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Applied Finite Mathematics HW Solutions 95

Applied Finite Mathematics HW Solutions 95 - EERCISES 5.3...

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Unformatted text preview: EERCISES 5.3 93 -13._ With 77 = 7.5, m m 12, n = 360, and A350 2 800 000, formula 5.4 gives 7.5 360 . P 1 + 100- 12) ._ 1] 1207.5 36" 7.5(800 000) Therefore, P = i = 593.72. 1207.5 36" .. 1 1200 14. With P = 2000, 'i = 10, m m 1, and 71. = 8, the value of the annuity after 8 years is 10 3 2000 (H1004) 4] 10/100 To calculate what this accumulates to over the next 36 years, we use formula 5.3 with P m 22871.?8,1’= 10,171 m 1, and 71 z 36, A3 a 2 22871.73. . ' 10 36 :2 . ——~ 2 2 . . C35 2871 78(1-1— 100.1) 7070 803 15. With P = 1000, 7‘. .—. 7.5, m e 4, and a = 32, the value of the annuity after 8 yeata is 7.6 32 1000[(1+100_4) .4] 44 W = 3 89.8?- To calculate what this accumulates to over the next 18.5 years, we use formula 5.3 with P = 43 489.87, 77 = 7.6, m = 4, and n = 74, A32: . 7.5 7“ 07., .—. 13489.87 (1 + m) =_ 175 096.61. With P = 200, i = 7.6, m = 4, and a = 74-, the value of the 13.5 year annuity is 200 7.6 74 (1+ 100-4) “ ] 711/400 They have therefore \$206,950.79 in the account. 16. With P = 50, 2' = 6, m = 12, and 71 m 12, the value of the annuity on April 1, 2005 is An = = 31 854.18. 6 1‘2 50 (1+ 100.12) —1 6/1200 This amount collects one month’s interest from April 1, 2005 until May 1, 2005. Using formula 5.2 with P = 616.79, 7'. = 6, m = 12, and n = 1, the accumulated value is 6 - 1 100- 12 A12 2 = 616.79. SI = 616.79 (I + ) = 619.87. .! .-_. 1-... I... .L- : I..:........:o-.. Arumanu... undue“... ﬂurlknr Wham-13M :n nnrl' nr in ﬁt“ it: stridlv mhihited. ...
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