This problem is better understood with a given figure. Assuming
that the flight is in a perfect northwest direction such that the angle is 45°,
therefore I believe I have the correct figure to simulate the situation (see
attached).
Now we are asked to find for the value of the hypotenuse
(flight speed) given the angle and the side opposite to the angle. In this
case, we use the sin function:
sin θ = opposite side / hypotenuse
sin 45 = 68 miles per hr / flight
flight = 68 miles per hr / sin 45
<span>flight = 96.17 miles per hr</span>
Answer:
where are the expressions??
If Marcus can type 40 words in half a minute, then 40 times 2 equals 80. 80 words a minute for Marcus. If Rhys can type 100 words in 1 1/2 minute, 100/ 1.5 = 66.66 and 80>66.66 so Marcus can type a greater rate of words per minute.
1) Set up an equation for the total, with T representing Trevor and M representing Marissa:
T + M = 26
2) Use the provided information related Marissa's t-shirts to Trevor's to solve for one variable at a time (in this case, T):
M = T-6
3) Substitute M with this equation into the initial, first equation:
T + (T-6) = 26
4) Combine like terms and isolate T:
T+T = 26 + 6
2T = 32
T = 16
5) Therefore, you now know that Trevor has 16 t-shirts. To find the number of t-shirts that Marissa has, simply plug in this amount for T into the equation that was already set up for M:
M = T - 6
M = 16 - 6
M = 10
Therefore, Trevor has 16 t-shirts and Marissa has 10 t-shirts.
Hope this helps, and don't forget to mark my answer as the Brainliest! :)
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)