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
Answer:
The required heat flux = 12682.268 W/m²
Explanation:
From the given information:
The initial = 25°C
The final = 75°C
The volume of the fluid = 0.2 m/s
The diameter of the steel tube = 12.7 mm = 0.0127 m
The fluid properties for density 
= 1000 kg/m³
The mass flow rate of the fluid can be calculated as:
To estimate the amount of the heat by using the expression:
q = 0.0253 × 4000(75-25)
q = 101.2 (50)
q = 5060 W
Finally, the required heat of the flux is determined by using the formula:
q" = 12682.268 W/m²
The required heat flux = 12682.268 W/m²
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