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Scilla [17]
3 years ago
5

the base of a triangle and a parallelogram are the same length there Heights are also the same if the area of the parallelogram

is 48 m 2 what is the area of the triangle​
Mathematics
1 answer:
Sphinxa [80]3 years ago
7 0

Answer:

24

Step-by-step explanation:

The area of a parallelogram is A = b*h. The area of a triangle is A = 1/2 b*h. This means if the area of the parallelogram is 48 square meters then the area of the triangle is 24 square meters.

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Which graph represents the function f(x) = x1 - 22​
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A student takes ten exams during a semester and receives the following grades: 90, 85, 97, 76, 89, 58, 82, 102, 70, and 67. Find
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Formula: Minimum = first element of set First Quartile = (n+1)/4 Median = (n+1)/2 Third Quartile = 3(n+1)/4 Maximum = Last element of set. Solution:<span> The five number summary of ( 90, 85, 97, 76, 89, 58, 82, 102, 70, and 67) is, Minimum = 58 First Quartile = 70 Median: = 83.5 Third Quartile = 90 Maximum = 102</span>
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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.)

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