The correct answer is Choice E: 1 milligram.
The prefix milli means "1/1000". That means a milligram is 1/1000th the size of a gram. All of the other weights in the list are larger.
2/18 is the multiple of 1/9
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:
What is 0.42857142857 as a fraction?
To write 0.42857142857 as a fraction you have to write 0.42857142857 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.
0.42857142857 = 0.42857142857/1 = 4.2857142857/10 = 42.857142857/100 = 428.57142857/1000 = 4285.7142857/10000 = 42857.142857/100000 = 428571.42857/1000000 = 4285714.2857/10000000 = 42857142.857/100000000 = 428571428.57/1000000000 = 4285714285.7/10000000000 = 42857142857/100000000000
And finally we have:
0.42857142857 as a fraction equals 42857142857/100000000000
<span>The goal in simplifying numbers
is to reduce it that can be best seen in the eye. Also in
converting units, you must take note of the units given in the problem. If you
don’t, you would get a lot of errors. Length is the measure of distance and can
be best represented by either in SI units or English units. Example of an SI unit is meter, m, and English
unit is feet, ft. You are given 3 inches and 2 feet. You can convert any of the
numbers into a number that would feel comfortable to you. I will take 3 inches.
Note that there are 12 inches in 1 feet. So divide 3 by 12 and you will get ¼ feet.
Then divide 2 feet to ¼ feet and you will get 8.</span>