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motikmotik
3 years ago
8

A light flashes every 8 seconds how many times will it flash in 3 minutes

Mathematics
1 answer:
pogonyaev3 years ago
3 0

Answer:

22 times.

Step-by-step explanation:

3 minutes = 180 seconds.

180 divided by 8 = 22.5

The light only flashed 22 times.

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To create a unique house paint color, Melton mixes together a sample that is 12 gallons of red, 2.5 gallons of yellow and 0.5 ga
Mariana [72]

Answer:

180

Step-by-step explanation:

total ratio is 15

gallons of red =

\frac{12}{15}  \times 6 \times  \frac{15}{0.5 }  \\  \\  = 144

gallons of yellow

30

gallons of blue

\frac{0.5}{15}  \times 30 \times  \frac{15}{2.5}  \\  \\  = 6

total =144+30+6

=180

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Answer:

-3/2

Step-by-step explanation:

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(6-9) / (5-3)

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The graphs of f(x) =-2x and g(x)=(1/2)^z are shown
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Answer:

f(x)-3x+5 and g(x)=4+5 is it true ?Step-by-step explanation:

6 0
3 years ago
((PLEASE HELP ASAP!! D:)) Natasha bought a tennis racket that was marked down in price, p, by 20%. Write an expression with TWO
kumpel [21]

Answer: p - 0.2p

Step-by-step explanation:

Given the following :

Original Price of tennis racket = p

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Amount after discount = Original price - Discount

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7 0
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
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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