Answer:
[tex]F(t) = 18000(0.6666)^{t}[/tex]
Step-by-step explanation:
The fish population after t years can be modeled by the following equation:
[tex]F(t) = F(0)(1-r)^{t}[/tex]
In which F(0) is the initial population and r is the constant rate of decay.
Year one the fish population was 18000
This means that [tex]F(0) = 18000[/tex]
In year three the fish population was 8000 fish.
Two years later, so [tex]F(2) = 8000[/tex]
[tex]F(t) = F(0)(1-r)^{t}[/tex]
[tex]8000 = 18000(1-r)^{2}[/tex]
[tex](1-r)^{2} = 0.4444[/tex]
[tex]\sqrt{(1-r)^{2}} = \sqrt{0.4444}[/tex]
[tex]1 - r = 0.6666[/tex]
[tex]r = 0.3334[/tex]
So
[tex]F(t) = 18000(0.6666)^{t}[/tex]