Answer:
7.3% of the bearings produced will not be acceptable
Step-by-step explanation:
Problems of normally distributed samples can be 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:

Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.
So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.
Larger than 0.504
1 subtracted by the pvalue of Z when X = 0.504.



has a pvalue of 0.9938
1 - 0.9938= 0.0062
Smaller than 0.496
pvalue of Z when X = -1.5



has a pvalue of 0.0668
0.0668 + 0.0062 = 0.073
7.3% of the bearings produced will not be acceptable
A relation is a function, if and only if, the x- values do not repeat in a given set.
1 mile=1,609.34 meters
because each lap is equal to 100m it might be 16 laps
16 laps =1600m which is almost equal to 1609.34m
I think there's not a exact number of laps. you can try with your own calculations you'll end up with some decimal remaining.
I hope this will help.
So measure the pen in meters
Ok so do 100 divided by 77 to get about 1.3% per fruit. then do 1.3 times 53 to get the current percent of oranges which is about 69% then do 84% minus 69% to get the difference then divide by 1.3 to get about 11.5 oranges so you round to 12 oranges. So 12+53 to get 65 total oranges to equal 84% of the 77 fruit in the box.