Let's solve this step by step.
### Part (a)
To model the population of foxes [tex]$t$[/tex] years after the year 2000, we use an exponential growth function. The general form of an exponential growth function is:
[tex]\[ P(t) = P_0 \cdot e^{rt} \][/tex]
where:
- \( P(t) \) is the population at time \( t \),
- \( P_0 \) is the initial population,
- \( r \) is the continuous growth rate,
- \( t \) is the time in years after the initial time.
We know the following:
- The initial population \( P_0 \) in the year 2000 is 14900.
- The continuous growth rate \( r \) is 5 percent per year, which we write as \( r = 0.05 \).
Putting these values into the function, we get:
[tex]\[ P(t) = 14900 \cdot e^{0.05t} \][/tex]
So, the function that models the population \( t \) years after 2000 is:
[tex]\[ P(t) = 14900 \cdot e^{0.05t} \][/tex]
### Part (b)
Now, we use this function to estimate the fox population in the year 2008. The year 2008 is 8 years after the year 2000, so \( t = 8 \).
We substitute \( t = 8 \) into the function \( P(t) \):
[tex]\[ P(8) = 14900 \cdot e^{0.05 \cdot 8} \][/tex]
Calculating this, we get:
[tex]\[ P(8) = 14900 \cdot e^{0.4} \][/tex]
The value of \( e^{0.4} \) is approximately 1.49182. So:
[tex]\[ P(8) = 14900 \cdot 1.49182 \][/tex]
[tex]\[ P(8) \approx 22228.08 \][/tex]
Since we need the population as an integer:
The estimated fox population in the year 2008 is [tex]\( \boxed{22228} \)[/tex].
