Any number 1-26 should work. the expression would me x<$26.
I think it is 8/17...
hope this helps!
First you start off by setting up a Subtraction problem there anything common denominators of 7/8 and two over three verse 7/8 you should get 21/24 invert to over three you should get 16/24 then you would subtract those two anyway to get 5/24 I hope that was helpful get it
Answer:
The volume of the solid is 
Step-by-step explanation:
In this case, the washer method seems to be easier and thus, it is the one I will use.
Since the rotation is around the y-axis we need to change de dependency of our variables to have 
. Thus, our functions with 
as independent variable are:
For the washer method, we need to find the area function, which is given by:
![A=\pi\cdot [(\rm{outer\ radius)^2 -(\rm{inner\ radius)^2 ]](https://tex.z-dn.net/?f=A%3D%5Cpi%5Ccdot%20%5B%28%5Crm%7Bouter%5C%20radius%29%5E2%20-%28%5Crm%7Binner%5C%20radius%29%5E2%20%5D)
By taking a look at the plot I attached, one can easily see that for a rotation around the y-axis the outer radius is given by the function 
and the inner one by 
. Thus, the area function is:
![A(y)=\pi\cdot [(\sqrt{y} )^2-(y^2)^2]\\A(y)=\pi\cdot (y-y^4)](https://tex.z-dn.net/?f=A%28y%29%3D%5Cpi%5Ccdot%20%5B%28%5Csqrt%7By%7D%20%29%5E2-%28y%5E2%29%5E2%5D%5C%5CA%28y%29%3D%5Cpi%5Ccdot%20%28y-y%5E4%29)
Now we just need to integrate. The integration limits are easy to find by just solving the equation 
, which has two solutions 
and 
. These are then, our integration limits.
The probability that 5 would not be working is 0.18665, the probability that at least one machine would be working is 0.00602 and the probability that all would be working is 1.
Given a company has 200 machines. Each machine has a 12% probability of not working.
If we working pick 40 machines randomly then we have to find the probability that 5 would not be working, the probability that at least one machine would be working, and the probability that all would be working.
So
1) probability that 5 would not be working
C(40,5)·0.12⁵·0.88³⁵= 40!/(5!(40-5)!)·0.12⁵·0.88³⁵
≈ 0.18665
2) probability that at least one machine would be working
0.88⁴⁰ ≈ 0.00602
3) probability that all would be working
1 - 0.12⁴⁰ ≈ 1.0000
Learn more about probability here: brainly.com/question/24756209
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