Question

The fish population of Lake Collins is decreasing at a rate of 5% per year. In 2004 there were about 1,350 fish. Write an exponential decay function that models this situation. Then find the population in 2010.

Answer 1

Answer:

Part 1) The exponential function is equal to $y=1,350(0.95)^{x}$

Part 2) The population in 2010 was  $992\ fish$

Step-by-step explanation:

Part 1) Write an exponential decay function that models this situation

we know that

In this problem we have a exponential function of the form

$y=a(b)^{x}$

where

y ----> the fish population of Lake Collins since 2004

x ----> the time in years

a is the initial value

b is the base

we have

$a=1,350\ fish$

$b=(100\%-5\%)=95\%=0.95$

substitute

$y=1,350(0.95)^{x}$ ----> exponential function that represent this scenario

Part 2) Find the population in 2010

we have

$y=1,350(0.95)^{x}$

so

For $x=(2010-2004)=6\ years$

substitute

$y=1,350(0.95)^{6}=992\ fish$