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)
<h3>
Answer: 12</h3>
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Explanation:
You can use the AAS (angle angle side) theorem to prove that triangle ABD is congruent to triangle CBD.
From there, we can then say that AD and DC are the same length
AD = DC
3y+6 = 5y-18
3y-5y = -18-6
-2y = -24
y = (-24)/(-2)
y = 12
Answer:
I'm not completely sure but try either 3x - 7
or 3 - 7x
Step-by-step explanation:
Hope one of them helps!
In this question, both tickets cost 2$ per ticket.
The answer to this question would be: $0
In WinOne scenario, you need to match a ticket that has to pick from A-J(10 possibilities) and 0-9 (10 possibilities). The chance to win would be: 1/10* 1/10= 1/100
The expected value must be:
E= chance to win * win amount - ticket price
E= 1//100*$200 - $2= $2-$2= 0
