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yawa3891 [41]
2 years ago
8

Please help. I don't know any theorems that relate to this

Mathematics
1 answer:
Tema [17]2 years ago
4 0

Answer:

125°

Step-by-step explanation:

In circle with center O.

  • BC is tangent at point B.
  • Draw segment OB
  • OB is radius
  • \implies OB\perp AB (Tangent theorem)
  • \implies\red{\bold{ m\angle OBC = 90\degree}}
  • m\angle AOB= 90\degree + 20\degree (By exterior angle theorem)
  • \implies m\angle AOB= 110\degree
  • OA = OB (radii of same circle)
  • \implies m\angle OAB = m\angle OBA = x \: (say) (By isosceles triangle theorem)
  • m\angle AOB + m\angle OAB +m\angle OBA=180\degree (By interior angle sum theorem of a triangle)
  • \implies 110\degree+ x +x=180\degree
  • \implies 2x =180\degree-110\degree
  • \implies 2x =70\degree
  • \implies x =35\degree
  • \implies \purple{\bold{m\angle OBA =35\degree}}
  • m\angle CBA=\red{\bold{m\angle OBC}} +\purple{\bold{m\angle OBA }}
  • m\angle CBA=90\degree + 35\degree
  • \huge{\orange{m\angle CBA=125\degree}}
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Approximate the stationary matrix S for the transition matrix P by computing powers of the transition matrix P.
Scrat [10]

Answer:

S = [0.2069,0.7931]

Step-by-step explanation:

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P=\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

Stationary matrix S for the transition matrix P is obtained by computing powers of the transition matrix P ( k powers ) until all the two rows of transition matrix p are equal or identical.

Transition matrix P raised to the power 2 (at k = 2)

P^{2} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{2} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right]

Transition matrix P raised to the power 3 (at k = 3)

P^{3} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{3} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

  P^{3} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right]

Transition matrix P raised to the power 4 (at k = 4)

P^{4} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{4} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{4} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]

Transition matrix P raised to the power 5 (at k = 5)

P^{5} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{5} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{5} =\left[\begin{array}{ccc}0.2069&0.7931\\0.2069&0.7931\end{array}\right]

P⁵ at k = 5 both the rows identical. Hence the stationary matrix S is:

S = [ 0.2069 , 0.7931 ]

6 0
3 years ago
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