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2 years ago

(2x+3)+(5x17)+90=180

Mathematics

2 answers:

klasskru [66]2 years ago

5 0

Answer:

x=10

Step-by-step explanation:

Alex73 [517]2 years ago

3 0

Answer:

x is 10 degrees

Step-by-step explanation:

180-90=90

(2x+3)+(5x+17) = 90

7x+20=90

7x=70

x=10

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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

3 years ago

E-mails do not require professionalism; it's acceptable to be informal. True False

Artyom0805 [142]

The answer is - False

3 0

2 years ago

Read 2 more answers

Math is not my strong point..

Leya [2.2K]

Answer: 16 in

Explanation:

PQ/QR = PT/TS

PQ/4 = 12/3
PQ x 3 = 12 x 4 (cross multiply)
PQ x 3 = 48
PQ = 48/3
PQ = 16

6 0

3 years ago

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7 = -x+3. Consider the line y = Find the equation of the line that is parallel to this line and passes through the point (-5, 6)

artcher [175]

Answer with explanation:

The equation of line is, y= -x +3

→x+y-3=0---------(1)

⇒Equation of line Parallel to Line , ax +by +c=0 is given by, ax + by +K=0.

Equation of Line Parallel to Line 1 is

x+y+k=0

The Line passes through , (-5,6).

→ -5+6+k=0

→ k+1=0

→k= -1

So, equation of Line Parallel to line 1 is

x+y-1=0

⇒Equation of line Perpendicular to Line , ax +by +c=0 is given by, bx - a y +K=0.

Equation of Line Perpendicular to Line 1 is

x-y+k=0

The Line passes through , (-5,6).

→ -5-6+k=0

→ k-11=0

→k= 11

So, equation of Line Parallel to line 1 is

x-y+11=0

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2 years ago

The length of a rectangle is 4 cm longer than the width of the rectangle. The perimeter of the rectangle is 88 Centimeters.What

Tcecarenko [31]

Im not sure but I think the answer is 9.4cm2

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3 years ago