Answer:
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Step-by-step explanation:
can be simplified to:

Make them have common denominators so:
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Simplify:

I don't think you can simplify that fraction further so that's the answer.
If we draw the contingency table of x (vertical) against y (horiz.), we have a square.
For n=4, we have (legend: < : x<y = : x=y > : x>y
y 1 2 3 4
x
1 = < < <
2 > = < <
3 > > = <
4 > > > =
We see that there are n(n-1)/2 cases of x<y out of n^2.
Therefore,
p(x<y)=n(n-1)/(2n^2)=(n-1)/(2n)
However, if the sample space is continuous, it will be simply p(x<y)=1/2.
If you need to you can write it out. the answer is X=14
3.15 addd all of them! Then divide
Two angles<span> are said to be supplementary when the sum of the two </span>angles<span> is </span>180<span>°. ... All </span>angles<span> that are either exterior </span>angles<span>, </span>interior angles<span>, </span>alternate angles<span> or corresponding </span>angles<span> are all congruent.</span>