Answer:
If it counts as a question, lol sure, I'd love to help
Answer:
69.14% probability that the diameter of a selected bearing is greater than 84 millimeters
Step-by-step explanation:
According to the Question,
Given That, The diameters of ball bearings are distributed normally. The mean diameter is 87 millimeters and the standard deviation is 6 millimeters. Find the probability that the diameter of a selected bearing is greater than 84 millimeters.
- In a set with mean and standard deviation, the Z score of a measure X is given by Z = (X-μ)/σ
we have μ=87 , σ=6 & X=84
- Find the probability that the diameter of a selected bearing is greater than 84 millimeters
This is 1 subtracted by the p-value of Z when X = 84.
So, Z = (84-87)/6
Z = -3/6
Z = -0.5 has a p-value of 0.30854.
⇒1 - 0.30854 = 0.69146
- 0.69146 = 69.14% probability that the diameter of a selected bearing is greater than 84 millimeters.
Note- (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 p-value, we get the probability that the value of the measure is greater than X)
Answer:
x = 9 ft
Step-by-step explanation:
The volume of a cylinder is given by
V = pi r^2 h
We know the radius is x
4069.44 = pi x^2 16
Divide each side by 16 pi
4069.44 / 16 pi = x^2
Using 3.14 for pi
81 = x^2
Take the square root of each side
sqrt(81) = sqrt(x^2)
x = 9
A=1/3h(b1-b2). So h=3/(b1-b2)
