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.)
300 cannot be converted to a rational number. It has to stay and irrational number because it is not a perfect square
Y=x+70
Slope-intercept form
Answer:
B
Step-by-step explanation:
If you arrange a statistics in order from least to greatest, the middle number is the median.
If there is even number of numbers in a list, let it be n numbers, then we take the average of n/2th and n/2 + 1 th terms.
Here, all of them have 6 numbers, so 6/2 = 3 and 4th, we take average of 3rd and 4th number to find the median.
Since they are arrange in order we check each:
Jon = average of 6 and 7, (6+7)/2 = 6.5
Leroy = average of 6 and 8, (6+8)/2 = 7
Simon = average of 5 and 6, (5+6)/2 = 5.5
Leroy's median is the greatest (7).
Answer:
We are given an area and three different widths and we need to determine the corresponding length and perimeter.
The first width that is provided is 4 yards and to get an area of 100 we need to multiply it by 25 yards. This would mean that our length is 25 yards and our perimeter would be 2(l + w) which is 2(25 + 4) = 58 yards.
The second width that is given is 5 yards and in order to get an area of 100 yards we need to multiply by 20 yards. This would mean that our length is 20 yards and our perimeter would be 2(l + w) which is 2(20 + 5) = 50 yards.
The final width that is given is 10 yards and in order to get an area of 100 yards we need to multiply by 10. This would mean that our length is 10 yards and our perimeter would be 2(l + w) which is 2(10 + 10) = 40 yards.
Therefore the field that would require the least amount of fencing (the smallest perimeter) is option C, field #3.
<u><em>Hope this helps!</em></u>
