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kakasveta [241]
2 years ago
14

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.
Engineering
1 answer:
LenKa [72]2 years ago
8 0

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.

You might be interested in
Twenty distinct cars park in the same parking lot every day. Ten of these cars are US-made, while the other ten are foreign-made
Zina [86]

Answer:

Total no. of ways to line up cars is 20! = 2.43 c 10^18

Probability that the cars alternate is 0.00001 or 0.001%

Explanation:

Since, the position of a car is random.Therefore, number ways in which cars can line up is given as:

<u>No. of ways = 20! = 2.43 x 10^18</u>

For the probability that cars alternate, two groups will be formed, one consisting of US-made 10 cars and other containing 10 foreign made. The number of favorable outcomes for this can be found out as the arrangements of 2! between these groups multiplied by the arrangements of 10! for each group, due to the arrangements among the groups themselves.

Favorable Outcomes = 2! x 10! x 10!

Thus the probability of event will be:

Probability = Favorable Outcomes/Total No. of Ways

Probability = (2! x 10! x 10!)/20!

<u>Probability = 0.00001 = 0.001%</u>

4 0
2 years ago
Bananas are to be cooled from 28°C to 12°C at a rate of 1140 kg/h by a refrigerator that operates on a vapor-compression refrige
Lera25 [3.4K]

Answer:

A) COP = \frac{16.97}{9.8} = 1.731

B) P_{IN} = 0.4763

C) Second law efficiency 4.85%

exergy destruction for the cycle = 9.3237 kW

Explanation:

Given data:

T_1 = 28 degree celcius

T_2 = 12 degree celcius

\dot m = 1140 kg/h

Power to refrigerator = 9.8 kW

Cp = 3.35 kJ/kg degree C

A) Q = \dot m Cp \Delta T

        = 1140 \times 3.35\times (28-12) = 61,104 kJ/h

Q_{abs} = 61,104 kJ/h = 16.97 kJ/sec

COP = \frac{16.97}{9.8} = 1.731

b)

COP ∝ \frac{1}{P_{in}}

P_{in} wil be max when COP maximum

taking surrounding temperature T_H = 20 degree celcius

COP_{max} = \frac{T_L}{T_H- T_L} = \frac{285}{293 - 285} = 35.625

we know that

COP = \frac{heat\ obsorbed}{P_{in}}

P_{IN} = \frac{16.97}{35.62} = 0.4763

c) second law efficiency

\eta_{11} = \frac{COP_R}{(COP)_max} = \frac{1.731}{35.625} = 4.85\%

exergy destruction os given as X = W_{IN} - X_{Q2}

                                                         = 9.8 - 0.473 = 9.3237 kW

8 0
2 years ago
Consider a 1.80-m-tall man standing vertically in water and completely submerged in a pool. Determine the difference between the
defon

Answer:

17.658 kPa

Explanation:

The hydrostatic pressure of a fluid is the weight of a column of that fluid divided by the base of that column.

P = \frac{weight}{base}

Also, the weight of a column is its volume multiplied by it's density and the acceleration of gravity:

weight = \delta * v * g

Meanwhile, the volume of a column is the area of the base multiplied by the height:

V = base * h

Replacing:

P = \frac{\delta * base * h * g}{base}

The base cancels out, so:

P = \delta * h * g

The pressure depends only on the height of the fluid column, the density of the fluid and the gravity.

If you have two point at different heights (or depths in the case of objects submerged in water) each point will have its own column of fluid exerting pressure on it. Since the density of the fluid and the acceleration of gravity are the same for both points (in the case of hydrostatics density is about constant for all points, it is not the case in the atmosphere), we can write:

\Delta P = \rho * g * (h1 - h2)

We do not know at what depth the man of this problem is, but it doesn't matter, because we know the difference in height of the two points of interes (h1 - h2) = 1.8 m. So:

\Delta P = 9.81 \frac{m}{s^{2} } * 1000 \frac{kg}{m^3} * 1.8 m = 17658 Pa = 17.658 kPa

4 0
2 years ago
2.5 kg of air at 150 kPa and 12°C is contained in a gas-tight, frictionless piston-cylinder device. The air is now compressed to
Korolek [52]

Answer:

Work input =283.47 KJ

Explanation:

Given that

P_1=150\ KPa

P_2=600\ KPa

T=12°C=285 K

m= 2.5 kg

Given that this is the constant temperature process.

e know that work for isothermal process  

W=P_1V_1\ln \dfrac{P_1}{P_2}

W=mRT\ln \dfrac{P_1}{P_2}

So now putting the values

W=mRT\ln \dfrac{P_1}{P_2}

W=2.5\times 0.287\times 285\ln \dfrac{150}{600}

W=-283.47 KJ

Negative sign indicates that work is done on the system.

So work input =283.47 KJ

8 0
2 years ago
1. Consider the steady flow in a water pipe joint shown in the diagram. The areas are:
marshall27 [118]

Answer:

-4.5 m/s

Explanation:

Assuming steady and incompressible flow and uniform properties at each section

V_1A_1+V_2A_2+A_3V_3+Q_4=0

Here V is velocity of flow and A is area, Q is flow rate out of the leak, subscript 1-4 represent different sections

At the surface,  is negative hence the equation above will be

-V_1A_1+V_2A_2+A_3V_3+Q_4=0

Making  the subject of the formula then

V_2=\frac {V_1A_1-A_3V_3-Q_4}{A_2}

Substituting the given values then

V_2=\frac {(5\times 0.2)-(12\times 0.15)-0.1}{0.2}=-4.5 m/s\\V_2=-4.5m/s

8 0
2 years ago
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