This question is incomplete because it lacks the diagram of the right angles shaped charging stand.
Find attached to this answer, the appropriate diagram.
Answer:
3(tan 50°)
Step-by-step explanation:
To solve this question, we would be using the Trigonometric function called Tangent.
tan θ = opposite side/ adjacent side
θ = 50°
Opposite side (Height)= Side AC = Unknown
Adjacent side = Side BC = 3 inches
Hence,
tan 50° = Side AC/ Side BC
tan 50° = Side AC / 3
Cross multiply
tan 50° × 3 = Side AC
3(tan 50°) = Side AC
Therefore, the expression shows the height, AC, of the charging stand is
3(tan 50°).
Answer:
Her answer should have been x=17/34
Step-by-step explanation:
Sum of n terms = a1 (r^n-1/ (r-1)
Sum of first one = 3 * (4^5 - 1) / 3 = 1023
.. 2nd = 5 * (6^2 - 1) / 5 = 35
.. 3rd = 5 ^4 - 1 / 4 = 156
.. 4th = 4 * (5^4 - 1) / 4 = 624
So in increasing order its 2nd, 3rd, 4th and first.
Just times everything like if you had
Lenght-3
widght-4
you times them so
3 x 4 = 12
so your answer would be 12 your area is 12
hope that helped you!
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)
