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2 answers:
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Answer:
C)
region C
Step-by-step explanation:
We have to use what is called the zero-interval test [test point] in order to figure out which portion of the graph these inequalities share:
0 ≤ 2 ☑ [We shade the portion of the graph that CONTAIN THE ORIGIN, which is the bottom portion.]
0 ≥ −3 ☑ [We shade the portion of the graph that CONTAINS THE ORIGIN, which is the left side.]
So, now that we got that all cleared up, we can tell that both graphs share a region where the ORIGIN IS VISIBLE. Therefore region C matches the above inequalities.
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Answer:
(a)
(b)
Step-by-step explanation:
Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.
(a)We want to show that E[X + Y ] = E[X] + E[Y ].
When we have two random variables instead of one, we consider their joint distribution function.
For a function f(X,Y) of discrete variables X and Y, we can define
Since f(X,Y)=X+Y
Let us look at the first of these sums.
Similarly,
Combining these two gives the formula:
Therefore:
(b)We want to show that if X and Y are independent random variables, then:
By definition of Variance, we have that:
Since X and Y are independent, Cov(X,Y)=0
Therefore as required: