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2 years ago

If M and m represents the number of sexagesimal and centesimal minute of any angle respectively, prove that M/27=m/50.

Mathematics

1 answer:

riadik2000 [5.3K]2 years ago

7 0

Answer:

if ma and m id black life matters if balck

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A child's toy is in the shape of a cone on top of a

sammy [17]

Answer:

1. 167.55

2. 10.77

Step-by-step explanation:

For #1 use the formula for volume of cone: $V=\pi r^2\frac{h}{3}$ , and for #2 use the Pythagorean Theorem: $a^2+b^2=c^2$

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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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2 years ago

1.The data set shows the October 1 noon temperatures in degrees Fahrenheit for a particular city in each of the past 6 years.

Paladinen [302]

Did you figure it out?

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3 years ago

Will mark brainliest!!! Two trains leave towns 663 kilometers apart at the same time and travel toward each other. One train tra

topjm [15]

Answer: rate of train A = 120 km/h

rate of train B = 101 km/h

Step-by-step explanation:

Let x = the speed of train A

Let y = the speed of train B

One train travels 19 km/h slower than the other. Let train B be the slower train. Therefore,

x = y + 19 - - - - - - - - - -1

Recall

Speed = distance/time

Distance = speed × time

If they meet in 3 hours, it means that train A has travelled a distance of

x × 3 = 3x km

And train B has travelled a distance of y × 3 = 3y km

Train A and Train B were 663 kilometers apart when they left their respective destinations. They left at the same time and travelled towards each other. This means that if they meet in 3hours time, they must have covered a total of 663 km. Therefore,

3x + 3y = 663 - - - - - - - - - 2

Substituting x = y + 19 into equation 2, it becomes

3(y + 19) + 3y = 663

3y + 57 + 3y = 663

6y + 57 = 663

6y = 663 - 57 = 606

y = 606/6 = 101 km/h

x = y + 19 = 101 + 19

x = 120 km/h

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3 years ago

PLEASE HELP ME!!!!!!!!!!

Svetlanka [38]

V = r + at (solve for a).

You want to isolate a. First subtract r.

v - r = at

Divide by t.

(v - r)/t = a

The answer is a = $\frac{v - r}{t}$ .

6 0

2 years ago

If M and m represents the number of sexagesimal and centesimal minute of any angle respectively, prove that M/27=m/50.​

If M and m represents the number of sexagesimal and centesimal minute of any angle respectively, prove that M/27=m/50.