Answer:
The probability that a randomly selected call time will be less than 30 seconds is 0.7443.
Step-by-step explanation:
We are given that the caller times at a customer service center has an exponential distribution with an average of 22 seconds.
Let X = caller times at a customer service center
The probability distribution (pdf) of the exponential distribution is given by;
![f(x) = \lambda e^{-\lambda x} ; x > 0](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Clambda%20e%5E%7B-%5Clambda%20x%7D%20%3B%20x%20%3E%200)
Here,
= exponential parameter
Now, the mean of the exponential distribution is given by;
Mean =
So,
⇒
SO, X ~ Exp(
)
To find the given probability we will use cumulative distribution function (cdf) of the exponential distribution, i.e;
; x > 0
Now, the probability that a randomly selected call time will be less than 30 seconds is given by = P(X < 30 seconds)
P(X < 30) =
= 1 - 0.2557
= 0.7443
Answer:
![p(q(x)) = 2x^2 - 16x + 30](https://tex.z-dn.net/?f=p%28q%28x%29%29%20%3D%202x%5E2%20-%2016x%20%2B%2030)
Step-by-step explanation:
The notation (p•q)(x) means p(q(x)) or p of q of x. To find it, substitute q(x) into p(x) and simplify:
![p(q(x)) = 2(x-3)^2 - 4(x-3)\\p(q(x))=2(x^2-6x+9) - 4x + 12\\p(q(x)) = 2x^2-12x+18-4x+12\\p(q(x)) = 2x^2 - 16x + 30](https://tex.z-dn.net/?f=p%28q%28x%29%29%20%3D%202%28x-3%29%5E2%20-%204%28x-3%29%5C%5Cp%28q%28x%29%29%3D2%28x%5E2-6x%2B9%29%20-%204x%20%2B%2012%5C%5Cp%28q%28x%29%29%20%3D%202x%5E2-12x%2B18-4x%2B12%5C%5Cp%28q%28x%29%29%20%3D%202x%5E2%20-%2016x%20%2B%2030)
Answer:
x=4
y=7 there tou go that is your answer
2 x 10 = 20 apples and plums that Charlie should gather :)