-6x^4y^5-15x^3y^2+9x^2y^3
The GCF OF -6, -15, & 9, is -3
The GCF OF x's is x^2
The GCF OF y's is y^2
These are the number of x's and y's that each term has.
-3x^2y^2(2x^2y^3+5x-3y)
LETTER B
Answer:
0.1507 or 15.07%.
Step-by-step explanation:
We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.
First of all, we will find z-scores for data points using z-score formula.
, where,
z = z-score,
x = Sample score,
= Mean,
= Standard deviation.
![z=\frac{21.97-22}{0.016}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B21.97-22%7D%7B0.016%7D)
![z=\frac{-0.03}{0.016}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B-0.03%7D%7B0.016%7D)
![z=-0.1875](https://tex.z-dn.net/?f=z%3D-0.1875)
Let us find z-score of data point 22.03.
![z=\frac{22.03-22}{0.016}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B22.03-22%7D%7B0.016%7D)
![z=\frac{0.03}{0.016}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B0.03%7D%7B0.016%7D)
![z=0.1875](https://tex.z-dn.net/?f=z%3D0.1875)
Using probability formula
, we will get:
![P(-0.1875](https://tex.z-dn.net/?f=P%28-0.1875%3Cz%3C0.1875%29%3DP%28z%3C0.1875%29-P%28z%3C-0.1875%29)
Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.
Answer:
Percent, Part, whole, I am not sure the rest hope this helped though
Step-by-step explanation:
Answer:
64 cups of sweet tea
Step-by-step explanation: