Answer
$P \approx 6283.17$
Work Step by Step
To find the present value for an interest rate $r=0.05$ compounded continuously for $t=3$ years
$A=Pe^{rt}$
$7300=Pe^{0.05(3)}$
$P = \frac{7300}{e^{0.05(3)}} \approx 6283.17$
Calculus with Applications (10th Edition)
by Lial, Margaret L.; Greenwell, Raymond N.; Ritchey, Nathan P.
Published by
Pearson
ISBN 10:
0321749006
ISBN 13:
978-0-32174-900-0
Chapter 2 - Nonlinear Functions - 2.6 Applications: Growth and Decay; Mathematics of Finance - 2.6 Exercises - Page 108: 13
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