<u>Answer:</u>
X and Y are stochastically dependent RVs .
<u>Step-by-step explanation:</u>
Let ,
X = sum of the values that come up after throwing n (≥ 1) fare dice.
Y = number of times an odd number come up.
Let, n = 3
then, P(X =6) = p (say) clearly 0 < p < 1
and P (Y = 3) = 
And,
P( X = 6, Y = 3) = 0 ≠ 
Hence, X and Y are stochastically dependent RVs
Answer:
x=7
Step-by-step explanation:
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
Answer:
C. 2x^2-7x+7
Step-by-step explanation:
(3x^2-2x+2)-(x^2+5x-5) <em>Given</em>
From the given you will subtract like terms from both parenthesis.
3x^2-x^2=2x^2
-2x-5x=-7x
2-(-5)=7
