Solve the problem. The data in the table represent the amount of raw material (in tons) put into an injection molding machine each day (x), and the amount of scrap plastic (in tons) that is collected from the machine every four weeks (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, not x and y. State the slope and intercept of the prediction equation. Round all calculations to the nearest hundredth. X y 2.71 3.61 2.33 2.80 2.33 2.77 2.21 2.34 2.11 2.15 2.08 2.06 ▬▬▬▬▬▬▬▬▬▬▬▬ 1.98 2.02 1.95 1.95 1.84 1.84 1.73 1.68 LinReg y = a + bx a = -2.376991175 b=2.192699333 r.9806834986 r = .9902946524 Intercept X Variable 1 Coefficients -2.376991175 2.192699333 scrap = 2.19 -2.38(raw material); slope = 2.19 and the intercept is -2.38., scrap = -2.38 + 2.19(raw material); slope = -2.38 and the intercept is 2.19. scrap = -2.38 + 2.19(raw material); slope = 2.19 and the intercept is -2.38. scrap = 2.19 2.38(raw material); slope = -2.38 and the intercept is 2.19.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter4: Equations Of Linear Functions
Section4.6: Regression And Median-fit Lines
Problem 6PPS
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Solve the problem.
The data in the table represent the amount of raw material (in tons) put into an injection
molding machine each day (x), and the amount of scrap plastic (in tons) that is collected from
the machine every four weeks (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, not x and y. State the slope and intercept of the prediction
equation. Round all calculations to the nearest hundredth.
X
y
2.71
3.61
2.33 2.80
2.33 2.77
2.21
2.34
2.11 2.15
2.08
2.06
1.98
1.95
1.84
1.84
2.02
1.95
1.73
1.68
LinReg
y=a+bx
a = -2.376991175
b=2.192699333
r² =.9806834986
P = .9902946524
Intercept
X Variable 1
Coefficients
-2.376991175
2.192699333
scrap = 2.19 -2.38(raw material); slope = 2.19 and the intercept is -2.38.,
scrap = -2.38 + 2.19(raw material); slope = -2.38 and the intercept is 2.19.
scrap = -2.38 + 2.19(raw material); slope = 2.19 and the intercept is -2.38.
scrap = 2.19 2.38(raw material); slope = -2.38 and the intercept is 2.19.
Transcribed Image Text:Solve the problem. The data in the table represent the amount of raw material (in tons) put into an injection molding machine each day (x), and the amount of scrap plastic (in tons) that is collected from the machine every four weeks (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, not x and y. State the slope and intercept of the prediction equation. Round all calculations to the nearest hundredth. X y 2.71 3.61 2.33 2.80 2.33 2.77 2.21 2.34 2.11 2.15 2.08 2.06 1.98 1.95 1.84 1.84 2.02 1.95 1.73 1.68 LinReg y=a+bx a = -2.376991175 b=2.192699333 r² =.9806834986 P = .9902946524 Intercept X Variable 1 Coefficients -2.376991175 2.192699333 scrap = 2.19 -2.38(raw material); slope = 2.19 and the intercept is -2.38., scrap = -2.38 + 2.19(raw material); slope = -2.38 and the intercept is 2.19. scrap = -2.38 + 2.19(raw material); slope = 2.19 and the intercept is -2.38. scrap = 2.19 2.38(raw material); slope = -2.38 and the intercept is 2.19.
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