Answer:
7.6 in
Step-by-step explanation:
c = 2πr Divide each side by 2π
r = C/2π Insert values
r = 48/(2×3.14)
r = 48/6.28
r = 7.6 in
Answer:
Warranty of 66 months.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

If the company wants no more than 2% of the components to wear out before they reach the warranty date, what number of months should be used for the warranty?
Only the lowest 2% will be replaced, so the warranty is the value of X when Z has a pvalue of 0.02. So it is X when Z = -2.055.




Warranty of 66 months.
Answer:
a. proportions have not changed significantly
Step-by-step explanation:
Given
Business College= 35 %
Arts College= 35 %
Education College = 30%
Calculated
Business College = 90/300= 9/30= 0.3 or 30%
Arts College= 120/300= 12/30= 2/5= 0.4 or 40%
Education College= 90/300= 9/30 = 0.3 or 30%
First we find the mean and variance of the three colleges using the formulas :
Mean = np
Standard Deviation= s= 
Business College
Mean = np =300*0.3= 90
Standard Deviation= s=
=
= 7.94
Arts College
Mean = np =300*0.4= 120
Standard Deviation= s=
=
= 8.49
Education College
Mean = np =300*0.3= 90
Standard Deviation= s=
=
= 7.94
Now calculating the previous means with the same number of students
Business College
Mean = np =300*0.35= 105
Arts College
Mean = np =300*0.35= 105
Education College:
Mean = np =300*0.3= 90
Now formulate the null and alternative hypothesis
Business College
90≤ Mean≥105
Arts College
105 ≤ Mean≥ 120
Education College
U0 : mean= 90 U1: mean ≠ 90
From these we conclude that the proportions have not changed significantly meaning that it falls outside the critical region.
The greatest common factor of 11m^3 and 13m^2 is m^2. 11 and 13 don't really have a greatest common factor (they're both prime numbers), and m^2 goes into each of them.
80 minutes because 8•10=80