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Applied Finite Mathematics HW Solutions 97

Applied Finite Mathematics HW Solutions 97 - EXERCISES 5.3...

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Unformatted text preview: EXERCISES 5.3 95 22. With P = 300, 1;: 7, m =4, and A“ = 4000, fornmla 5.4 gives .7 1'1 . 300[(1+—-—) — J _ 100 4 407 _ 4000(7) _ 4000_~———m—-~—7/400 =1- 300K4—00) 11— 400 _70. Thus, 407 n 70 7 407 37 (m) "17370-5 =" (5)” 73—0-1233 A calculator shows that 12 13 (%) =1.231 and (%) =1.253. The annuity will have at least $4000 after the 13th deposit. This is June 1, 2007. 23. With P = 500, 71:: 5,711. = 12, and A,1 = 10 000, formula 5.4 gives ' n 500 (1+ ) —1] _ -_ 100-12 241 _10000(5) 125 10000_ =- 500[(— ) —1]_ 1200 3. 5/1200 240 3 Thus, ' 241 “ 125 1 241 13 (m) ‘1“3‘5—mfl—2 =’" (EY “54°83 A calculator shows that 19 20 (g) -1082 .11 (22,3) 41.087. The annuity will have at least $10,000 after the 20th deposit. This is December 1, 2005. 24. With P = 100, 1' = 3, m = 12, and A,1 = 10 000, formula 5.4 gives 3 n 100[(1+ ) -1] . 0(3 10000=__100_13_ =1. 100K411) —1]=10—9{)——( )— =.25 3/1200 400 1200 Thus, _ 401 “ 25 1 401 ”#5 (as) "1—1001 ='" (470) ‘1‘1'25' A calculator shoe/5 that 89 90 (2%) - = 1.249 and (fig—é) = 1.252. _ _The annuity will have at least $10,000 after the £30th deposit. 25. With P = 2000, 7; = 6, m = 2, and n. = 20, the value of the annuity immediately after payment number 20 is . 6 20 2000 [(1 + 100-2) — ] 6/200 To find the value of this annuity 4 months later, we use formula 5.2 with P = 53, 740.75, 0: 6,711: 12, andn=4, A20 = = 53, 740.75. ..-I_ _ ____=_L. |_...|_..u... n..:.......:+.. Aprlnnhnlm umbdm I'q‘anl-mrrnmvfilr‘lfnn in heart or in fin". is stricthvnrol'llbi'led. ...
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