Answer
$8.013$ grams of carbon-14
Work Step by Step
$A=A_{0}e^{kt}$
For $A=16e^{-0.000121t}, t=5715$
$A=16e^{-0.000121\times 5715}$
$A=8.013$ grams of carbon-14 will be present in $5715$ years.
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 740: 15
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