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Valentin [98]
3 years ago
14

A and B are bounded non-empty subsets of R. For inf(A) to be less than or equal to inf(B), which of the following conditions mus

t be met?
a) For every b in B and epsilon > 0, there exists a in A, such that a < b + epsilon.

b) There exists a in A, and b in B such that a < b.

If neither of these conditions are appropriate, what would be appropriate conditions for inf(A) to be less than or equal to inf(B)?
Mathematics
1 answer:
nirvana33 [79]3 years ago
6 0

Answer:

a) must be met

Step-by-step explanation:

We have two conditions:

a) For every b\in B and \epsilon>0, there exists a\in A, such that a.

b) There exists a\in A and b\in B such that  a.

We will prove that conditon a) is equivalent to inf(A)\leq inf(B)

If a) is not satisfied, then it would exist b\in B and \epsilon >0 such that, for every a\in A, a\geq b+\epsilon. This implies that b+\epsilon is a lower bound for A and in consequence

inf(A)\geq b+\epsilon > b\geq inf(B)

Then, inf(A) \leq inf(B) implies a).

If inf(A) \leq inf(B) is not satisfied then, inf(A) > inf(B) and in consequence exists b\inB such that b-inf(A)=\epsilon >0. Then b-\epsilon=inf(A) and, for every a\in A,

b-\epsilon =inf(A)\leq a.

So, a) is not satisfied.

In conclusion, a) is equivalent to inf(A)\leq inf(B)

Finally, observe that condition b) is not an appropiate condition to determine if inf(A)\leq inf(B) or not. For example:

  • <u>A={0}, B={1}</u>. b) is satisfied and inf(A)=0
  • <u>A={0}. B={-1,1}</u>. b) is satisfied and inf(A)=0>-1=inf(B)

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MissTica
You need to use PEMDAS
So first you do the parenthesis.
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So now re-write the problem.
50+10*8+2
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A sample of 47 observations is selected from a normal population. The sample mean is 30, and the population standard deviation i
erma4kov [3.2K]

Answer:

We accept the null hypothesis and reject  the alternate hypothesis. There is no evidence to conclude that the population mean is greater than 29. The population mean is less than or equal to 29.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 29

Sample mean, \bar{x} = 30

Sample size, n = 47

Alpha, α = 0.05

Population standard deviation, σ = 5

First, we design the null and the alternate hypothesis

H_{0}: \mu \leq 29\\H_A: \mu > 29

a) This is a one-tailed test because the alternate hypothesis is in greater than direction.

We use One-tailed z test to perform this hypothesis.

b) z_{stat} > z_{critical} , we reject the null hypothesis and accept the alternate hypothesis and if z_{stat} < z_{critical} , we accept the null hypothesis and reject  the alternate hypothesis.

c) Formula:

z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

z_{stat} = \displaystyle\frac{30 - 29}{\frac{5}{\sqrt{47}} } = 1.37

d) Now, z_{critical} \text{ at 0.05 level of significance } = 1.64

Since,  

z_{stat} < z_{critical}

We accept the null hypothesis and reject  the alternate hypothesis. There is no evidence to conclude that the population mean is greater than 29. The population mean is less than or equal to 29.

e) P-value is 0.0853

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Let's the length to be x,

then the width is 8x - 2.

Perimeter of a rectangle = 2*Width + 2*Length

Perimeter of the rectangle = 2(8x-2) + 2x = 16x - 4 + 2x = 18x - 4

Ar the same time, Perimeter of the rectangle = 88.

So, we can write

18x - 4 = 88

18x = 84

9x=42

x= 42/9 cm (length) or ≈ 4.67 cm (length)

8x - 2 = 8*42/9 - 2 ≈ 35.33 cm (width)


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[[ 20 POINTS IF YOU ANSWER THIS QUESTION CORRECTLY. ]]
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Answer: the height of the kite is 106.065 ft

Step-by-step explanation:

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To determine x, the height of the kite, we would apply the sine trigonometric ratio which is expressed as

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-8-(-1)-5

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