# math homework that i just can't solve.

• May 3rd 2010, 07:28 PM
lsk91
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.
• May 3rd 2010, 08:08 PM
Soroban
Hello, lsk91!

Quote:

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: .$\displaystyle A \;=\;\left(\begin{array}{ccc}0&1&0 \\ 0&0&1 \\ \frac{1}{2} & \frac{1}{2} & 0 \end{array}\right)$

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

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

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

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

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

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

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

• May 4th 2010, 04:35 PM
lsk91
thank you (Bow) (Bow) (Bow)