The Solution:
Given the equation that modeled the population of fish after time (t) in years as below:
[tex]p(t)=\frac{1250}{2+3e^{-0.6t}}[/tex]part 1:
We are to find the initial population of fish. This means we are to find the value of p when t = 0.
[tex]p(0)=\frac{1250}{2+3e^{-0.6(0)}}=\frac{1250}{2+3e^0}=\frac{1250}{2+3}=\frac{1250}{5}=250\text{ fish}[/tex]Part 2:
We are required to find the doubling time (t) for this population (Round the answer to the nearest tenth). This means we are to find t when P(t) = 500.
Note: double of 250 fish is 2x250 = 500 fish.
[tex]\begin{gathered} p(t)=\frac{1250}{2+3e^{-0.6t}} \\ \\ \text{Where P(t)=500, and t=?} \end{gathered}[/tex][tex]\begin{gathered} 500=\frac{1250}{2+3e^{-0.6t}} \\ \text{Cross multiplying, we get} \\ \\ 500(2+3e^{-0.6t})=1250 \\ \text{ Dividing both sides by 500, we get} \\ \\ \frac{500(2+3e^{-0.6t})}{500}=\frac{1250}{500} \end{gathered}[/tex][tex]\begin{gathered} 2+3e^{-0.6t}=2.5 \\ \text{ Collecting the like terms, we get} \\ 3e^{-0.6t}=2.5-2 \\ 3e^{-0.6t}=0.5 \end{gathered}[/tex]Dividing both sides by 3, we get
[tex]\begin{gathered} \frac{3e}{3}^{-0.6t}=\frac{0.5}{3} \\ \\ e^{-0.6t}=\frac{1}{6} \end{gathered}[/tex]Taking the natural log of both sides, we get
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