Answer:
![\frac{4}{9}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B9%7D)
Step-by-step explanation:
First, we gather the data:
The model is a single
landing strip with ![\mu = \frac{2}{3}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B2%7D%7B3%7D)
This based on the assumption that service times are exponential.
Then, we must find the maximum rate such that ![w_{q} \leq 3](https://tex.z-dn.net/?f=w_%7Bq%7D%20%5Cleq%203)
Using the
table, we find that:
![w_{p} = \frac{\rho }{(\mu (1 -\rho) } \\ \frac{\lambda }{\mu (1 - \lambda } \leq 3](https://tex.z-dn.net/?f=w_%7Bp%7D%20%20%3D%20%5Cfrac%7B%5Crho%20%7D%7B%28%5Cmu%20%281%20-%5Crho%29%20%20%7D%20%5C%5C%20%5Cfrac%7B%5Clambda%20%7D%7B%5Cmu%20%281%20-%20%5Clambda%20%20%7D%20%5Cleq%203)
if and only if
![\lambda\leq \frac{3\mu ^{2} }{(1 + 3\mu) }](https://tex.z-dn.net/?f=%5Clambda%5Cleq%20%5Cfrac%7B3%5Cmu%20%5E%7B2%7D%20%7D%7B%281%20%2B%203%5Cmu%29%20%7D)
=![\frac{4}{9}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B9%7D)
Answer:
a = 80
b = 40
c = 58
d = 19
e = 34
f = 36
Pls mark brainliest. Thanks!
13.45 + 9.5 = 22.95
42.6 - 22.95 = 19.65
19.65 : 3 = 6.55m did he fence each day wednesday thursday & friday
Factor 6x^5−18x^3+54x^2
6x^5−18x^3+54x^2
=6x^2(x^3−3x+9)
Answer:
6x^2(x^3−3x+9)