a small lake is stocked with a certain species of fish. the fish population is modeled by the function p =…

Explanation:

Step1: Substitute $t = 4$ into the formula

We have $P=\frac{10}{1 + 4e^{-0.4\times4}}$. First, calculate the exponent: $-0.4\times4=-1.6$. Then find $e^{-1.6}\approx0.2019$. Next, $4e^{-1.6}\approx4\times0.2019 = 0.8076$. And $1+4e^{-1.6}\approx1 + 0.8076=1.8076$. So $P=\frac{10}{1.8076}\approx5.531$. Since $P$ is in thousands, the number of fish is $5.531\times1000 = 5531$.

Step2: Set $P = 5$ (since $P$ is in thousands) and solve for $t$

Set $\frac{10}{1 + 4e^{-0.4t}}=5$. Cross - multiply to get $10 = 5(1 + 4e^{-0.4t})$. Then $10=5 + 20e^{-0.4t}$. Subtract 5 from both sides: $5=20e^{-0.4t}$. Divide both sides by 20: $e^{-0.4t}=\frac{5}{20}=0.25$. Take the natural logarithm of both sides: $\ln(e^{-0.4t})=\ln(0.25)$. Using the property $\ln(e^{x})=x$, we have $-0.4t=\ln(0.25)$. Since $\ln(0.25)\approx - 1.3863$, then $t=\frac{\ln(0.25)}{-0.4}=\frac{-1.3863}{-0.4}\approx3.47$.

Answer:

(a) 5531 fish (b) 3.47 yr