From what I gather from your latest comments, the PDF is given to be
and in particular, <em>f(x, y)</em> = <em>cxy</em> over the unit square [0, 1]², meaning for 0 ≤ <em>x</em> ≤ 1 and 0 ≤ <em>y</em> ≤ 1. (As opposed to the unbounded domain, <em>x</em> ≤ 0 *and* <em>y</em> ≤ 1.)
(a) Find <em>c</em> such that <em>f</em> is a proper density function. This would require
(b) Get the marginal density of <em>X</em> by integrating the joint density with respect to <em>y</em> :
(c) Get the marginal density of <em>Y</em> by integrating with respect to <em>x</em> instead:
(d) The conditional distribution of <em>X</em> given <em>Y</em> can obtained by dividing the joint density by the marginal density of <em>Y</em> (which follows directly from the definition of conditional probability):
(e) From the definition of expectation:
![E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_0%5E1%5Cint_0%5E1%20x%5C%2Cf_%7BX%2CY%7D%28x%2Cy%29%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D4%5Cleft%28%5Cint_0%5E1x%5E2%5C%2C%5Cmathrm%20dx%5Cright%29%5Cleft%28%5Cint_0%5E1y%5C%2C%5Cmathrm%20dy%5Cright%29%3D%5Cboxed%7B%5Cfrac23%7D)
![E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}](https://tex.z-dn.net/?f=E%5BY%5D%3D%5Cdisplaystyle%5Cint_0%5E1%5Cint_0%5E1%20y%5C%2Cf_%7BX%2CY%7D%28x%2Cy%29%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D4%5Cleft%28%5Cint_0%5E1x%5C%2C%5Cmathrm%20dx%5Cright%29%5Cleft%28%5Cint_0%5E1y%5E2%5C%2C%5Cmathrm%20dy%5Cright%29%3D%5Cboxed%7B%5Cfrac23%7D)
![E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}](https://tex.z-dn.net/?f=E%5BXY%5D%3D%5Cdisplaystyle%5Cint_0%5E1%5Cint_0%5E1xy%5C%2Cf_%7BX%2CY%7D%28x%2Cy%29%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D4%5Cleft%28%5Cint_0%5E1x%5E2%5C%2C%5Cmathrm%20dx%5Cright%29%5Cleft%28%5Cint_0%5E1y%5E2%5C%2C%5Cmathrm%20dy%5Cright%29%3D%5Cboxed%7B%5Cfrac49%7D)
(f) Note that the density of <em>X</em> | <em>Y</em> in part (d) identical to the marginal density of <em>X</em> found in (b), so yes, <em>X</em> and <em>Y</em> are indeed independent.
The result in (e) agrees with this conclusion, since E[<em>XY</em>] = E[<em>X</em>] E[<em>Y</em>] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)
Answer:
Part A: T(x) = $49.99x + $499.00
Part B: $748.95
Part C: T(x) = $49.99x + $24.99y + $499.00
Part D: $773.93
Step-by-step explanation:
Part A:
Let x represent the amount of games bought. Let T(x) represent total cost.
T(x) = $49.99x + $499.00
Part B:
T(x) = $49.99(5) + $499.00
T(x) = $249.95 + $499.00
T(x) = $748.95
Part C:
Let y represent the amount of controllers bought.
T(x) = $49.99x + $24.99y + $499.00
Part D:
T(x) = $49.99(4) + $24.99(3) + $499.00
T(x) = $199.96 + $74.97 + $499.00
T(x) = $773.93
Answer: 6/12 are white, 3/12 are colored and 3/12 are albino.
Step-by-step explanation: If the horses are white and their parents are ccww (albino) and CCWw (white horse), according to Mendel's premises, they both must be CcWw, since the crossing provides one C from one parent and other c from the other parent, one W and the other w. Using Mendel's chess and the principle of independent segregation, the crossing between CcWw results in the following fenotypical ratio:
1/16 CCWW (lethal)
2/16 CCWw (white)
2/16 CcWW (lethal)
4/16 CcWw (white)
1/16 CCww (normal)
2/16 Ccww (normal)
2/16 ccWw (albino)
1/16 ccWW (lethal)
1/16 ccww (albino)
Excluding the 4 individuals that have the lethal locus, we have 6/12 that are white (2/12 + 4/12) and 3/12 (1/12 + 2/12) that are colored. Also, there are 3/12 of albino individuals as well.
3 (3/1) represents the slope and -4 is the y-intercept
D) (X,Y) <span> → (x + 4, y - 5).
HOPE THIS HELPED :-)</span>
