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puteri [66]
3 years ago
15

The following sets of numbers are possible side lengths of a triangle. Which of the following sets forms a Pythagorean triple?

Mathematics
2 answers:
Inga [223]3 years ago
4 0

I would say C as the answer

Sphinxa [80]3 years ago
4 0
The first choice is not a triangle  because 42 = 18 + 24 You cant draw a tangle with these measurements  = In fact they make  a straight line.

The  same goes for choice 3 (45 = 30 + 15)

The second one forms a Pythagorean Triple  53^2 = 28^2 + 45^2

The last one does nt form a triangle  because  77 > 22 + 33
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Answer:

Step-by-step explanation:

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3 years ago
I have tried this like 10 diffrent ways<br> -3(r+9)=-32
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Distribute -3 to parenthasees
-3(r+9)=-3r-27

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-3r-27=-32
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3 years ago
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Y: 2x+3+9x+2<br> Simplify this
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y=16x ??

Step-by-step explanation:

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g A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of pa
adoni [48]

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:

The value required is  \sigma =  0.0133

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

Now from the question we are asked to evaluated the value of  standard deviation that will produce a  capability index of 2

Now let assuming that

         \frac{\mu - LSL  }{ 3 *  \sigma } =  2

So

         \frac{ 1.60 -  1.52  }{ 3 *  \sigma } =  2

=>    0.08 = 6 \sigma

=>     \sigma =  0.0133

So

        \frac{ 1.68  - 1.60 }{ 3 *  0.0133 }

=>      2

Hence

      Cpk  =  min[ 2, 2 ]

So

    Cpk  = 2

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

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