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Math Help - math homework that i just can't solve.

  1. #1
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    math homework that i just can't solve.

    Yes so i have some math assighments i need to solve and i am almost done (yes it is for a grade) but i just have problems with it. I need answers for these 2 problems:

    1. Cody, Ray, and Amanda are kicking a soccerball. Cody always kicks to Ray, and Ray always kicks to Amanda, but Amanda is equally likely to kick the ball to either Cody or Ray. Find the stable-state vector and tell what it means.

    2. The B & M Company produces batteries and motors. The battery division requires 3 batteries from the battery division and 1 motor from the motor division to produce 100 batteries. The motor division requires 4 motors from the motor division and 8 batteries from the battery division to produce 100 motors. B & M has received an order for 900 batteries and 400 motors. Determine the total production needed from each division to fill the order.
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  2. #2
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    Hello, lsk91!

    1. Cody, Ray, and Amanda are kicking a soccerball.
    Cody always kicks to Ray, and Ray always kicks to Amanda,
    but Amanda is equally likely to kick the ball to either Cody or Ray.

    Find the stable-state vector and tell what it means.
    The Markov process is designated by: . A \;=\;\left(\begin{array}{ccc}0&1&0 \\ 0&0&1 \\ \frac{1}{2} & \frac{1}{2} & 0 \end{array}\right)


    The steady-state vector is: . \vec v \:=\:( p,q,r)

    . . where: . (p,q,r )\cdot A \:=\: (p,q,r) \:\text{ and }\:p+q+r\:=\:1


    We have: . (p,q,r)\cdot\left(\begin{array}{ccc}0&1&0 \\ 0&0&1 \\ \frac{1}{2}&\frac{1}{2}&0 \end{array}\right) \;=\;(p.q.r)

    . . . . . . . \begin{Bmatrix}\frac{1}{2}r &=& p \\ p+\frac{1}{2}r &=& q \\ q &=& r \end{Bmatrix}\;\;\text{ and: }\;p+q+r\:=\:1


    Solve the system: . \boxed{(p,q,r) \;=\;\left(\frac{1}{5},\:\frac{2}{5},\:\frac{2}{5}  \right)}



    After a very large number of consecutive kicks,
    . . there are the probabilities of who has the ball.

    . . \begin{array}{ccc}P(\text{Cody has the ball}) &=& \dfrac{1}{5} \\ \\[-3mm]<br />
P(\text{Ray has the ball}) &=& \dfrac{2}{5} \\ \\[-3mm]<br />
P(\text{Amanda has the ball}) &=& \dfrac{2}{5} \end{array}

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  3. #3
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    thank you
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