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Margarita [4]
2 years ago
8

The GFC of 84 and 56 is ?

Mathematics
2 answers:
Semmy [17]2 years ago
8 0

Answer:

28

Step-by-step explanation:

Express the numbers as a product of their prime factors

84 = 2 × 2 × 3 × 7

56 = 2 × 2 × 2 × 7

Identify the prime factors hat both numbers have in common and multiply them.

GCF = 2 × 2 × 7 = 28

arsen [322]2 years ago
6 0

Answer:

28

What is the GFC of 56 and 84? The gcf of 56 and 84 is 28.

Step-by-step explanation:

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9966 [12]

Answer:

C

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Because c has the variables in the right spot and the other ones don't make scene

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3 years ago
4 + -5 * -1 + 8=? <br> i need help so yeah
Natasha2012 [34]

Answer:

1

Step-by-step explanation:

4+(-5*-1)-8

4+5-8

9-8

pemdas

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2 years ago
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If the length of side a is 16 centimeters, the length of side b is 10 centimeters, and m = 42°, what is the measure of ? Round y
kupik [55]

Answer:Substitute the given values in the Law of Sines and find that the measure is 24.72°

5 0
3 years ago
The town of Madison has a population of 25000 The population is increasing by a factor of 1.12 each year.
Fudgin [204]

Answer:

P(t) = 25,000*1.12^t

Step-by-step explanation:

The populational growth is exponential with a factor of 1.12 each year. An exponential function has the following general equation:

y(x) = ab^x

Where 'a' is the initial population (25,000 people), 'b' is the growth factor (1.12 per year), 'x' is the time elapsed, in years, and 'y(x)' is the population after 'x' years.

Therefore, the function P(t) that models the population in Madison t years from now is:

P(t) = 25,000*1.12^t

8 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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