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icang [17]
3 years ago
8

Can someone help me with 23?

Mathematics
1 answer:
maxonik [38]3 years ago
3 0
2.5 pounds of apples * $1.20 = $3

He spent total $4.50

So he had to have spent 1.50 on oranges.

So $1.50=$1.20 x where x will be the pound of oranges bought.

1.50/1.20 = 1.25 or 1 1/4 pounds of oranges.
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Write an experession that shows 10 ^4
Licemer1 [7]
10^4+6x=8 that’s one equation I made up :)
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2 years ago
What is the area of a<br> parallelogram with a base<br> of 2.5 cm and a height of<br> 6 cm?
zloy xaker [14]

2.5 multiplied by 6 = 15

15 square centimeters

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3 years ago
What is an extraneous solution? How does it arise in solving radical equations?
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fdsdfsdfsdfsdf

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2 years ago
2) X and Y are jointly continuous with joint pdf
Nady [450]

From what I gather from your latest comments, the PDF is given to be

f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}

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

\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}

(b) Get the marginal density of <em>X</em> by integrating the joint density with respect to <em>y</em> :

f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}

(c) Get the marginal density of <em>Y</em> by integrating with respect to <em>x</em> instead:

f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}

(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):

f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}

(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}

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}

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}

(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.)

8 0
3 years ago
If the revenue for the week is $2000, and labor consists of two workers earning $8.00 per hourwho worked 40 each, what is labor
julsineya [31]
Labor costs of the workers at $8 per hour working 40 hours each=$640
Revenue for the week=$2000
therefore labor cost as a percentage of revenue would be=32%
8 0
4 years ago
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