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alexandr1967 [171]
3 years ago
8

In a normal distribution, a data value located 0.8 standard deviations below the mean has Standard Score: z =

Mathematics
1 answer:
V125BC [204]3 years ago
4 0

Answer:

answer cuzzz euevaksn

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Solve the equation by finding square roots. To start, remember to isolate x^2. Equation= x^2+4=20
prohojiy [21]
Get the equation simplified. Subtract 4 by each side.
20-4= 16
and the square root of 16, is 4. 
Plug it in to check. 
(4)^2+4=20
16+4=20
20=20

Ultimately your final answer is X=4

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How many solutions does the equation sin(5x) = 1/2 have on the interval (0,2pi]?
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1/2 has two solutions for 5x

Step-by-step explanation:

But if you add 2pi, 4pi, 6pi, 8pi to the solutions then divide by 5, you get 10 solutions for x

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y = -x^2 − 2.5x + 350

Step-by-step explanation:

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4 years ago
A sample of 66 obese adults was put on a low-carb diet for a year. The average weight loss was 11 pounds and the standard deviat
Illusion [34]

Answer:

The 99% lower confidence bound for the true average weight loss is 3.98 pounds.

This means that we can be 99% sure that the mean weight loss for all obese adults on a low-carb diet is positive.

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.

So it 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{19}{\sqrt{66}} = 6.02

The lower end of the interval is the sample mean subtracted by M. So it is 11 - 6.02 = 3.98 pounds

The 99% lower confidence bound for the true average weight loss is 3.98 pounds.

This means that we can be 99% sure that the mean weight loss for all obese adults on a low-carb diet is positive.

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