(1 point) Monday's lecture went over a logistic equation: \( f(t)=\frac{1000}{1+999 e^{-0.603 t}} \) which models how many people \( f \) have had the flu \( t \) days after an initial infection in a community of 1000 people. According to the model, how many people will have had the flu after 10 days? \( \square \) According to the model, how many people will have had the flu after 20 days? \( \square \) According to the model, how many days until 500 people have had the flu? \( \square \) According to the model, how many days until 700 people have had the flu? \( \square \) According to the model, how many days until 2000 people have had the flu? \( \square \) Hint: One of your answers should be "never"

To solve the logistic equation \( f(t) = \frac{1000}{1 + 999 e^{-0.603 t}} \), we will calculate the number of people who have had the flu after 10 days and 20 days, as well as determine how many days it will take for 500 and 700 people to have had the flu. We will also check how many days it will take for 2000 people to have had the flu. 1. **Number of people after 10 days:** \[ f(10) = \frac{1000}{1 + 999 e^{-0.603 \cdot 10}} \] First, calculate \( e^{-0.603 \cdot 10} \): \[ e^{-6.03} \approx 0.0024 \] Now substitute this value back into the equation: \[ f(10) = \frac{1000}{1 + 999 \cdot 0.0024} = \frac{1000}{1 + 2.398} = \frac{1000}{3.398} \approx 294.4 \] So, approximately **294 people** will have had the flu after 10 days. 2. **Number of people after 20 days:** \[ f(20) = \frac{1000}{1 + 999 e^{-0.603 \cdot 20}} \] Calculate \( e^{-0.603 \cdot 20} \): \[ e^{-12.06} \approx 0.0006 \] Now substitute this value back into the equation: \[ f(20) = \frac{1000}{1 + 999 \cdot 0.0006} = \frac{1000}{1 + 0.5994} = \frac{1000}{1.5994} \approx 624.5 \] So, approximately **625 people** will have had the flu after 20 days. 3. **Days until 500 people have had the flu:** Set \( f(t) = 500 \): \[ 500 = \frac{1000}{1 + 999 e^{-0.603 t}} \] Rearranging gives: \[ 1 + 999 e^{-0.603 t} = 2 \implies 999 e^{-0.603 t} = 1 \implies e^{-0.603 t} = \frac{1}{999} \] Taking the natural logarithm: \[ -0.603 t = \ln\left(\frac{1}{999}\right) \implies t = -\frac{\ln(1/999)}{0.603} \approx \frac{6.9078}{0.603} \approx 11.45 \] So, it will take approximately **11.45 days** for 500 people to have had the flu. 4. **Days until 700 people have had the flu:** Set \( f(t) = 700 \): \[ 700 = \frac{1000}{1 + 999 e^{-0.603 t}} \] Rearranging gives: \[ 1 + 999 e^{-0.603 t} = \frac{1000}{700} \implies 999 e^{-0.603 t} = \frac{300}{700} = \frac{3}{7} \] Thus: \[ e^{-0.603 t} = \frac{3}{7 \cdot 999} \] Taking the natural logarithm: \[ -0.603 t = \ln\left(\frac{3}{6993}\right) \implies t = -\frac{\ln(3) - \ln(6993)}{0.603} \approx \frac{8.524}{0.603} \approx 14.15 \] So, it will take approximately **14.15 days** for 700 people to have had the flu. 5. **Days until 2000 people have had the flu:** Since the maximum number of people in the community is 1000, it is impossible for 2000 people to have had the flu. Therefore, the answer is **never**. ### Summary of Answers: - After 10 days: **294** - After 20 days: **625** - Days until 500 people: **11.45** - Days until 700 people: **14.15** - Days until 2000 people: **never**