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3 years ago

What are the primary colors?

Mathematics

2 answers:

Zigmanuir [339]3 years ago

7 0

Technically none, but if you have to chose red,blue yellow

lawyer [7]3 years ago

6 0

Red, yellow, and blue.

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Which of the following appear in the diagram below?

vladimir1956 [14]

Answer:

A , B, D

Step-by-step explanation:

YX is the segment joining points Y and X and is defined (A)

∠ XYZ is the angle between XY and YZ and is defined (B)

∠ YXZ is the angle between YX and XZ

However there is no segment joining XZ ← not defined

YW is the segment joining points Y and W and is defined (D)

7 0

2 years ago

Find a parametrization of the line in which the planes x + y + z = -6 and y + z = -8 intersect.

rewona [7]

Answer:

L(x,y) = (2,-8,0) + (0,-1,1)*t

Step-by-step explanation:

for the planes

x + y + z = -6 and y + z = -8

the intersection can be found subtracting the equation of the planes

x + y + z - ( y + z ) = -6 - (-8)

x= 2

therefore

x=2

z=z

y= -8 - z

using z as parameter t and the point (2,-8,0) as reference point , then

x= 2

y= -8 - t

z= 0 + t

another way of writing it is

L(x,y) = (2,-8,0) + (0,-1,1)*t

6 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

3 years ago

How many continents are there

9966 [12]

Answer:

(7) seven continents

By most standards, there is a maximum of seven continents - Africa, Antarctica, Asia, Australia/Oceania, Europe, North America, and South America

7 0

3 years ago

Read 2 more answers

How can I solve this?

kotegsom [21]

I completely forgot how to do this but maybe multiply the angles I’m so sorry

4 0

3 years ago