Solve the equation by using the Properties of Equality. 9x

Jane addresses twice as many envelopes as Louis in the same amount of time. Together their dress 360 envelopes each day. How man

Flauer [41]

Louis = x
Jane = 2x (Twice as many as Louis)
Together they have 360
x + 2x = 360
3x= 360
x=120
2(120)= 240
Jane addresses 240 envelopes

7 0

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, 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

2 years ago

Help me pleaseee urgenttt

marusya05 [52]

A. 4x b. 3y c. 8p is the answer

6 0

2 years ago

Read 2 more answers

Solve the following equation for x. 0.1(x + 250) = 5,000 h

yan [13]

Answer:

x=49750

Step-by-step explanation:

0.1(x + 250) = 5,000

Divide each side by .1

0.1(x + 250)/.1 = 5,000/.1

x+250 =50000

Subtract 250 from each side

x+250-250 = 50000-250

x=49750

3 0

2 years ago

Read 2 more answers

The seating capacity of a theater is 250. Tickets are $3 for children and $5 for adults. The theater must take in at least $1100

maksim [4K]

Answer:

x + y ≤

3x + 5y ≥ 1100

Step-by-step explanation:

Given:

Seating capacity of theater = 250

Cost of each child ticket = $3

Cost of each adult ticket = $5

Cost per performance = $1100 at least

Find:

System of inequalities

Computation:

Let;

x = Number of children's tickets

y = Number of adult tickets

So

x + y ≤

3x + 5y ≥ 1100

5 0

2 years ago

Solve the equation by using the Properties of Equality. 9x – 3 = 42