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2 years ago

Which fraction equals the ratio of rise to run between the points (-3,-4) and (0, 0)?

Mathematics

2 answers:

Marizza181 [45]2 years ago

6 0

Answer:

4/3

Step-by-step explanation:

Aleksandr [31]2 years ago

5 0

Answer:

THe answer is C

Step-by-step explanation:

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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

koban [17]

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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2 years ago

For a magic trick , Ashley has a classmate pick one card from a standard deck of cards . what is the probability of Ashley 's cl

Ludmilka [50]

Answer:

1/4

Step-by-step explanation:

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Answer :
Positive

(Unless there is any information or problem on there)

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Let f(x)=x2+6x+11.<br><br> What is the average rate of change from x = 5 to x = 8?

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Plug 8 into the equation and divide it by that equation but with 5 plugged in

$\frac{ {8}^{2} + 6(8) + 11}{ {5}^{2} + 6(5) + 11 }$

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three, twelve, five, and six

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