Suive the problem. The data in the table represent the amount of pressure (psi) exerted by a stamping machine (x), and the amount of scrap brass shavings (in pounds) that are collected from the machine each hour (y). Also shown below are the outputs from two different statistical technologies (TI-83/84/94 Calculator and Excel). A scatterplot of the data confirms that there is a linear association. Report the equation for predicting scrap brass shavings using words such as scrap, not x and y. State the slope and intercept of the prediction equation. Round all calculations to the nearest thousandth. y LinReg 2.00 2.30 7.80 15.14 14.51 28.65 2.80 4.15 槽。 4.01 6.35 6.21 10.52 11.84 24.05 5.11 8.75 11.67 22.22 8.70 17.02 y=ax + b a=2.134464237 b=-2.018775528 72=0.996074796 7= 0.998035467 Intercept X Variable 1 Coefficients -2.018775528 2.134464237 O scrap = -2.019 +2.134(pressure); slope = -2.019 and the intercept is 2.134. O scrap = -2.019 +2.134(pressure); slope = 2.134 and the intercept is -2.019. O scrap = 2.134 - 2.019(pressure); slope = -2.019 and the intercept is 2.134. scrap = 2.134 - 2.019(pressure); slope = 2.134 and the intercept is -2.019.