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%.
Answer:
[(a)^2-(4^2)](a+4)
Step-by-step explanation:
Answer:
615.44cm
Step by Step Answer:
A=πr^{2}=π·14^{2}≈615.75216
Answer:
Step-by-step explanation:
7*(7 + x+1) = 6*(x + 5 + 6) {Intersecting secant theorem}
7 *(x + 8) = 6*(x + 11)
Distributive property,
7x + 56 = 6x + 66
Subtract 56 from both sides
7x = 6x + 66 - 56
7x = 6x + 10
Subtract 6x from both sides
7x - 6x = 10
x = 10