Sure, let's address each part of the question step by step.
### Part a: Choosing the correct graph
Given the model [tex]\(N(t) = \frac{2900}{1 + 21.8 e^{-0.5t}}\)[/tex], we need to determine the correct behavior of the graph.
Key characteristics of the graph:
1. At [tex]\(t = 0\)[/tex], the population starts with a specific number of infected people.
2. As [tex]\(t\)[/tex] approaches infinity (a long time after the disease starts), the number of infected people should approach the total population of 2900 because the denominator will approach 1 (since [tex]\(e^{-0.5t}\)[/tex] approaches 0).
The correct graph will show:
- A rapid increase in the number of infected people initially.
- A plateauing effect as [tex]\(t\)[/tex] increases, approaching the total population of 2900.
Without visual aids here, let's infer that you should choose the graph that matches this exponential growth pattern nearing the population as [tex]\(t\)[/tex] grows.
### Part b: Calculating the initial number of infected people (at [tex]\(t = 0\)[/tex])
To find how many people are initially infected with the disease when [tex]\(t = 0\)[/tex]:
Given the function:
[tex]\[ N(t) = \frac{2900}{1 + 21.8 e^{-0.5 t}} \][/tex]
Plugging [tex]\(t = 0\)[/tex] into the equation:
[tex]\[ N(0) = \frac{2900}{1 + 21.8 e^{0}} \][/tex]
Since [tex]\( e^0 = 1 \)[/tex]:
[tex]\[ N(0) = \frac{2900}{1 + 21.8 \cdot 1} \][/tex]
[tex]\[ N(0) = \frac{2900}{1 + 21.8} \][/tex]
[tex]\[ N(0) = \frac{2900}{22.8} \][/tex]
Now, performing the division:
[tex]\[ N(0) \approx 127.19 \][/tex]
Rounding to the nearest whole number:
[tex]\[ N(0) \approx 127 \][/tex]
So, initially, 127 people are infected with the disease. Therefore, the answer to part b is:
[tex]\[ \boxed{127} \][/tex]
Feel free to ask if you have any more questions or need further assistance!
