9514 1404 393
Answer:
- central: XWR, VWU
- inscribed: VST
Step-by-step explanation:
Point W is the center of the circle. Any angle with W as its vertex is a central angle:
angles XWR and VWU are central angles
Any angle with its vertex on the circle and rays that intersect the circle is an inscribed angle.
angle VST is an inscribed angle
Answer:
Step-by-step explanation:
the formula for a circle which would be C = 2 \pi r
The 9.5*3.14 divide by two so 14.915
and this rounded would be 15.0 or 14.9 not too sure just pick whatever
Answer:
264
Step-by-step explanation:
well you could have just used a calculator!
Answer: -5
Step-by-step explanation:
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
Step-by-step explanation:
From the question we are told that
The upper specification is ![USL = 1.68 \ mm](https://tex.z-dn.net/?f=USL%20%20%3D%20%201.68%20%5C%20mm)
The lower specification is
The sample mean is
The standard deviation is ![\sigma = 0.03 \ mm](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%200.03%20%5C%20mm)
Generally the capability index in mathematically represented as
![Cpk = min[ \frac{USL - \mu }{ 3 * \sigma } , \frac{\mu - LSL }{ 3 * \sigma } ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%20%5Cfrac%7BUSL%20-%20%20%5Cmu%20%7D%7B%203%20%2A%20%20%5Csigma%20%7D%20%20%2C%20%20%5Cfrac%7B%5Cmu%20-%20LSL%20%7D%7B%203%20%2A%20%20%5Csigma%20%7D%20%5D)
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} ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%20%5Cfrac%7B1.68%20-%20%201.6%20%7D%7B%203%20%2A%20%200.03%20%7D%20%20%2C%20%20%5Cfrac%7B1.60%20-%20%201.52%20%7D%7B%203%20%2A%20%200.03%7D%20%5D)
=> ![Cpk = min[ 0.88 , 0.88 ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%200.88%20%2C%200.88%20%20%5D)
So
![Cpk = 0.88](https://tex.z-dn.net/?f=Cpk%20%20%3D%200.88)
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](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmu%20-%20LSL%20%20%7D%7B%203%20%2A%20%20%5Csigma%20%7D%20%3D%20%202)
So
![\frac{ 1.60 - 1.52 }{ 3 * \sigma } = 2](https://tex.z-dn.net/?f=%5Cfrac%7B%201.60%20-%20%201.52%20%20%7D%7B%203%20%2A%20%20%5Csigma%20%7D%20%3D%20%202)
=> ![0.08 = 6 \sigma](https://tex.z-dn.net/?f=0.08%20%3D%206%20%5Csigma)
=> ![\sigma = 0.0133](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%200.0133)
So
![\frac{ 1.68 - 1.60 }{ 3 * 0.0133 }](https://tex.z-dn.net/?f=%5Cfrac%7B%201.68%20%20-%201.60%20%7D%7B%203%20%2A%20%200.0133%20%7D)
=> ![2](https://tex.z-dn.net/?f=2)
Hence
![Cpk = min[ 2, 2 ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%202%2C%202%20%5D)
So
![Cpk = 2](https://tex.z-dn.net/?f=Cpk%20%20%3D%202)
So
is the value of standard deviation required