How to find the x intercept of 3y=2x-6: (when y is 0)
3(0) = 2x - 6
0 = 2x - 6
+6 + 6
-------------------------
6 = 2x
------ -----
2 2
x = 3
How to find the y intercept of 3y=2x-6: (when x is 0)
3y = 2(0) - 6
y = 0 - 6
3y = -6
------ -----
3 3
y = -2
The probability that the train will be there when Alex arrives is 5/18
If Alex arrives at any time after 1.20pm the chances that train will be there is 1/3.
However if alex arrives at 1.00pm exactly there is no chance the train will be arrive there.
The probability that the train will be there increase linearly to 1/3 as alex's arrival time moves from 1.00pm to 1.20pm.
By arranging the probabilities over the first 20 minutes to get a 1/6 chance the train will be there if alex arrives between 1.00pm to 1.20pm
we get the final answer by
=1/3( 1/6 + 1/3 + 1/3)
=5/18
So, the probability that the train will be there when Alex arrives is 5/18
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Answer:
7 1/6
Step-by-step explanation:
<span>solution of a system of linear equations</span>
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.
