled the number away from the origin, and rotated it clockwise about 45° z1 = iz + = i (2) + = + 2i 21 z = iz1 + = i (1 + 2i ) + = i − + = −1 + i z = iz + = i (−1 + i ) + = −i − + = −i + 3i 5+i real Solutions to Selected Exercises 427 z0 = z1 = z − 0.25 = − 0.25 = −0.25 23 z = z1 − 0.25 = ( −0.25) − 0.25 = −0.1875 z = z − 0.25 = (−0.1875) − 0.25 = −0.21484 z = z − 0.25 = (−0.21484) − 0.25 = −0.20384 25 attracted, to approximately -0.37766 + 0.14242i 27 periodic 2-cycle 29 Escaping 31 periodic 3-cycle 33 a) Yes, periodic 3-cycle b) Yes, periodic 3-cycle c) No Cryptography ZLU KZB WWS PLZ SHRED EVIDENCE O2H DO5 HDV MERGER ON MNB AET RTE HAT TLR EII YN 11 THE STASH IS HIDDEN AT MARVINS QNS 13 UEM IYN IOB WYL TTL N 15 HIRE THIRTY NEW EMPLOYEES MONDAY 17 ZMW NDG CDA YVK 19 a) b) c) 21 We test out all n from to 10 n 4n 4n mod 11 4 16 64 256 1024 4096 16384 65536 9 262144 10 1048576 Since we have repeats, and not all values from to 10 are produced (for example, there is no n is 4n mod 11 = 7), is not a generator mod 11 428 23 15710 mod = (157 mod 5)10 mod = 210 mod = 1024 mod = 25 37 mod 23 = 27 Bob would send 57 mod 33 = 14 Alice would decrypt it as 143 mod 33 = 31 a 678 mod 83 = (674 mod 23)2 mod 83 = 492 mod 83 = 2401 mod 83 = 77 6716 mod 83 = (678 mod 23)2 mod 83 = 772 mod 83 = 5929 mod 83 = 36 b 17000 mod 83 = (100 mod 83)*(170 mod 83) mod 83 = (17)(4) mod 83 = 68 c 675 mod 83= (674 mod 83)(67 mod 83) mod 83 = (49)(67) mod 83 = 3283 mod 83 = 46 d 677 mod 83= (674 mod 83) (672 mod 83)(67 mod 83) mod 83 = (49)(7)(67) mod 83 = 22981 mod 83 = 73 e 6724 = 6716678 so 6724 mod 83 = (6716 mod 83)(678 mod 83) mod 83 = (77)(36) mod 83 = 2272 mod 83 = 33 ... of how many calories mini-muffins will contain, we would want to know the number of calories in each mini-muffin To find the calories in each mini-muffin, we could first find the total calories... whole recipe 12 muffins ⋅ muffin 3000 calories gives 150 calories per mini-muffin 20 mini − muffins 150 calories totals 600 calories consumed 4 mini muffins ⋅ mini − muffin Example 27 You need... billion is not a cut in current spending, but a cut in future increases in spending, largely reducing future growth in health care payments In this case, at least the numerical claims in both statements