A lake is stocked with 569 fish of a new variety. The size of the lake, the availability of food, and the number of other fish restrict growth in the lake to a limiting value of 3,556 . The population of fish in the lake after time \( t \), in months, is given by the function \( P(t)=\frac{3,556}{1+5.3 e^{-0.27 t}} \). Find the population after 1 months.

To find the population after 1 month, we need to substitute \( t = 1 \) into the function \( P(t) \). Starting with the formula: \[ P(t) = \frac{3,556}{1 + 5.3 e^{-0.27 t}} \] Substituting \( t = 1 \): \[ P(1) = \frac{3,556}{1 + 5.3 e^{-0.27(1)}} \] Calculating \( e^{-0.27} \) using a calculator gives approximately \( 0.763 \): \[ P(1) = \frac{3,556}{1 + 5.3 \times 0.763} \] \[ P(1) = \frac{3,556}{1 + 4.0429} \] \[ P(1) = \frac{3,556}{5.0429} \] \[ P(1) \approx 704.51 \] So, the population of fish after 1 month is approximately 705 fish (rounding to the nearest whole fish).