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3 years ago

I will mark as BRAINLIEST

Mathematics

1 answer:

NARA [144]3 years ago

3 0

Answer:

the answer will be k^3

Step-by-step explanation:

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Identify the transformation that was applied to this boomerang.

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D. Since the dark part of the boomerang is on the different side, it had to have been rotated

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28 percent of 63 is what

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A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is

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The expected length of code for one encoded symbol is

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha$

where $p_\alpha$ is the probability of picking the letter $\alpha$ , and $\ell_\alpha$ is the length of code needed to encode $\alpha$ . $p_\alpha$ is given to us, and we have

$\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}$

so that we expect a contribution of

$\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375$

bits to the code per encoded letter. For a string of length $n$ , we would then expect $E[L]=1.375n$ .

By definition of variance, we have

$\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2$

For a string consisting of one letter, we have

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4$

so that the variance for the length such a string is

$\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859$

"squared" bits per encoded letter. For a string of length $n$ , we would get $\mathrm{Var}[L]=1.859n$ .

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3 years ago

Han and Andre were comparing the miles they ran during one week of cross country practice Hantan 12 which was three times as muc

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Answer:

I don't know.

Step-by-step explanation:

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3 years ago

Determine the value of x in the diagram

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Answer:

3x + x + 3x + x = 360

8x = 360

x = 45

3(45)= 135

Step-by-step explanation:

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3 years ago

I will mark as BRAINLIEST ​

I will mark as BRAINLIEST