please rewrite your question so that i can be able to give answer of it
Answer:
1/256
Step-by-step explanation:
(-16)^-2
= 1/(-16)^2
= 1/ (-16· -16)
= 1/ 256
Answer:
Step-by-step explanation:
The problem states that there are only two types of busses - M104 and M6 with probable occurence of 0.6 and 0.4 respectively.
If the average number of busses arriving per hour is λ, the average number of M6 busses per hour is 0.4λ
Now consider a set of 3 M6 busses as an event. The average number of such events per hour will be
μ = 0.4λ / 3
The expected number of hours for the event "THIRD M6 arrives", let's say X is
E[X] = 1 / μ ( exponential distribution) = 3 / 0.4λ
= 7.5 / λ
The variance of event X is =
![Var[x] = \frac{1}{U^2} = \frac{56.25}{\lambda ^2}](https://tex.z-dn.net/?f=Var%5Bx%5D%20%3D%20%5Cfrac%7B1%7D%7BU%5E2%7D%20%3D%20%5Cfrac%7B56.25%7D%7B%5Clambda%20%5E2%7D)