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The table below includes a caterer's data representing the number of people ([tex]$x$[/tex]) and the total cost to cater an event ([tex]$y$[/tex]).

[tex]\[
\begin{tabular}{|c|c|}
\hline
x & y \\
\hline
39 & 425 \\
\hline
45 & 350 \\
\hline
50 & 499 \\
\hline
56 & 530 \\
\hline
62 & 760 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|c|c|}
\hline
x & y \\
\hline
71 & 651 \\
\hline
84 & 770 \\
\hline
97 & 915 \\
\hline
115 & 1,100 \\
\hline
\end{tabular}
\][/tex]

Apply the median-fit method to write the equation of the line of best fit for the data. The summary points are (45, 425), (62, 651), and (97, 915).

Using approximate values for the slope and [tex]$y$[/tex]-intercept, what is the linear model?

[tex]y = \square \cdot x + \square[/tex]

Answer :

To apply the median-fit method, we use three summary points:

- Extreme points:
[tex]$$ (x_1,y_1)=(45,425) \quad \text{and} \quad (x_3,y_3)=(97,915) $$[/tex]
- Median point:
[tex]$$ (x_2,y_2)=(62,651) $$[/tex]

Step 1. Calculate the slope.

Using the extreme points, the slope [tex]$m$[/tex] is given by:
[tex]$$
m = \frac{y_3 - y_1}{x_3 - x_1} = \frac{915 - 425}{97 - 45} = \frac{490}{52} \approx 9.4231.
$$[/tex]

Step 2. Calculate the [tex]$y$[/tex]-intercept.

With the median point [tex]$(62,651)$[/tex] and the slope found, the [tex]$y$[/tex]-intercept [tex]$b$[/tex] is calculated by:
[tex]$$
b = y_2 - m \cdot x_2 = 651 - (9.4231 \times 62) \approx 651 - 584.230 \approx 66.7692.
$$[/tex]

Step 3. Write the linear model.

Thus, the equation of the line of best fit (rounded to two decimal places) is:
[tex]$$
y = 9.42x + 66.77.
$$[/tex]

This is the linear model representing the caterer's data.

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