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user100 [1]
3 years ago
9

20098765432 plus 34567890

Mathematics
1 answer:
Rom4ik [11]3 years ago
7 0

Answer: 20133333322

Step-by-step explanation: I just did it thx

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If we had a Pyramid with a hexagonal base, how many of these pyramids would it take to fill up a hexagonal Prism with a congruen
klemol [59]
Idk what is the answers
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3 years ago
I need to solve for X<br> X=
Gemiola [76]

\dfrac{2\frac{1}{4}}{\frac{3}{8}}=\dfrac{\frac{15}{16}}{x}\ \ \ \ |\text{cross multiply}\\\\2\dfrac{1}{4}x=\dfrac{3}{8}\cdot\dfrac{15}{16}\\\\\dfrac{2\cdot4+1}{4}x=\dfrac{45}{128}\\\\\dfrac{9}{4}x=\dfrac{45}{128}\ \ \ \ \ |\text{multiply both sides by 4}\\\\9x=\dfrac{45}{32}\ \ \ \ \ |\text{divide both sides by 9}\\\\\boxed{x=\dfrac{5}{32}}

6 0
3 years ago
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
Can someone pls help??
Vlad1618 [11]

Answer:

a. The y coordinates for A are either 10 or -14

b.  The y coordinates for A are either -11 or 7

Step-by-step explanation:

distance = sqrt ((x2-x1)^2+(y2-y1)^2)

We know Point B and the x coordinate of point A as well as the distance

15 = sqrt ((-4-5)^2+(-2-y1)^2)

15 = sqrt ((-9)^2+(-2-y1)^2)

15 = sqrt (81+(-2-y1)^2)

Square each side

15^2 =sqrt (81+(-2-y1)^2)^2

225 =  81+(-2-y1)^2

Subtract 81 from each side

225-81 = 81-81 +(-2-y1)^2

144 = (-2-y1)^2

Take the square root of each side

±sqrt 144 = sqrt(-2-y1)^2

±12 = (-2-y1)

Add 2 to each side

2±12 = 2-2-y1

2±12 = -y1

14 = -y1  or -10 = -y1

Multiply by -1

-14 = y1  or 10 = y1

The y coordinates for A are either 10 or -14

Now move B to -7,-2

15 = sqrt ((-7-5)^2+(-2-y1)^2)

15 = sqrt ((-12)^2+(-2-y1)^2)

15 = sqrt (144+(-2-y1)^2)

Square each side

15^2 =sqrt (144+(-2-y1)^2)^2

225 =  144+(-2-y1)^2

Subtract 144 from each side

225-144 = 144-144 +(-2-y1)^2

81 = (-2-y1)^2

Take the square root of each side

±sqrt 81 = sqrt(-2-y1)^2

±9 = (-2-y1)

Add 2 to each side

2±9 = 2-2-y1

2±9 = -y1

11 = -y1  or -7 = -y1

Multiply by -1

-11 = y1  or 7 = y1

The y coordinates for A are either -11 or 7

3 0
3 years ago
2x −3 y = 21 hmu wit da answer
bazaltina [42]

slope is 2/3 and y intercept is 7

5 0
3 years ago
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