3 is double 6 so the relation ship is that it is doubled
<em>Y</em>₁ and <em>Y</em>₂ are independent, so their joint density is

By definition of conditional probability,
P(<em>Y</em>₁ > <em>Y</em>₂ | <em>Y</em>₁ < 2 <em>Y</em>₂) = P((<em>Y</em>₁ > <em>Y</em>₂) and (<em>Y</em>₁ < 2 <em>Y</em>₂)) / P(<em>Y</em>₁ < 2 <em>Y</em>₂)
Use the joint density to compute the component probabilities:
• numerator:






• denominator:

(I leave the details of the second integral to you)
Then you should end up with
P(<em>Y</em>₁ > <em>Y</em>₂ | <em>Y</em>₁ < 2 <em>Y</em>₂) = (1/6) / (2/3) = 1/4
Answer:
4z^2+7z
you have to combine the like terms
7z+4z^2+6-6
then becomes
(4z^2) + (7z) + (6-6)
gets you to the simplified version which is
4z^2 +7z
9x-6+6x-1=98
15x+7=98
15x=105
x=7
hope this helps :)
16 Ounces, I hope this helps, Mark me brainliest