Answer:
0.0668
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 123, \sigma = 4](https://tex.z-dn.net/?f=%5Cmu%20%3D%20123%2C%20%5Csigma%20%3D%204)
Find the probability that the diameter of a selected bearing is greater than 129 millimeters.
This is 1 subtracted by the pvalue of Z when X = 129. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{129 - 123}{4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B129%20-%20123%7D%7B4%7D)
![Z = 1.5](https://tex.z-dn.net/?f=Z%20%3D%201.5)
has a pvalue of 0.9332
1 - 0.9332 = 0.0668
0.0668 = 6.68% probability that the diameter of a selected bearing is greater than 129 millimeters.
You can conclude that AE = DE because the triangles are about equal
Answer:
9, 4.5 x 2 = 9
4.5 = 45 tenths and 0.5 = 5 tenths
5 tenths x 9 = 45 tenths
4.5/0.5 = 9
6.3/0.9 = 7, 0.9 x 7 = 63, 63 / 0.9 = 7
Step-by-step explanation: