Answer
$k=-0.006134$ decay rate or $0.6134\%$ decay per hour.
Work Step by Step
$A=A_{0}e^{kt}$
For $A=0.5, A_{0}=1, t=113, $ hour
$0.5=1e^{113k} $
$\ln0.5=113k $
$k=-0.006134$ decay rate or $0.6134\%$ decay per hour.
Intermediate Algebra for College Students (7th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13417-894-7
ISBN 13:
978-0-13417-894-3
Chapter 9 - Section 9.6 - Exponential Growth and Decay; Modeling Data - Exercise Set - Page 741: 26
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