Answer
$Q(t)\approx Q_{0}e^{-0.0248t}$
Work Step by Step
$Q(t)=Q_{0}e^{-kt}$
The decay constant k and half-life $t_{h}$ are related by
$t_{h}k=\ln 2$
$28k=\ln 2$
$k=\displaystyle \frac{\ln 2}{28}\approx 0.0248$
$Q(t)\approx Q_{0}e^{-0.0248t}$
Finite Math and Applied Calculus (6th Edition)
by Waner, Stefan; Costenoble, Steven
Published by
Brooks Cole
ISBN 10:
1133607705
ISBN 13:
978-1-13360-770-0
Chapter 9 - Section 9.3 - Logarithmic Functions and Models - Exercises - Page 658: 54a
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