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Stells [14]
3 years ago
7

Dawn earned $97.50 for 10 hours of work. Amy earned $120 for 12 hours of work.How much did each person earn per hour? How can yo

u use this information to compare their earnings? EXPLIAN​
Mathematics
2 answers:
netineya [11]3 years ago
8 0

Answer:

Step-by-step explanation:

dawn earned less for how many he worked but amy got more cause she work for a greater value

RUDIKE [14]3 years ago
6 0

Answer:

Dawn earned 9.75 an hour

Amy earned 10 an hour

Amy makes 0.25 more than Dawn

Step-by-step explanation:

97.50/10=9.75

120/12=10

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What is 9 out of 100 as a percent ?
UkoKoshka [18]

Answer:

9%

Step-by-step explanation:

9/100 = 0.09

To convert to a percent, multiply by 100 and add a percent sign:

0.09 * 100 = 9%

5 0
3 years ago
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PLEASE HELP IN ONE MINUTE
Marysya12 [62]

Answer:

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2 years ago
Write an equation of the line that passes through (2,−5) and is parallel to the line 2y=3x+10
LenaWriter [7]

Step-by-step explanation:

Divide two on both sides to get rid of it and the make the equation in the form y = mx + c

\frac{2}{2} y = \frac{3}{2} x +  \frac{10}{2}

y = 3/2x + 5

since both lines are parallel they must have the same gradient which is 3/2

y = 3/2x + c

All you have to do now is to replace x and y with (2, -5) to find c

x = 2

y = -5

-5 = 3/2 × 2 + c

-5 = 3 + c

c = -5-3

c = -8

y  = \frac{3}{2} x  - 8

8 0
3 years ago
Round 60488 to the nearest ten thousand
ANTONII [103]
60,000 would be the answer because it stays the same if under 5 it stays if above 5 rounds up once
4 0
3 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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