Answer:
Step-by-step explanation:
Given that a professor sets a standard examination at the end of each semester for all sections of a course. The variance of the scores on this test is typically very close to 300.

(Two tailed test for variance )
Sample variance =480
We can use chi square test for testing of hypothesis
Test statistic = 
p value = 0.0100
Since p <0.05 our significance level, we reject H0.
The sample variance cannot be claimed as equal to 300.
Answer:
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Step-by-step explanation:

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Answer:
A≈1075.21
d Diameter
37
d
r
r
r
d
d
C
A
Using the formulas
A=
π
r
2
d=
2
r
Solving forA
A=
1
4
π
d
2
=
1
4
π
37
2
≈
1075.21009
Step-by-step explanation: