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3 years ago

Hey could anyone explain how to answer this kind of question? Or just solve and show what you did! Thank you!

Mathematics

1 answer:

STALIN [3.7K]3 years ago

4 0

Step-by-step explanation:

9x-3=60°{each angle of equilateral triangle is 60°}
9x=60+3
x=63/9
x=7

so,

9x-3
9×7-3
63-3
60°

hope it helps.

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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

ABC has coordinates of A(-8,-8) B(4,-2) C(2,2). Find the coordinates of its image after the dilation centered at the origin with

garri49 [273]

Answer:

A' ( -12 , -12 )

B' ( 6 , -3 )

C' ( 3 , 3 )

Step-by-step explanation:

To find the coordinates of a point after a dilation simply multiply the x and y values of the points by the scale factor

Points: A(-8,-8) B(4,-2) C(2,2)

Scale factor: 1.5

Coordinates after the dilation

A' = (-8,-8) --------> (-8 * 1.5 , -8 * 1.5 ) ------------> (-12 , -12)

B' = (4,-2) ---------> (4 * 1.5 , -2 * 1.5) -----------> (6 , -3 )

C' = (2 , 2) ----------> (2 * 1.5 , 2 * 1.5) -----------> (3, 3)

So inclusion the coordinates of ABC after a dilation centered at the origin with a scale factor of 1.5 are A' ( -12 , -12 ) B' ( 6 , -3 ) C' ( 3 , 3 )

4 0

3 years ago

Prove that a line parallel to one side of a triangle divides the other two sides proportionally. Be sure to create and name the

ryzh [129]

Answer:

in steps

Step-by-step explanation:

DE // BC

m∠ADE = m∠ABC and m∠AED = m∠ACB

∴ ΔADE similar to ΔABC

AB/AD = AC/AE

(AD + DB) / AD = (AE + EC) / AE

AD/AD + DB/AD = AE/AE + EC/AE

1 + DB/AD = 1 + EC/AE

DB/AD = EC/AE (AD/DB = AE/EC)

7 0

3 years ago

Find the value of variable x. If your answer is not an integer, write it in simplest radical form with the denominator rationali

Otrada [13]

Answer:

the answer is 7

Step-by-step explanation:

angle 30°

x=14/2

x=7

8 0

3 years ago

Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c

GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

$f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0$

The value of E (X) is 10.

The parameter λ is:

$\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10$

(a)

Compute the value of P (X > 10) as follows:

$P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679$

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

$P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353$