It's called the endpoint lol
R=S*0.5^(t/8)
<span>R is the remaining amount </span>
<span>S is the starting amount (500) </span>
<span>0.5^ is for the HALF in half-life </span>
<span>t/8 show that every 8 ts (every 8 hours), it will be halved once </span>
<span>...so plug in 500mg for the general solution... </span>
<span>R=(500)*(0.5)^(t/8) </span>
<span>... plug in 24h to solve for after 24h </span>
<span>R=(500)*(0.5)^(24/8) </span>
<span>R=(500)*(0.5)^(3) </span>
<span>R=(500)*(0.125) </span>
<span>R=(0.0625) </span>
<span>...therefore there with be 0.0625 mg of the dose remaining</span>
The sporting goods store receives the shipment of 124 golf bags in total.
Let us assume that the number of full-sized bags = X
Then the number of collapsible golf bags = 124 - X
Cost of each full sized bags = $38.50
Cost of each collapsible bags = $22.50
Total cost of the 124 bags received by the sporting goods store = $3430
Then
(38.50)X + (124 - X)22.50 = 3430
(38.50X - 22.50X) + 2790 = 3430
16X = 3430 - 2790
16X = 640
X = 640/16
= 40
So the number of full-sized golf bags received = 40
Then the number of collapsible golf bags received = 124 - 40
= 84
Answer:
(a)0.16
(b)0.588
(c)![[s_1$ 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)