We are given a logistic growth model for the fox population, described by the function:

$N(t) = \frac{37.5}{0.25 + 0.76t}$

where $t$ is time in years, and $N(t)$ gives the number of foxes at time $t$. Let's tackle each part of the question:

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(a) How many foxes were introduced into the protected area?

To find the number of foxes initially introduced, we evaluate $N(0)$, which represents the population at $t = 0$.

$N(0) = \frac{37.5}{0.25 + 0.76 \cdot 0} = \frac{37.5}{0.25} = 150$

Answer: 150 foxes were introduced into the protected area.

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(b) Calculate $N(2)$.

To calculate the number of foxes after 2 years, we evaluate $N(2)$:

$N(2) = \frac{37.5}{0.25 + 0.76 \cdot 2} = \frac{37.5}{0.25 + 1.52} = \frac{37.5}{1.77} \approx 21.19$

Rounding to the nearest whole number:

$N(2) \approx 21 \text{ foxes}$

Explanation: After 2 years, there were about 21 foxes in the protected area.

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(c) Explain how the population varies with time.

The population follows a logistic growth model, which typically means that the population grows rapidly at first and then slows as it approaches the carrying capacity (the maximum population that the environment can sustain).

• In the early years, the population grows rapidly.
• As time goes on, the population growth slows down and approaches the carrying capacity.

To understand the population's growth rates over specific intervals:

Average Rate of Increase (First 10 Years):

We compute $N(10)$ and compare it with the initial population $N(0)$.

$N(10) = \frac{37.5}{0.25 + 0.76 \cdot 10} = \frac{37.5}{7.85} \approx 4.78$

The average rate of increase in the first 10 years is:

$\text{Rate (First 10 years)} = \frac{N(10) - N(0)}{10} = \frac{4.78 - 150}{10} \approx -14.52$

Thus, in the first 10 years, the population decreases at an average rate of about 14.5 foxes per year.

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(d) Find the carrying capacity for foxes in the protected area.

The carrying capacity of a population in a logistic growth model is the horizontal asymptote of the function as $t \to \infty$. Mathematically, this is the limit of $N(t)$ as $t \to \infty$:

$\lim_{t \to \infty} N(t) = \lim_{t \to \infty} \frac{37.5}{0.25 + 0.76t} = 0$

Thus