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3 years ago

4 , 2 9 5

Mathematics

1 answer:

Bumek [7]3 years ago

5 0

4,295+___=8,329

Answer:4034

Hope this helps, plz make brainly-est!!!

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Please I need help ASAP

artcher [175]

This is an exponential table

4 0

3 years ago

Kevin, Michael and Darren all do their own laundry at home. Kevin does his laundry every 9 days, Michael every 10 days, and Darr

VladimirAG [237]

Answer:

In 180 days the three of them will do their laundry again on the same day.

Step-by-step explanation:

we need to find the least common factor or 9, 10 and 12:

LCM(9, 10, 12) = 180

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3 years ago

Read 2 more answers

In the diagram below of triangle ABC, CD is the bisector of angle BCA, AE is the bisector of angle CAB and BG is drawn.

pashok25 [27]

There are two solutions
AG =BG
mesDBG = mesEBG

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3 years ago

Select the correct answer.<br><br>which function is continuous across its domain?

SSSSS [86.1K]

Answer:

Step-by-step explanation:

3 0

3 years ago

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 $\mathrm{Var}[L]=1.859n$ .

5 0

3 years ago