On a coordinate gride what is the distance between (1,3) and (6,15)

Use the diagram and equations. below to match the following angles with their correct measures

Lorico [155]

So you would set angle y plus angle z equal to 90 degrees.
2b + 6 +3b - 1 = 90
5b + 5 = 90
5b = 85
b = 17
so for y 3(17)-1 = 50 degrees
so for z 2(17) + 6 = 40 degrees
angle y and w form a straight line equal to 180 degreew so w = 180 - 50 = 130 degrees which leaves x being 140 degrees

5 0

3 years ago

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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, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

(a) Find c such that f 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 X by integrating the joint density with respect to y :

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

(c) Get the marginal density of Y by integrating with respect to x 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 X given Y can obtained by dividing the joint density by the marginal density of Y (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 X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.

The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

8 0

3 years ago

Can someone help me please ???????

Inessa [10]

Answer:

d=1

Step-by-step explanation:

Factor d^2 - 2d - 8 into (d-4)(d+2)

Move -2/d+2 onto the other side, changing it into 2/d+2.

$\frac{-3d}{(d-4)(d+2)} +\frac{3}{d-4}+\frac{2}{d+2} = 0$

Let the empty side equal zero.

Add 3/d-4 and 2/d+2

$\frac{5d-2}{(d+2)(d-4)}$

Then add that to -3d.

$\frac{2d-2}{(d+2)(d+4)}$

Multiply 0 by d+2 and d-4 to get 2d-2 by itself.

2d=2

d=1

I can't edit the equation, but it's d-4 althroughout the question. Sorry for being so slow.

6 0

2 years ago

A recipe for cookies requires 2/3 cup of butter.Rama wants to make 3/4 of the recipe.how many cups of butter does she need to us

notsponge [240]

1 recipe = 2/3 cup of butter.

3/4 recipe = 3/4 x 2/3 = 1/2 cup of butter

Answer: 1/2 cup of butter

5 0

3 years ago

How many teaspoons are in 15 milliliters? 1 tsp = 5 mL

zhannawk [14.2K]

Answer:

Milliliters to teaspoons (mL to tsp) converter, formulas and conversion table. ... 1 mL = 0.202884 tsp; 5 mL = 1.014 tsp; 10 mL = 2.028 tsp; 15 mL = 3.04

Step-by-step explanation:

hope it helps

8 0

2 years ago

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