What is the LCD of 6/7 , 3/5 and 1/4 ? =^=<br> A. 100<br> B. 140

Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

nadya68 [22]

I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

and so the standard deviation is $\sqrt{\dfrac{350}3}\approx\boxed{116.67}$

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

8 0

3 years ago

Louise is walking 12 miles to raise funds for a local cause. If she has completed

jok3333 [9.3K]

12 x 3/4 = 36/4 = 9

she has walked 9 miles

4 0

3 years ago

Need help pronto! Will give brainliest!

irina1246 [14]

156°

D is a 24° angle, and is placed in a 90° angle, so 90-24=66

C is a right angle, so 90+66=156°

Hope this helped!

6 0

2 years ago

Read 2 more answers

Applying the Third Corollary to the Inscribed Angles Theorem A circle is shown. Points A, B, C, and D are on the circle. Lines c

Marat540 [252]

Answer:

∠B = 110°
∠D = 70°

Step-by-step explanation:

Opposite angles in an inscribed quadrilateral are supplementary, so ...

∠B +∠D = 180°

(3x -4) +(2x -6) = 180

5x = 190 . . . . . . . . . . . add 10, collect terms

x = 38 . . . . . . . . . . . . . divide by 5

∠D = 2(38) -6 = 70

∠B = 180 -70 = 110

The measures of angles B and D are 110° and 70°, respectively.

8 0

3 years ago

A photograph of sides 35cm by 22cm is mounted onto a

trapecia [35]

<h3>Answer: 580 square cm</h3>

===================================================

Work Shown:

A = area of smaller rectangle

A = length*width

A = 35*22

A = 770

B = area of larger rectangle

B = length*width

B = 45*30

B = 1350

C = area of the border only

C = B - A

C = 1350-770

C = 580 square cm

7 0

3 years ago

Read 2 more answers

What is the LCD of 6/7 , 3/5 and 1/4 ? =^= A. 100 B. 140