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nlexa [21]
2 years ago
6

Please help me out!​

Mathematics
1 answer:
gavmur [86]2 years ago
5 0

Answer:

B is correct because it value become 15^9

Step-by-step explanation:

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What is the Y 2(7y−1)=40
MissTica

Answer:

3

Step-by-step explanation:

2(7y−1)=40

14y-2=40

14y=40+2

14y=42

y=3

please mark as brainliest

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3 years ago
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Q = 15<br> a = 16<br> What would you do with q and a to get 240?
ch4aika [34]

You multiply Q and a to get 240, because 15 x 16 = 240.

6 0
3 years ago
The fraction 4/5 is equivalent to which of the following decimals?
Sophie [7]

Answer:

0.8(Decimal Form) and 80%(Percent Form)

Step-by-step explanation:

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3 years ago
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The table represents a linear function.<br> What is the slope of the function?
TEA [102]

Answer:

-4

Step-by-step explanation:

The function is linear; therefore, the slope of the line joining any two points is the same.

We take any two points from the table and compute the slope or  rise/ run between them— let is take points (-4, -2) and (-2, -10).

The slope m of the line joining the points is

m=\dfrac{\Delta y }{\Delta x } =\dfrac{-2-(-10)}{-4-(-2)} \\\\\boxed{m=-4}

The slope of the function is -4.

3 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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