The given question is incomplete, the complete question is as follows:
Our text describes a trade-off that we must make as engineers between our confidence in the value of a parameter versus the precision with which we know the value of that parameter. That trade-off might be affected by whether we are looking at a two-sided or bounded (one-sided) interval.
Question: Discuss your interpretation of the confidence-precision trade-off, and provide a few examples of how you might make a choice in one direction or the other in an engineering situation.
Answer: A balancing point is required to be reached to obtain a better confidence level in the predicted values.
Explanation:
The confidence interval and precision are the two terms that aims at providing the accurate estimation of the measurability of an object. If the precision increases, we can compromise on the confidence level and if the confidence level increases, then the precision of the predicted value also dilutes.
Thus a balance point is required to be reached between these two variables so that we get better confidence in the values being predicted without losing the correct estimation on precision. Ensuring that both the confidence and precision are maintained.
Answer:
The depth to which the cage will drop for each single rotation of the drum is approximately 18.85 m.
Explanation:
The mine winch drum consists of a drum around around which the haulage rope is attached
Therefore, given that the diameter of the winch drum = 6 m
The length of rope unwound by each rotation = How far the cage will drop for each single rotation of the drum
The length of the rope unwound by each rotation = The rope that goes round the circumference of the winch drum, once
∴ Since the rope that goes round the circumference of the winch drum, once = The circumference of the winch drum, we have;
The length of the rope unwound by each rotation = The circumference of the winch drum = π × The diameter of the winch drum
The length of the rope unwound by each rotation = π × 6 m = 6·π m
The length of rope unwound by each rotation = How far the cage will drop for each single rotation of the drum = 6·π m
How far the cage will drop for each single rotation of the drum = 6·π m ≈ 18.85 m
The depth to which the cage will drop for each single rotation of the drum ≈ 18.85 m.
Answer:
=>> 167.3 kpa.
=>> 60° from horizontal face.
Explanation:
So, we are given the following data or parameters or information which is going to assist us in solving this kind of question;
=>> "A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. "
=>>"If the deviator stress is 120 kPa and the shear strength parameters are a friction angle of 30° and a cohesion of 5 kPa."
The orientation of this plane with respect to the major principle stress plane = 50 tan^2 (45 + 30/2) + 10 tan ( 45 + 30/2).
magnitude of the stresses on the failure plane = 167.3 kpa.
The orientation of this plane with respect to the major principle stress plane => x = 60 cos 60° = 30kpa.
y = 60 sin 60° = 30√3 = sheer stress.
the orientation of this plane with respect to the major principle stress plane.
Theta = 45 + 15 = 60°.
Answer:
where P is density of air and v flow is volume flow rate, so in order to find velocity of air and mass flow rate formula is given by :
so velocity is:
