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pantera1 [17]
2 years ago
5

How do I solve numbers 8 and 9?

Mathematics
1 answer:
Tresset [83]2 years ago
7 0
8: 20
9: 90
I am so sorry if this is wrong. I am not good at this type of things
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Here are 2 right triangles.
Eva8 [605]

Answer

Part A: The right triangles are smiliar because If the lengths of the hypotenuse and a leg of a right triangle are proportional to the corresponding parts of another right triangle, then the triangles are similar.

Part B: 2z:2x or 2z/2x. You can write z/x as z:x. Then just multiply both sides by the same number to get an equivalent ratio. Let's multiply by 2. then the answer will be: 2z:2x or 2z/2x

Step-by-step explanation:

8 0
2 years ago
Solve for x <br> Need help on this
enyata [817]
Mid segment of triangle = 1/2 base
base = x here, so
1/2x = 15
multiply both sides by 2,
x=30

answer: x=30
5 0
2 years ago
What is 3x+4x-7=8-3x+11<br> need step by step explanation
Nadya [2.5K]

Answer:

exact form: 13/5

Decimal form: 2.6

x= 2 3/5

Step-by-step explanation:

3x+4x-7=8-3x+11

7x-7=8-3+11

7x-19-3x

7x-7 + 7 =-3x+19+7

7x=-3+26

10x=26

x= 13/5

6 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
2 years ago
In the problem 4 x 12 = 48 which numbers are the factors
Volgvan
C 4and 12 are the factors of 48
3 0
3 years ago
Read 2 more answers
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