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nalin [4]
3 years ago
6

Give an example of each of the following or explain why you think such a set could not exist.

Mathematics
1 answer:
Oksanka [162]3 years ago
6 0

Answer:

a.No

b.No

c.No

Step-by-step explanation:

a.No,Such set does not exist .A set of natural numbers is N

Every point of this set is an isolated point but no accumulation point

Accumulation point:It is defined as that point a of set Swhich every neighborhood contains infinitely many distinct point of set

(a-\epsilon,e+\epsilon)\cap S-{a}\neq\phi

Isolated point : it is defined as that point a of set S which neighborhood   does not contain any other point of set except itself

(a-\epsilon0,a+\epsilon)\cap S-{a}=\phi

Interior point of set :Let a\in S .Then a is called interior point of set when its neighborhood is a subset of set S.

a\in(a-\epsilon,e+\epsilon)\subset S

When a set is uncountable then interior point exist it is  necessary for interior points existance .

Boundary points :Let a\in S .If every non empty neighborhood of a  intersect S and complement of S.

Every member of  a set is a boundary point

b.No, such set does not exist .A non empty set with isolated point then the set have no interior points .By definition of interior point and isolated point .For example.set of natural numbers

c.No, Such set does not exist ,for example set of natural every point is an isolated point and boundary point.By definition of  boundary point and isolated point

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For this case we have the following equation:

2v ^ 2-12 = -12v

Rewriting we have:

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Dividing by 2 to both sides of the equation:

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We apply the quadratic formula:

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Substituting:

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Thus, we have two roots:

x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}

ANswer:

x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}

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Complete Question

A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?

Answer:

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Step-by-step explanation:

From the question we are told that

   The upper specification is  USL  =  1.68 \ mm

    The lower specification is  LSL  = 1.52  \  mm

     The sample mean is  \mu =  1.6 \  mm

     The standard deviation is  \sigma =  0.03 \ mm

Generally the capability index in mathematically represented as

             Cpk  =  min[ \frac{USL -  \mu }{ 3 *  \sigma }  ,  \frac{\mu - LSL }{ 3 *  \sigma } ]

Now what min means is that the value of  CPk is the minimum between the value is the bracket

          substituting value given in the question

           Cpk  =  min[ \frac{1.68 -  1.6 }{ 3 *  0.03 }  ,  \frac{1.60 -  1.52 }{ 3 *  0.03} ]

=>      Cpk  =  min[ 0.88 , 0.88  ]

So

         Cpk  = 0.88

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Now let assuming that

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So

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So

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=>      2

Hence

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So

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So    \sigma =  0.0133 is  the value of standard deviation required

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3 years ago
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