Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Review - Page 112: 43b

Answer

$\$3,700.81$

Work Step by Step

Using $A=P\left( 1+\dfrac{r}{n} \right)^{nt}$ or the compound interest formula, then \begin{array}{l}\require{cancel} A=3000\left( 1+\dfrac{0.03}{52} \right)^{52(7)} \\\\ A=3000\left( 1.0005769230769230769230769230769 \right)^{364} \\\\ A=3700.8100755850663837034463072889 \\\\ A\approx3700.81 .\end{array} Hence, with the given values, the amount will be $ \$3,700.81 .$
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