Answer:
The correct answer is
(0.0128, 0.0532)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence interval
, we have the following confidence interval of proportions.

In which
Z is the zscore that has a pvalue of 
For this problem, we have that:
In a random sample of 300 circuits, 10 are defective. This means that
and 
Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.
So
= 0.05, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The correct answer is
(0.0128, 0.0532)
The statement that is true is option A that is Line segment TU is parallel to line segment RS because StartFraction 32 Over 36 EndFraction = StartFraction 40 Over 45 EndFraction.
<h3>What is the line segment about?</h3>
The law of the side- splitter connote that when f the line is parallel to a side of the triangle and if that said line do intersects the other 2 sides, tt is said to divides those given sides proportionally.
So the converse means that when the sides are proportional so therefore, the side TU is said to be parallel to the side RS.
Thus;
= 

Therefore, the ratios are equal and as such TU is parallel to RS
Hence, The statement that is true is option A that is Line segment TU is parallel to line segment RS because StartFraction 32 Over 36 EndFraction = StartFraction 40 Over 45 EndFraction.
Learn more about Line segment from
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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)