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Aleonysh [2.5K]
2 years ago
12

Find the number of paths of length 4 between two different vertices in K5​

Mathematics
1 answer:
Mkey [24]2 years ago
4 0

K₅ is the 5-complete graph, with adjacency matrix

A = \begin{bmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{bmatrix}

The (i, j)-th entry of the matrix A⁴ gives the number of length-4 paths from vertex i to vertex j. Computing A⁴ isn't so bad:

A^2 =  \begin{bmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{bmatrix}  \begin{bmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{bmatrix} =  \begin{bmatrix} 4 & 3 & 3 & 3 & 3 \\ 3 & 4 & 3 & 3 & 3 \\ 3 & 3 & 4 & 3 & 3 \\ 3 & 3 & 3 & 4 & 3 \\ 3 & 3 & 3 & 3 & 4 \end{bmatrix}

A^4 = \begin{bmatrix} 4 & 3 & 3 & 3 & 3 \\ 3 & 4 & 3 & 3 & 3 \\ 3 & 3 & 4 & 3 & 3 \\ 3 & 3 & 3 & 4 & 3 \\ 3 & 3 & 3 & 3 & 4 \end{bmatrix} \begin{bmatrix} 4 & 3 & 3 & 3 & 3 \\ 3 & 4 & 3 & 3 & 3 \\ 3 & 3 & 4 & 3 & 3 \\ 3 & 3 & 3 & 4 & 3 \\ 3 & 3 & 3 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 52 & 51 & 51 & 51 & 51 \\ 51 & 52 & 51 & 51 & 51 \\ 51 & 51 & 52 & 51 & 51 \\ 51 & 51 & 51 & 52 & 51 \\ 51 & 51 & 51 & 51 & 52 \end{bmatrix}

We want the paths between two distinct vertices, so we ignore the entries on the diagonal and take the total of the non-diagonal entries, 20 • 51 = 1020.

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(2x-10) °= °<br><br><br> Please help
Marizza181 [45]

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Step-by-step explanation:

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3 years ago
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A random sample of the costs of repair jobs at a large muffler repair shop produces a mean of $127.95. and a standard deviation
lubasha [3.4K]

Answer:

127.95-1.685\frac{24.03}{\sqrt{40}}=121.55    

127.95 +1.685\frac{24.03}{\sqrt{40}}=134.35    

So on this case the 90% confidence interval would be given by (121.55;134.35)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=127.95 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=24.03 represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=40-1=39

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,39)".And we see that t_{\alpha/2}=1.685

Now we have everything in order to replace into formula (1):

127.95-1.685\frac{24.03}{\sqrt{40}}=121.55    

127.95 +1.685\frac{24.03}{\sqrt{40}}=134.35    

So on this case the 90% confidence interval would be given by (121.55;134.35)    

8 0
2 years ago
Help me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ahrayia [7]

Answer:

9. x=-6, y=0

10. x=-9, y=-10

11. x=3, y=-1

Step-by-step explanation:

2x-9y=-12\\3x+18y=-18

Multiply the first equation by 2.

4x-18y=-24\\3x+18y=-18

Add the equations.

7x=-42

x=-6

We figured out x now we have to figure out y. To figure out y plug in x into any equation.

2(-6)-9y=-12

-12-9y=-12

-9y=0

y=0

Your answer is x = -6 and y = 0 for number nine.

6x+3y=24\\2x+3y=-12

Change the signs of the second equation becuase 3y and 3y are both positive. We need one of them to be negative.

6x+3y=24\\-2x-3y=12

Add them.

4x = 36

x=9

Now we have to figure out y.

2(9)+3y=-12

18+3y=-12

3y=-30

y=-10

x=-9 and y=-10 is your answer to number ten.

10x+5y=25\\-7x-3y=-18

30x+15y=75\\-35x-15y=-90

-5x=-15

x=3

-7(3)-3y=-18

-21-3y=-18

-3y=3

y=-1

x=3 and y=-1 is the answer to number eleven. I hope this helps! Let me know if I got anything wrong.

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