Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 7 Exponential and Logarithmic Functions - 7.1 Graph Exponential Growth Functions - 7.1 Exercises - Problem Solving - Page 484: 38b

Answer

$\approx2701.4$

Work Step by Step

The amount can be obtained by the formula $A=P(1+\frac{r}{n})^{nt}$ where $P$ is the initial amount, and $r$ is the annual interest rate which is compounded $n$ times per year over $t$ years. Hence here: $A=P(1+\frac{r}{n})^{nt}=P(1+\frac{0.035}{12})^{12\cdot3}=3000\\P=\frac{3000}{(1+\frac{0.035}{12})^{12\cdot3}}\approx2701.4$
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