Show using the definition of expected values for discrete random variables (Eq. 5.1) that for a discrete random variable, X, E(g
(X)) = ∑g(x) p(x) where g(x) is linear. That is, show the formula is true assuming that g(x) = ax + b. You are showing that a special case of Eq. 5.2 is true; therefore, you many not use that formula in the derivation. Submission type: Submit entries below as answer to receive credit Save entries below (not submitted for credit yet)
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Answer:
Step-by-step explanation:
When solving a quadratic inequality, first solve it normally like you would for a normal quadratic equation. We have:
Ignore the less than sign and replace it with an equal sign and solve the quadratic for its zeros:
Subtract 6 from both sides:
Divide everything by 3:
Factor. Find two numbers that equal (3)(-2)=-6 that add up to -5.
-6 and 1 works. Thus:
Find the x using the Zero Product Property:
Now, we need to replace the equal signs with symbols again. To do so, we need to test which symbol to place. Let's do the first zero first.
So, the first zero is:
Assume that the correct symbol is >. Thus,
Now, pick any number that is greater than -1/3. I'll pick 0 since it's the easiest. Now, plug 0 back into the very original inequality. If it works, then the sign is correct, if it doesn't, then simply use the opposite one. Therefore:
0 is indeed less than six, so our first correct solution is:
For the second one, do the same thing. We have:
Assume that the correct symbol is <. Thus:
Again, pick any number less than 2. I'm going to use 0. Plug 0 back into the original equation
Again, this is correct. Therefore, x<2 is also the correct inequality.
So together, we have:
Together, we can write them as:
(Note that we don't need to worry about the "or equal to" part since the original inequality didn't have it.)