I Believe it is 80% but I am not sure
Six nights.
1. Saturday night
2. Sunday night
3. Monday night
4. Tuesday night
5. Wednesday night
6. Thursday night
Answer:
(3, - 3)
Step-by-step explanation: that might be it
Answer:
The correct option is;
A. A number line that goes from 95 to 103. A closed circle is at 98.6. Everything to the right is shaded.
Step-by-step explanation:
To represent x greater-than-or-equal-to 98.6 on the number line we note that an open circle stands for either a less than or a greater than, while an open circle stands for either less than or equal to or greater than or equal to
The correct way to represent x greater than or equal to 98.6 on the number line, we indicate the point 98.6 with a closed circle and we shade everything in the region of the solution which is to the right of the closed circle
Therefore the correct option is A. A number line that goes from 95 to 103. A closed circle at is at 98.6. Everyting to the right is shaded.
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.

(a)To determine the probability of the system being down or running after any k hours, we determine the kth state matrix
.
(a)


If the system is initially running, the probability of the system being down in the next hour of operation is the 
The probability of the system being down in the next hour of operation = 0.16
(b)After two(periods) hours, the transition matrix is:

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)