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]