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3 years ago

PLS HELP ME ASAP!! WILL MARK BRAINLIEST

Mathematics

2 answers:

SIZIF [17.4K]3 years ago

7 0

Answer: 10

Step-by-step explanation:

Lana71 [14]3 years ago

3 0

Answer:

Step-by-step explanation:

The sqrt of 81 equals to 9, the sqrt of 100 equals to 10

So, sqrt of 92 is between 9 and 10, but closer to 10

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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

$Cpk = min[ \frac{1.68 - 1.6 }{ 3 * 0.03 } , \frac{1.60 - 1.52 }{ 3 * 0.03} ]$

=> $Cpk = min[ 0.88 , 0.88 ]$

$Cpk = 0.88$

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

$\frac{\mu - LSL }{ 3 * \sigma } = 2$

$\frac{ 1.60 - 1.52 }{ 3 * \sigma } = 2$

=> $0.08 = 6 \sigma$

=> $\sigma = 0.0133$

$\frac{ 1.68 - 1.60 }{ 3 * 0.0133 }$

=> $2$

Hence

$Cpk = min[ 2, 2 ]$

$Cpk = 2$

So $\sigma = 0.0133$ is the value of standard deviation required

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3 years ago

What is the distance between the points (80,55) and (20,44)

kaheart [24]

Distance formula : d = sqrt (x2 - x1)^2 + (y2 - y1)^2
(20,44)...x1 = 20 and y1 = 44
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d = sqrt (80 - 20)^2 + (55 - 44)^2
d = sqrt ((60^2) + (11^2))
d = sqrt (3600 + 121)
d = sqrt 3721
d = 61 <==

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