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)
Ratio triangle height to square side 7:8
ratio triangle base to square side 1:2
Area of the square is 64 in² then the sides = 8 in
ratios give
h = 7(8)/8
h = 7
b = 8/2
b = 4
Area Triangle = 0.5(7)(4)
= 14
Then shaded region is the Area of the square minus the Area of the triangle.
A = 64 - 14
A = 50 in²
Y = 9x
y = 5x + 24
9x = 5x + 24
9x - 5x = 24
4x = 24
x = 24/4
x = 6
y = 9x
y = 9(6)
y = 54
solution (6,54)....x = 6 and y = 54
The function is given
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Now let's solve for
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Rule of negative exponent:
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Rule for sum of fractions:
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And the result is:
