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)
Answer:
hey there, the answer is 8.9429 × 10^8
Answer:
8
Step-by-step explanation:
Answer:
13
Step-by-step explanation:
formula: a^2 + b^2 = c^2
5^2 + 12^2 = c^2
25 + 144 = c^2
169 = c^2
13 x 13 = 169
13
3,500 meters bc 3.5 times 1000 meters bc since 1 kilo = 1000 so 3.5 times 1000 = 3,500