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malfutka [58]
3 years ago
6

Find the value of the test statistic to test for a difference in the areas. Round your answer to two decimal places, if necessar

y.

Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

hello your question has some missing parts attached below is a picture of the complete question

Answer : 3.59

Step-by-step explanation:

Calculating the standard deviation, mean and standard error of the hourly wages

Area 1 : mean = 12.75 , std = 4.9497 , std error = 1.75

Area 2 : mean = 18.25, std = 4.3671,  std error = 1.54399

Area 3 : mean = 16.25, std = 2.8660, std error = 1.01330

mean = sum of terms / number of terms

std = \sqrt{} (X − μ)2 / n

std error = std / \sqrt{n}

The value of the test statistic to test for a difference in the areas is

3.59 ( using anova table attached below )

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13. The slope for a line that is perpendicular to the line y=4x+8 is  -1/4

14. The slope for a line that is perpendicular to the line x=-6 is 0

15. The equation for a line that is perpendicular to the line 8x-4y=12 and passes through the origin is y = - (1/2) x

16. The equation for a line that is perpendicular to the line y=-(1/3)x and passes through the point (0,-10) is y=3x-10

Step-by-step explanation:

13. What is the slope for a line that is perpendicular to the line y=4x+8?

y=mx+b

Comparing with the form slope-intercept, the slope of the given line is the coefficient of x, then the slope of the given line is 4.

A line perpendicular to y=4x+8 must have a slope opposite and inverse, then:

Slope of the perpendicular = - 1/4


14. What is the slope for a line that is perpendicular to the line x=-6?

The line x=-6 is a vertical line. A line perpendicular to the line x=-6 must be a horizontal line (Angle=0°), then:

Slope of the perpendicular = tan Angle = tan 0° = 0


15. Write the equation for a line that is perpendicular to the line 8x-4y=12 and passes through the origin.

8x-4y=12

Isolating y: Subtracting 8x both sides of the equation:

8x-4y-8x=12-8x

-4y=-8x+12

Dividing all the terms by -4:

-4y/(-4)= -8x/(-4)+12/(-4)

y=2x-3

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The perpendicular passes through the origin:

P1=(0,0)=(x1,y1)→x1=0, y1=0

Using the equation point - slope:

y-y1=m(x-x1)

Replacing the known values:

y-0=(-1/2)(x-0)

y=(-1/2)x


16. Write an equation for a line that is perpendicular to the line y=-(1/3)x and passes through the point (0,-10).

The slope of the given line is -(1/3)

The slope of the perpendicular is m=3/1→m=3 (opposite and inverse to the slope of the given line)

The perpendicular passes through the point:

P1=(0,-10)=(x1,y1)→x1=0, y1=-10

Using the equation point - slope:

y-y1=m(x-x1)

Replacing the known values:

y-(-10)=3(x-0)

y+10=3x

Isolating y: Subtracting 10 both sides of the equation:

y+10-10=3x-10

y=3x-10


 

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