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2 years ago

What is the slope of the line on the graph? Enter your answer in the box.

Mathematics

2 answers:

pshichka [43]2 years ago

8 0

I know this is WAY late.. but this is for whoever finds this in the future.

The answer is: -2

:)

HACTEHA [7]2 years ago

7 0

Answer:

-2

Step-by-step explanation:

i took the test

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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

2 years ago

Name the regular polygon that has 12 sides

AnnZ [28]

Answer:

dodecagon

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Does anyone know this? Also, can you explain how you get the answer? cuz I have 10 other questions same as this.

Nezavi [6.7K]

The answer is 36

Multiply each side by 9

9 times for and 9 times n/9

The n/9 will cancel out so you are left with n= 9 times 4

This comes to 36

7 0

2 years ago

according to the fundamental theorem of algebra , how many zeros does the polynomial below have x^4 +5x^3+10x^2+20x+24

Vinil7 [7]

Answer:

Step-by-step explanation:

the Fundamental Theorem of Algebra states that for any polynomial of degree n, there are n roots, some of which may be complex

The polynomial shown is of degree 4 ( highest exponent of x )

Hence the polynomial has 4 roots/ zeros

8 0

2 years ago

Read 2 more answers

A rectangular painting is 24 inches wide and 20 inches tall without the frame. With the frame, it is 28 inches wide and 24 inche

evablogger [386]

The area of the frame not covered by the painting is 192 square inches

Solution:

Given that, A rectangular painting is 24 inches wide and 20 inches tall without the frame

The area of rectangle is given as:

$Area = length \times width$

Find the area of painting without frame:

Length = 20 inches

Width = 24 inches

$Area\ Without\ Frame = 20 \times 24 = 480$

Thus area of painting without frame is 480 square inches

With the frame, it is 28 inches wide and 24 inches tall

Find the area of painting with frame:

Length = 24 inches

Width = 28 inches

$Area\ with\ frame = 24 \times 28 = 672$

Thus area of painting with frame is 672 square inches

What is the area of the frame not covered by the painting

Area of the frame not covered by the painting = Area with farme - Area without frame

Area of the frame not covered by the painting = 672 - 480 = 192

Thus the area of the frame not covered by the painting is 192 square inches

4 0

2 years ago