Answer:
It is a many-to-one relation
Step-by-step explanation:
Given
See attachment for relation
Required
What type of function is it?
The relation can be represented as:
![\left[\begin{array}{c}x\\ \\-5\\-2\\1\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5C%20%5C%5C-5%5C%5C-2%5C%5C1%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}y\\ \\10\\11\\4\\10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dy%5C%5C%20%5C%5C10%5C%5C11%5C%5C4%5C%5C10%5Cend%7Barray%7D%5Cright%5D)
Where
and 
Notice that the range has an occurrence of 10 (twice)
i.e.
and 
In function and relations, when two different values in the domain point to the same value in the range implies that, <em>the relation is many to one.</em>
Answer:
I would have 360000.
Step-by-step explanation:
250 (the amount of money I earned) x 60(as in 60 mins in a hour) and 24(as in 24 hours a day) equals 360000 total.
Y-5= 3(x-0)
Y-5= 3x
Y= 3x + 5
Hope this helps!
The valid probability distributions are the ones in options C and D.
<h3>
Which of the following are valid probability distributions?</h3>
For discrete random variables with probabilities p₁, p₂, ..., pₙ, there are two rules:
- All of these probabilities are numbers between 0 and 1.
- p₁ + p₂ + ... + pₙ = 1.
So, for the first rule we can discard the first option, where we have negative probabilities.
To check the other 4 options, just add the probabilities and see if the addition gives 1.
The options that add up to 1 are C and D, so these two are the correct options.
D: 1/5 + 1/10 + 1/10 + 1/10 + 1/5 + 1/10 + 1/10 + 1/10 = 1
C: 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1
If you want to learn more about probability:
brainly.com/question/25870256
#SPJ1
Hey, we just need to find the gcf for them.
A) 3:4
B) 14:9
C) 3:11
D) 2:3
Hope these helped!
