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)
When height turns to 3 times height or 3h
we can replace h with 3h and see what happens
A=1/2b(3h)
A=3/2bh
A=3(1/2bh)
well, the area is trippled
Answer: A. "Segment AD bisects angle CAB." is the right answer.
Step-by-step explanation:
Given : In ΔABC ,AC≅AB.
⇒∠ACB=∠CBA....(1) (∵ angles opposite to equal sides of a triangle are equal )
Now in ΔACD and ΔABD
AD=AD (common)....(2)
Here we need one more statement to prove the triangles congruent that is only statement (A) fits in it.
If AD bisects ∠CAB then ∠CAD=∠BAD..(3)
Now again Now in ΔACD and ΔABD
∠ACB=∠CBA [from (1)]
AD=AD [common]
∠CAD=∠BAD [from (3)]
So by ASA congruency criteria ΔADC≅ΔABD.
It’s a Trapezium (also called Trapezoid)
