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3 years ago

Plz help me with this

Mathematics

2 answers:

Makovka662 [10]3 years ago

7 0

Answer:

117° I believe.

Step-by-step explanation:

It would make sense. As it would equal the same as the other side

Olegator [25]3 years ago

7 0

Answer:

117° (Degrees)

Step-by-step explanation:

If you look at it like two triangles, you will see that the two triangles are equal and they are opposite so the answer would be 117° (Degrees)

Hope this helped :3

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What is the check digits (d) for the master card number 5130 7201 2608 209d A)1 B)3 C)5 D)7

LenKa [72]

I hope this is just a maths problem coz if this is your card number I wouldn't be putting it on here for all to see.

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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

Please help! Simplify. Your answer should contain only positive exponents.

maxonik [38]

Answer:

$\frac{1}{4k^3}$

Step-by-step explanation:

It's already in a form that only ha positive exponents. Just to state it though you could rewrite it as $k*(4k^{4})^{-1}$ or $k*4^{-1}*k^{-4}$ The question asks for positive exponents though so yu don't need to do that.

You also need to know $\frac{x^a}{x^b}=x^{a-b}$ so in this case $\frac{k}{k^4}=\frac{k^1}{k^4} =k^{1-4}=k^{-3}=\frac{1}{k^3}$

so $\frac{k}{4k^4}=\frac{1}{4}*\frac{k}{k^4}=\frac{1}{4}*\frac{1}{k^3}=\frac{1}{4k^3}$

3 0

3 years ago

What is the common denominator for 1 2/3 and 3/4

Elena L [17]

Answer:

The answer is 12

Step-by-step explanation:

Multiply 1*3

Add 2 to the number

Change the first number to 5/3

Multiply 3*4

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3 years ago

Rotation 180 degrees about the origin

WITCHER [35]

Answer:

Rule for rotation 180° about the origin:

(x, y) → (-x, -y)

Given coordinates:

U(1, 0), V(4, -1), T(1, -3)

Coordinates of the image, apply the rule above:

U'(-1, 0), V'(-4, 1), T'(-1, 3)

4 0

3 years ago