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.)
-Make variables
-Write equation given using variables
-Plugin the equation for L with 2w to solve for W.
-plugin W value to solve for L
Answer:
x-intercepts = (-3/4, 0)
y-intercepts = (0, 3/7)
Step-by-step explanation:
Thirty two point eight seven five
Answer:
d. each trial has exactly two outcomes whose probabilities do not change
Step-by-step explanation:
A binomial experiment is one where there are exactly two outcomes for each trial and probability for getting success is constant in each trial.
In other words, each trial is independent of the other.
The trials need not be continuous nor time between trials to be constant.
Since trials are to be independent, each trial cannot influence the next.
Only option d is right.
d. each trial has exactly two outcomes whose probabilities do not change
Examples are tossing of coins, throwing dice, drawing cards or balls with replacement, etc