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:
Step-by-step explanation:
2(5m+4)=2(3m-10)
10m+8= 6m-20
-6m -6m
4m+8= -20
-8 -8
4m= -28
m = -7
The equation is y=-2/3x-1/3 i think
Answer:
1000002
Step-by-step explanation: