We can solve this problem using the formula for exponential growth:

$P(t) = P_0 e^{kt}$

Where:

• $P(t)$ is the population at time $t$,
• $P_0$ is the initial population,
• $k$ is the growth rate constant,
• $t$ is the time in years (we will adjust as needed).

Step 1: Define known values

• Initial population $P_0 = 149149$,
• Population after 33 years $P(33) = 11921192$,
• Time $t = 33$ years.

Step 2: Solve for the growth constant $k$

Using the exponential growth formula:

$11921192 = 149149 e^{33k}$

To isolate $k$, divide both sides by 149149:

$\frac{11921192}{149149} = e^{33k}$

$79.9 \approx e^{33k}$

Now take the natural logarithm of both sides:

$\ln(79.9) = 33k$

$\ln(79.9) \approx 4.381$

So,

$33k = 4.381$

Now solve for $k$:

$k = \frac{4.381}{33} \approx 0.132$

Step 3: Calculate the population after 66 months

66 months is $\frac{66}{12} = 5.5$ years.

Using the exponential growth formula again, we find the population after 5.5 years:

$P(5.5) = 149149 e^{0.132 \times 5.5}$

$P(5.5) = 149149 e^{0.726}$

$P(5.5) = 149149 \times 2.067 \approx 308657$

Final Answer:

After another 66 months, the population is approximately 308,657 rabbits (rounded to the nearest rabbit).

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Would you like more details or further clarification?

Here are some additional related questions:

1. How do you determine the growth rate constant $k$ more generally from population data?
2. What happens to the population if the environment starts to limit growth?
3. How would the population calculation change if a predator were introduced?
4. How do exponential and logistic growth models differ in real-world scenarios?
5. What would the population be after 10 years under the same growth conditions?

Tip: When dealing with exponential growth problems, it’s important to pay attention to the units of time and make sure they match up (in this case, years and months).