Answer:
The true solution is x=4/9
EXPLANATION
The logarithmic equation given to us is
We need to use the quotient rule of logarithms.
When we apply this law the expression becomes
We now take the antilogarithm of both sides to get
We square both sides to get,
We evaluate to obtain,
This simplifies to
We divide both sides by 36 to get
We simplify to get,
Its parallel so they are equal
7x-40=5x+10
2x=50
x=25
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)
<span>First, we write an equation to represent that the fencing lengths add up to 568 feet. we call the side of the fence that has three segments of its length x and the side with only two segments y. We write 3x + 2y = 568. We also know that the area of the rectangle is equal to xy, so area = xy. We put y in terms of x using our first equation and find that y = (568 - 3x)/2. We plug this into our area equation and find that area = (568x - 3x^2)/2. To find the maximum we set the derivative equal to 0 and end up with 0 = 284 - 3x. We solve for x and get 94 and 2/3. We then put that into our first equation to find y = 142. So the dimensions that maximize the area are 94 2/3 x 142.</span>
