Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 5 - Exponential Functions - 5.5 Exponential Growth and Decay Rates and Compounding Interest - 5.5 Exercises - Page 470: 15

Answer

$P(t)=40(1.03)^{t}$

Work Step by Step

We define our variables as $P(t)=\text{Population of animals}$ $t=\text{Time in years}$ Since the growth started with $40$ animals, in our model, $a=40$. The growth rate $r$ is $3\%=0.03$ We can find the base using the formula $b=1+r$ $b=1+0.03=1.03$ Exponential model is of the form $P(t)=a\cdot b^{t}$ Therefore, our model is $P(t)=40(1.03)^{t}$
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