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erica [24]
3 years ago
5

Y equals 4x to the 3rd power minus 3 plus 2x to the second power

Mathematics
2 answers:
Nikolay [14]3 years ago
8 0
It equals 3,969x........,,,.
Marina CMI [18]3 years ago
7 0
Y=4x^3 - (3+2x)^2 hope this helped! I wrote it out in the picture

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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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Which of the following is a non-prime factor of 70? Select one: A. 5 B. 7 C. 10 D. 15
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Answer:

C. 10

Step-by-step explanation:

by deduction, 5 and 7 are prime numbers and 15 is not a multiple of 70, therefore the answer must be 10 as it is a non-prime number and is a factor of 70 (I.e. 10 and 7)

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Answer:

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Step-by-step explanation:

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