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liraira [26]
3 years ago
15

For a certain​ candy, 15​% of the pieces are​ yellow, 5​% are​ red, 20​% are​ blue, 20​% are​ green, and the rest are brown. ​a)

if you pick a piece at​ random, what is the probability that it is​ brown? it is yellow or​ blue? it is not​ green? it is​ striped?
Mathematics
1 answer:
LuckyWell [14K]3 years ago
3 0
P(Brown) = 100 - 15 -5- 20 - 20 = 40%

P(yellow or blue) = 15 + 20 = 35%

P(not green) = 100 - 20 = 80%

P(striped) = 0%
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Answer:

AB = \left[\begin{array}{ccc}-3&4&6\\-6&3&5\\5&0&-4\end{array}\right]

Each column of AB is written as a linear combination of columns of Matrix A in the explanation below.

Step-by-step explanation:

A = \left[\begin{array}{ccc}-2&2&1\\-3&1&1\\2&0&-1\end{array}\right]

B= \left[\begin{array}{ccc}2&-1&0\\1&2&1\\-1&-2&4\end{array}\right]

We need to write each column of AB as a linear combination of the columns of A so we will multiply each column of A with each column element of B to get the column of AB. So,

AB Column 1 = 2 * \left[\begin{array}{ccc}-2\\-3\\2\end{array}\right]  + 1 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + (-1) \left[\begin{array}{ccc}1\\1\\-1\end{array}\right] = \left[\begin{array}{ccc}-3\\-6\\5\end{array}\right]

AB Column 2 = (-1)\left[\begin{array}{ccc}-2\\-3\\2\end{array}\right] + 2\left[\begin{array}{ccc}2\\1\\0\end{array}\right] + (-2)\left[\begin{array}{ccc}1\\1\\-1\end{array}\right] = \left[\begin{array}{ccc}4\\3\\0\end{array}\right]

AB Column 3 = (0)\left[\begin{array}{ccc}-2\\-3\\2\end{array}\right] + (1)\left[\begin{array}{ccc}2\\1\\0\end{array}\right] + 4\left[\begin{array}{ccc}1\\1\\-1\end{array}\right] = \left[\begin{array}{ccc}6\\5\\-4\end{array}\right]

Finally, we can combine all three columns of AB to form the 3x3 matrix AB.

AB = \left[\begin{array}{ccc}-3&4&6\\-6&3&5\\5&0&-4\end{array}\right]

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