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xxTIMURxx [149]
2 years ago
10

52.85 - 9.09 = what rounded to the nearest one

Mathematics
1 answer:
bonufazy [111]2 years ago
6 0
52.85 - 9.09 rounded to the nearest one is 44
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Coral is sometimes used to replace human bone
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PLSSS HURRY ILL GIVE YOU A BRAINLYEST
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Answer:

36 inches

Step-by-step explanation:

lets think of it as a ratio,

80:w

20:9

so we want 20 to get to 80

and 9 to get to w right?

so if we just move the smaller rectangle to the corner, then 20*4=80 right?

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2 years ago
14x=x^2-15<br><br> find the value of x
hram777 [196]

Answer: x=-1, 15

Step-by-step explanation:

x^2 -14x-15=0\\\\(x-15)(x+1)=0\\\\x=-1, 15

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1 year ago
A card is Chosen at random from a deck of 52 cards it is replaced and a second card is chosen what is the probability both cards
ankoles [38]
There are 4 aces in a deck of 52 cards. the odds of drawing one is 1/13. the probability of doing it again is still 1/13. the odds of doing it twice, as described, is then 1/13 * 1/13, or (1/13)², or 1/169 option A
6 0
3 years ago
Read 2 more answers
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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