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2 years ago

Several students made comments about this image.

Mathematics

1 answer:

yawa3891 [41]2 years ago

3 0

Answer:

The third comment is the true one: "If I know the measures of angles C and B, I can find the measures of A and D."

This is because you can use the angle B to find angle A by subtracting it from 180, and you can find angle D by adding together C and B and subtracting that from 180.

I hope this helps!

Step-by-step explanation:

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3 years ago

Approximate the stationary matrix S for the transition matrix P by computing powers of the transition matrix P.

Scrat [10]

Answer:

S = [0.2069,0.7931]

Step-by-step explanation:

Transition Matrix:

$P=\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

Stationary matrix S for the transition matrix P is obtained by computing powers of the transition matrix P ( k powers ) until all the two rows of transition matrix p are equal or identical.

Transition matrix P raised to the power 2 (at k = 2)

$P^{2} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{2} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right]$

Transition matrix P raised to the power 3 (at k = 3)

$P^{3} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{3} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{3} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right]$

Transition matrix P raised to the power 4 (at k = 4)

$P^{4} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]$

Transition matrix P raised to the power 5 (at k = 5)

$P^{5} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{5} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{5} =\left[\begin{array}{ccc}0.2069&0.7931\\0.2069&0.7931\end{array}\right]$

P⁵ at k = 5 both the rows identical. Hence the stationary matrix S is:

S = [ 0.2069 , 0.7931 ]

6 0

3 years ago