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adelina 88 [10]
2 years ago
8

PLEASE HELP!!! In a set of 1,000 integers from 1 to 1,000, an integer chosen at random on a single trial should be an integer fr

om 1 to 25 about 25 out of every 1,000 trials, or one out of every 40 integers selected. A sample of 5 integers selected is shown. Does this sample represent the general rule for picking an integer from 1 to 25 in the population of integers from 1 to 1,000?
A. Yes - the sample is representative of the expected number of integers from 1 to 25 in a sample of 5 integers, which would be none or zero.
B. No - the sample is not representative of the expected number of integers from 1 to 25 in a sample of 5 integers, which would be at least 10.

Mathematics
1 answer:
Mumz [18]2 years ago
3 0

Answer:

Step-by-step explanation:

Use math-way go quick and easy answers without step by step explanations, and use math-soup-calculator for step by step explanation

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The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 manag
VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

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

\alpha = \frac{1 - 0.95}{2} = 0.025

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 p-value of 1 - 0.025 = 0.975, so Z = 1.96.

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 = 1.96\frac{2050}{\sqrt{49}} = 574

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

5 0
3 years ago
Simplify. 2³+4⋅8 please I beg you
dalvyx [7]
2^3=2*2*2
2*2*2+4*8
8+32
=> 40
8 0
3 years ago
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I'll mark as Brainliest, Please answer ASAP, Ihave MCAS tomorrow. How could this be true?? Please Explain.​
Vaselesa [24]

Answer:

See below

Step-by-step explanation:

The number of loaves of bread remaining can only go as low as ZERO (you can't have negative loaves) ...and can only go as high as when h = 0

     so    0 = 42-3.5 h    

             3.5 h = 42

                h = 12       and   min is  h = 0   ( when there are 42 loaves to start)

So:

<u>0≤ h ≤ 12 </u>

5 0
2 years ago
When a trend line is drawn on a scatterplot, there are 4 points above the trend line. About how many points should be below the
scoundrel [369]
There is no exact rule for lines of best fit. However, in general, there should be roughly the same amount above as below. So, if we are following this rule, there should be about 4 below as well. 
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3 years ago
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Help!!!!!!!!!!!!!!!!!!!!!!!!!!
7nadin3 [17]
The answer is the blue one (the one your mouse is hovering over)
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