Answer:
x=4x-16 is what you get after you distribute
Step-by-step explanation:
2 by 2 is 4
2 by 8 is 16
There is no picture to be solved
The volume of a cone is

where r = radius and h = height. If the cone has a volume of 94.2 cm³ (I assume you didn't mean m³ because that would be ridiculously huge) and a height of 10 cm, we can plug these values into the formula to find the radius. Don't do any rounding.

Now we know that's going to be the radius of our <em>new </em>cone as well since we're keeping the diameter the same. The volume is going to be double 94.2 which is 188.4. Let's solve for the height.
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:
7
Step-by-step explanation: