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Vera_Pavlovna [14]
3 years ago
10

Help, please????????

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
5 0

9514 1404 393

Answer:

  E)  25.5 units²

Step-by-step explanation:

The figure is not carefully drawn, so it is difficult to tell that it is a trapezoid. The "height" of it is BC, the diagonal of a rectangle that is 1 unit by 4 units. The Pythagorean theorem tells you that length is ...

  BC = √(1² +4²) = √17

We notice that the "bases" of the trapezoid (CD and AB) are 1 and 2 times this length. The trapezoid area formula tells us the area is ...

  A = 1/2(b1 +b2)h

  A = 1/2((√17)(1 +2))(√17) = 3/2(17) = 25.5 . . . . square units

_____

<em>Alternate solution</em>

<em>Pick's Theorem</em> is a theorem that can help you find the area of a polygon when its vertices are all on grid points. It tells you the area is ...

  A = i + b/2 -1

where i is the number of grid points interior to the figure, and b is the number of grid points on the boundary. Here, the boundary grid points are those that are labeled plus the one at (-2, 1). Counting the interior grid points gives a total of 24, so the area by Pick's theorem is ...

  A = 24 + 5/2 -1 = 25.5 . . . square units

__

<em>Additional comment</em>

Once you're familiar with Pick's theorem (and/or any of a few other methods of computing area from coordinates), you see that <em>the area of any polygon whose vertices are grid points will be an integer multiple of 1/2</em>. The only such answer choice here is choice E.

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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
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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:

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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