Answer:
![F(t) = 18000(0.6666)^(t)](https://img.qammunity.org/2021/formulas/mathematics/high-school/huux0gv534b4p9ghcd1ghe96va99efa6id.png)
Explanation:
The fish population after t years can be modeled by the following equation:
![F(t) = F(0)(1-r)^(t)](https://img.qammunity.org/2021/formulas/mathematics/high-school/p02vbdlt0jb1gnex7d80xzxpkz51dpxjfy.png)
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
![F(0) = 18000](https://img.qammunity.org/2021/formulas/mathematics/high-school/93bse1lreq177ihe5ai53cj5608t3dn57z.png)
In year three the fish population was 8000 fish.
Two years later, so
![F(2) = 8000](https://img.qammunity.org/2021/formulas/mathematics/high-school/fr60cdrwqd3bmjvi1rqryugkictpn4g7hl.png)
![F(t) = F(0)(1-r)^(t)](https://img.qammunity.org/2021/formulas/mathematics/high-school/p02vbdlt0jb1gnex7d80xzxpkz51dpxjfy.png)
![8000 = 18000(1-r)^(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/1ywjejeplclpgxv8vs2goansy8b1llnotx.png)
![(1-r)^(2) = 0.4444](https://img.qammunity.org/2021/formulas/mathematics/high-school/gs3oi36amxasldpaghhyeqwet2ai8dqby4.png)
![\sqrt{(1-r)^(2)} = √(0.4444)](https://img.qammunity.org/2021/formulas/mathematics/high-school/9tw9xuqsvpejlxeehc6kxgj45156libo2u.png)
![1 - r = 0.6666](https://img.qammunity.org/2021/formulas/mathematics/high-school/ttydai4w492r20wwokh4qzyhi26qex077b.png)
![r = 0.3334](https://img.qammunity.org/2021/formulas/mathematics/high-school/neyy6jmejmcbohcyg925relc5wsedoa7xq.png)
So
![F(t) = 18000(0.6666)^(t)](https://img.qammunity.org/2021/formulas/mathematics/high-school/huux0gv534b4p9ghcd1ghe96va99efa6id.png)