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Wittaler [7]
3 years ago
13

Producers will make 1000 refrigerators available when the unit price is $280.00. At a unit price of $400, 5000 refrigerators wil

l be marketed . Find the equation relating the unit price p of a refrigerator to the quantity supplied x if the equation is known to be linear.
p =

How many refrigerators will be marketed when the unit price is $440?______ refrigerators

What is the lowest price at which a refrigerator will be marketed? $_________ (lowest price)
Mathematics
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

For linear equations we use:

p=mx+b   ------  (1)

Now we have the following coordinates:

(x1,p1)= (1000,280) and (x2, p2)=(5000,400)

First we need slope (m)

m= (400-280)/(5000-1000)

= 120/4000=0.03

Now we will plug the value of m in the first equation

280=0.03(1000)+b

=> 280=30+b

=> b = 250

Now plug into p=mx+b using only m=0.03 and b=250

p=0.03x+250

When the unit price is $440, we can plug in 440 in for p;

440=0.03x+250

=> 0.03x=440-250

=> 0.03x=190

=> x = 6333 refrigerators

The lowest price at which  a refrigerator will be marketed, we can find this by plugging x = 0 in p=mx+b.

p=0.03(0)+250

=> p = $250

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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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