Answer:
7
O
⋅
(
24
h
O
)
⋅
(
25
O
)
⋅
31
.
X
⋅
7
=
7
X
24
=
7
X
7
O
⋅
(
24
h
O
)
⋅
(
25
O
)
⋅
31
=
130200
O
3
h
First you must add the same numbers together, for instance, you wouldn't add 4y with 4 you have to add it with another number that is just like it, for instance 4y and 9y they are the same, so you can add them. So let's get to the problem now:
5y(2y-3)=(2y-3)
5y*2y=10y
(10y-3)=(2y-3)
You can do nothing to the other side because there are no numbers, that are the same.
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:
Yes.
Step-by-step explanation:
Let's first convert his speed to km/h.
>
. Therefore, Rob is traveling within the speed limit.