Answer:
0.1507 or 15.07%.
Step-by-step explanation:
We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.
First of all, we will find z-scores for data points using z-score formula.
, where,
z = z-score,
x = Sample score,
= Mean,
= Standard deviation.



Let us find z-score of data point 22.03.



Using probability formula
, we will get:

Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.
Yes this is the right answer
Answer:
a = 3
Step-by-step explanation:
1) Simplify 5a - 9 + a to 6a - 9.

2) Add 9 to both sides.

3) Simplify 9 + 9 to 18.

4) Divide both sides by 6.

5) Simplify 18/6 to 3.

Therefor, the answer is, a = 3.
4m+28=44
4m=44-28
4m=16
m=16/4
m=4
2(5*4)+2(4*3)+2(3*5)=94 sq.m (Orginial SA)
2(4*5)+2(5*6)+2(6*4)=148 sq.m (Total SA)
148-94=54 (New SA)