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Katena32 [7]
3 years ago
15

Please answer

Mathematics
1 answer:
beks73 [17]3 years ago
8 0

Answer:

(0,8)

Step-by-step explanation:

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Assume the population has a normal distribution. A sample of 25 randomly selected students Has a mean test score of 81.5 With a
Natasha_Volkova [10]

Answer:

The 90% confidence interval for the mean test score is between 77.29 and 85.71.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution 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 = 25 - 1 = 24

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.95. So we have T = 2.064

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.064\frac{10.2}{\sqrt{25}} = 4.21

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 81.5 - 4.21 = 77.29

The upper end of the interval is the sample mean added to M. So it is 81.5 + 4.21 = 85.71.

The 90% confidence interval for the mean test score is between 77.29 and 85.71.

6 0
3 years ago
URGENT PLEASE I AM TIMED URGENT SOLVE THE PROBLEM AND PLEASE, PLEASE EXPLAIN HOW YOU GOT YOUR ANSWER, GIVE DETAIL'S I WILL MARK
Neko [114]

I'm pretty sure you just multiply the 1 1/4 by 3 because if 1 1/4 makes 10 muffins you need to make 3 more dozens so 1 1/4 times 3 equals 3.75

3 0
3 years ago
What is y=1/3x+2 on a graph
Alika [10]

hope you find this helpful

8 0
3 years ago
Read 2 more answers
Solve for a. Assume that a, b, c are not zero.<br> abc=c
inn [45]
It is C because a and b are zeros
6 0
3 years ago
Read 2 more answers
Can anyone help me please.
EleoNora [17]
2x +9 <0

x > -12

2x<-9

2x<-9/2  (9/2 = 4.5)


x >-12
 the answer is all integers between -12 and -4.5

-11, -10, -9, -8, -7, -6, -5

8 0
4 years ago
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