Answer:
The first one is 79
The second one is 26
I hope this helped! :D
Step-by-step explanation:
For the first one you just plug in 23 into x so it would be 3(23) + 10 = x
For the second one you just plug in 88 into g so it would be 88 = 3x + 10
Answer: a) 4.6798, and b) 19.8%.
Step-by-step explanation:
Since we have given that
P(n) = ![\dfrac{15}{120}=0.125](https://tex.z-dn.net/?f=%5Cdfrac%7B15%7D%7B120%7D%3D0.125)
As we know the poisson process, we get that
![P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda](https://tex.z-dn.net/?f=P%28n%29%3D%5Cdfrac%7B%28%5Clambda%20t%29%5En%5Ctimes%20e%5E%7B-%5Clambda%20t%7D%7D%7Bn%21%7D%5C%5C%5C%5CP%28n%3D0%29%3D0.125%3D%5Cdfrac%7B%28%5Clambda%20%5Ctimes%2014%29%5E0%5Ctimes%20e%5E%7B-14%5Clambda%7D%7D%7B0%21%7D%5C%5C%5C%5C0.125%3De%5E%7B-14%5Clambda%7D%5C%5C%5C%5C%5Cln%200.125%3D-14%5Clambda%5C%5C%5C%5C-2.079%3D-14%5Clambda%5C%5C%5C%5C%5Clambda%3D%5Cdfrac%7B2.079%7D%7B14%7D%5C%5C%5C%5C0.1485%3D%5Clambda)
So, for exactly one car would be
P(n=1) is given by
![=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B%280.1485%5Ctimes%2014%29%5E1%5Ctimes%20e%5E%7B-0.1485%5Ctimes%2014%7D%7D%7B1%21%7D%5C%5C%5C%5C%3D0.2599)
Hence, our required probability is 0.2599.
a. Approximate the number of these intervals in which exactly one car arrives
Number of these intervals in which exactly one car arrives is given by
![0.2599\times 18=4.6798](https://tex.z-dn.net/?f=0.2599%5Ctimes%2018%3D4.6798)
We will find the traffic flow q such that
![P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr](https://tex.z-dn.net/?f=P%280%29%3De%5E%7B%5Cfrac%7B-qt%7D%7B3600%7D%7D%5C%5C%5C%5C0.125%3De%5E%7B%5Cfrac%7B-18q%7D%7B3600%7D%7D%5C%5C%5C%5C0.125%3De%5E%7B-0.005q%7D%5C%5C%5C%5C%5Cln%200.125%3D-0.005q%5C%5C%5C%5C-2.079%3D-0.005q%5C%5C%5C%5Cq%3D%5Cdfrac%7B-2.079%7D%7B-0.005%7D%3D415.88%5C%20veh%2Fhr)
b. Estimate the percentage of time headways that will be 14 seconds or greater.
so, it becomes,
![P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%](https://tex.z-dn.net/?f=P%28h%5Cgeq%2014%29%3De%5E%7B%5Cfrac%7B-qt%7D%7B3600%7D%7D%5C%5C%5C%5CP%28h%5Cgeq%2014%29%3De%5E%7B%5Cfrac%7B-415.88%5Ctimes%2014%7D%7B3600%7D%7D%5C%5C%5C%5CP%28h%5Cgeq%2014%29%3D0.198%5C%5C%5C%5CP%28h%5Cgeq%2014%29%3D19.8%5C%25)
Hence, a) 4.6798, and b) 19.8%.
On what do u Need help on?
Answer:
yes wait im doing it. do I put them un order
Answer:
0.1306
Step-by-step explanation:
This is a binomial distribution problem
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = number of free throws
x = Number of successes required = 6
p = probability of success = 0.60
q = probability of failure = 1 - 0.60 = 0.40
P(X = 6) = ⁷C₆ (0.6)⁶ (0.4)⁷⁻⁶ = 0.1306368 = 0.1306
Hope this Helps!!!