Answer:
#3. d = -10.25
Step-by-step explanation:
#3. d = -6.5 + -3.75
(1) Looks like the joint density is

In order for this to be a proper density function, integrating it over its support should evaluate to 1. The support is a triangle with vertices at (0, 0), (4, 0), and (4, 4) (see attached shaded region), so the integral is


(2) The region in which <em>X</em> > 2 and <em>Y</em> < 1 corresponds to a 2x1 rectangle (see second attached shaded region), so the desired probability is

(3) Are you supposed to find the marginal density of <em>X</em>, or the conditional density of <em>X</em> given <em>Y</em>?
In the first case, you simply integrate the joint density with respect to <em>y</em>:

In the second case, we instead first find the marginal density of <em>Y</em>:

Then use the marginal density to compute the conditional density of <em>X</em> given <em>Y</em>:

10.75,11.50,12.25,13.00,13.75,14.50.
So 6
Answer:
$11.78
Step-by-step explanation:
14.89+9.33=24.22
36-24.22=11.78
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http://www.mathopenref.com/quadvertexexplorer.html