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An equation was created for the line of best fit from the actual enrollment data. It was used to predict the dance studio enrollment values shown in the table below:

\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
\multirow{2}{*}{Enrollment} & \multicolumn{6}{|c|}{Month} \\
\cline{2-7}
& January & February & March & April & May & June \\
\hline
Actual & 120 & 140 & 150 & 140 & 150 & 130 \\
\hline
Predicted & 80 & 150 & 110 & 150 & 110 & 150 \\
\hline
Residual & 40 & -10 & 40 & -10 & 40 & -20 \\
\hline
\end{tabular}

Analyze the data. Determine whether the equation that produced the predicted values represents a good line of best fit.

A. No, the equation is not a good fit because the residuals are all far from zero.
B. No, the equation is not a good fit because the sum of the residuals is a large number.
C. Yes, the equation is a good fit because the residuals are not all far from zero.
D. Yes, the equation is a good fit because the sum of the residuals is a small number.

Sagot :

GinnyAnsweridn
To determine whether the equation for the line of best fit provides a good prediction of the dance studio enrollment, let's analyze the given data step by step.

### Step 1: Residuals
The residuals are calculated as the difference between the actual values and the predicted values for each month:
- January: Actual = 120, Predicted = 80, Residual = [tex]\(120 - 80 = 40\)[/tex]
- February: Actual = 140, Predicted = 150, Residual = [tex]\(140 - 150 = -10\)[/tex]
- March: Actual = 150, Predicted = 110, Residual = [tex]\(150 - 110 = 40\)[/tex]
- April: Actual = 140, Predicted = 150, Residual = [tex]\(140 - 150 = -10\)[/tex]
- May: Actual = 150, Predicted = 110, Residual = [tex]\(150 - 110 = 40\)[/tex]
- June: Actual = 130, Predicted = 150, Residual = [tex]\(130 - 150 = -20\)[/tex]

Summarizing the residuals:
[tex]\[ [40, -10, 40, -10, 40, -20] \][/tex]

### Step 2: Sum of Residuals
The sum of the residuals is:
[tex]\[ 40 + (-10) + 40 + (-10) + 40 + (-20) = 80 \][/tex]

### Step 3: Mean of Residuals
To find the mean of the residuals:
[tex]\[ \text{Mean of Residuals} = \frac{\text{Sum of Residuals}}{\text{Number of Residuals}} = \frac{80}{6} \approx 13.33 \][/tex]

### Step 4: Analyzing Residuals
1. Residuals Far from Zero:
We check if all the residuals are far from zero. Typically, a residual is considered "far from zero" if its absolute value is greater than 5 or another threshold defined by the context.
[tex]\[ |40| > 5, |-10| > 5, |40| > 5, |-10| > 5, |40| > 5, |-20| > 5 \][/tex]
All these residuals have an absolute value greater than 5, hence they are all considered far from zero.

2. Large Sum of Residuals:
A large sum of residuals indicates that the total error across all predictions is significant. Generally, we might define "large" based on the context or a fixed threshold, but given the sum here (80), it is quite clear that it is not small.

### Conclusion
Based on our analysis:
- The residuals are all far from zero.
- The sum of the residuals is a large number.

Therefore, both criteria indicate that the equation that produced the predicted values does not represent a good line of best fit. The correct conclusion is:
- No, the equation is not a good fit because the residuals are all far from zero.
- No, the equation is not a good fit because the sum of the residuals is a large number.
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