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3 years ago

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The sum of three numbers is 855. One of the numbers, x, is 50% more than the sum of the other two numbers. What is the value of

x?

1 answer:

Snowcat [4.5K]3 years ago

8 0

Answer:

The value of x = 513

Step-by-step explanation:

Let x, y, and z be the three numbers

Given

$x+y+z=855$

Assuming x be the number which is 50% more than the sum of the other two numbers.

i.e.

(150)% of (y+z) = 1.5(y+z)

y + z = x/1.5

As we know that

$x + y + z = 855$

substituting y + z = x / 1.5 in the equation

$x+\frac{x}{1.5}=855$

Multiply both sides by 1.5

$x\cdot \:1.5+\frac{x}{1.5}\cdot \:1.5=855\cdot \:1.5$

$1.5x+x=1282.5$

$2.5x=1282.5$

Divide both sides by 2.5

$\frac{2.5x}{2.5}=\frac{1282.5}{2.5}$

$x = 513$

Therefore, the value of x = 513

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2 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:

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

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