\left[x _{2}\right] = \left[ 6+\sqrt{33}\right][x2]=[6+√33]
<span>−4.0 is an integer.
hope it helps</span>
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
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Answer:
Infintely many solutions
Step-by-step explanation:
I'm going to assume that the capital y is equal to the lowerase y
if you subtract y and two from both sides in the second equation you get
-y=2x-2
you then divide by -1 to get it into a normal form
y= -2x+2
this is the same as the first equation, these lines are the same
