Step-by-step explanation:
Given equation is
2y = 3x + 10
3x - 2y + 10 = 0 .....i)
Any line parallel to line I) is
3x - 2y + k = 0 ......ii)
As the line two passes through the point ( 2 , - 5 ) Now substituting the values
3 * 2 - 2 * ( - 5) + k = 0
6 + 10 + k = 0
16 + k = 0
k = - 16
Now putting the value of k in equation two
3x - 2y + 16 = 0 is the required equation.
Hope it will help :)❤
Answer:
14.36446281 is the answer, you probably have to round
Answer:
New area is 175 42/45 feet ²
Step-by-step explanation:
Given data
Length l= 15 1/4
To proper fraction = 61/4 feet
Width w= 8 11/15
To proper fraction = 131/15 feet
Extention = 1 2/3
To proper fraction = 5/3
Dimensions of new rectangle
Length =61/4+5/3 = 183+20/12
LCM = 12 = 203/12
Width = 131/15+5/3= 131+25/15
LCM = 15 = 156/15
Area = 203/12*156/15= =31668/180
=31668/180
= 7917/45
= 175 42/45 feet ²
Answer:
xit the problem
Step-by-step explanation:
Answer:
(a)0.16
(b)0.588
(c)![[s_1$ s_2]=[0.75,$ 0.25]](https://tex.z-dn.net/?f=%5Bs_1%24%20s_2%5D%3D%5B0.75%2C%24%20%200.25%5D)
Step-by-step explanation:
The matrix below shows the transition probabilities of the state of the system.
![\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7Ccc%7D%26%24Running%26%24Down%5C%5C---%26---%26---%5C%5C%24Running%260.90%260.10%5C%5C%24Down%260.30%260.70%5Cend%7Barray%7D%5Cright%29)
(a)To determine the probability of the system being down or running after any k hours, we determine the kth state matrix
.
(a)
![P^1=\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)](https://tex.z-dn.net/?f=P%5E1%3D%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7Ccc%7D%26%24Running%26%24Down%5C%5C---%26---%26---%5C%5C%24Running%260.90%260.10%5C%5C%24Down%260.30%260.70%5Cend%7Barray%7D%5Cright%29)
![P^2=\begin{pmatrix}0.84&0.16\\ 0.48&0.52\end{pmatrix}](https://tex.z-dn.net/?f=P%5E2%3D%5Cbegin%7Bpmatrix%7D0.84%260.16%5C%5C%200.48%260.52%5Cend%7Bpmatrix%7D)
If the system is initially running, the probability of the system being down in the next hour of operation is the ![(a_{12})th$ entry of the P^2$ matrix.](https://tex.z-dn.net/?f=%28a_%7B12%7D%29th%24%20entry%20of%20the%20P%5E2%24%20matrix.)
The probability of the system being down in the next hour of operation = 0.16
(b)After two(periods) hours, the transition matrix is:
![P^3=\begin{pmatrix}0.804&0.196\\ 0.588&0.412\end{pmatrix}](https://tex.z-dn.net/?f=P%5E3%3D%5Cbegin%7Bpmatrix%7D0.804%260.196%5C%5C%200.588%260.412%5Cend%7Bpmatrix%7D)
Therefore, the probability that a system initially in the down-state is running
is 0.588.
(c)The steady-state probability of a Markov Chain is a matrix S such that SP=S.
Since we have two states, ![S=[s_1$ s_2]](https://tex.z-dn.net/?f=S%3D%5Bs_1%24%20%20s_2%5D)
![[s_1$ s_2]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[s_1$ s_2]](https://tex.z-dn.net/?f=%5Bs_1%24%20%20s_2%5D%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D0.90%260.10%5C%5C0.30%260.70%5Cend%7Barray%7D%5Cright%29%3D%5Bs_1%24%20%20s_2%5D)
Using a calculator to raise matrix P to large numbers, we find that the value of
approaches [0.75 0.25]:
Furthermore,
![[0.75$ 0.25]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[0.75$ 0.25]](https://tex.z-dn.net/?f=%5B0.75%24%20%200.25%5D%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D0.90%260.10%5C%5C0.30%260.70%5Cend%7Barray%7D%5Cright%29%3D%5B0.75%24%20%200.25%5D)
The steady-state probabilities of the system being in the running state and in the down-state is therefore:
![[s_1$ s_2]=[0.75$ 0.25]](https://tex.z-dn.net/?f=%5Bs_1%24%20s_2%5D%3D%5B0.75%24%20%200.25%5D)