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lutik1710 [3]
3 years ago
8

Determine the value for b assuming that the two horizontal lines displayed are parallel.

Mathematics
1 answer:
Neko [114]3 years ago
8 0

Answer:

29

Step-by-step explanation:

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Hey can anyone pls help me out in dis real quick!!!!!!!!!!
Kazeer [188]

Answer:

It looks the same but it can be smaller or big and they both look the same shape.

Step-by-step explanation:

4 0
3 years ago
6) Solve for x. 2/3 (x + 7) = 10 <br><br> A)22 <br><br> B)8 <br><br> C)- 1 /3 <br><br> D)41/ 3
Amiraneli [1.4K]

For this case we must find the value of the variable "x" of the following equation:

\frac {2} {3} (x + 7) = 10

We multiply by 3 on both sides of the equation:

2 (x + 7) = 30

We divide between 2 on both sides of the equation:

(x + 7) = 15\\x + 7 = 15

We subtract 7 on both sides of the equation:

x = 8

Answer:

Option B

7 0
2 years ago
The scatter plot shows the number of cars and trucks sold by 10 different employees at a car and truck dealership during a month
Hitman42 [59]
1 employee
I think....
3 0
2 years ago
Read 2 more answers
Please I need answer fast will give brainliest
Sonbull [250]
Point slope is y-y1=m(x-x1)
Keep y and x as they are (do not plug anything in for those).
Variable m is the slope.
x1 and y1 are part of the point, which is in (x, y) form.
With the information given, plug each into the equation.
m= 2 (It's the slope)
x1=-8
y1=5
Put these variables into the equation like this-
y-5=2(x--8)
The two negatives cancel out (between x and 8) so your final answer is-
y-5=2(x+8)
6 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
2 years ago
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