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.

Respuesta :

Answer:

Part 1) The exponential function is equal to [tex]y=1,350(0.95)^{x}[/tex]

Part 2) The population in 2010 was  [tex]992\ fish[/tex]

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

[tex]y=a(b)^{x}[/tex]

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

[tex]a=1,350\ fish[/tex]

[tex]b=(100\%-5\%)=95\%=0.95[/tex]

substitute

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

Part 2) Find the population in 2010

we have

[tex]y=1,350(0.95)^{x}[/tex]

so

For [tex]x=(2010-2004)=6\ years[/tex]

substitute

[tex]y=1,350(0.95)^{6}=992\ fish[/tex]

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