I believe that would be 0.4691g
Answer:
0.025
Step-by-step explanation:
Given that a statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 46.0 and 56.0 minutes
Let X be the length of the class of statistics professor
X is U(46,56)
We are to find the probability that a given class period runs between 51.25 and 51.5 minutes.
The pdf of X is = ![\frac{1}{56-46} =0.1, 46\leq x\leq 56](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B56-46%7D%20%3D0.1%2C%2046%5Cleq%20x%5Cleq%2056)
Using this we can find probability using integration
![P(51.25\leq x\leq 51.5)](https://tex.z-dn.net/?f=P%2851.25%5Cleq%20x%5Cleq%2051.5%29)
=![\int\limits^{51.5} _{51.25} 0.1 \, dx \\=0.1(51.5-51.25)\\= 0.025](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B51.5%7D%20_%7B51.25%7D%20%200.1%20%5C%2C%20dx%20%5C%5C%3D0.1%2851.5-51.25%29%5C%5C%3D%200.025)
Look at the picture.
Domain: -4 ≤ x ≤ 4
Range: -1 ≤ y ≤ 1
<h3>Answer: {x | -4 ≤ x ≤ 4}</h3>
Step-by-step explanation:
the most common measure of how much sample mean differ from each other is the standard deviation of the sampling distribution of the mean. this standard deviation is called the standard error of the mean
Answer:
The chart below shows Mrs. Thompson's grocery bill. What is the average amount Mrs. Thompson spent on groceries?
A. $58
B. $68
C. $78
D. $340