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Lorico [155]
2 years ago
9

Need some help here

Mathematics
1 answer:
Natalija [7]2 years ago
5 0

Answer:

bottom right

Step-by-step explanation:

it goes down consitantly by 2

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Help me please. I don’t understand this math. My math teacher is mean. :(((((
Verdich [7]

Answer:

The height of the given rectangular prism is 5cm.

Step-by-step explanation:

Given that the volume of the rectangular prism is 5cm^3

And length is \frac{3}{4} and width is 1\frac{1}{3}

That is V=5cm^3[/tex] , l=\frac{3}{4}cm and w=1\frac{1}{3}cm

We have the  volume of the rectangular prism V=lwh cubic centimeter

V=lwhcm^3

Substituting the values in above formula

5=\frac{3}{4}\times 1\frac{1}{3}\times h

5=\frac{3}{4}\times \frac{4}{3}\times h

5=1\times h

5=h

Rewritting as below

h=5cm

Therefore the height is 5cm

5 0
3 years ago
Read 2 more answers
Trinity collects rainwater in a cylinder rain barrel. The area of the base of the
Fittoniya [83]

Answer:

suggestions use algebra calculator or mathaway they will help u and give you the answer

8 0
2 years ago
HELP ASAP PLZ HELP<br> I will give a brain list if it is right !!! <br> just help me plz
lilavasa [31]
Answer: A

Explanation:
First you have to write the model in an equation form.

6x+5=15

Subtract 5 from both sides

6x=10

Divide 6 from both sides

X=10/6

Simplify

X=5/3 or 1.67
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2 years ago
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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
2 years ago
Please help me solve for x =
Sladkaya [172]

Answer:

x = 13

Step-by-step explanation:

Given that Δ NML and Δ PST are similar right triangles, we can set up the following proportional statement to establish their relationship:

\frac{ML}{NM} = \frac{ST}{PS}

\frac{8}{10} = \frac{x - 1}{x + 2}

Cross multiply:

8(x + 2) = 10 (x - 1)

8x + 16 = 10x - 10

Subtract 8x from both sides:

8x - 8x + 16 = 10x - 8x - 10

16 = 2x - 10

Add 10 to both sides:

16 + 10 = 2x - 10 + 10

26 = 2x

Divide both sides by 2:

\frac{26}{2} = \frac{2x}{2}

13 = x

Verify whether x = 13 is the correct value:

\frac{ML}{NM} = \frac{ST}{PS}

\frac{8}{10} = \frac{x - 1}{x + 2}

\frac{8/2}{10/2} = \frac{4}{5}

\frac{x -1}{x+2} =  \frac{13 - 1}{13 + 2} = \frac{12}{15} = \frac{12 / 3}{15 /3} = \frac{4}{5}

This shows the proportional relationship between \frac{ML}{NM} = \frac{ST}{PS}, and that ΔNML and ΔPST are indeed similar right triangles.

Therefore, the correct answer is x = 13.

4 0
3 years ago
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