Problem
6. $[-/ 9.09$ Points] DETAILS OSCOLALG1 6.7.392. MY NOTES ASK YOUR TEACHER Use a graphing calculator and the following scenario. The population $P$ of a fish farm in $t$ years is modeled by the equation $P(t)=\frac{1200}{1+9 e^{-0.8 t}}$. To the nearest whole number, what will the fish population be after 2 years? fish Additional Materials eBook Example Video
Solution
Step 1 ：Substitute \(t=2\) into the equation to get \(P(2)=\frac{1200}{1+9 e^{-0.8 * 2}}\)
Step 2 ：Simplify the exponent to get \(P(2)=\frac{1200}{1+9 e^{-1.6}}\)
Step 3 ：Calculate the value of the exponent to get \(P(2)=\frac{1200}{1+9 * 0.201896517995}\)
Step 4 ：Multiply inside the denominator to get \(P(2)=\frac{1200}{1+1.81706866195}\)
Step 5 ：Add inside the denominator to get \(P(2)=\frac{1200}{2.81706866195}\)
Step 6 ：Divide to find the population to get \(P(2)=426.26\)
Step 7 ：So, to the nearest whole number, the fish population after 2 years will be \(\boxed{426}\)
