Length:Width
530:1
30:x
X/30= 1/530
Multiply 30 on both sides
X= 3/53 =0.05660377
5C2 * (2x)^3 * y^2
5 * 4 / 2 * 8 * x ^ 3 * y ^ 2
20 / 2 * 8 * x ^ 3 * y ^ 2
80 * x^3 y^2
Answer = 80
Answer:
3/4
Step-by-step explanation:
Answer:
![P(x\geq 3)=0.3233](https://tex.z-dn.net/?f=P%28x%5Cgeq%203%29%3D0.3233)
Step-by-step explanation:
If the number of defects in poured metal follows a Poisson distribution, the probability that x defects occurs is:
![P(x)=\frac{e^{-m}*(m)^{x}}{x!}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7Be%5E%7B-m%7D%2A%28m%29%5E%7Bx%7D%7D%7Bx%21%7D)
Where x is bigger or equal to zero and m is the average. So replacing m by 2, we get that the probability is equal to:
![P(x)=\frac{e^{-2}*(2)^{x}}{x!}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7Bx%7D%7D%7Bx%21%7D)
Finally, the probability that there will be at least three defects in a randomly selected cubic millimeter of this metal is equal to:
![P(x\geq 3)=1-p(x\leq 2)\\](https://tex.z-dn.net/?f=P%28x%5Cgeq%203%29%3D1-p%28x%5Cleq%202%29%5C%5C)
Where ![P(x\leq 2)=P(0)+P(1)+P(2)](https://tex.z-dn.net/?f=P%28x%5Cleq%202%29%3DP%280%29%2BP%281%29%2BP%282%29)
So, P(0), P(1) and P(2) are equal to:
![P(0)=\frac{e^{-2}*(2)^{0}}{0!}=0.1353\\P(1)=\frac{e^{-2}*(2)^{1}}{1!}=0.2707\\P(2)=\frac{e^{-2}*(2)^{2}}{2!}=0.2707](https://tex.z-dn.net/?f=P%280%29%3D%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7B0%7D%7D%7B0%21%7D%3D0.1353%5C%5CP%281%29%3D%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7B1%7D%7D%7B1%21%7D%3D0.2707%5C%5CP%282%29%3D%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7B2%7D%7D%7B2%21%7D%3D0.2707)
Finally,
and
are equal to:
![P(x\leq 2)=0.1353+0.2707+0.2707=0.6767\\P(x\geq 3)=1-0.6767=0.3233](https://tex.z-dn.net/?f=P%28x%5Cleq%202%29%3D0.1353%2B0.2707%2B0.2707%3D0.6767%5C%5CP%28x%5Cgeq%203%29%3D1-0.6767%3D0.3233)
I think it would be the second one but i am not sure hope this helps! :D