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
To solve this problem you must apply the proccedure shown below:
1. You have that the ellipse given as a vertical major axis (a=13), therefore, taking the ellipse with its center at the origin, you have the following equation:
(y^2/a^2)+(x^2/b^2)=1
2. You have the distance from the center of the ellipse to the focus:
c=12, therefore, you can calculate the value of b, the minor radius:
c^2=a^2-b^2
b=√(13^3-12^2)
b=5
3. Therefore, the equation is:
a^2=169
b^2=25
(y^2/169)+(x^2/25)=1
The answer is: (y^2/169)+(x^2/25)=1
Answer:
-6/5
Step-by-step explanation:
-5-1 / 0-(-5)
S=<span>−<span>1.546727
hope this helps.</span></span>
Answer:
yes
Step-by-step explanation:
yes that is true you are correct
