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3 years ago

11

What is the value of (7 + 2i)(3 - i)?

1 answer:

Harrizon [31]3 years ago

7 0

There are two ways to work this out: normal variables or using "imaginary" numbers.

Normal variables:
$(7+2i)(3-i)\\(7*3)+[7*(-i)]+(3*2i)+[2i*(-i)]\\21-7i+6i-2i^{2}\\\\21-i-2i^{2}$

Imaginary numbers:
Using the result from earlier:
$21-i-2i^{2}$
Now since $i = \sqrt{-1}$ , then the expression becomes:
$21-i-2i^{2}\\21-i-2(\sqrt{-1})^{2}\\21-i+(-2)(-1)\\21-i+2\\\\23-i$

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Andrej [43]

Step-by-step explanation:

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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

$\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859$

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2 years ago

In a two-player game, five cards, numbered 1 through 5, are placed in a bag. A card is drawn at random, and the players look at

IceJOKER [234]

Answer:

If it is less than 3, Player 1 earns 3 points.

If not, Player 2 earns 2 points.

Step-by-step explanation:

<u>Player 1</u> :

p(N < 3) = p(N = 1 or N = 2) = 2/5

<u>Player 2</u> :

p(N ≥ 3) = p(N = 3 or N = 4 or N = 5) = 3/5

<u>We notice that</u> :

p(N < 3) × 3 = (2/5) × 3 = 6/5

On the other hand,

p(N ≥ 3) × 2 = (3/5) × 2 = 6/5

since ,the probability player 1 win multiplied by the associated number of points (3)

is equal to

the probability player 2 win multiplied by the associated number of points (2).

Then the game is fair.

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3 years ago

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