1. **Problem statement:** We are given the fruit fly population model $$P(t) = \frac{291}{1+565e^{-0.37t}}$$ where $t$ is the number of days. We need to find: a. The initial population $P(0)$. b. The population after 2 days $P(2)$. c. The time $t$ when the population reaches 220. --- 2. **Formula and rules:** The population function is a logistic model. To find values, substitute $t$ into the formula. To solve for $t$ when $P(t)$ is known, rearrange the equation and solve the exponential. --- 3. **Part a: Initial population $P(0)$** $$P(0) = \frac{291}{1 + 565e^{-0.37 \times 0}} = \frac{291}{1 + 565e^0} = \frac{291}{1 + 565 \times 1} = \frac{291}{566}$$ Calculate: $$P(0) \approx 0.514$$ Rounded to nearest whole number: $$P(0) = 1$$ --- 4. **Part b: Population after 2 days $P(2)$** Substitute $t=2$: $$P(2) = \frac{291}{1 + 565e^{-0.37 \times 2}} = \frac{291}{1 + 565e^{-0.74}}$$ Calculate $e^{-0.74} \approx 0.477$: $$P(2) = \frac{291}{1 + 565 \times 0.477} = \frac{291}{1 + 269.6} = \frac{291}{270.6} \approx 1.076$$ Rounded: $$P(2) = 1$$ --- 5. **Part c: Find $t$ when $P(t) = 220$** Start with: $$220 = \frac{291}{1 + 565e^{-0.37t}}$$ Rearrange: $$1 + 565e^{-0.37t} = \frac{291}{220}$$ Calculate right side: $$\frac{291}{220} \approx 1.3227$$ Subtract 1: $$565e^{-0.37t} = 0.3227$$ Divide both sides by 565: $$e^{-0.37t} = \frac{0.3227}{565} \approx 0.000571$$ Take natural logarithm: $$-0.37t = \ln(0.000571)$$ Calculate: $$\ln(0.000571) \approx -7.466$$ Divide both sides by -0.37: $$t = \frac{-7.466}{-0.37} = 20.18$$ Rounded: $$t = 20$$ days --- **Final answers:** - Initial population: 1 - Population after 2 days: 1 - Time to reach population 220: 20 days