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Orlov [11]
2 years ago
11

16. Let W be the set of all vectors in R3 of the form a+2b b -3a Find a basis for Wand state the dimension of W.

Mathematics
1 answer:
Yuki888 [10]2 years ago
7 0

Answer:

W=\{\left[\begin{array}{ccc}a+2b\\b\\-3a\end{array}\right]: a,b\in\mathbb{R} \}

Observe that if the vector x=\left[\begin{array}{ccc}x\\y\\z\end{array}\right] is in W then it satisfies:

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

This means that each vector in W can be expressed as a linear combination of the vectors \left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]

Also we can see that those vectors are linear independent. Then the set

\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]\} is a basis for W and the dimension of W is 2.

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The class average on a math test was an 82 with a standard deviation of 5. If the 
Alex787 [66]

Answer:

Option 3 and 4.

Step-by-step explanation:

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation

Given that mean (μ) = 82, standard deviation (σ) = 5

1) For x = 74:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{74-82}{5} =-1.6

Option 1 is incorrect

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z=\frac{x-\mu}{\sigma} \\\\z=\frac{87-82}{5} =1

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Option 3 is correct

4) For x < 77

z=\frac{x-\mu}{\sigma} \\\\z=\frac{77-82}{5} =-1

From the normal distribution table, P(x < 77) = P(z < -1) = 0.16 = 16%

Option 4 is correct

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Option 5 is incorrect

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2 years ago
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