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pav-90 [236]
4 years ago
11

_______ variation is a variation whose cause can be identified. assignable controllable random statistical

Mathematics
1 answer:
Tom [10]4 years ago
4 0
Hey there :)

Assigned variation is a variation whose cause can be identified, since assigned variation has a special cause
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OFFERING 15 POINTS PLEASE HELP ASAP
enot [183]
Let L be the length and W be the width. 
"the width of a rectangle is half as long as the length"
W = (1/2)·L
"the rectangle has an area of 18 square feet"
A = L·W = 18
Substitute, 
L·W = 18
L·(1/2)·L = 18
(L²)/2 = 18
L² = 36
L = 6 feet
then W = (1/2) · L
W = (1/2) · 6
W = 3 feet
4 0
3 years ago
How do you write 900.5 and expanded form and word form?
Anna11 [10]

(9 x 100) + (5 x 0.1)

OR

900 + .5

Nine hundred and five tenths

Hope this helped!

4 0
4 years ago
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Which expression represents a negative integer?
olchik [2.2K]
The answer is D to the question Because the result is -1
8 0
3 years ago
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Use 913,256. write the digits in the ten thousand place
disa [49]
1 is in the 'Ten Thousand's' place.

8 0
4 years ago
Suppose a 90% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$38,737, $50,46
JulsSmile [24]

Answer:

a. The point estimate was of $44,600.

b. The sample size was of 16.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the difference between the two bounds, divided by 2.

a. What is the point estimate of the mean salary for all college graduates in this town?

Mean of the bounds, so:

(38737+50463)/2 = 44600.

The point estimate was of $44,600.

b. Determine the sample size used for the analysis.

First we need to find the margin of error, so:

M = \frac{50463-38737}{2} = 5863

Relating the margin of error with the sample size:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.9}{2} = 0.05

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.64.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

For this problem, we have that \sigma = 14300, M = 5863. So

M = z\frac{\sigma}{\sqrt{n}}

5863 = 1.645\frac{14300}{\sqrt{n}}

5863\sqrt{n} = 1.645*14300

\sqrt{n} = \frac{1.645*14300}{5863}

(\sqrt{n})^2 = (\frac{1.645*14300}{5863})^2

n = 16

The sample size was of 16.

7 0
3 years ago
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