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)
The answer is b, B cause why not
Cross Multiply by 'x - 5'
Hence
y(x - 5) = 4 -3x
xy - 5y = 4 - 3x
Move all the terms containing an 'x' to one side of the equals and all the others to the opposite side.
xy + 3x = 4 + 5y
Factor out 'x'
x(y + 3) = 4 + 5y
'Cross' divide by 'y + 3'
x = ( 4 + 5y) / (y + 3 )
'x' is now the subject.
Answer:
A and D
Step-by-step explanation:
A) 10+2^3*4-1 D) 10+(2^3*4)-1
10+8*4-1 10+(8*4)-1
10+32-1 10+32-1
42-1 42-1
41 41