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fomenos
2 years ago
5

A researcher is using a repeated-measures study to evaluate the difference between two treatments. If the difference between the

treatments is consistent from one participant to another, then the data should produce ______.
A. a small variance for the difference scores and a small standard error
B. a small variance for the difference scores and a large standard error
C. a large variance for the difference scores and a small standard error
D. a large variance for the difference scores and a large standard error
Mathematics
1 answer:
Leya [2.2K]2 years ago
8 0

Answer:

A) a small variance for the difference scores and a small standard error

Step-by-step explanation:

Since the difference scores are obtained by subtracting one variable form another, if the difference scores are consistent between treatments, then the variance will be small. The higher the variance, the higher the standard error. So if the variance is small, then the standard error will also be small.

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Step-by-step explanation:

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1 year ago
a car rental service rently purchased all new vehicles .to help cover the cost ,rental rates have to increase 26% from the norma
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Answer:

The new rate should be $56.67 per day

Step-by-step explanation:

Proportion states that the two fractions or ratios are equal.

As per the statement:

Normal rate per day = $45

To find the new rate:

Let new rate be x per day

By definition of proportion:

\frac{126}{100} = \frac{x}{45}

By cross multiply we have;

100x = 126 \times 45

Divide both sides by 100 we get;

x = \frac{126 \times 45}{100}

Simplify:

x = $56.7

Therefore, the new rate should be $56.7 per day




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2 years ago
How can patterns be used to determine products of a number and a power of 10?
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2 years ago
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Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.Tiles
Ugo [173]

By definition, the average rate of change is given by:

AVR = \frac{f(x2)-f(x1)}{x2-x1}

We evaluate each of the functions in the given interval.

We have then:

For f (x) = x ^ 2 + 3x:

Evaluating for x = -2:

f (-2) = (-2) ^ 2 + 3 (-2)\\f (-2) = 4 - 6\\f (-2) = - 2

Evaluating for x = 3:

f (3) = (3) ^ 2 + 3 (3)\\f (3) = 9 + 9\\f (3) = 18

Then, the AVR is:

AVR = \frac{18-(-2)}{3-(-2)}

AVR = \frac{18+2}{3+2}

AVR = \frac{20}{5}

AVR = 4


For f (x) = 3x - 8:

Evaluating for x =4:

f (4) = 3 (4) - 8\\f (4) = 12 - 8\\f (4) = 4

Evaluating for x = 5:

f (5) = 3 (5) - 8\\f (5) = 15 - 8\\f (5) = 7

Then, the AVR is:

AVR = \frac{7-4}{5-4}

AVR = \frac{3}{1}

AVR = 3


For f (x) = x ^ 2 - 2x:

Evaluating for x = -3:

f (-3) = (-3) ^ 2 - 2 (-3)\\f (-3) = 9 + 6\\f (-3) = 15

Evaluating for x = 4:

f (4) = (4) ^ 2 - 2 (4)\\f (4) = 16 - 8\\f (4) = 8

Then, the AVR is:

AVR = \frac{8-15}{4-(-3)}

AVR = \frac{-7}{4+3}

AVR = \frac{-7}{7}

AVR = -1


For f (x) = x ^ 2 - 5:

Evaluating for x = -1:

f (-1) = (-1) ^ 2 - 5\\f (-1) = 1 - 5\\f (-1) = - 4

Evaluating for x = 1:

f (1) = (1) ^ 2 - 5\\f (1) = 1 - 5\\f (1) = - 4

Then, the AVR is:

AVR = \frac{-4-(-4)}{1-(-1)}

AVR = \frac{-4+4}{1+1}

AVR = \frac{0}{2}

AVR = 0


Answer:

from the greatest to the least value based on the average rate of change in the specified interval:


f(x) = x^2 + 3x interval: [-2, 3]

f(x) = 3x - 8 interval: [4, 5]

f(x) = x^2 - 5 interval: [-1, 1]

f(x) = x^2 - 2x interval: [-3, 4]


4 0
1 year ago
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