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adoni [48]
3 years ago
7

If u, v, and w are nonzero vectors in r 2 , is w a linear combination of u and v?

Mathematics
1 answer:
Tju [1.3M]3 years ago
8 0
Not necessarily. \mathbf u and \mathbf v may be linearly dependent, so that their span forms a subspace of \mathbb R^2 that does not contain every vector in \mathbb R^2.

For example, we could have \mathbf u=(0,1) and \mathbf v=(0,-1). Any vector \mathbf w of the form (r,0), where r\neq0, is impossible to obtain as a linear combination of these \mathbf u and \mathbf v, since

c_1\mathbf u+c_2\mathbf v=(0,c_1)+(0,-c_2)=(0,c_1-c_2)\neq(r,0)

unless r=0 and c_1=c_2.
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Answer:

Step-by-step explanation:

Hello!

X: Cholesterol level of a woman aged 30-39. (mg/dl)

This variable has an approximately normal distribution with mean μ= 190.14 mg/dl

1. You need to find the corresponding Z-value that corresponds to the top 9.3% of the distribution, i.e. is the value of the standard normal distribution that has above it 0.093 of the distribution and below it is 0.907, symbolically:

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P(Z≤z₀)= 0.907

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Using the Z value from 1., the mean Cholesterol level (μ= 190.14 mg/dl) and the Medical guideline that indicates that 9.3% of the women have levels above 240 mg/dl you can clear the standard deviation of the distribution from the Z-formula:

Z= (X- μ)/δ ~N(0;1)

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δ=(X- μ)/Z

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I hope it helps!

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