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
Answer:
-1.575, or -1 575/1000, or -1 23/40
Step-by-step explanation:
Simplify the problem slightly by prefacing it with " - " as follows:
334
- ----------- = -1.575 = -1 23/40
212
Step-by-step explanation:
An=A1. R raised to the power n-1
From the given table, the points can be written as : (6,7.55) , (12,7.85), (18, 8.15) and (24, 8.45)
Slope 
m = 
m = 0.05
slope intercept form of line = 
We plug m = 0.05, x = 6 and y = 7.55 and find c.

c= 7.25
Required equation:
... Answer