Answer
See below.
Work Step by Step
We know that $x=2xe^{k\cdot5.27}\\0.5=e^{5.27k}\\5.27k=\ln0.5\\k=\frac{\ln0.5}{5.27}\approx-0.131527$
Thus $N(20)=100e^{-0.131527\cdot20}\approx7.204$
$N(40)=100e^{-0.131527\cdot40}\approx0.519$
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 6 - Review Exercises - Page 500: 54
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