The data in the table represent the amount of raw material (in tons) put into an injection moldingmachine each day (x), and the amount of scrap plastic (in tons) that is collected from the machineevery fourweeks (y).Also shown below are the outputs from two different statistical technologies(TI-83/84 Calculator and Excel). A scatterplot of the data confirms that there is a linear association.Report the equation for predicting scrap from raw material using words such as scrap, notxandy.State the slope and intercept of the prediction equation. Round all calculations to the nearesthundredth

\begin{array}{|c|c|}\hline x & y \\\hline 2.71 & 3.61 \\\hline 2.33 & 2.80 \\\hline 2.33 & 2.77 \\\hline 2.21 & 2.34 \\\hline 2.11 & 2.15 \\\hline 2.08 & 2.06 \\\hline 1.98 & 2.02 \\\hline 1.95 & 1.95 \\\hline 1.84 & 1.73 \\\hline 1.84 & 1.68 \\\hline\end{array}

\begin{array}{l}\text { LinReg }\\y=a+b x \\a=-2.376991175 \\b=2.192699333 \\r^{2}=.9806834986 \\r=.9902946524\end{array}

\begin{array}{|l|r|}\hline & \text { Coefficients } \\\hline \text { Intercept } & -2.376991175 \\\hline \text { X Variable 1 } & 2.192699333 \\\hline\end{array}

A) scrap

= - 2.38 + 2.19

(raw material); slope

= 2.19

and the intercept is

- 2.38

.
B) scrap

= 2.19 - 2.38

(raw material); slope

= - 2.38

and the intercept is

2.19

.
C) scrap

= 2.19 - 2.38

(raw material); slope

= 2.19

and the intercept is

- 2.38

.
D) scrap

= - 2.38 + 2.19

(raw material); slope

= - 2.38

and the intercept is

2.19

The Following Regression Equation Was Found to Model Commute Distance

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