The easiest way to go about problems like these is to divide the complex shape into simpler shapes.
for instance, you could imagine the 20x9 and 4x20 rectangles separate from the 9x7 section in the middle.
it would be 9x7 because 9+4+x = 22 and 11++y+2 = 20
Step-by-step explanation:
The first one has x going up by 2's and y going up by 1's. The second has x going up by 1's and y going up by 2's.
Answer: 371 mm
46 mm
Step-by-step explanation:
Answer:
![a. \ P(X=0)=5.574\times10^{-7}](https://tex.z-dn.net/?f=a.%20%5C%20P%28X%3D0%29%3D5.574%5Ctimes10%5E%7B-7%7D)
![b. \ \ P(X\leq 1)=8.584\times10^{-6}](https://tex.z-dn.net/?f=b.%20%5C%20%5C%20P%28X%5Cleq%201%29%3D8.584%5Ctimes10%5E%7B-6%7D)
![c.\ \ \ P(X\geq 4)=0.9997](https://tex.z-dn.net/?f=c.%5C%20%5C%20%5C%20P%28X%5Cgeq%204%29%3D0.9997)
Step-by-step explanation:
a. #We notice this is a Poisson probability function expressed as:
![f(x)=\frac{\mu^xe^{-\mu}}{x!}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B%5Cmu%5Exe%5E%7B-%5Cmu%7D%7D%7Bx%21%7D)
x-number of occurrences in a given interval.
-mean occurrences of the event
-The mean is calculated as:
![\mu=\frac{14.4}{4}=3.6](https://tex.z-dn.net/?f=%5Cmu%3D%5Cfrac%7B14.4%7D%7B4%7D%3D3.6)
#the probability of no accidents in a 15-minute period is :
![P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X=0)=\frac{14.4^0e^{-14.4}}{0!}\\\\=5.574\times10^{-7}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7B%5Cmu%5Exe%5E%7B-%5Cmu%7D%7D%7Bx%21%7D%5C%5C%5C%5CP%28X%3D0%29%3D%5Cfrac%7B14.4%5E0e%5E%7B-14.4%7D%7D%7B0%21%7D%5C%5C%5C%5C%3D5.574%5Ctimes10%5E%7B-7%7D)
Hence, the probability of no accident in a 15-min period is ![=5.574\times10^{-7}](https://tex.z-dn.net/?f=%3D5.574%5Ctimes10%5E%7B-7%7D)
b. The the probability of at least one accident in a 15-minute period. is calculated as:
![P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X\leq 1)=\frac{14.4^0e^{-14.4}}{0!}+\frac{14.4^1e^{-14.4}}{1!}\\\\=5.574\times10^{-7}+8.026\times 10^{-6}\\\\=8.584\times 10^{-6}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7B%5Cmu%5Exe%5E%7B-%5Cmu%7D%7D%7Bx%21%7D%5C%5C%5C%5CP%28X%5Cleq%201%29%3D%5Cfrac%7B14.4%5E0e%5E%7B-14.4%7D%7D%7B0%21%7D%2B%5Cfrac%7B14.4%5E1e%5E%7B-14.4%7D%7D%7B1%21%7D%5C%5C%5C%5C%3D5.574%5Ctimes10%5E%7B-7%7D%2B8.026%5Ctimes%2010%5E%7B-6%7D%5C%5C%5C%5C%3D8.584%5Ctimes%2010%5E%7B-6%7D)
Hence, the probability of at least one accident in a 15-minute period is ![8.584\times10^{-6}](https://tex.z-dn.net/?f=8.584%5Ctimes10%5E%7B-6%7D)
c. The probability of four or more accidents in a 15-minute period is calculated as:
![P(X\geq 4)=1-P(X\leq 3)=1-[P(X=0)+P(X1)+P(X=2)+P(X=3)]\\\\=1-[5.574\times10^{-7}+a. \ 8.026\times10^{-6}+\frac{14.4^2e^{-14.4}}{2!}+\frac{14.4^3e^{-14.4}}{3!}]\\\\=1-[8.584\times 10^{-6}+5.779\times10^{-5}+2.774\times10^{-4}]\\\\=0.9997](https://tex.z-dn.net/?f=P%28X%5Cgeq%204%29%3D1-P%28X%5Cleq%203%29%3D1-%5BP%28X%3D0%29%2BP%28X1%29%2BP%28X%3D2%29%2BP%28X%3D3%29%5D%5C%5C%5C%5C%3D1-%5B5.574%5Ctimes10%5E%7B-7%7D%2Ba.%20%5C%208.026%5Ctimes10%5E%7B-6%7D%2B%5Cfrac%7B14.4%5E2e%5E%7B-14.4%7D%7D%7B2%21%7D%2B%5Cfrac%7B14.4%5E3e%5E%7B-14.4%7D%7D%7B3%21%7D%5D%5C%5C%5C%5C%3D1-%5B8.584%5Ctimes%2010%5E%7B-6%7D%2B5.779%5Ctimes10%5E%7B-5%7D%2B2.774%5Ctimes10%5E%7B-4%7D%5D%5C%5C%5C%5C%3D0.9997)
Hence,the probability of four or more accidents in a 15-minute period. is 0.9997