radius=diameter/2
therefore r=3÷2 ie 1.5m
area=pi (r^2)m^2
=1.5^2 pi m^2
2.25 pi m^2 is required area
In order to find x-axis interception coordinates you need to set other two variables (y and z) to zero and solve the remaining equation.
So the following simple equations need to be solved:
2x = 40
5y = 40
-4z = 40
It intercepts x-axis at coordinates (20;0;0), y-axis at (0;8;0) and z-axis at (0;0;-10).
Answer:
a)
USL = 6.2 inches
LSL = 5.8 inches
b) Cp = 1.33
Cpk = 0.67
c)
Yes it meets all specifications
Step-by-step explanation:
The specification for a plastic handle calls for a length of 6.0 inches ± .2 inches. The standard deviation of the process is estimated to be 0.05 inches. What are the upper and lower specification limits for this product? The process is known to operate at a mean thickness of 6.1 inches. What is the Cp and Cpk for this process? Is this process capable of producing the desired part?
Given that:
Mean (μ) = 6.1 inches, Standard deviation (σ) = 0.05 inches and the length of the plastic handle is 6.0 inches ± .2
a) Since the length of the plastic handle is 6.0 inches ± .2 = (6 - 0.2, 6 + 0.2)
The Upper specification limits (USL) = 6 inches + 0.2 inches = 6.2 inches
The lower specification limits (LSL) = 6 inches - 0.2 inches = 5.8 inches
b) The Cp is given by the formula:

The Cpk is given by the formula:
c)
The upper specification limit lies about 3 standard deviations from the centerline, and the lower specification limit is further away, so practically all units will meet specifications
