Answer
$\approx 4.01$ grams
Work Step by Step
In the model $A=16e^{-0.000121t}$ ,
substitute t=11,430 and calculate A
$A=16e^{-0.000121(11,430)}\approx 4.01$
In 11,430 years.
approximately $4.01$ 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: 16
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