Calculus with Applications (10th Edition)

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


$P \approx 6283.17$

Work Step by Step

To find the present value for an interest rate $r=0.05$ compounded continuously for $t=3$ years $A=Pe^{rt}$ $7300=Pe^{0.05(3)}$ $P = \frac{7300}{e^{0.05(3)}} \approx 6283.17$
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