Which statement is not true for goodness-of-fit tests? Question content area bottom Part 1 A. Expected frequencies must be whol
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1/3x ≥ y + 2
You’ll have to transform the equation into its slope-intercept form, y = mx + b, so that you could graph the line. Set y to “=“
1/3x = y + 2
Subtract 2 from both sides
1/3x -2 = y + 2- 2
1/3x - 2 = y or y = 1/3x - 2
Next, to graph the solid line, you need at least two points to plot. We can start with the y-intercept, (0, -2). Another point to use is the x-intercept. To solve for the x-intercept, (a, 0), set y = 0:
0 = 1/3x - 2
Add 2 on both sides:
0 + 2 = 1/3x - 2 + 2
2 = 1/3x
Multiply both sides by 3 to solve for x:
2(3) = (3) 1/3x
6 = x
Therefore, the x-intercept = (6,0).
Plot the intercepts on the graph and connect those two points to create line.
Next, to identify which half-plane region to shade, choose a convenient test point (not on the line) to see whether it part of the solution. We could use the point of origin, (0, 0). Substitute its values into the linear inequality:
1/3x ≥ y + 2
1/3(0) ≥ (0) + 2
0 ≥ 2 (false statement). Therefore, you must shade the half-plane region where it doesn’t contain the test point, (0, 0).
Attached is the graph where it shows that the shaded lower half-plane region.
Please mark my answers as the Brainliest if you find my explanations helpful :)
You’ll have to transform the equation into its slope-intercept form, y = mx + b, so that you could graph the line. Set y to “=“
1/3x = y + 2
Subtract 2 from both sides
1/3x -2 = y + 2- 2
1/3x - 2 = y or y = 1/3x - 2
Next, to graph the solid line, you need at least two points to plot. We can start with the y-intercept, (0, -2). Another point to use is the x-intercept. To solve for the x-intercept, (a, 0), set y = 0:
0 = 1/3x - 2
Add 2 on both sides:
0 + 2 = 1/3x - 2 + 2
2 = 1/3x
Multiply both sides by 3 to solve for x:
2(3) = (3) 1/3x
6 = x
Therefore, the x-intercept = (6,0).
Plot the intercepts on the graph and connect those two points to create line.
Next, to identify which half-plane region to shade, choose a convenient test point (not on the line) to see whether it part of the solution. We could use the point of origin, (0, 0). Substitute its values into the linear inequality:
1/3x ≥ y + 2
1/3(0) ≥ (0) + 2
0 ≥ 2 (false statement). Therefore, you must shade the half-plane region where it doesn’t contain the test point, (0, 0).
Attached is the graph where it shows that the shaded lower half-plane region.
Please mark my answers as the Brainliest if you find my explanations helpful :)