To answer the given question, let's follow the steps one by one:
### 1. Find the regression equation for the rabbit population as a function of time \( x \).
We are given the data for the number of rabbits at different times in months:
- Time (months): [0, 3, 6, 9, 12]
- No. of Rabbits: [6, 32, 107, 309, 770]
We need to find the regression equation that models the population growth. Given the nature of the data, an exponential growth model will be suitable:
[tex]\[ P(t) = a \cdot e^{bt} \][/tex]
Where:
- \( P(t) \) is the population at time \( t \),
- \( a \) and \( b \) are constants to be determined.
Through regression analysis, we determine the values of \( a \) and \( b \). The optimal parameters for this model are found to be approximately:
[tex]\[ a \approx 16.70809436 \][/tex]
[tex]\[ b \approx 0.31952958 \][/tex]
### 2. Write the regression equation in terms of base \( e \).
Using the values of \( a \) and \( b \) derived from the regression analysis, the regression equation for the rabbit population as a function of time \( t \) is:
[tex]\[ P(t) = 16.70809436 \cdot e^{0.31952958t} \][/tex]
### 3. Use the equation from part (2) to estimate the time for the rabbits to exceed 10,000.
We need to find the time \( t \) when the rabbit population \( P(t) \) exceeds 10,000. So, we set up the equation:
[tex]\[ 10000 = 16.70809436 \cdot e^{0.31952958t} \][/tex]
First, isolate the exponential term:
[tex]\[ \frac{10000}{16.70809436} = e^{0.31952958t} \][/tex]
[tex]\[ 598.422 \approx e^{0.31952958t} \][/tex]
Next, take the natural logarithm of both sides to solve for \( t \):
[tex]\[ \ln(598.422) = 0.31952958t \][/tex]
[tex]\[ t = \frac{\ln(598.422)}{0.31952958} \][/tex]
Estimating the values:
[tex]\[ t \approx \frac{6.395}{0.31952958} \][/tex]
[tex]\[ t \approx 20.01206603670371 \][/tex]
Therefore, the time for the rabbit population to exceed 10,000 is approximately 20 months.
