The answer is —2
Each number is multiplied by —2 to get the following number in the pattern
Hope this helps and good luck you can do thissss :)
Answer:
There is a 99.44% probability that the squad will have at most 2 calls in an hour.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given time interval.
In this problem, we have that:
2.83 rescues every eight hours.
What is the probability that the squad will have at most 2 calls in an hour?
This is
![P = P(X = 0) + P(X = 1) + P(X = 2)](https://tex.z-dn.net/?f=P%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29)
We have 2.83 rescues every 8 hours. So for an hour, we have ![\mu = \frac{2.83}{8} = 0.354](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B2.83%7D%7B8%7D%20%3D%200.354)
So
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-0.354}*(0.354)^{0}}{(0)!} = 0.7019](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-0.354%7D%2A%280.354%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.7019)
![P(X = 1) = \frac{e^{-0.354}*(0.354)^{1}}{(1)!} = 0.2485](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-0.354%7D%2A%280.354%29%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.2485)
![P(X = 2) = \frac{e^{-0.354}*(0.354)^{2}}{(2)!} = 0.0440](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-0.354%7D%2A%280.354%29%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.0440)
So
![P = P(X = 0) + P(X = 1) + P(X = 2) = 0.7019 + 0.2485 + 0.0440 = 0.9944](https://tex.z-dn.net/?f=P%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%3D%200.7019%20%2B%200.2485%20%2B%200.0440%20%3D%200.9944)
There is a 99.44% probability that the squad will have at most 2 calls in an hour.
Answer:
Below.
Step-by-step explanation:
12. That is a semicircle
The central angle is 180 degrees.
answer in bold
4L-12=68
4L=80
L=20
W= 14inches
77/100, 81%, 0.84 because it equals 77% than 81% than 84%