Let's break down the reasoning to determine if all 2900 people will be infected according to the given model [tex]\( N(t) = \frac{2900}{1 + 21.8 e^{-0.5t}} \)[/tex].
Given the nature of the function [tex]\( N(t) \)[/tex]:
1. The denominator [tex]\( 1 + 21.8 e^{-0.5t} \)[/tex] involves an exponential decay term [tex]\( e^{-0.5t} \)[/tex].
2. As [tex]\( t \)[/tex] increases, the term [tex]\( e^{-0.5t} \)[/tex] gets smaller and smaller, approaching 0.
3. As [tex]\( e^{-0.5t} \)[/tex] approaches 0, the denominator of the function approaches 1.
So, let's analyze the behavior of [tex]\( N(t) \)[/tex] as [tex]\( t \)[/tex] approaches infinity:
- When [tex]\( t = 0 \)[/tex], the exponential term [tex]\( e^{-0.5 \cdot 0} = e^0 = 1 \)[/tex]. Plugging this into the function, we get:
[tex]\[
N(0) = \frac{2900}{1 + 21.8 \cdot 1} = \frac{2900}{22.8} \approx 127
\][/tex]
- As [tex]\( t \)[/tex] increases, [tex]\( e^{-0.5t} \)[/tex] decreases:
- For very large [tex]\( t \)[/tex], [tex]\( e^{-0.5t} \)[/tex] becomes very small, approaching 0.
- When [tex]\( e^{-0.5t} \to 0 \)[/tex], the denominator [tex]\( 1 + 21.8e^{-0.5t} \to 1 \)[/tex].
- This makes the function approach:
[tex]\[
N(t) = \frac{2900}{1 + 0} = 2900
\][/tex]
So, as [tex]\( t \)[/tex] approaches infinity, the function [tex]\( N(t) \)[/tex] approaches 2900.
However, it never exactly reaches 2900, it just gets arbitrarily close to it. Thus, we can conclude that the number of people infected will approach 2900, but never actually reach it.
This analysis corresponds to choice C:
C. As [tex]\( t \rightarrow \infty \)[/tex], [tex]\( N(t) \rightarrow 2900 \)[/tex], so the number approaches 2900, but never actually reaches it.
