College Algebra (10th Edition)

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

Chapter 6 - Section 6.3 - Exponential Functions - 6.3 Assess Your Understanding - Page 437: 112



Work Step by Step

By plugging in $P=1000,d=3$ into the formula: $N=1000(1-e^{-0.15\cdot3})=1000(1-e^{-0.45})\approx362.37$
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