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Unformatted text preview: 41 I= P*N*i I= $10,000*4.25*0.1 I= $4,250 The amount of interest paid on November 1, 2012 would be $4,250 42 I= P*N*i I= 200*6*0.01 I= 12 I would owe $212 after 6 months and $12 would be the interest owed 43 F= P(1 + i)^N F= $1000(1+ 0.08) ^ 2.5 F= $1,212.16 The future equivalence of the $1,000 after 2.5 years is $1,212.16. 44 I= P*N*i I= P* 47 EOYK 1 2 3 Interest Paid $800 $553.60 $287.49 Principal Repayment $3,080 $3,326.40 $3,592.51 b. The amount of principle owed at the beginning of year 3 would be $3,593.60 48 F= P(1 + i)^N F= $10,000(1 + 0.06) ^ 5 F= $13,382.3 Juanita will owe Jim $13,382.3 if she repays the entire loan at the end of five years 415 72/10 = 7.2 Ashlea's money would double every 7.2 years 428 F= A * [(((1+i)^N) 1)/ i ] F= $730 * [(((1 + 0.06) ^ 35) 1) / 0.06 ] F= $117,732 I will accumulate $117,732 by saving $2 a day for 35 years if the annual interest rate is 7% ...
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This note was uploaded on 09/09/2009 for the course ISE 13715 taught by Professor Koelling during the Fall '09 term at Virginia Tech.
 Fall '09
 KOELLING

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