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.
If you divide 27 by 17 you get 0.6296 and if you move the decimal and round it would be about 63%
Answer:
The answer is 1.5
Step-by-step explanation:
Hope it helps :)
Answer:
D. 2019
Step-by-step explanation:
So I started by rewriting the function f(x)
1000 = 10^3 so
f(x)=10^3 times 1.03^x
1.03 = 103/100 so
f(x)= 10^3 times (103/100)^x
to raise a fraction to a power you just raise the numerator and denominator to that power so
f(x)= 10^3 times 103^x/100^x
change it to exponential form with a base of 10
f(x)=10^3 times 103^x/10^2x
then you can reduce it with the 10^3
f(x)=103^x/10^2x-3
so now that thats in more of a standard form you can find the intersection of the two functions
which is at (9.012,1305.244)
x is the number of years after 2010, so they will be equal ~9 years after 2010, which is 2019.
Y = 3x + 5
(0,5) (1,8)
y = -2x + 20
(0,20) (1,18)