Simplify completely. (w15/w5)4 Enter your answer in the box.

We expect that students who do well on the midterm exam in a course will usually also do well on the final exam. Gary Smith of P

dybincka [34]

Using the model equation, the predicted mean score on the final given a score of 10 points above the class mean in the mid term exam is 50.7

The Least - Square Regression equation which models the relationship between midterm and final exam score is :

γ=46.6 + 0.41x

x = 10 points ; substitute the value of x = 10 into the regression equation ;

γ=46.6 + 0.41(10)

γ=46.6 + 4.1

γ = 50.7

The number of points above the mean he'll score in the final exam is predicted to be 50.7

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What’s the property of 45 times 6

lidiya [134]

Answer:

270

Step-by-step explanation:

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3 years ago

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Find the solution -1.2x-8.2>-9.7

andriy [413]

The solution to the equation is x < 1.25

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3 years ago

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Xelga [282]

Answer:

16. Mst = 104°

17. The value of mPL is 31°

18. 129°

Step-by-step explanation:

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2 years ago

Sample annual salaries (in thousands of dollars) for employees at a company are listed. 42 36 48 51 39 39 42 36 48 33 39 42 45 (

Sonja [21]

Answer:

a) Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45

$\bar X = 41.538$

$s = 5.317$

b) 44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25

$\bar X = 43.615$

$s= 5.583$

c) 3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75

$\bar X= 3.462$

$s = 0.443$

d) As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3.462).

And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.

And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.

Step-by-step explanation:

For this case we can use the following formulas for the mean and standard deviation:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Part a

Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45

And if we calculate the mean we got:

$\bar X = 41.538$

$s = 5.317$

Part b

For this case we know that each value present a 5% of rise so we just need to multiply each value bu 1.05 and we have this new dataset:

44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25

And if we calculate the new mean and deviation we got:

$\bar X = 43.615$

$s= 5.583$

Part c

The new dataset would be each value divided by 12 so we have:

3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75

And the new mean and deviation would be:

$\bar X= 3.462$

$s = 0.443$

Part d

As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3,462).

And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.

And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.

5 0

3 years ago