Let X be the number of lightning strikes in a year at the top of particular mountain.
X follows Poisson distribution with mean μ = 3.8
We have to find here the probability that in randomly selected year the number of lightning strikes is 0
The Poisson probability is given by,
P(X=k) = ![\frac{e^{-mean} mean^{x}}{x!}](https://tex.z-dn.net/?f=%20%5Cfrac%7Be%5E%7B-mean%7D%20mean%5E%7Bx%7D%7D%7Bx%21%7D%20%20%20%20)
Here we have X=0, mean =3.8
Hence probability that X=0 is given by
P(X=0) = ![\frac{e^{-3.8} 3.8^{0}}{0!}](https://tex.z-dn.net/?f=%20%5Cfrac%7Be%5E%7B-3.8%7D%203.8%5E%7B0%7D%7D%7B0%21%7D%20%20%20%20)
P(X=0) = ![\frac{0.02237 * 1}{1}](https://tex.z-dn.net/?f=%20%5Cfrac%7B0.02237%20%2A%201%7D%7B1%7D%20%20)
P(X=0) = 0.0224
The probability that in a randomly selected year, the number of lightning strikes is 0 is 0.0224
Answer:
Step-by-step explanation:
Io
Answer: 1.25 or 5/4
Steps:
Move the -3x to the other side by adding it which gives you 4x=5. Then divide by 4 to have the x alone which gives you the answer. Hopes this helps :)
The gcf out of those three numbers should be 8
Answer:
3
Step-by-step explanation:
![\frac{7-(-2)}{5-2} =\frac{9}{3} =3](https://tex.z-dn.net/?f=%5Cfrac%7B7-%28-2%29%7D%7B5-2%7D%20%3D%5Cfrac%7B9%7D%7B3%7D%20%3D3)