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Alisiya [41]
4 years ago
12

If you have 21 cats and give 2 away how much will you have

Mathematics
2 answers:
Rzqust [24]4 years ago
7 0

Answer:

19 Cats

Step-by-step explanation:

21 (Cats) - 2 (Cats) = 19 (Cats)

nexus9112 [7]4 years ago
5 0

Answer:

19

Step-by-step explanation:

21 - 2 is 19

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Todd paid $0.08 less per gallon of gasoline than Hope.

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What are the numerical measures of each angle in the diagram? 1 and 3 measure degrees. 2 and 4 measure degrees. (3x - 1)º (2x +
Tanya [424]

Answer:

Step-by-step explanation:

In the diagram shown, the measure of angle 1 is oppositely directed to angle 2 and oppositely directed angles are equal.

Hence <1 = <3

Given < 1 = 3x-1 and <3 = 2x+9

Hence 3x-1 =  2x+9

collect like terms

3x-2x = 9+1

x = 10°

Since <1 = 3x-1

on substituting x = 10

<1 = 3(10)-1

<1 = 30-1

<1 = 29°

<1+<2 = 180 (angle on a straight line)

29+<2 = 180

<2 = 180-29

<2 = 151°

Similarly, on substituting x = 10 into <3

<3 = 2x+9

<3 = 2(10)+9

<3 = 20+9

<3 = 29°

<3+<4 = 180 (angle on a straight line)

29+<4 = 180

<4 = 180-29

<4 = 151°

3 0
4 years ago
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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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