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Applied Finite Mathematics HW Solutions 96

Applied Finite Mathematics HW Solutions 96 - 94 EXERCISES...

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Unformatted text preview: ' 94 EXERCISES 5.3 17' 18. 19. 20. 21. T1131: mania-in] I'll-us luau mam-lunar! :n swan-loam un'lh mun-ink! In... In: il-u I Ink-mun ni‘ll’anflnl‘n Dnnbnlm 'L‘...t-'l.....— With P = 100, 2‘ = 5, m = 12, and n = 120, the value of the annuity on May 1, 2014 is 5 120 100[(1+ 100.12) —1] 5/1200 This amount collects one month's interest from May 1, 2014 until June 1, 2014. Using formula 5.2 with P = 15 528.23, 1' = 5, m = 12, and n = 1, the accumulated value is 5 100-12 With P : 200, 2' = 6, m 2. 12, and n = 43, formula 5.4 gives the value of the annuity on October 1, 2014 ' 6 43 200 l(1+100_12) —1] 6/1200 With P =‘500, i = 5, m = 4, and n = 28, formula 5.4 gives the value of the annuity on 5 23 500 (”100-4) — A120 = = 15 523.23. 01 = 15,528.23 (1 + ) = 15 592.93. A43 = = 9567.91. October 1, 2010 A23 = 5,400 = 16 639.69. With P = 50, 1: = 5, m = 12, and An = 10 000, formula 5.4_ gives 5 n 50 [(1 + 100- 12) _ 1] 1205 " 10 000(5) 125 10000=———————— :5 50 — —1=_=—. 5/1200 1200 1200 3 Thus, (a 1,1,1. E -2 240 "50 3 "6 A calculator shows that 145 146 (g) :15. 1.. (3.33.) 21.335. You-will have at least $10,000 after the 146‘” deposit. This is on June 1, 2016. With P = 500, 3': 6, m = 12, and An = 500 000, formula 5.4 gives 241 n 11 (E) —-F—-1.833. 6 n ' . 500 [(1 + — 1.] n _ 100 - 12 201 _ 500 000(6) _ 500000— 6/1200 ==> 500 [(200) —1] —- 1200 —25[}[}. Thus, 2—06 - 500 E A calculator shows that 359 I 380 (%) = 5.99 and (-33%) = 6.02. You will have $500,000 after the 360“1 deposit. This is when you are 55. 1| 11 (201) _1:2500:5 :2} (201) =6. Judi” :n a...“ M L. A." :. .|_'.al-. .._I..:l.:¢...I r‘"\ I'M-.0 ...
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