Looks like a badly encoded/decoded symbol. It's supposed to be a minus sign, so you're asked to find the expectation of 2<em>X </em>² - <em>Y</em>.
If you don't know how <em>X</em> or <em>Y</em> are distributed, but you know E[<em>X</em> ²] and E[<em>Y</em>], then it's as simple as distributing the expectation over the sum:
E[2<em>X </em>² - <em>Y</em>] = 2 E[<em>X </em>²] - E[<em>Y</em>]
Or, if you're given the expectation and variance of <em>X</em>, you have
Var[<em>X</em>] = E[<em>X</em> ²] - E[<em>X</em>]²
→ E[2<em>X </em>² - <em>Y</em>] = 2 (Var[<em>X</em>] + E[<em>X</em>]²) - E[<em>Y</em>]
Otherwise, you may be given the density function, or joint density, in which case you can determine the expectations by computing an integral or sum.
Answer:
0 in²
Step-by-step explanation:
Maybe it would help to show some lines or something shape-like.
we are given

we can see that
there is no value of x for which g(x) is not defined
so, no vertical asymptote exists
now, we will find horizontal asymptote



so, we get
horizontal asymptote as
............Answer
H+20=35-4H
first, combine your like terms, which are the Hs
add the 4H to both sides, and you will get
5H+20=35
now subtract 20 from both sides, that will cancel out the 20 on the left and leave you with
5H=15
finally, divide both sides by 5 and you will get H=3.