Answer: D) The linear model shows a strong fit to the data
The actual strength of the relationship is unknown unless we have the actual values of each data point (so we can compute the correlation coefficient r), but the residuals are randomly scattered about both above and below the horizontal axis. This means we have a fairly good linear fit. If all of the points were above the line, or all below the line, or all residuals fit a certain pattern (eg: parabola), then it wouldn't be a good linear fit.
Answer:
x^3 + y-3
Step-by-step explanation:
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Answer:
Distance: 8.5440037453175
Step-by-step explanation:
Distance: 8.5440037453175
Steps:
Distance (d) = √(-7 - -10)2 + (-1 - -9)2
= √(3)2 + (8)2
= √73
= 8.5440037453175
Slope and Angle:
ΔX = -7 – -10 = 3
ΔY = -1 – -9 = 8
Slope (m) =
ΔY
ΔX
=
8
3
= 2.6666666666667
θ =
arctan( ΔY )
ΔX
= 69.443954780417°
Equation of the line:
y = 2.6666666666667x + 17.666666666667
or
y =
8 x
3
+ 53
3
When x=0, y = 17.666666666667
When y=0, x = -6.625
Answer:
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