What is 7/20 in decimals
2 answers:
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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
Answer:
Approximately 59 stacked cups.
Step-by-step explanation:
Given,
Height of a cup = 12.5 cm,
Two stacked cups = 14 cm,
Three stacked cups = 15.5 cm,
........, so on,....
Thus, there is an AP that represents the given situation,
12.5, 14, 15.5,....
First term is, a = 12.5,
Common difference, d = 1.5 cm,
Thus, the height of x cups is,
According to the question,
h(x) = 200
⇒ 1.5x + 11 = 200
⇒ 1.5x = 189
⇒ x = 59.3333333333 ≈ 59,
Hence, approximately 59 stacked cups will need.