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saw5 [17]
3 years ago
14

If p = .8 and n = 50, then we can conclude that the sampling distribution of pˆ p ^ is approximately a normal distribution.

Engineering
1 answer:
defon3 years ago
4 0

Answer:

It is true that the sampling distribution of the proportions is approximately a normal distribution.

Explanation:

Solution

Given that:

N = 50

P= 0.8

Thus

p ^ =√p(1-p)/n

p ^=√0.8 (1-0.8)/50

p ^= 0.0566

Therefore,we conclude that the sampling distribution of the proportions is approximately a normal distribution.

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The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
Sedaia [141]

Answer:

a) 254.6 GPa

b) 140.86 GPa

Explanation:

a) Considering the expression of rule of mixtures for upper-bound and calculating the modulus of elasticity for upper bound;

Ec(u) = EmVm + EpVp

To calculate the volume fraction of matrix, 0.63 is substituted for Vp in the equation below,

Vm + Vp = 1

Vm = 1 - 0.63

Vm = 0.37

In the first equation,

Where

Em = 68 GPa, Ep = 380 GPa, Vm = 0.37 and Vp = 0.63,

The modulus of elasticity upper-bound is,

Ec(u) = EmVm + EpVp

Ec(u) = (68 x 0.37) + (380 x 0.63)

Ec(u) = 254.6 GPa.

b) Considering the express of rule of mixtures for lower bound;

Ec(l) = (EmEp)/(VmEp + VpEm)

Substituting values into the equation,

Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)

Ec(l) = 25840/183.44

Ec(l) = 140.86 GPa

6 0
4 years ago
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of wid
dmitriy555 [2]

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

T_{min} = 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, d_c = 0.5 mm

Number of passes necessary for this reduction, n = \frac{\triangle d}{d_c}

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec

minimum time required to reduce the depth of the plate by 20 mm:

T_{min} = number of passes * Time/pass

T_{min} = n * Time/pass

T_{min} = 40 * 40

T_{min} =  1600 = 26 mins 40 secs

3 0
4 years ago
Read 2 more answers
primitive transportation and storage systems that make local distribution ineffective if not impossible, the lack of clean water
Vera_Pavlovna [14]

Primitive transportation and storage systems that make local distribution ineffective if not impossible, and the lack of clean water and the lack of effective sewer systems are all examples of the barrier called physical & environmental barriers.

<h3>What are the problems with having physical & environmental barriers?</h3>

Time, Place, Space, Climate, and Noise are the key environmental/physical constraints. Some are simple to change, while others may prove to be difficult.

Physical barriers are essentially defined by three primary factors: environment, distance, and medium unawareness. Environmental barriers, on the other hand, are connected to variables that occur in the current environment.

Learn more about transportation:
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5 0
1 year ago
PLEASE HELP, TEST MULTIPLE CHOICE QUESTIONS
Rina8888 [55]

Answer:

c

Explanation:

dbfjex vadamhqrmtwg

8 0
3 years ago
Read 2 more answers
Water enters the pump of a steam power plant as saturated liquid at 20 kPa at a rate of 45 kg/s and exits at 6 MPa. Neglecting t
Natasha2012 [34]

Answer:

\dot W_{in} = 273.69\,kW

Explanation:

The pump is modelled after the First Law of Thermodynamics. A reversible process means that fluid does not report any positive change in entropy:

\dot W_{in} + \dot m \cdot (h_{in}-h_{out}) = 0

The properties of the fluid at entrance and exit are, respectively:

Inlet (Saturated Liquid)

P = 20\,kPa

T = 60.06\,^{\textdegree}C

h = 251.42\,\frac{kJ}{kg}

s = 0.8320\,\frac{kJ}{kg\cdot K}

Outlet (Subcooled Liquid)

P = 6000\,kPa

T = 60.06\,^{\textdegree}C

h = 257.502\,\frac{kJ}{kg}

s = 0.8320\,\frac{kJ}{kg\cdot K}

The power input to the pump is computed hereafter:

\dot W_{in} = \left(45\,\frac{kg}{s} \right)\cdot \left(257.502\,\frac{kJ}{kg} -251.42\,\frac{kJ}{kg} \right)

\dot W_{in} = 273.69\,kW

8 0
4 years ago
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