evenans56 - $31,058.48 (b) 9.69 years 10. (a) $37,552.21...

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Even-Numbered Answers to Exercise Set 5.6: Applications 2. A(t) = the total amount of money in the investment after t years P = the principal, i.e. the original amount of money invested r = the interest rate, expressed as a decimal (rather than a percent) t = the number of years for which the money is invested 4. N(t) = the population (number of people, bacteria, animals, etc.) after time t . N 0 = the initial population r = the relative growth rate, expressed as a decimal (rather than a percent) t = the amount of time ( t may be measured in hours, days, years, etc. This information is contained somewhere in the problem.) 6. (a) $8,881.47 (b) 8.84 years 8. (a)
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Unformatted text preview: $31,058.48 (b) 9.69 years 10. (a) $37,552.21 (b) 17.45 years 12. (a) $32,416.98 (b) $32,604.51 (c) $32,700.54 (d) $32,765.43 (e) $ 32,798.14 14. $9,974.89 16. 4.07% 18. Option (b) is a better investment (7.97%, compounded continuously). 20. (a) 1,350 mosquitoes (b) 0.1% per hour (c) 1,358 mosquitoes (d) 1,416 mosquitoes (e) 1,098.61 hours 45.78 days 22. (a) t e t N 041 . 000 , 35 ) ( = (b) 46,635 people (c) 4.45 years 24. 4,578 ladybugs 26. (a) 672 bacteria (b) 127 million bacteria (c) 5.41 hours 28. (a) 12 grams (b) 2.02 grams (c) 16.86% (d) 7.79 days 30. (a) t e t m 0058 . 850 ) ( = (b) 46.77 milligrams (c) 249.47 years...
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