True or False: The relation shown is a function. 20 POINTS!!
1 answer:
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Answer:
see explanation
Step-by-step explanation:
The excluded values are any values of x that make the function undefined
Given
The denominator of the rational function cannot be zero as this would make it undefined. Equating the denominator to zero and solving gives the values that x cannot be.
(x + 4)(x - 2)(x - 5) = 0
Equate each factor to zero and solve for x
x + 4 = 0 ⇒ x = - 4
x - 2 = 0 ⇒ x = 2
x - 5 = 0 ⇒ x = 5
x = - 4, x = 2, x = 5 are the excluded values
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: