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:
Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.
Explanation:
Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow. You can also see that 
represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:
- Wall Thickness. More thickness, more thermal resistance.
- Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
- Cross-section Area. More cross-section area, less thermal resistance.
A expression to define the thermal resistance for the wall is as follows: 
, where l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.
Answer:
ambages. pitiless or without compassion; cruel; merciless. winding, roundabout paths or ways.
Explanation:
Answer: hello some aspects of your question is missing below is the missing information
The gas tank is made from A-36 steel and has an inner diameter of 1.50 m.
answer:
≈ 22.5 mm
Explanation:
Given data:
Inner diameter = 1.5 m
pressure = 5 MPa
factor of safety = 1.5
<u>Calculate the required minimum wall thickness</u>
maximum-shear-stress theory ( σ allow ) = σγ / FS
= 250(10)^6 / 1.5 = 166.67 (10^6) Pa
given that |σ| = σ allow
3.75 (10^6) / t = 166.67 (10^6)
∴ t ( wall thickness ) = 0.0225 m ≈ 22.5 mm
