4 years ago

Which expressions are equivalent to the expression...

Mathematics

1 answer:

Ulleksa [173]4 years ago

3 0

Most likely C............... yea C

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A triangle has side lengths of 34in. 28in. and 42in. Is the triangle acute, right or obtuse?

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I believe the answer is acute. Hope this helps!!

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

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3 years ago

G(t)= t-3 <br> h(x)= 3t-2 <br> find (g-h)(t)

Marina86 [1]

Answer:

g(t) - h(t) = -2t - 1

Step-by-step explanation:

g(t) - h(t) = t-3-3t+2

= -2t -1

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4 years ago

What percent of 95 is 60.8

vampirchik [111]

Answer:

57.76 :)

Step-by-step explanation:

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3 years ago

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BaLLatris [955]

Y^2 my teach taught me this last year

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3 years ago