Ask question

Ask question

All categories

3 years ago

6

PLEASE HELP ME DO THIS

1 answer:

gladu [14]3 years ago

8 0

Answer:

Perimeter = 8x

Step-by-step explanation:

If I can see correctly, the area of the squares are x². Area is just s², so we can find one side by finding the square root of x², which is x. Now that we know that x is one side, we can count the amount of sides and multiply that number by x.

There are 8 sides, so the answer is P = 8x

You might be interested in

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

In a triangle, the midsegment is ___________ to and ____________ of the non-intersecting side.

erastova [34]

In a triangle, the midsegment is parallel to and half of non-intersecting side.

7 0

3 years ago

Read 2 more answers

Please answer this question for me

makkiz [27]

I think the answer is par

4 0

3 years ago

15 POINTS! PLEASE HELP! BRAINLIEST!

lina2011 [118]

Answer:

D. 15.3%

Step-by-step explanation:

<u>Total number of outcomes:</u>

2¹⁵ = 32768

<u>Number of combinations of getting 6 heads:</u>

15C6 = 15!/6!(15-6)! = 5005

<u>Required probability is:</u>

P(6 heads out of 15 flips) = 5005/32768 = 0.1527... ≈ 15.3%

Correct choice is D

7 0

3 years ago

Read 2 more answers

Someone please help me will give BRAILIEST!!!!!!

asambeis [7]

The answer to this problem is 2/5

6 0

3 years ago

Read 2 more answers