10.2: Appendix B - Solutions to Selected Exercises
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1: Mathematics and Problem-Solving
1. 18/230 = 0.07826 = about 7.8%
3. €250(0.23) =€ 57.50 in VAT
5. $15000(5.57) = $83,550
7. absolute increase: 1050. Relative: 1050/3250 = 0.323 = 32.3% increase
9. a. 2200 – 2200(0.15) = 2200(0.85) = $1870
b. Yes, their goal was to decrease by at least 15%. They exceeded their goal.
11. Dropping by 6% is the same as keeping 94%. a(0.94) = 300. a = 319.15. Attendance was about 319 before the drop.
13. a. Kaplan’s enrollment was 64.3% larger than Walden’s. 30510
b. Walden’s enrollment was 39.1% smaller than Kaplan’s.
c. Walden’s enrollment was 60.9% of Kaplan’s.
15. If the original price was $100, the basic clearance price would be $100 – $100(0.60) = $40. The additional markdown would bring it to $40 - $40(0.30) = $28. This is 28% of the original price.
17. These are not comparable; “a” is using a base of all Americans and is talking about health insurance from any source, while “b” is using a base of adults and is talking specifically about health insurance provided by employers.
21. These statements are equivalent, if we assume the claim in “a” is a percentage point increase, not a relative change. Certainly these messages are phrased to convey different opinions of the levy. We are told the new rate will be $9.33 per $1000, which is 0.933% tax rate. If the original rate was 0.833% (0.1 percentage point lower), then this would indeed be a 12% relative increase.
23. 20% of 30% is 30%(0.20) = 6%, a 6 percentage point decrease.
25. Probably not, unless the final is worth 50% of the overall class grade. If the final was worth 25% of the overall grade, then a 100% would only raise her average to 77.5%
27. $4/10 pounds = $0.40 per pound (or 10 pounds/$4 = 2.5 pounds per dollar)
29. x = 15
31. 2.5 cups
33. 74 turbines
35. 96 inches
37. $6000
39. 55.6 meters
43. The population density of the US is 84 people per square mile. The density of India is about 933 people per square mile. The density of India is about 11 times greater than that of the U.S.
49. The oil in the spill could produce 93.1 million gallons of gasoline. Each car uses about 600 gallons a year. That would fuel 155,167 cars for a year.
53. An answer around 100-300 gallons would be reasonable
57. 156 million miles
59. The time it takes the light to reach you is so tiny for any reasonable distance that we can safely ignore it. 750 miles/hr is about 0.21 miles/sec. If the sound takes 4 seconds to reach you, the lightning is about 0.84 miles away. In general, the lightning will be 0.21 n miles away, which is often approximated by dividing the number of seconds by 5.
61. About 8.2 minutes
63. Four cubic yards (or 3.7 if they sell partial cubic yards)
3: Mathematics and Finance
1. A = 200 + .05(200) = $210
3. I=200. t = 13/52 (13 weeks out of 52 in a year). \(P_0\) = 9800
200 = 9800(r)(13/52) r = 0.0816 = 8.16% annual rate
5. \(P_{10}=300(1+.05 / 1)^{10(1)}=\$ 488.67\)
7. a. \(P_{20}=2000(1+.03 / 12)^{20(12)}=\$ 3641.51\) in 20 years
b. 3641.51 – 2000 = $1641.51 in interest
9. \(P_{8}=P_{0}(1+.06 / 12)^{8(12)}=6000 \). \(P_{0}=\$ 3717.14\) would be needed
11. a. \(P_{30}=\dfrac{200\left((1+0.03 / 12)^{30(12)}-1\right)}{0.03 / 12}=\$ 116,547.38\)
b. 200(12)(30) = $72,000
c. $116,547.40 - $72,000 = $44,547.38 of interest
13. a. \(P_{30}=800,000=\dfrac{d\left((1+0.06 / 12)^{30(12)}-1\right)}{0.06 / 12}\) d = $796.40 each month
b. $796.40(12)(30) = $286,704
c. $800,000 - $286,704 = $513,296 in interest
15. a. \(P_{0}=\dfrac{30000\left(1-(1+0.08 / 1)^{-25(1)}\right)}{0.08 / 1}=\$ 320,243.29\)
b. 30000(25) = $750,000
c. $750,000 - $320,243.29 = $429,756.71
17. \(P_{0}=500,000=\dfrac{d\left(1-(1+0.06 / 12)^{-20(12)}\right)}{0.06 / 12}\) d = $3582.16 each month
19. a. \(P_{0}=\dfrac{700\left(1-(1+0.05 / 12)^{-30(12)}\right)}{0.05 / 12}=\) a $130,397.13 loan
b. 700(12)(30) = $252,000
c. $252,200 - $130,397.13 = $121,602.87 in interest
21. \(P_{0}=25,000=\dfrac{d\left(1-(1+0.02 / 12)^{-48}\right)}{0.02 / 12}=\$ 542.38\) a month
23. a. Down payment of 10% is $20,000, leaving $180,000 as the loan amount
b. \(P_{0}=180,000=\dfrac{d\left(1-(1+0.05 / 12)^{-30(12)}\right)}{0.05 / 12}\), d = $966.28 a month
c. \(P_{0}=180,000=\dfrac{d\left(1-(1+0.06 / 12)^{-30(12)}\right)}{0.06 / 12}\), d = $1079.19 a month
25. First we find the monthly payments:
\(P_{0}=24,000=\dfrac{d\left(1-(1+0.03 / 12)^{-5(12)}\right)}{0.03 / 12}\), d = $431.25
Remaining balance: \(P_{0}=\dfrac{431.25\left(1-(1+0.03 / 12)^{-2(12)}\right)}{0.03 / 12}=\$ 10,033.45\)
27. \(\begin{aligned}
6000(1+0.04 / 12)^{12 N}&=10000\\
(1.00333)^{12 N}&=1.667\\
\log \left((1.00333)^{12 N}\right)&=\log (1.667)\\
12 N \log (1.00333)&=\log (1.667)\\
N&=\dfrac{\log (1.667)}{12 \log (1.00333)}=\text { about } 12.8 \text { year }
\end{aligned} \)
29. \(\begin{aligned}
3000&=\dfrac{60\left(1-(1+0.14 / 12)^{-12 N}\right)}{0.14 / 12}\\
3000(0.14 / 12)&=60\left(1-(1.0117)^{-12 N}\right)\\
\dfrac{3000(0.14 / 12)}{60}&=0.5833=1-(1.0117)^{-12 N}\\
0.5833-1&=-(1.0117)^{-12 N}\\
-(0.5833-1)&=(1.0117)^{-12 N}\\
\log (0.4167)&=\log \left((1.0117)^{-12 N}\right)\\
\log (0.4167)&=-12 N \log (1.0117)\\
N&=\dfrac{\log (0.4167)}{-12 \log (1.0117)}=\text { about } 6.3 \text { years }
\end{aligned}\)
31. First 5 years: \(P_{5}=\dfrac{50\left((1+0.08 / 12)^{5(12)}-1\right)}{0.08 / 12}\) = $3673.84
Next 25 years: \(3673.84(1+.08 / 12)^{25(12)}=\$ 26,966.65\)
33. Working backwards, \(P_{0}=\dfrac{10000\left(1-(1+0.08 / 4)^{-10(4)}\right)}{0.08 / 4}=\$ 273,554.79\) needed at retirement. To end up with that amount of money, \(273,554.70=\dfrac{d\left((1+0.08 / 4)^{15(4)}-1\right)}{0.08 / 4}\). He’ll need to contribute d = $2398.52 a quarter.
5: Data and Statistics
1. 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
3. This suffers from leading question bias
5. This question would likely suffer from a perceived lack of anonymity
7. This suffers from leading question bias
9. Quantitative
11. Observational study
13. Stratified sample
15. a. Group 1, receiving the vaccine
b. Group 2 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
6: Growth and Decay
1. a. \(P_{0}=20\). \(P_{n}=P_{n-1}+5\)
b. \(P_{n}=20+5 n\)
3. a. \(P_{1}=P_{0}+15=40+15=55\). \(P_{2}=55+15=70\)
b. \(P_{n}=40+15 n\)
c. \(P_{10}=40+15(10)=190\) thousand dollars
d. \(40 + 15n = 100\) when \(n = 4\) years.
5. Grew 64 in 8 weeks: 8 per week
a. \(P_{n}=3+8 n\)
b. \(187 = 3 + 8n\). \(n = 23\) weeks
7. a. \(P_0 = 200\) (thousand), \(P_{n}=(1+.09) P_{n-1}\) where \(n\) is years after 2000
b. \(P_{n}=200(1.09)^{n}\)
c. \(P_{16}=200(1.09)^{16}=794.061\) (thousand) = 794,061
d. \(200(1.09)^{n}=400\). \(n=\log (2) / \log (1.09)=8.043\). In 2008.
9. 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.
11. If \(n\) is in hours, better to start with the explicit form. \(P_{0}=300\). \(P_{4}=500=300(1+r)^{4}\)
\(500 / 300=(1+r)^{4}\). \(1+r=1.136\). \(r=0.136\)
a. \(P_{0}=300\). \(P_{n}=(1.136) P_{n-1}\)
b. \(P_{n}=300(1.136)^{n}\)
c. \(P_{24}=300(1.136)^{24}=6400\) bacteria
d. \(300(1.136)^{n}=900\). \(n=\log (3) / \log (1.136)=\) about 8.62 hours
13. a. \(P_{0}=100 \quad P_{n}=P_{n-1}+0.70\left(1-P_{n-1} / 2000\right) P_{n-1}\)
b. \(P_{1}=100+0.70(1-100 / 2000)(100)=166.5\)
c. \(P_{2}=166.5+0.70(1-166.5 / 2000)(166.5)=273.3\)
15. To find the growth rate, suppose \(n=0\) was 1968. Then \(P_0\) would be 1.60 and \(P_{8}=2.30=1.60(1+r)^{8}\), \(r = 0.0464\). Since we want \(n=0\) to correspond to 1960, then we don’t know \(P_0\), but \(P_8\) would \(1.60 = P_{0}(1.0464)^{8}\). \(P_{0}=1.113\)
a. \(P_{n}=1.113(1.0464)^{n}\)
b. \(P_{0}=\$ 1.113\), or about $1.11
c. 1996 would be \(n=36\). \(P_{36}=1.113(1.0464)^{36}=\$ 5.697\). Actual is slightly lower.
17. The population in the town was 4000 in 2005, and is growing by 4% per year.
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A logarithm is an exponent. Specifically, it is the exponent to which a base \(b\) is raised to produce a given value. In the expressions given, the base \(b\) has the same value. The exponent, \(y\), in the expression by can also be written as the logarithm, \(\log _{b} x\), and the value of \(x\) is the result of raising \(b\) to the power of \(y\).
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Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation \(b^{y}=x\), and then properties of exponents can be applied to solve for \(x\).
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The natural logarithm is a special case of the logarithm with base b in that the natural log always has base \(e\). Rather than notating the natural logarithm as \(\log _{e}(x)\), the notation used is \(\ln (x)\).
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\(a^c=b \)
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\(x^y=64\)
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\(x^y=64\)
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\(15^b=a\)
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\(13^a=142\)
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\(e^n=w\)
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8: Mathematics and Politics
Apportionment
1. a. Math: 6, English: 5, Chemistry: 3, Biology: 1
b. Math: 7, English: 5, Chemistry: 2, Biology: 1
c. Math: 6, English: 5, Chemistry: 3, Biology: 1
d. Math: 6, English: 5, Chemistry: 3, Biology: 1
e. Math: 6, English: 5, Chemistry: 2, Biology: 2
3. a. Morning: 1, Midday: 5, Afternoon: 6, Evening: 8
b. Morning: 1, Midday: 4, Afternoon: 7, Evening: 8
c. Morning: 1, Midday: 5, Afternoon: 6, Evening: 8
d. Morning: 1, Midday: 5, Afternoon: 6, Evening: 8
e. Morning: 2, Midday: 5, Afternoon: 6, Evening: 7
5. a. Alice: 18, Ben: 14, Carlos: 4
b. Alice: 19, Ben: 14, Carlos: 3
c. Alice: 19, Ben: 14, Carlos: 3
d. Alice: 19, Ben: 14, Carlos: 3
e. Alice: 18, Ben: 14, Carlos: 4
7. a. A: 40, B: 24, C: 15, D: 30, E: 10
b. A: 41, B: 24, C: 14, D: 30, E: 10
c. A: 40, B: 24, C: 15, D: 30, E: 10
d. A: 40, B: 24, C: 15, D: 30, E: 10
e. A: 40, B: 24, C: 15, D: 29, E: 11
Describing Data
1. a. Different tables are possible
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Score |
Frequency |
|---|---|
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30 |
1 |
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40 |
0 |
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50 |
4 |
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60 |
3 |
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70 |
6 |
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80 |
5 |
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90 |
2 |
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100 |
3 |
b. This is technically a bar graph, not a histogram:
c.
3. a. 5+3+4+2+1 = 15
b. 5/15 = 0.3333 = 33.33%
5. Bar is at 25%. 25% of 20 = 5 students earned an A
7. 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 = 2. Q1 is average of 2nd and 3rd data values: $7.375
0.75*8 = 6. Q3 is average of 6th and 7th data values: $8.125
5-number summary: $7.00, $7.375, $7.75, $8.125, $9.00
d. 0.637
9. a. (5*0 + 3*1 + 4*2 + 2*3 + 1*5)/15 = 1.4667
b. Median is 8th data value: 1 child
c. 0.25*15 = 3.75. Q1 is 4th data value: 0 children
0.75*15 = 11.25. Q3 is 12th data value: 2 children
5-number summary: 0, 0, 1, 2, 5
d.
11. Kendra makes $90,000. Kelsey makes $40,000. Kendra makes $50,000 more.
Fractals
3.
5.
7.
11. Four copies of the Koch curve are needed to create a curve scaled by 3. \[D=\dfrac{\log (4)}{\log (3)} \approx 1.262 \nonumber \]
13. Eight copies of the shape are needed to make a copy scaled by 3.
Voting Theory
1.
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Number of voters |
3 |
3 |
1 |
3 |
2 |
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1st choice |
A |
A |
B |
B |
C |
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2nd choice |
B |
C |
A |
C |
A |
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3rd choice |
C |
B |
C |
A |
B |
3. a. 9+19+11+8 = 47
b. 24 for majority; 16 for plurality (though a choice would need a minimum of 17 votes to actually win under the Plurality method)
c. Atlanta, with 19 first-choice votes
d. Atlanta 94, Buffalo 111, Chicago 77. Winner: Buffalo
e. Chicago eliminated, 11 votes go to Buffalo. Winner: Buffalo
f. A vs B: B. A vs C: A. B vs C: B. B gets 2 pts, A 1 pt. Buffalo wins.
5. a. 120+50+40+90+60+100 = 460
b. 231 for majority; 116 for plurality
c. A with 150 first choice votes
d. A 1140, B 1060, C 1160, D 1240. Winner: D
e. B eliminated, votes to C. D eliminated, votes to A. Winner: A
f. A vs B: B. A vs C: A. A vs D: D. B vs C: C. B vs D: D. C vs D: C
A 1pt, B 1pt, C 2pt, D 2pt. Tie between C and D.
Winner would probably be C since C was preferred over D
7. a. 33
b. 17
9. Yes, B
11. B, with 17 approvals
13. Independence of Irrelevant Alternatives Criterion
15. Condorcet Criterion
Weighted Voting
1. a. 9 players
b. 10+9+9+5+4+4+3+2+2 = 48
c. 47
3. a. 9, a majority of votes
b. 17, the total number of votes
c. 12, which is 2/3 of 17, rounded up
5. a. P1 is a dictator (can reach quota by themselves)
b. P1, since dictators also have veto power
c. P2, P3, P4
7. a. none
b. P1
c. none
9. a. 11+7+2 = 20
b. P1 and P2 are critical
11. Winning coalitions, with critical players underlined:
{ P1 , P2 } { P1 ,P2,P3} { P1,P2 ,P4} { P1 ,P2,P3,P4} { P1,P3 } { P1,P3 ,P4}
P1: 6 times, P2: 2 times, P3: 2 times, P4: 0 times. Total: 10 times
Power: P1: 6/10 = 60%, P2: 2/10 = 20%, P3: 2/10 = 20%, P4: 0/10 = 0%
13. a. { P1 } { P1 ,P2} { P1 ,P3} { P1 ,P4} { P1 ,P2,P3} { P1 ,P2,P4} { P1 ,P3,P4} { P1 ,P2,P3,P4}
P1: 100%, P2: 0%, P3: 0%, P4: 0%
b. { P1,P2 } { P1,P3 } { P1,P4 } { P1 ,P2,P3} { P1 ,P2,P4} { P1 ,P3,P4} { P1 ,P2,P3,P4}
P1: 7/10 = 70%, P2: 1/10 = 10%, P3: 1/10 = 10%, P4: 1/10 = 10%
c. { P1,P2 } { P1,P3 } { P1 ,P2,P3} { P1 , P2 ,P4} { P1 , P3 ,P4} { P1 ,P2,P3,P4}
P1: 6/10 = 60%, P2: 2/10 = 20%, P3: 2/10 = 20%, P4: 0/10 = 0%
15. P3 = 5. P3+P2 = 14. P3+P2+P1 = 27, reaching quota. P1 is critical.
17. Sequential coalitions with pivotal player underlined
<P1, P2 ,P3> <P1, P3 ,P2> <P2, P1 ,P3> <P2, P3 ,P1> <P3, P1 ,P2> <P3, P2 ,P1>
P1: 2/6 = 33.3%, P2: 2/6 = 33.3%, P3: 2/6 = 33.3%
19. a. 6, 7
b. 8, given P1 veto power
c. 9, given P1 and P2 veto power
21. If adding a player to a coalition could cause it to reach quota, that player would also be critical in that coalition, which means they are not a dummy. So a dummy cannot be pivotal.
23. We know P2+P3 can’t reach quota, or else P1 wouldn’t have veto power.
P1 can’t reach quota alone.
P1+P2 and P1+P3 must reach quota or else P2/P3 would be dummy.
a. { P1 , P2 } { P1 , P3 } { P1 ,P2,P3}. P1: 3/5, P2: 1/5, P3: 1/5
b. <P1, P2 ,P3> <P1, P3 ,P2> <P2, P1 ,P3> <P2,P3, P1 > <P3, P1 ,P2> <P3,P2, P1 >
P1: 4/6, P2: 1/6, P3: 1/6
25. [4: 2, 1, 1, 1] is one of many possibilities
27. [56: 30, 30, 20, 20, 10]
29. [54: 10, 10, 10, 10, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] is one of many possibilities
Contributors and Attributions
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Saburo Matsumoto
CC-BY-4.0