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AysviL [449]
3 years ago
8

The ANSWER IS A, C,and D

Mathematics
2 answers:
cupoosta [38]3 years ago
7 0

wowww your totally right
WITCHER [35]3 years ago
7 0
Where is the question?
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Step-by-step explanation:

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Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confide
ioda

Answer:

The question is incomplete, but the step-by-step procedures are given to solve the question.

Step-by-step explanation:

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 - 0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.005 = 0.995, so Z = 2.575.

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.

M = 2.575\frac{\sigma}{\sqrt{n}}

The lower end of the interval is the sample mean subtracted by M.

The upper end of the interval is the sample mean added to M.

The 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (lower end, upper end).

7 0
3 years ago
PLEASE HELP ME!
salantis [7]

Answer:

the last one

Step-by-step explanation:

3 0
2 years ago
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