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:
perimeter = 19.42 ft
Step-by-step explanation:
perimeter of semi circle = 6π/2 = 9.42
perimeter of cone = 5 + 5 = 10
10 + 9.42 = 19.42 ft
Answer:
Kara incorrectly factored 2(16x^2 + 24x + 9) she did not find the correct factors when she factored out. The correct factored expression would be 2(4x+3)^2
Step-by-step explanation:
32x^2 + 48x + 18
2(16x^2 + 24x + 9)
2(16x^2 + 12x) (12x + 9)
2 4x(4x + 3) 3(4x + 3)
2(4x + 3)(4x + 3)
2(4x + 3)^2
X > -8
The line would be on -8 going to the left, the circle would not be shaded
1. You are given PR and RS, so simply add the two measurements to get PS.
28.2+30.7= 58.9
2. Subtract QV from QT to get TV.
78-56= 22
3. Again, Subtract QV from QT to get TV.
74-36= 38
4. Similar to question 1, add PR and RS to get PS.
40+53= 93
