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LenKa [72]
3 years ago
13

William cycles at a sped of 15 miles per hour. He cycles 12 miles from home to school. If he increases his cycling speed ny 5 mi

les per hour how much faster will he arrive his school
Mathematics
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:he would arrive 0.2 miles faster.

Step-by-step explanation:

Distance = speed × time

Time = distance/speed

William cycles at a sped of 15 miles per hour. He cycles 12 miles from home to school. This means that the time it takes William to get to school from home would be

12/15 = 0.8 hours

If he increases his cycling speed by 5 miles per hour, his new speed becomes

15 + 5 = 20 miles per hour

Therefore, the new time it takes William to get to school from home would be

12/20 = 0.6 hours

The difference in both times is

0.8 - 0.6 = 0.2 hours

Therefore, he would arrive 0.2 miles faster.

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3 years ago
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:

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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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