Answer
$\approx 8.01$ grams
Work Step by Step
In the model $A=16e^{-0.000121t}$ ,
substitute t=5715 and calculate A
$A=16e^{-0.000121(5715)}\approx 8.01$
In 5715 years,
approximately $8.01$ grams of carbon-14 will be present.
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.5 - Page 504: 15
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