If you draw one card from a standard deck, what is the probability that you draw a king or queen?
2 answers:
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The matrix equation that represents this situation is
![\left[\begin{array}{ccc}3&2&0\\1&0&4\\3&1&1\end{array}\right]*\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}13.50\\16.50\\14.00\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%260%5C%5C1%260%264%5C%5C3%261%261%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D13.50%5C%5C16.50%5C%5C14.00%5Cend%7Barray%7D%5Cright%5D%20)
Use technology to find the inverse of matrix A:
![A^{-1}= \left[\begin{array}{ccc}-\frac{2}{5}&-\frac{1}{5}&\frac{4}{5}\\ \frac{11}{10}&\frac{3}{10}&-\frac{6}{5}\\\frac{1}{10}&\frac{3}{10}&-\frac{1}{5}\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B2%7D%7B5%7D%26-%5Cfrac%7B1%7D%7B5%7D%26%5Cfrac%7B4%7D%7B5%7D%5C%5C%0A%5Cfrac%7B11%7D%7B10%7D%26%5Cfrac%7B3%7D%7B10%7D%26-%5Cfrac%7B6%7D%7B5%7D%5C%5C%5Cfrac%7B1%7D%7B10%7D%26%5Cfrac%7B3%7D%7B10%7D%26-%5Cfrac%7B1%7D%7B5%7D%5Cend%7Barray%7D%5Cright%5D%20)
Multiplying A inverse by B, we get the solution matrix
[tex]\left[\begin{array}{ccc}2.50\\3\\3.50\end{array}\right][\tex]
Use technology to find the inverse of matrix A:
Multiplying A inverse by B, we get the solution matrix
[tex]\left[\begin{array}{ccc}2.50\\3\\3.50\end{array}\right][\tex]
bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.
we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.
we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.