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3 years ago

HelpI don’t understand graphs and I will mark brainliest for the correct answer

Mathematics

1 answer:

qwelly [4]3 years ago

5 0

Answer:

G(x)=x^3-1

Step-by-step explanation:

Надеюсь тебе это поможет

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The equation y=0.002x-0.20 can be used to determine the approximate profit,y in dollars,of producing x items.solve for x. How ma

lakkis [162]

Answer:

1966600 items must be produced in other to profit $3933

Step-by-step explanation:

from the equation y = 0.002x -0.20

where y is the profit in dollars

and x is the number of items

then to get the number of item to be produced in other to profit $3933 ?

will be by substituting $3933 for y in the equation and solving for x,

3933 = 0.002x - 0.20

Firstly you will add 0.20 to both sides, which will be

3933 + 0.20 = 0.002x - 0.20 + 0.20

3933.20 = 0.002x

then we will divide both sides by 0.002

3933.20 / 0.002 = 0.002x / 0.002

therefore, x = 1966600

5 0

3 years ago

Someone pls help me i do not understand !!

pantera1 [17]

Yea me ether did you ever find the answer

3 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

Find a irrational number between 5.2 and 5.3 .. Explain why its irrational. Include the decimal approximation of thr nearest hun

mihalych1998 [28]

$\sqrt{28}$
It is in between
$\sqrt{25}$
And
$\sqrt{36}$
There for it is in between 5 and 6
Additionally, it is closer to root 25, meaning that it is closer to 5

Rounded to the nearest hundredth is

5.29

4 0

3 years ago

Find x of the triangle

sukhopar [10]

Answer:

Step-by-step explanation:

altitude = sqrt(6*8) = 4*sqrt(3)

Use the Pythagorean Theorem to derive x

altitude^2 + 18^2 = x^2

48 + 18^2 = x^2

x^2 = 342

x = 3*sqrt(38)

x = 18.49

5 0

2 years ago