Answer
$9.797$
Work Step by Step
We know that $x=2xe^{k\cdot1690}\\0.5=e^{1690k}\\1690k=\ln0.5\\k=\frac{\ln0.5}{1690}\approx-0.000410146$
Thus $N(50)=10e^{-0.000410146\cdot50}\approx9.797$
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 6 - Section 6.8 - Exponential Growth and Decay Models; Newton's Law: Logistic Growth and Decay Models - 6.8 Assess Your Understanding - Page 486: 9
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