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To solve the logistic model problem, we need to determine the parameters \( c \), \( a \), and \( b \) using the given data.

The logistic model is given by:
\[ P(t) = \frac{c}{1 + ae^{-bt}} \]

### Step-by-Step Solution:

1. **Identify the carrying capacity \( c \):**
   - The total predicted population to get flu: 7500
   - Therefore, \( c = 7500 \)

2. **Set up the initial conditions:**
   - Initial number of cases at \( t = 0 \) is 50.
   - One week (7 days) later, \( t = 7 \), the number of cases is 200.

Using the initial condition at \( t = 0 \):
\[ P(0) = \frac{c}{1 + ae^{0}} = \frac{c}{1 + a} \]
\[ 50 = \frac{7500}{1 + a} \]
\[ 1 + a = \frac{7500}{50} \]
\[ 1 + a = 150 \]
\[ a = 149 \]

3. **Use the second condition to find \( b \):**
\[ P(7) = \frac{7500}{1 + 149e^{-7b}} = 200 \]
\[ 200 = \frac{7500}{1 + 149e^{-7b}} \]
\[ 1 + 149e^{-7b} = \frac{7500}{200} \]
\[ 1 + 149e^{-7b} = 37.5 \]
\[ 149e^{-7b} = 36.5 \]
\[ e^{-7b} = \frac{36.5}{149} \]
\[ e^{-7b} = 0.245 \]
\[ -7b = \ln(0.245) \]
\[ b = -\frac{\ln(0.245)}{7} \]
\[ b \approx 0.188 \]

### Results:

- \( c = 7500 \)
- \( a = 149 \)
- \( b \approx 0.188 \)

### Summary:

The logistic model is:
\[ P(t) = \frac{7500}{1 + 149e^{-0.188t}} \]

Would you like any further details or have any questions?

Here are 5 questions you might be interested in asking next:
1. Can you explain how to solve another type of logistic growth problem?
2. How do you solve exponential growth problems?
3. What is the difference between logistic and exponential growth?
4. Can you help me with a problem involving differential equations?
5. How do you solve problems involving decay models?

**Tip:** When solving logistic models, always ensure you accurately identify the initial conditions and carrying capacity, as these are critical for finding the correct parameters.

Learn how to solve a logistic growth model problem step-by-step. Determine parameters c, a, and b using given data and apply the logistic model formula P(t) = 7500 / (1 + 149e^(-0.188t)). Suitable for advanced high school students and above.

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