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lawyer [7]
2 years ago
13

If jordan runs 16 miles and Jake ran 18 times as many how many miles did jake ran​

Mathematics
2 answers:
zheka24 [161]2 years ago
8 0
288 miles




Yghhhhhhhhhhhh
QveST [7]2 years ago
5 0

Answer:

288 miles

Step-by-step explanation:

18 times 16

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I REALLLLLY NEED HELP WITH THAT!!!! Plz help
Inga [223]

Answer:

If it says (10a + 4): 40a + 16

If it says (10a - 4): 40a - 16

Step-by-step explanation:

If it's an equilateral quadrilateral, that means all sides are equal. So you would multiply the given side by the number of sides. Then simplify you equation.

I can't read it if it says (10a + 4) or (10a - 4) so i'll just do both and you can use which one it says.

If it says (10a + 4):

4(10a + 4)

40a + 16

If it says (10a - 4)

4(10a - 4)

40a - 16

8 0
3 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
Ben and Carmella want to go to a basketball game. Carmella proposes they use a bag of 90 lettered tiles to decide who will pay f
navik [9.2K]

Answer:

72.1%

2

Step-by-step explanation:

Please give me brainliest, I really need it.

4 0
3 years ago
Find the number of tiles required to be put in a floor with dimensions 25 m x 20 m. Each tile is a rhombus with side 25 cm altit
Romashka-Z-Leto [24]

Answer:

8000

Step-by-step explanation:

Area of floor=25×20m

=500²

Area of each tile=area if each rhombus

=25×25=625cm²

No. of tiles required= area of floor/ area of each tile

=500×104⁴/ 625

=4/5 ×10×1000

=8000

hope it helps!

6 0
2 years ago
What’s the solution of x+4&lt;8
aniked [119]

Answer:

x<4

Move all terms not containing  x  to the right side of the inequality.

7 0
3 years ago
Read 2 more answers
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