College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.7 - Financial Models - 6.7 Assess Your Understanding - Page 477: 69

Answer

See below.

Work Step by Step

a) Using the formula the equation is: $233=215.3(1+0.01r)^5\\\frac{233}{215.3}=(1+0.01r)^5\\\sqrt[5] {\frac{233}{215.3}}=1+0.01r\\\sqrt[5] {\frac{233}{215.3}}-1=0.01r\\r=100(\sqrt[5] {\frac{233}{215.3}}-1)\approx1.59$ b) Using the formula the equation is: $300=215.3(1+0.01\cdot1.59)^n\\\frac{300}{215.3}=(1+0.01\cdot1.59)^n\\n=\log_{1+0.01\cdot1.59}\frac{300}{215.3}\approx21.03$
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