Yup it is C. Integers and Positive Integers are under positive numbers so it has to be, becuase B is wrong.
Answer:
56.65% probability that more than 3 customers arrive during 15 minutes
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 interval, which is the same as the variance. The standard deviation is the square root of the variance.
Standard deviation of 2 customers per 15-minutes.
So ![\mu = 2^{2} = 4](https://tex.z-dn.net/?f=%5Cmu%20%3D%202%5E%7B2%7D%20%3D%204)
What is the probability that more than 3 customers arrive during 15 minutes
Either three or less customers arrive, or more than 3 do. The sum of these probabilities is decimal 1. Mathematically, we have that:
![P(X \leq 3) + P(X > 3) = 1](https://tex.z-dn.net/?f=P%28X%20%5Cleq%203%29%20%2B%20P%28X%20%3E%203%29%20%3D%201)
We want P(X > 3). So
![P(X > 3) = 1 - P(X \leq 3)](https://tex.z-dn.net/?f=P%28X%20%3E%203%29%20%3D%201%20-%20P%28X%20%5Cleq%203%29)
In which
![P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%203%29%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%2B%20P%28X%20%3D%203%29)
![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^{-4}*(4)^{0}}{(0)!} = 0.0183](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.0183)
![P(X = 1) = \frac{e^{-4}*(4)^{1}}{(1)!} = 0.0733](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.0733)
![P(X = 2) = \frac{e^{-4}*(4)^{2}}{(2)!} = 0.1465](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.1465)
![P(X = 3) = \frac{e^{-4}*(4)^{3}}{(3)!} = 0.1954](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.1954)
![P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0183 + 0.0733 + 0.1465 + 0.1954 = 0.4335](https://tex.z-dn.net/?f=P%28X%20%5Cleq%203%29%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%2B%20P%28X%20%3D%203%29%20%3D%200.0183%20%2B%200.0733%20%2B%200.1465%20%2B%200.1954%20%3D%200.4335)
![P(X > 3) = 1 - P(X \leq 3) = 1 - 0.4335 = 0.5665](https://tex.z-dn.net/?f=P%28X%20%3E%203%29%20%3D%201%20-%20P%28X%20%5Cleq%203%29%20%3D%201%20-%200.4335%20%3D%200.5665)
56.65% probability that more than 3 customers arrive during 15 minutes
Answer: 34,450 tonnes
Step-by-step explanation:
35/100 of 53000 food was spoilt= 18, 550 tonnes
Subtract 18,550 from 53,000
C)50 I think because it goes up 50 units each time
Answer b: the system of equations does not have a solution
Suppose the systems consits of two first grade equations with two varaibles. If the two equations are parallel you would obtaing a result like Vinnet.
This is an example:
x = y + 1
7x = 7y + 9
substitue to get:
7(y+1) = 7y +9
7y + 7 = 7y + 9
7=9
There is not mistake, the system has no solutions, because the two equations represent parallel lines, which do no intercept one to each other.