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alexandr402 [8]
3 years ago
12

Hlpppppppppppppppppppppppppppppppppp

Mathematics
2 answers:
denpristay [2]3 years ago
8 0

Answer:

It is the 4th quadrant!

Step-by-step explanation:

Hope it helps! ;D

goblinko [34]3 years ago
7 0

Answer:

4th quadrant

Step-by-step explanation:

quadrant 4 is (+,-)

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Please help. Could you explan your answer also? thanks
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Answer: Choice B) -4

The lower quartile, also known as Q1 or the first quartile, is the left edge of the box. In this case, that is at -4. We can drop a vertical line from the left edge of the box until we hit -4 on the number line. 

Side Note: 25% of the data values are below Q1, while 75% of the data values are above Q1
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As one once said Another one
aleksandrvk [35]

Answer:

f

Step-by-step explanation:

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Help me please in my module​
Ghella [55]

Answer:

1.(b)

2.(b)

3. (d) 4cm

4.(a)

Explanation :

The polygons will have the same shape and size, but one may be a rotated, or be the mirror image of the other.Polygons are congruent when they have the same number of sides, and all corresponding sides and interior angles are congruent.

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2 years ago
State if the triangle is acute obtuse or right​
Dmitrij [34]

Answer:

It should be obtuse.

Step-by-step explanation: if i'm not mistaken it has to be obtuse. sorry i'm still learning i'm a beginner but, i'm volunteering to help everyone anyways because i'm a very nice person.

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2 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

3 0
3 years ago
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