What is 88 over 55 in long division

Mathematics

1 answer:

8090 [49]4 years ago

8 0

Dividing 55 into 88

= 1 /33/55 = 1 3/5 or 1.6

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

5 0

3 years ago

a box contains 5 plain pencils and 5 pens. a second box contains 7 color pencils and 5 crayons. one item from each box is chosen

Ket [755]

First box: $\dfrac{pen}{total}=\dfrac{5}{10}=\dfrac{1}{2}$

Second box: $\dfrac{crayon}{total}=\dfrac{5}{12}$

Multiply them:

$\dfrac{1}{2}*\dfrac{5}{12}=\dfrac{5}{24}$

The probabilityis $\dfrac{5}{24}$