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natulia [17]
3 years ago
14

Solve for r. -7.61 - 4.3r = -8.1r - 3.81

Mathematics
2 answers:
marysya [2.9K]3 years ago
7 0

Answer:

r = 1

Step-by-step explanation:

-7.61 - 4.3r = -8.1r - 3.81

-7.61 = -3.8r - 3.81

-3.8 = -3.8r

r = 1

Aleonysh [2.5K]3 years ago
6 0

Answer:

r=1

Step-by-step explanation:

-7.61 - 4.3r = -8.1r - 3.81

(bring like terms to one side)

-4.3 r = -8.1r -3.81 + 7.61

-4.3r + 8.1 r =-3.81 + 7.61

(consolodate)

3.8r = 3.8

(divide right by # of r)

r = 3.8/3.8

r = 1

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The correct answer is

(0.0128, 0.0532)

Step-by-step explanation:

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For this problem, we have that:

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So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 - 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0128

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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.

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So the converse means that when the sides are proportional so therefore, the side TU is said to be parallel to the side RS.

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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.

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The computer center at Dong-A University has been experiencing computer down time. Let us assume that the trials of an associate
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\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)

(a)To determine the probability of the system being down or running after any k hours, we determine the kth state matrix P^k.

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P^1=\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)

P^2=\begin{pmatrix}0.84&0.16\\ 0.48&0.52\end{pmatrix}

If the system is initially running, the probability of the system being down in the next hour of operation is the (a_{12})th$ entry of the P^2$ matrix.

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(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]

[s_1$  s_2]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[s_1$  s_2]

Using a calculator to raise matrix P to large numbers, we find that the value of P^k 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]

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]

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