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:
i think photomath will help you with those
Step-by-step explanation:
go to your appstore and download photomath
A because the length is greater
Answer:
<u>If we remove 61 from the data set, the median changes from 87.5 to 93.</u>
Step-by-step explanation:
1. Let's calculate the median of the original data set:
Median = (3rd term + 4th term)/2 because the number of terms are even and our median mark is the average of the two middle marks, in this case, 82 and 93.
Median = (82 + 93)/2
Median = 87.5
2. Let's calculate the median of the data set removing 61:
Median = 3rd term because our median mark is the middle mark, in this case, 93. It is the middle mark because there are 2 scores before it (80 and 82) and 2 scores (94 and 98) after it.
Median = 93
Answer:
7
Step-by-step explanation:
The left hand limit is when we approach zero from left. We use the function on this domain in finding the limit.
The right hand limit is
Since the left hand limit equals the right hand limit;
