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Zielflug [23.3K]
3 years ago
15

What is the slope of the line through the points (6,-8)And (3,- 4)​

Mathematics
1 answer:
Nina [5.8K]3 years ago
7 0

Answer:

y=(-4/3)x

Step-by-step explanation:

you put the points into this equation to find the slope

(y2-y1)/(x2-x1)

now lets label the points

(6,-8)

x1=6 and y1=-8

(3,-4)

x2=3 and y2=-4

Input into equation

(-4-(-8))/(3-6)=

(-4+8)/(3-6)

4/-3

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The answer is 6*4*4 square inches where 4*4 gives us the area of one square and multiplying by 6 gives you the total surface area.
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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.

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2 years ago
Question number 8. Prove that 3 + √5 is an irrational number.​
Marina86 [1]

Answer :

Let's assume the opposite of the statement i.e., 3 + √5 is a rational number.

\\  \sf \: 3 +  \sqrt{5}  =  \frac{a}{b}   \\  \\   \qquad \: \tiny \sf{(where \:  \: a \:  \: and \:  \: b \:  \: are \:  \: integers \:  \: and \:  \: b \:  \neq \: 0)} \\

\\  \sf \:  \sqrt{5}  =  \frac{a}{b}  - 3 =  \frac{a - 3b}{b}  \\

Since, a, b and 3 are integers. So,

\\ \sf \:  \frac{p - 3b}{b}  \\  \\  \qquad \tiny \sf{ \: (is \:  \: a \:  \: rational \:  \: number \:) } \\

Here, it contradicts that √5 is an irrational number.

because of the wrong assumption that 3 + √5 is a rational number.

\\

Hence, 3 + √5 is an irrational number.

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3 years ago
What is the answer to this?​
myrzilka [38]

Answer:r=3cm,diameter =6cm and C=18.84cm

Step-by-step explanation:

To find the radius,diameter and circumference according to the diagram drawn above

r=3cm

Diameter=2×3cm

d=6cm

To find the circumference

C=2πr

C=2×3.14×3cm

C=18.84cm

4 0
3 years ago
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