Answer:
The stationary matrix is:
S = [0.2791, 0.7209]
Step-by-step explanation:
The transition matrix, <em>P</em> is:
![P=\left[\begin{array}{cc}0.38&0.62\\0.24&0.76\end{array}\right]](https://tex.z-dn.net/?f=P%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.38%260.62%5C%5C0.24%260.76%5Cend%7Barray%7D%5Cright%5D)
The stationary matrix S for the transition matrix P would be obtained by computing <em>k</em> powers of <em>P</em> until all the two rows of <em>P</em> are identical.
Compute P² as follows:
![P^{2}=\left[\begin{array}{cc}0.38&0.62\\0.24&0.76\end{array}\right]\times \left[\begin{array}{cc}0.38&0.62\\0.24&0.76\end{array}\right]](https://tex.z-dn.net/?f=P%5E%7B2%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.38%260.62%5C%5C0.24%260.76%5Cend%7Barray%7D%5Cright%5D%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.38%260.62%5C%5C0.24%260.76%5Cend%7Barray%7D%5Cright%5D)
![=\left[\begin{array}{cc}0.2932&0.7068\\0.2736&0.7264\end{array}\right]](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.2932%260.7068%5C%5C0.2736%260.7264%5Cend%7Barray%7D%5Cright%5D)
Compute P³ as follows:
![P^{3}=P^{2}\times P](https://tex.z-dn.net/?f=P%5E%7B3%7D%3DP%5E%7B2%7D%5Ctimes%20P)
![=\left[\begin{array}{cc}0.2932&0.7068\\0.2736&0.7264\end{array}\right]\times \left[\begin{array}{cc}0.38&0.62\\0.24&0.76\end{array}\right]\\\\=\left[\begin{array}{cc}0.2810&0.7190\\0.2783&0.7217\end{array}\right]](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.2932%260.7068%5C%5C0.2736%260.7264%5Cend%7Barray%7D%5Cright%5D%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.38%260.62%5C%5C0.24%260.76%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.2810%260.7190%5C%5C0.2783%260.7217%5Cend%7Barray%7D%5Cright%5D)
Compute P⁴ as follows:
![P^{4}=P^{3}\times P](https://tex.z-dn.net/?f=P%5E%7B4%7D%3DP%5E%7B3%7D%5Ctimes%20P)
![=\left[\begin{array}{cc}0.2810&0.7190\\0.2783&0.7217\end{array}\right]\times \left[\begin{array}{cc}0.38&0.62\\0.24&0.76\end{array}\right]\\\\=\left[\begin{array}{cc}0.2793&0.7207\\0.2790&0.7210\end{array}\right]](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.2810%260.7190%5C%5C0.2783%260.7217%5Cend%7Barray%7D%5Cright%5D%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.38%260.62%5C%5C0.24%260.76%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.2793%260.7207%5C%5C0.2790%260.7210%5Cend%7Barray%7D%5Cright%5D)
Compute P⁵ as follows:
![P^{5}=P^{4}\times P](https://tex.z-dn.net/?f=P%5E%7B5%7D%3DP%5E%7B4%7D%5Ctimes%20P)
![=\left[\begin{array}{cc}0.2793&0.7207\\0.2790&0.7210\end{array}\right]\times \left[\begin{array}{cc}0.38&0.62\\0.24&0.76\end{array}\right]\\\\=\left[\begin{array}{cc}0.2791&0.7209\\0.2791&0.7209\end{array}\right]](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.2793%260.7207%5C%5C0.2790%260.7210%5Cend%7Barray%7D%5Cright%5D%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.38%260.62%5C%5C0.24%260.76%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.2791%260.7209%5C%5C0.2791%260.7209%5Cend%7Barray%7D%5Cright%5D)
For <em>k</em> = 5, we get both the rows identical.
The stationary matrix is:
S = [0.2791, 0.7209]
Answer:
Step-by-step explanation:
Uui
7m=t is the correct answer and you can check by plugging 1 into m and you will get 7 for t.
Given:
The vertices of the rectangle ABCD are A(0,1), B(2,4), C(6,0), D(4,-3).
To find:
The area of the rectangle.
Solution:
Distance formula:
![D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=D%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
Using the distance formula, we get
![AB=\sqrt{(2-0)^2+(4-1)^2}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%282-0%29%5E2%2B%284-1%29%5E2%7D)
![AB=\sqrt{(2)^2+(3)^2}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%282%29%5E2%2B%283%29%5E2%7D)
![AB=\sqrt{4+9}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B4%2B9%7D)
![AB=\sqrt{13}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B13%7D)
Similarly,
![BC=\sqrt{(6-2)^2+(0-4)^2}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B%286-2%29%5E2%2B%280-4%29%5E2%7D)
![BC=\sqrt{(4)^2+(-4)^2}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B%284%29%5E2%2B%28-4%29%5E2%7D)
![BC=\sqrt{16+16}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B16%2B16%7D)
![BC=\sqrt{32}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B32%7D)
![BC=4\sqrt{2}](https://tex.z-dn.net/?f=BC%3D4%5Csqrt%7B2%7D)
Now, the length of the rectangle is
and the width of the rectangle is
. So, the area of the rectangle is:
![A=length \times width](https://tex.z-dn.net/?f=A%3Dlength%20%5Ctimes%20width)
![A=\sqrt{13}\times 4\sqrt{2}](https://tex.z-dn.net/?f=A%3D%5Csqrt%7B13%7D%5Ctimes%204%5Csqrt%7B2%7D)
![A=4\sqrt{26}](https://tex.z-dn.net/?f=A%3D4%5Csqrt%7B26%7D)
![A\approx 20](https://tex.z-dn.net/?f=A%5Capprox%2020)
Therefore, the area of the rectangle is 20 square units.
Answer:
slope= 0.2; intercept=-1; slope-interception form: y=0.2x-1.
Step-by-step explanation:
1) if to re-write the given equation, then
![y=\frac{1}{5}x-1;](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B5%7Dx-1%3B)
or y=0.2x-1;
this equation is slope-interception form;
2) according to the new view of the equation,
the required slope is 1/5;
the required y-intercept is -1.