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Minchanka [31]
3 years ago
13

A small school with 606060 total students records how many of their students attend school on each of the 180180180 days in a sc

hool year. The mean number of students in attendance daily is 555555 students and the standard deviation is 444 students. Suppose that we take random samples of 555 school days and calculate the mean number of students \bar x x ˉ x, with, \bar, on top in attendance on those days in each sample.
Mathematics
2 answers:
charle [14.2K]3 years ago
8 0

Answer:

bbbcadacdababcada

Step-by-step explanation:

egoroff_w [7]3 years ago
5 0

Answer: 180,842,239

Step-by-step explanation:

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A Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of
GenaCL600 [577]

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 1.989.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

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3 years ago
The product of 5 and a number is at least 20.
svp [43]

Answer:

Step-by-step explanation:

the answer is 5x\geq 20

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