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Daniel [21]
3 years ago
13

Need number 6.) With answer and graphing chart included!

Mathematics
1 answer:
Aloiza [94]3 years ago
6 0
#4 is B, #5 is A, and #6 is C
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(4xy^2+6y)dx+(5yx^2+8x)dy=0<br> solve by using integrating factor
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Are u sure about the second arrow?BTW
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4 years ago
An overdraft is a check that cannot be covered by funds in the account.
Degger [83]

Answer:

True

Step-by-step explanation:

7 0
4 years ago
I need help again :/
Kobotan [32]

Answer:

Hola

Step-by-step explanation:

Well if this is based off the first thing I answered, then 1. C

as I told you before, every time the x axis values go up one, the y values go up by x2 (aka times 2). Therefore it’s C.

2. It is linear because the unit change (amount it goes up aka slope) is the same since it’s alway x2. Hence the line on a graph would look linear

8 0
3 years ago
Please help ASAP! Only one question!!
levacccp [35]
It is negative because the plot is going down not up 
5 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
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