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]](https://tex.z-dn.net/?f=P%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=P%5E%7B2%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D%20X%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D)
![P^{2} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right]](https://tex.z-dn.net/?f=P%5E%7B2%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.2203%260.7797%5C%5C0.2034%260.7966%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=P%5E%7B3%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D%20X%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5DX%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D)
![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]](https://tex.z-dn.net/?f=P%5E%7B3%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.2203%260.7797%5C%5C0.2034%260.7966%5Cend%7Barray%7D%5Cright%5D%20X%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D)
![P^{3} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right]](https://tex.z-dn.net/?f=P%5E%7B3%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.2086%260.7914%5C%5C0.2064%260.7936%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=P%5E%7B4%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D%20X%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5DX%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5DX%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D)
![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]](https://tex.z-dn.net/?f=P%5E%7B4%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.2086%260.7914%5C%5C0.2064%260.7936%5Cend%7Barray%7D%5Cright%5D%20X%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D)
![P^{4} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]](https://tex.z-dn.net/?f=P%5E%7B4%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.2071%260.7929%5C%5C0.2068%260.7932%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=P%5E%7B5%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D%20X%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5DX%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5DX%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5DX%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D)
![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]](https://tex.z-dn.net/?f=P%5E%7B5%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.2071%260.7929%5C%5C0.2068%260.7932%5Cend%7Barray%7D%5Cright%5D%20X%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.31%260.69%5C%5C0.18%260.82%5Cend%7Barray%7D%5Cright%5D)
![P^{5} =\left[\begin{array}{ccc}0.2069&0.7931\\0.2069&0.7931\end{array}\right]](https://tex.z-dn.net/?f=P%5E%7B5%7D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.2069%260.7931%5C%5C0.2069%260.7931%5Cend%7Barray%7D%5Cright%5D)
P⁵ at k = 5 both the rows identical. Hence the stationary matrix S is:
S = [ 0.2069 , 0.7931 ]
Answer:
its easy bro
Step-by-step explanation:
the answer is
x=125
y=68
pls mark brainliest
Answer:
4/13
Step-by-step explanation:
NOT 2010s and NOT spruce wood (4) over total (13)
4/13 is in its reduced form
I was gonna ask for a new bike for a week and he was going camping and then after he got a bike he was a bike and a half bike bike and bike bike he was riding bike ride bike and he got a
I’ll teach you how to solve 3.26d+9.75d-2.65
——————————————-
3.26d+9.75d-2.65
Collect like terms by adding their coefficients:
(3.26+9.75)d-2.65
Add the numbers:
3.26+9.75= 13.01
13.01d-2.65
Your Answer Is 13.01d-2.65
Plz mark me as brainliest if this helped :)
