All categories

3 years ago

Hey guy pls help explain answer no links

Mathematics

1 answer:

laiz [17]3 years ago

8 0

Answer:

4, 5

Step-by-step explanation:

the mode is 4 since mode is the number/s that appear most. Median is the number in the middle when the range is in order. That is 5

You might be interested in

What interval notation represents the data graphed below?

Aliun [14]

Answer: Choice D

$(-\infty, -2) \cup [4, \infty)$

=====================================================

Explanation:

The left portion is the interval (-∞, -2)

This is a shorthand way of saying $-\infty < x < -2$

The curved parenthesis says "do not include this endpoint as part of the solution set". Note the open hole at x = -2 in the diagram.

In contrast, the value x = 4 is included (due to the filled in circle), so we use a square bracket for this endpoint. Therefore, the right-hand portion is represented by [4, ∞) which translates to $4 \le x < \infty$

Negative and positive infinity will always use a parenthesis, and never a square bracket. This is because we can only approach infinity but never reach it, so we cannot include it as an endpoint.

All of this builds up to the full interval notation to be $(-\infty, -2) \cup [4, \infty)$

The only square bracket is near the 4; everything else is a curved parenthesis. This is why choice D is the final answer.

6 0

2 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 is 0.0008235 written in scientific notation

jeka57 [31]

Okay, move the decimal behind the 8. There are 4 places before the decimal so it is nagitive. If there are places Infront of the decimal is nagitive. So the answer is 8.235×10-4

8 0

3 years ago

Please, help me with my homework?

ivann1987 [24]

Answer:

yea what do you need help with?

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Write an absolute value equation or inequality to describe each graph

Rina8888 [55]