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Zielflug [23.3K]
3 years ago
15

Which quadratic equation is equivalent to (x+2)2+5(x+2)-6=0

Mathematics
1 answer:
professor190 [17]3 years ago
6 0
(x+2)^2+5(x+2)-6=0\\\\x^2+2\cdot x\cdot2+2^2+5\cdot x+5\cdot2-6=0\\\\x^2+4x+4+5x+4-6=0\\\\x^2+9x+2=0
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Let P = 0.50.30.50.7 be the transition matrix for a Markov chain with two states. Let x0 = 0.50.5 be the initial state vector fo
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Answer:

Probability distribution vector = \left(\begin{array}{c}0.375\\ 0.625 \end{array} \right)

Step-By-Step Explanation

If P=\left(\begin{array}{cc}0.5&0.3\\ 0.5&0.7 \end{array} \right)  is the transition matrix for a Markov chain with two states.  

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X_{1}=P x_{0}=\left(\begin{array}{cc}0.5&0.3\\ 0.5&0.7 \end{array} \right) \left(\begin{array}{c}0.5\\ 0.5 \end{array} \right) =\left(\begin{array}{c}0.4\\ 0.6 \end{array} \right)  

X_{2}=P^{2} x_{0}=\left(\begin{array}{c}0.38\\ 0.62 \end{array} \right)  

X_{3}=P^{3} x_{0}=\left(\begin{array}{c}0.38\\ 0.62 \end{array} \right)  

X_{30}=P^{30} x_{0}=\left(\begin{array}{c}0.37499\\ 0.625 \end{array} \right)  

In the long run, the probability distribution vector Xm approaches the probability distribution vector \left(\begin{array}{c}0.375\\ 0.625 \end{array} \right) .

This is called the steady-state (or limiting,) distribution vector.

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