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garik1379 [7]
3 years ago
15

Find an example for each of vectors x, y ∈ V in R.

Mathematics
1 answer:
rjkz [21]3 years ago
8 0

(a) Both conditions are satisfied with <em>x</em> = (1, 0) for \mathbb R^2 and <em>x</em> = (1, 0, 0) for \mathbb R^3:

||(1, 0)|| = √(1² + 0²) = 1

max{1, 0} = 1

||(1, 0, 0)|| = √(1² + 0² + 0²) = 1

max{1, 0, 0} = 1

(b) This is the well-known triangle inequality. Equality holds if one of <em>x</em> or <em>y</em> is the zero vector, or if <em>x</em> = <em>y</em>. For example, in \mathbb R^2, take <em>x</em> = (0, 0) and <em>y</em> = (1, 1). Then

||<em>x</em> + <em>y</em>|| = ||(0, 0) + (1, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2

||<em>x</em>|| + ||<em>y</em>|| = ||(0, 0)|| + ||(1, 1)|| = √(0² + 0²) + √(1² + 1²) = √2

The left side is strictly smaller if both vectors are non-zero and not equal. For example, if <em>x</em> = (1, 0) and <em>y</em> = (0, 1), then

||<em>x</em> + <em>y</em>|| = ||(1, 0) + (0, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2

||<em>x</em>|| + ||<em>y</em>|| = ||(1, 0)|| + ||(0, 1)|| = √(1² + 0²) + √(0² + 1²) = 2

and of course √2 < 2.

Similarly, in \mathbb R^3 you can use <em>x</em> = (0, 0, 0) and <em>y</em> = (1, 1, 1) for the equality, and <em>x</em> = (1, 0, 0) and <em>y</em> = (0, 1, 0) for the inequality.

(c) Recall the dot product identity,

<em>x</em> • <em>y</em> = ||<em>x</em>|| ||<em>y</em>|| cos(<em>θ</em>),

where <em>θ</em> is the angle between the vectors <em>x</em> and <em>y</em>. Both sides are scalar, so taking the norm gives

||<em>x</em> • <em>y</em>|| = ||(||<em>x</em>|| ||<em>y</em>|| cos(<em>θ</em>)|| = ||<em>x</em>|| ||<em>y</em>|| |cos(<em>θ</em>)|

Suppose <em>x</em> = (0, 0) and <em>y</em> = (1, 1). Then

||<em>x</em> • <em>y</em>|| = |(0, 0) • (1, 1)| = 0

||<em>x</em>|| • ||<em>y</em>|| = ||(0, 0)|| • ||(1, 1)|| = 0 • √2 = 0

For the inequality, recall that cos(<em>θ</em>) is bounded between -1 and 1, so 0 ≤ |cos(<em>θ</em>)| ≤ 1, with |cos(<em>θ</em>)| = 0 if <em>x</em> and <em>y</em> are perpendicular to one another, and |cos(<em>θ</em>)| = 1 if <em>x</em> and <em>y</em> are (anti-)parallel. You get everything in between for any acute angle <em>θ</em>. So take <em>x</em> = (1, 0) and <em>y</em> = (1, 1). Then

||<em>x</em> • <em>y</em>|| = |(1, 0) • (1, 1)| = |1| = 1

||<em>x</em>|| • ||<em>y</em>|| = ||(1, 0)|| • ||(1, 1)|| = 1 • √2 = √2

In \mathbb R^3, you can use the vectors <em>x</em> = (1, 0, 0) and <em>y</em> = (1, 1, 1).

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Step-by-step explanation:

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According to Masterfoods, the company that manufactures M&amp;M's, 12% of peanut M&amp;M's are brown, 15% are yellow, 12% are re
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Answer:

The questions asked are

If you randomly select 4 peanuts

1. Compute the probability that exactly three of the four M&M’s are brown

2. Compute the probability that two or three of the four M&M’s are brown.

3. Compute the probability that at most three of the four M&M’s are brown.

4. Compute the probability that at least three of the four M&M’s are brown.

Step-by-step explanation:

Given the following information

Brown=12%. P(B)=0.12

Yellow=15%. P(Y)=0.15

Red=12%. P(R), =0.12

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Question 1.

They are independent events

If there are exactly three brown and the last is not brown

P(B n B n B n B')

P(B)×P(B)×P(B)×P(B')

0.12×0.12×0.12×(1-P(B))

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0.001728×0.88

0.00152.

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2. If two or three are brown

I.e we are going to two brown and two none brown or three brown and one not brown. (P(B)×P(B)×P(B')×P(B'))+ (P(B)×P(B)×P(B'))

(0.12×0.12×0.88×0.88)+(0.12×0.12×0.12×0.88)

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3. At most three brown out of four then we are going to have

BBBB', BBB'B', BB'B'B', B'B'B'B'

These are the cases of at most three brown.

P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B')×P(B') + P(B)×P(B')×P(B')×PB')+ P(B')×P(B')×P(B')×P(B')=

0.12×0.12×0.12×0.88+ 0.12×0.12×0.88×0.88+ 0.12×0.88×0.88×0.88+ 0.88×0.88×0.88×0.88=0.694

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4. At least 3 brown out of four selection

I.e BBBB', BBBB

These are the two options

P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B)×P(B)=

0.12×0.12×0.12×0.88 + 0.12×0.12×0.12×0.12

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