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:
30-2x
Step-by-step explanation:
yuson has to finish am total of 30 hours in community service
she does 2 hours each day
in x days, hour of service done=2x
after x days hours of service left=30-2x
therefore the linear equation representing the hours yuson has left after x days is:
30-2x
Answer:
5091 Km/hr and 505 km/hr
Step-by-step explanation:
Speed = Distance / Time
Let the speed of first automobile be 'x' and that of the second be 'y'
Since speed of one is 10 times greater than the other. therefore;
⇒ x = 10 y
also let time for faster automobile be 'T' and time for slower auto mobile be 't'
Since first arrive one hour earlier than second, therefore;
⇒ t = T + 1
⇒ For first automobile;
; substituting for 'x' and 'T'. Therefore;
⇒ 
⇒ For Second automobile;
⇒ 
⇒ 
⇒ 5600 +
= 560
⇒ 5600 - 560 = - 
⇒ t = 1.11 hr
also ; T = 1.11 - 1 = 0.11 hr
Speed of 1st auto = 560/0.11 = 5091 km /hr
Speed of 2nd auto = 560/1.11 = 505 km/hr