Answer :
To determine the exponential regression equation that best fits the given data, we need to follow several steps. Here is a detailed step-by-step solution:
1. Identify the given data points:
[tex]\[ \begin{tabular}{|c|c|} \hline x & y \\ \hline 1 & 4 \\ 2 & 8 \\ 3 & 27 \\ 4 & 85 \\ 5 & 250 \\ 6 & 600 \\ \hline \end{tabular} \][/tex]
2. Determine the form of the exponential regression equation:
Typically, an exponential regression equation is given by:
[tex]\[ y = a \cdot (b^x) \][/tex]
where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the constants we need to determine.
3. Transform the data for linear regression:
Since exponential functions can be linearized using logarithms, we take the natural logarithm of the [tex]\(y\)[/tex]-values. Let:
[tex]\[ \ln(y) = \ln(a) + x \cdot \ln(b) \][/tex]
This transforms the problem into a linear regression model. Let:
[tex]\[ \text{let } Y = \ln(y) \quad \text{and} \quad B = \ln(b) \][/tex]
Now the equation becomes:
[tex]\[ Y = \ln(a) + B \cdot x \][/tex]
4. Fit the linear model:
Using statistical techniques, we can determine the coefficients [tex]\( \ln(a) \)[/tex] and [tex]\( \ln(b) \)[/tex].
5. Convert back to the exponential form:
With the coefficients determined, we exponentiate the results to find [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
Given the data and the methodology, the constants derived are:
[tex]\[ a \approx 1.22 \quad \text{and} \quad b \approx 2.84 \][/tex]
6. Write the final regression equation:
The exponential regression equation that fits the given data points is:
[tex]\[ y = 1.22 \left(2.84^x\right) \][/tex]
Therefore, the correct exponential regression equation from the provided choices is:
[tex]\[ \boxed{y = 1.22 \left(2.84^x\right)} \][/tex]
And the correct answer is:
[tex]\[ \text{B. } y = 1.22\left(2.84^x\right) \][/tex]
1. Identify the given data points:
[tex]\[ \begin{tabular}{|c|c|} \hline x & y \\ \hline 1 & 4 \\ 2 & 8 \\ 3 & 27 \\ 4 & 85 \\ 5 & 250 \\ 6 & 600 \\ \hline \end{tabular} \][/tex]
2. Determine the form of the exponential regression equation:
Typically, an exponential regression equation is given by:
[tex]\[ y = a \cdot (b^x) \][/tex]
where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the constants we need to determine.
3. Transform the data for linear regression:
Since exponential functions can be linearized using logarithms, we take the natural logarithm of the [tex]\(y\)[/tex]-values. Let:
[tex]\[ \ln(y) = \ln(a) + x \cdot \ln(b) \][/tex]
This transforms the problem into a linear regression model. Let:
[tex]\[ \text{let } Y = \ln(y) \quad \text{and} \quad B = \ln(b) \][/tex]
Now the equation becomes:
[tex]\[ Y = \ln(a) + B \cdot x \][/tex]
4. Fit the linear model:
Using statistical techniques, we can determine the coefficients [tex]\( \ln(a) \)[/tex] and [tex]\( \ln(b) \)[/tex].
5. Convert back to the exponential form:
With the coefficients determined, we exponentiate the results to find [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
Given the data and the methodology, the constants derived are:
[tex]\[ a \approx 1.22 \quad \text{and} \quad b \approx 2.84 \][/tex]
6. Write the final regression equation:
The exponential regression equation that fits the given data points is:
[tex]\[ y = 1.22 \left(2.84^x\right) \][/tex]
Therefore, the correct exponential regression equation from the provided choices is:
[tex]\[ \boxed{y = 1.22 \left(2.84^x\right)} \][/tex]
And the correct answer is:
[tex]\[ \text{B. } y = 1.22\left(2.84^x\right) \][/tex]
