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4 years ago

9

Compare 5.7145 and

ddle" class="latex-formula"> show your work

1 answer:

Lunna [17]4 years ago

4 0

To get the square root of 29, we can simply the calculator.

So the square root of 29 using the calculator will be;

sqrt 29 = 5.3852

So 5.7145 is greater than the square root of 29.

The difference will be 5.7145 – 5.3852 = 0.3293

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The perimeter of a playing field for a certain sport is 300ft. The field is a rectangle, and the length is 48ft longer than the

lara31 [8.8K]

Answer:

Width of the rectangular field = $102 feet$

Step-by-step explanation:

Perimeter of the field= $300feet$

Length of the field= $48 feet$

Width=?

Perimeter of a rectangular field= $2(Length+Width)$

$300= 2(48+Width)\\\\150=48+Width\\\\Width=150-48\\\\Width=102 feet$

The Width of the rectangular field is $102 feet$

4 0

3 years ago

Read 2 more answers

What is the answer too -6/7 x 3/8

IrinaK [193]

Answer:

-9/28

Step-by-step explanation:

<u>Step 1: Solve</u>

-6/7 * 3/8

-18/56

-18/2 / 56/2

<em>-9/28</em>

<em />

Answer: -9/28

6 0

3 years ago

In Miami the number of highway accidents increased by 20% over a four year period. How many accidents were there in 2013 if ther

wariber [46]

X/5120=20/100
x/5120=1/5
5120/5=1024
1024+5120=6144
6144 is the answer.

9 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

Can anyone help me with this ?

lorasvet [3.4K]

The square root of 24 simplifys to:

√24
2√6

Since √6 is inbetween √4 and √9, it is somewhere in between 2 and 3. Really, the best option is to use a calculator.√24 is approximately 4.9. <span />

4 0

4 years ago