Answer:
6
Step-by-step explanation:
the mean is 6 so is the median, can I have brainliest plz and thanks if yes
Answer:
The charge for admission is $6 and the charge for each ride is $2
Step-by-step explanation:
Let
x ----> the charge for admission
y ----> the charge for each ride
we have that
-----> equation A
-----> equation B
Solve the system by elimination
Subtract equation B from equation A
![x+2y=10\\-(x+5y)=-16\\---------\\2y-5y=10-16\\-3y=-6\\y=2](https://tex.z-dn.net/?f=x%2B2y%3D10%5C%5C-%28x%2B5y%29%3D-16%5C%5C---------%5C%5C2y-5y%3D10-16%5C%5C-3y%3D-6%5C%5Cy%3D2)
Find the value of x
substitute the value of y in any equation
![x+2(2)=10](https://tex.z-dn.net/?f=x%2B2%282%29%3D10)
![x+4=10](https://tex.z-dn.net/?f=x%2B4%3D10)
![x=10-4](https://tex.z-dn.net/?f=x%3D10-4)
![x=6](https://tex.z-dn.net/?f=x%3D6)
therefore
The charge for admission is $6 and the charge for each ride is $2
It’s J, 3 - X would equal 3 - (-2) which turns into 3+2 which equals 5 but 5 is not greater than 10
Answer:
There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.
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
is the Euler number
is the mean in the given time interval.
The problem states that:
The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and mean 0.2 each day during the weekend.
To find the mean during the time interval, we have to find the weighed mean of calls he receives per day.
There are 5 weekdays, with a mean of 0.1 calls per day.
The weekend is 2 days long, with a mean of 0.2 calls per day.
So:
![\mu = \frac{5(0.1) + 2(0.2)}{7} = 0.1286](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B5%280.1%29%20%2B%202%280.2%29%7D%7B7%7D%20%3D%200.1286)
If today is Monday, what is the probability that Ben receives a total of 2 phone calls in a week?
This is
. 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 = 2) = \frac{e^{-0.1286}*0.1286^{2}}{(2)!} = 0.0073](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-0.1286%7D%2A0.1286%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.0073)
There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.
Answer:
it is a it the first one. 7/13