Answer:
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it fails, or it does not. The components are independent. We want to know how many outcomes until r failures. The expected value is given by
In which r is the number of failures we want and p is the probability of a failure.
In this problem, we have that:
r = 1 because we want the first failed unit.
![p = 0.4[\tex]So[tex]E = \frac{r}{p} = \frac{1}{0.4} = 2.5](https://tex.z-dn.net/?f=p%20%3D%200.4%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3ESo%3C%2Fp%3E%3Cp%3E%5Btex%5DE%20%3D%20%5Cfrac%7Br%7D%7Bp%7D%20%3D%20%5Cfrac%7B1%7D%7B0.4%7D%20%3D%202.5)
The expected number of systems inspected until the first failed unit is 2.5
Answer:
30(x+1)/(x+7)
Step-by-step explanation:
This is not a rational expression because denominator has a domain of which it can be undefined.
To rationalize multiply by the reciprocal of the denominator (1/6+1/x+1)
___<u>5___ </u>*(6)(x+1) --> _____<u>30(x+1)_____</u> --><u> 30x+1</u>_ --> <u>30x+1</u>
1/6+1/x+1 *(6)(x+1)--> 6(x+1)/6 + 6(x+1)/(x+1) --> (x+1)+6 --> x+7
Final Answer: 30(x+1)/(x+7)
I'm pretty sure this one is super easy
you just take 144 and divide it by 6 which would look like
144/6 then the answer is 24
It’s casey. when compared to other fractions, it stands supreme.
Answer:
7723/10000
Step-by-step explanation:
