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nordsb [41]
3 years ago
11

Find the length of an arc made by an 80° central angle in a circle with a 10 ft. radius.

Mathematics
2 answers:
docker41 [41]3 years ago
4 0

I'm not an expert at this but the answer should be 13.9 or 14

valentina_108 [34]3 years ago
4 0

Answer:

6.9778

Step-by-step explanation:

we use

theta/360 × 2pier

I.e,

theta=80

r=10

80/360 × 2× 3.14 ×10

62.8/9

6.9778

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Gillian eats 12 skittles. If there was originally 25 skittles in the bag, what percent of the skittles did Gillian eat? (NOT A S
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4 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
2 years ago
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