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Arada [10]
4 years ago
13

A bag has 3 yellow, 6 red, and 5 green marbles. You randomly choose a marble from the bag and replace it. What is the probabilit

y of choosing a yellow or red marble? Simplify

Mathematics
2 answers:
Naily [24]4 years ago
6 0

Answer:

9/14

Step-by-step explanation:

add them =14

= denominator

6+3=9

jek_recluse [69]4 years ago
4 0

Answer:

answer:

Step-by-step explanation:

3\17 for yellow and 6\17 for red.

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Consider the following two ordered bases of R3:
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Let A = (a_1, ..., a_n) and B = (b_1, ..., b_n) bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors b_1, ..., b_n at base A.

The, we case correspond to find the coordinates of vectors of C,

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

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1. We need to find a,b,c\in\mathbb{R} such that

\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]

Then we find these values solving the linear system

\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&-1\\0&-1&1&-1\end{array}\right]

Using rows operation we obtain the echelon form of the matrix

\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&1\end{array}\right]

now we use backward substitution

c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4

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2. We need to find a,b,c\in\mathbb{R} such that

\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]

Then we find these values solving the linear system

\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&0\\0&-1&1&-1\end{array}\right]

Using rows operation we obtain the echelon form of the matrix

\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&2\end{array}\right]

now we use backward substitutionc=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4

Then the coordinate vector of \left[\begin{array}{ccc}2\\0\\-1\end{array}\right] is \left[\begin{array}{ccc}4\\3\\2\end{array}\right]

3. We need to find a,b,c\in\mathbb{R} such that

\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]

Then we find these values solving the linear system

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

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

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Then the coordinate vector of \left[\begin{array}{ccc}-3\\1\\2\end{array}\right] is \left[\begin{array}{ccc}-2\\-4\\-2\end{array}\right]

Then the change of basis matrix from B to C is

\left[\begin{array}{ccc}4&4&-2\\2&3&-4\\1&2&-2\end{array}\right]

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