Answer:
Days: 6.9444 days
Production rate: 547.2035 ft²/s
Explanation:
the solution is attached in the Word file
Answer:
Option A
Chemical engineering
Explanation:
Chemical engineering mainly encompass the study of behavior of different particles such as petroleum, water, drugs and other products. When Anne is involved in a study with engineers who study flow of particles, the flow, viscosity and other properties are among the behavior that chemical engineers are involved in.
Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ =
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ =
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
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It is a criminal offense for any person to give unauthorized advance notice of an OSHA inspection.
Answer: the standard deviation STD of machine B is s (Lb) = 0.4557
Explanation:
from the given data, machine A and machine B produce half of the rods
Lt = 0.5La + 0.5Lb
so
s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)
but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)
so we substitute
s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)
0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)
0.64 = 0.25 + s²(Lb) + 0.4s(Lb)
s²(Lb) + 0.4s(Lb) - 0.39 = 0
s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2
s (Lb) = 0.4557
therefore the standard deviation STD of machine B is s (Lb) = 0.4557