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user100 [1]
3 years ago
10

Why is 24 a more convenient choice than 23 or 25

Mathematics
1 answer:
asambeis [7]3 years ago
6 0
Because it is easily divisible by 2, 3, 4, 6, 8, 12 and itself 
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Mel drives a bus 39 weeks in a year.
zlopas [31]

Answer:

11,622

Step-by-step explanation:

39 × 298 = 11,622

Have a good day!

7 0
4 years ago
Michael is doing an underwater handstand. His feet are sticking up 0.5 meters above the surface of the water Michael's hands are
Jlenok [28]
1.3 meters below the surface of the water
6 0
3 years ago
Read 2 more answers
The coffee shop owner explains that she will work 12 hours more than you do if you get the job. She will work 50 hours edch week
MrMuchimi

Answer:

The owner says that she will work 12 hours more than you if you get the job.

We know that she works 50 hours each week.

then, knowing that she works 50 hours per week and that she works 12 hours more than you, you have that, if H is the number of hours that you work per week, you have:

H + 12 = 50

Here says that the number of hours that you work, plus 12 hours, is equal to the number of hours that the owner works.

H = 50 - 12 = 38

This means that you would work 38 hours per week.

3 0
4 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
3 years ago
A square orange rug has a blue square in the center. The side length of the blue square is x inches. The width of the orange ban
Ivahew [28]

Answer:

(x+12)(x+12)-x^2

x^2 + 2(12)(x)+12^2-x^

=24x+144

Step-by-step explanation:

3 0
3 years ago
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