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)
27 divided by 40 is equal to 0.675.
A=p(1+i/m)^mn
P=3168
i=0.1275
M=12
n=3/12
A=3,168×(1+0.1275÷12)^(12*3/12)
A=3,270.06
Answer: -rad7, +rad7
Explanation:
- Divide the equation by 6 (x^2-7=0)
- Add 7 (x^2=7)
- Square root x to find solutions (x=rad7)
- Dont forget the radical means that there are two solutions!
First we need to find f(2) and f(5), which are 5 and 11 respectively; all you have to do is plug in 2 and 5.
Then, we use the following formula:
(f(b)-f(a))/(b-a),
where b and a are the largest and smallest x values respectively.
Finally, we plug our values in:
(11-5)/(5-2)=6/3=2
In fact, the average rate of change of any linear function is just the coefficient of the x term. Hope this helped!
