The possibilities are endless
If we're talking between 1-20
then the numbers divisible by 2 and 6 are
18,12,6
Answer: 0.75
Step-by-step explanation:
Given : Interval for uniform distribution : [0 minute, 5 minutes]
The probability density function will be :-

The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-
![P(x>1.25)=\int^{5}_{1.25}f(x)\ dx\\\\=(0.2)[x]^{5}_{1.25}\\\\=(0.2)(5-1.25)=0.75](https://tex.z-dn.net/?f=P%28x%3E1.25%29%3D%5Cint%5E%7B5%7D_%7B1.25%7Df%28x%29%5C%20dx%5C%5C%5C%5C%3D%280.2%29%5Bx%5D%5E%7B5%7D_%7B1.25%7D%5C%5C%5C%5C%3D%280.2%29%285-1.25%29%3D0.75)
Hence, the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes = 0.75
Is there any answer choices so I can choose?
(f+g) (n) = (–5n +1 ) + (- 6n +2) = –11n +3
(f+g) (n) = –11n +3
(f+g) (–2) = – 11 (–2) +3 = 22 +3 = 25
I hope I helped you^_^
Your answer should be C! Hope that helps