The confidence interval for the mean noise at locations is (127.09,148.51).
Given noise levels at 5 airports=152,154,139,124,120
We have to find the confidence interval for the mean noise and sample mean of the data. We know that mean is the sum divided by the number of items taken.
Mean = sum of X values/n
=137.8 (calculated in figure)
s=
=
=
=15.59
t value at 80% confidence level with degree of freedom 4 (5-1) is 1.533.
standard error=t value*s/
=1.533*15.59/
=10.71
Upper level= Mean + standard error
=137.8+10.71
=148.51
Lower level=Mean - standard error
=137.8-10.71
=127.09
Hence the confidence interval for the true mean noise is (127.09,148.51) and sample mean is 137.8.
Learn more about confidence interval at brainly.com/question/15712887
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Answer:
c. 3.2
Step-by-step explanation:
5(x+2)=7(4-x)
5(x+2)=7(-x+4)
5x+5*2=-7x+7*4
Answer:
0.4 and 0.2.
Step-by-step explanation:
For the first problem, you divide 5 by 2.

This gives you 0.4
For the second problem, we simplify the fraction, first.

Then, we divide 1 by 5.

This gives you 0.2.
Answer:
5 answer is A
6 answer is B
Hope I Helped
<em>P.S. I cant see the the #7 options but the answer is that it belongs to a rational number</em>
Answer:
All of the above.
Step-by-step explanation:
31%=.31=31/100.
even if there is 31 per 100, it is the same. what if there is twice the value. 62/200. This is still reduced to the same number 31/100. There is still 31% of that thing.