Answer:
Option C is the correct answer.
Step-by-step explanation:
Since a random number generator is used to select a single number between 1 and 28, inclusively;
For a fair decision to be made during the process, the number of persons in the set must be a divisor of 28.
Let's list the divisors of 28:
The divisors of 28 are {1, 2, 4, 7, 14, 28}
The only option that contains one of the divisors is option C which is 7.
<span>P= IRT solve for T
To solve for T, you must get T on it's own on one side of the equals sign, with all the other terms on the other side.
IRT means that I and R are multiplied by T.
To get rid of them we must do the opposite of multiply, which is divide.
But because this is an equation, anything we do on one side of the equals sign, must be repeated on the other side.
So if we divide the right side by IR, we must also divide the left side by IR</span>
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)
Answer:
(f+g)(x)=5x²-4x+3
(f-g)(x)=3x²-2x+3
(fg)(x)

Step-by-step explanation:
Given that,
f(x)=4x²-3x
g(x)=x²-x+3
(f+g)(x)
=f(x)+g(x)
=4x²-3x+x²-x+3
=(4x²+x²)+(-3x-x)+3 [ combined the like terms]
=5x²-4x+3
(f-g)(x)
=f(x)-g(x)
=4x²-3x-(x²-x+3)
=4x²-3x-x²+x-3
=(4x²-x²)+(-3x+x)-3 [ combined the like terms]
=3x²-2x+3
(fg)(x)
=f(x).g(x)
=(4x²-3x).(x²-x+3)
=4x²(x²-x+3)-3x(x²-x+3)





