Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.1 - Exponential Functions - Exercise Set - Page 672: 39

Answer

(a)$\$13,116.51$ . (b)$\$13,157.04$ . (c)$\$13,165.31$ .

Work Step by Step

Compounded interest formula is $A=P\left ( 1+\frac{r}{n} \right )^{nt}$. Where $A$ is the accumulated value. Given values are Investment $P=\$10,000$. Time $t=5$ years. Rate of interest $r=5.5\%=0.055$. (a.) Compounded semiannually $(n=2)$. Substitute all values into the formula. $A=10,000\left ( 1+\frac{0.055}{2} \right )^{2\cdot 5}$ $A=10,000\left ( \frac{2+0.055}{2} \right )^{10}$ $A=10,000\left ( \frac{2.055}{2} \right )^{10}$ $A=10,000\left ( 1.0275 \right )^{10}$ $A=10,000\left ( 1.31165103259 \right )$ $A=13116.5103259$ Rounded to the nearest cent. $A=\$13,116.51$ . (b.) Compounded monthly $(n=12)$ $A=10,000\left ( 1+\frac{0.055}{12} \right )^{12\cdot 5}$ $A=10,000\left ( \frac{12+0.055}{12} \right )^{60}$ $A=10,000\left ( \frac{12.055}{12} \right )^{60}$ $A=10,000\left ( 1.00458333333 \right )^{60}$ $A=10,000\left ( 1.31570377254 \right )$ $A=13157.0377254$ Rounded to the nearest cent. $A=\$13,157.04$ . (c.) Compounded continuously The formula is $A=Pe^{rt}$. Substitute all values. $A=10,000 e^{0.055\cdot 5}$ $A=13165.3067487$ Rounded to the nearest cent. $A=\$13,165.31$ .
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