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hoa [83]
2 years ago
6

This one is hard can some body help me

Mathematics
1 answer:
natali 33 [55]2 years ago
5 0
First option because 3 x 10^9 is 3,000,000,000
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Simplify the expression. Write the answer using scientific notation. (5 x 102)−2
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102 multiply 5 is equally to 610 so that's gonna be 6.10 I think
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The answer is 79.56.
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Custumers of a phone company can choose between two service plans for long distance calls. The first plan has a $9 monthly fee a
saul85 [17]
The answer is 350.
<em>
m= minutes</em>

Plan 1:  9+0.13<em>m</em>
Plan 2:  23+0.09<em>m</em>

9+0.13<em>m</em> = 23+0.09<em>m</em>
-9               -9
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3 years ago
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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
Complete the following sentence.
coldgirl [10]

Answer:arc lengths; radius

l guzz.. sry if wrong....

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