Answer:
Step-by-step explanation:
Let X(t) denote the number of machines breakdown at time t.
The givenn problem follows birth-death process with finite space
S={0, 1, 2, 3} with
![\lambda_0=\frac{3}{10}, \mu_1=\frac{1}{8}\\\\ \lambda_1=\frac{2}{10}, \mu_2=\frac{2}{8}\\\\ \lambda_2=\frac{1}{10}, \mu_3=\frac{2}{8}](https://tex.z-dn.net/?f=%20%5Clambda_0%3D%5Cfrac%7B3%7D%7B10%7D%2C%20%5Cmu_1%3D%5Cfrac%7B1%7D%7B8%7D%5C%5C%5C%5C%20%5Clambda_1%3D%5Cfrac%7B2%7D%7B10%7D%2C%20%5Cmu_2%3D%5Cfrac%7B2%7D%7B8%7D%5C%5C%5C%5C%20%5Clambda_2%3D%5Cfrac%7B1%7D%7B10%7D%2C%20%5Cmu_3%3D%5Cfrac%7B2%7D%7B8%7D)
The birth-death process having balance equations ![\lambda_sP_i=\mu_{s+1}P_{i+1},i=0,1,2](https://tex.z-dn.net/?f=%5Clambda_sP_i%3D%5Cmu_%7Bs%2B1%7DP_%7Bi%2B1%7D%2Ci%3D0%2C1%2C2)
since, state rate at which leave = rate at which enter
0 ![\lambda_0P_0=\mu_1P_1](https://tex.z-dn.net/?f=%5Clambda_0P_0%3D%5Cmu_1P_1)
1 ![(\lambda_1+\mu_1)P_1= \mu_2P_2 + \lambda_0P_0](https://tex.z-dn.net/?f=%28%5Clambda_1%2B%5Cmu_1%29P_1%3D%20%5Cmu_2P_2%20%2B%20%5Clambda_0P_0)
2 ![(\lambda_2+\mu_2)P_2= \mu_3P_3 + \lambda_1P_1](https://tex.z-dn.net/?f=%28%5Clambda_2%2B%5Cmu_2%29P_2%3D%20%5Cmu_3P_3%20%2B%20%5Clambda_1P_1)
![P_1=\frac{12}{5}P_0=P_0=\frac{5}{12}P_1\\\\P_2=\frac{48}{25}P_0=P_0=\frac{25}{48}P_2\\\\P_3=\frac{192}{250}P_0=P_0=\frac{250}{192}P_3](https://tex.z-dn.net/?f=P_1%3D%5Cfrac%7B12%7D%7B5%7DP_0%3DP_0%3D%5Cfrac%7B5%7D%7B12%7DP_1%5C%5C%5C%5CP_2%3D%5Cfrac%7B48%7D%7B25%7DP_0%3DP_0%3D%5Cfrac%7B25%7D%7B48%7DP_2%5C%5C%5C%5CP_3%3D%5Cfrac%7B192%7D%7B250%7DP_0%3DP_0%3D%5Cfrac%7B250%7D%7B192%7DP_3)
Since ![\sum\limits^3_{i=0} {P_i=1}\\\\p_0=[1+\frac{5}{12}+\frac{48}{25}+\frac{192}{250}]^{-1}=\frac{250}{1522}](https://tex.z-dn.net/?f=%5Csum%5Climits%5E3_%7Bi%3D0%7D%20%7BP_i%3D1%7D%5C%5C%5C%5Cp_0%3D%5B1%2B%5Cfrac%7B5%7D%7B12%7D%2B%5Cfrac%7B48%7D%7B25%7D%2B%5Cfrac%7B192%7D%7B250%7D%5D%5E%7B-1%7D%3D%5Cfrac%7B250%7D%7B1522%7D)
a)
Average number not in use equals the mean of the stationary distribution ![P_1+2P_2+3P_3=\frac{2136}{751}](https://tex.z-dn.net/?f=P_1%2B2P_2%2B3P_3%3D%5Cfrac%7B2136%7D%7B751%7D)
b)
Proportion of time both repairmen are busy ![P_2+P_3=\frac{672}{1522}=\frac{336}{761}](https://tex.z-dn.net/?f=P_2%2BP_3%3D%5Cfrac%7B672%7D%7B1522%7D%3D%5Cfrac%7B336%7D%7B761%7D)
Answer:
x=4.5,y=5,z=9
Step-by-step explanation:
First, we use the pythagorean theorem in the lower right triangle (with sides 4,x,6) and find x^2+4^2=6^2 and therefore x^2=6^2-4^2=20 so x=√20 which is approximately 4.5.
Also, we can see from the picture that z=y+4, and therefore the only option left is x=4.5,y=5,z=9.
Y=-13+1
y=-12
So the "range" is only a single number...-12
You are just multiplying each number by negative 4 so 1792*-4 would be -7168 and that would be your next number
Answer: c(x)=-2 x2 + 16x + 168
Step-by-step explanation: I hope this helps if im wrong im srry!