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
4x^2 + 16x-2= 5x
Minus 16x over
4x2-2= -11x
Add 11x over
4x^2 +11x -2
Use quadratic formula x= 2.9
I’m not sure here. I tried my best. Hope it helps!
<h2>
Answer:</h2><h2>917</h2>
<h3>what you do is that you find the area of the whole thing then you multiply the answer up to get 917</h3>
Answer:
(3, -3)
Step-by-step explanation:
The vertex of the most general absolute value function is y = |x|, whose vertex is at (0, 0). Focusing on the " |x - 3| " tells us that the graph of y = |x| is moved 3 units to the right. Next, focusing on the " -3 " as it tells us that the resulting graph is shifted 3 units down. Thus, the vertex of g(x)=|x-3|-3 is at (3, -3).
