Let's solve the problem step by step. Given: \[ P(t) = \frac{280}{1 + 4e^{-0.3t}} \] where \( t \) is time in years. ### **1. Initial Population Size (\( t = 0 \)):** Plug \( t = 0 \) into the equation: \[ P(0) = \frac{280}{1 + 4e^{-0.3 \times 0}} \] \[ P(0) = \frac{280}{1 + 4e^{0}} \] \[ e^0 = 1 \] \[ P(0) = \frac{280}{1 + 4 \times 1} \] \[ P(0) = \frac{280}{1 + 4} \] \[ P(0) = \frac{280}{5} \] \[ P(0) = 56 \] **Initial population size: 56 individuals** --- ### **2. Population Size After 10 Years (\( t = 10 \)):** Plug \( t = 10 \) into the equation: \[ P(10) = \frac{280}{1 + 4e^{-0.3 \times 10}} \] \[ P(10) = \frac{280}{1 + 4e^{-3}} \] First, calculate \( e^{-3} \): \[ e^{-3} \approx 0.0498 \] Now substitute: \[ P(10) = \frac{280}{1 + 4 \times 0.0498} \] \[ P(10) = \frac{280}{1 + 0.1992} \] \[ P(10) = \frac{280}{1.1992} \] \[ P(10) \approx 233.5 \] Round to the nearest whole number: **Population size after 10 years: 234 individuals** --- ## **Final Answers** - **Initial population size:** 56 individuals - **Population size after 10 years:** 234 individuals

The logistic growth model describes how a population initially grows exponentially but slows as it approaches a carrying capacity, represented by a sigmoidal (S-shaped) curve. The parameters in the model determine the initial population size and the long-term equilibrium.