1:1 is a square ratio rather than horizontally long rectangle, which would be similar to 720x1080
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:
See below
Step-by-step explanation:
<em>On the graphs we see transformations of exponential functions</em>
<h3>Graphic 1 = Horizontal shift </h3>
- f(x) = 2ˣ is the parent function
- g(x) = 2ˣ⁺³ indicates shift to the left by 3 units
- h(x) = 2ˣ⁻¹ indicates the shift to the right by 1 unit
<h3>Graphic 2 =Vertical shift</h3>
- p(x) = (1/3)ˣ is the parent function
- r(x) = (1/3)ˣ⁺³ indicates shift up by 3 units
- q(x) = (1/3)ˣ⁻² indicates the shift down by 2 units
Answer:
look in the explanation, it is in there
Step-by-step explanation:
i don't know if you are distributing or just using PEMDAS .
if you are distributing then: do in the equation 4(r + 1) + 3r.. do 4 x r and 4 x 1 , now you have 4r + 4 + 3r
7r +4
is what i got!
If you are doing PEMDAS then: first you do what is inside the perenthisis ()
4(r + 1) +3r changes into
4( r1 ) + 3r now you do multipulcation because that in the "M" in pemdas
4r + 3r, then if you are using like terms you would add them
7r
i hope this helps you out!
