An air-conditioner which uses R-134a operates on the ideal vapor compression refrigeration cycle with a given compressor efficiency.
--Given Values--
Evaporator Temperature: T1 (C) = 9
Condenser Temperature: T3 (C) = 39
Mass flow rate of refrigerant: mdot (kg/s) = 0.027
Compressor Efficiency: nc (%) = 90

a) Determine the specific enthalpy (kJ/kg) at the compressor inlet.
Your Answer =
b) Determine the specific entropy (kJ/kg-K) at the compressor inlet
Your Answer =
c) Determine the specific enthalpy (kJ/kg) at the compressor exit
Your Answer =
d) Determine the specific enthalpy (kJ/kg) at the condenser exit.
Your Answer =
e) Determine the specific enthalpy (kJ/kg) at the evaporator inlet.
Your Answer =
f) Determine the coefficient of performance for the system.
Your Answer =
g) Determine the cooling capacity (kW) of the system.
Your Answer =
h) Determine the power input (kW)to the compressor.
Your Answer =

Answer:

The rate of heat transfer has increased.

Explanation:

Heat transfer rate is the rate at which heat energy is dissipated to the ambient from a hot body. The rate of heat transfer is proportional to the available surface area for heat exchange. This means that the greater the exposed surface area for heat exchange, the greater the rate at which heat is lost to the ambient. In introducing the fins to the heat exchange system (fins have a large surface area to volume ratio for maximum exposure to the ambient), one maximizes the available surface area for heat exchange between the material and the ambient, increasing the rate of heat transfer.

Answer:

  Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Replace the 10.0-ohm resistor with a 20.0-ohm resistor. Measure the new current. Continue replacing the resistor with a different resistor of known resistance. Measure the current for each resistor. Record all data.

Explanation:

The only design that has resistance varying with everything else remaining the same is the first design. That would be what you'd want to do if you're exploring the effect of resistance on current.

Answer:

load speeds:

For V = 100 v  speed = 188 rad/sec

For V = 75 v   speed = 138 rad/sec

For V = 50 v   speed = 88 rad/sec

Explanation:

Given data

Ra = 0.3 Ω

Ke = Kt = 0.5

torque = 10 Nm

using  a constant torque = 10 Nm we can calculate the various load speed for the given values of 100 v , 75 v, 50 v

attached below is the detailed solutions and plot

Answer:

the rate of heat loss from the duct to the attic space = 1315.44 W

the pressure difference between the inlet and outlet sections of the duct  = 7.0045 N/m²

Explanation:

We know that properties of air 80⁰C  and 1atm  (from appendix table) are;

density p = 0.9994 kg/m³, Specifice heat Cp = 1008 J/kg.⁰C

Thermal conductivity k = 0.02953 W/m.⁰C, Prandtl number Pr = 0.7154,

Kinematic viscosity v = 2.097 × 10⁻⁵ m²/s

haven gotten that, we calculate the hydraulic diameter of square duct

Dh = 4Ac / P      { Ac = is cross sectional area of duct and P = perimeter}

now we substitute a² for Ac and 4a for P ( we know from the question that a = 0.2 m)

Dh = 4a² / 4a

Dh = 4(0.2)² / 4(0.2)

Dh = 0.2 m

Now we calculate the average velocity of air

Vₐ = Vˣ / Ac        { vˣ = volume flow rate of air}

Vₐ = Vˣ / a²      { Ac = a² }, we know that a = 0.2m₂, Vˣ = 0.15 m³

Vₐ = 0.15 / (0.2)²

Vₐ = 3.75 m/s

Next we calculate the Reynolds number

Re = Vₐ Dh / V

Re = (3.75 × 0.2) / 2.097× 10⁻⁵

Re = 35765.379

The  Reynolds number IS GREATER than 10,000

so the flow is turbulent and entry length in this case is nearly 10 times the hydraulic diameter

Lh ≈ Lt ≈ 10D

= 10 × 0.2

= 2m

As this length is quite small when compared to the total of tube, we assume fully developed flow for the entire tube length.

Now we calculate the Nusselt number from this relation;

Nu =  0.023 Re⁰'⁸ Pr⁰'³

so we substitute for Re and Pr

Nu = 0.023(35765.379)⁰'⁸ (0.7154)⁰'³

Nu = 91.4

Now calculate the convective heat transfer coefficient

h = Nu × K/ Dh

we substitute

h = 91.4 × 0.02953 W/m.°C / 0.2 m

h = 13.5 W/m².°C

We calculate the surface area of the square duct

Aₓ = 4aL       { L= length of duct}

we substitute

Aₓ = 4 × 0.2 × 8

Aₓ = 6.4 m²

Mass flow rate of air

m = pVˣ

we substitute again ( from our initials)

m = 0.9994 kg/m₃ × 0.15 m³/s

m= 0.150 kg/s

We calculate the exit temperature of the air from the duct

Te = Ts - (Ts -Ti) exp ( - hAₓ / mCp)

we know that

Ts = 60°C , Ti = 80°C, h = 13.5 W/m².°C , Aₓ = 6.4m², m = 0.150 kg/s , Cp = 1008 J/kg.°C

we substitute

Te = 60 - (60-80) exp(- ((13.5 × 6.4)/(0.15 × 1008))

Te = 71.3°

Now we calculate the rate of heat loss from the duct.

Q = mCp ( Ti -Te )

we substitute again

Q = 0.150 × 1008 × ( 80 - 71.3 )

Q = 1315.44 W

Next we calculate the estimated friction factors by using Haaland equation

1/√f = - 1.8log₁₀ [ 6.9/Re + (E/D)/3.7)¹'¹¹]

we know that E/D = relative roughness = 10⁻³

we substitute

so

1/√f = - 1.8log₁₀ [ (6.9/35765.379) + ( 10⁻³/3.7)¹'¹¹]

1/√f = - 1.8log₁₀ { 0.000192924 + 0.00010947}

1/√f = - 1.8log₁₀ 0.000302324  

√f =   1/6.334

f = (1/6.334)²

f = 0.02492

We calculate the pressure difference between inlet and outlet sections of the duct

ΔPl = fLPVa² / Dh × 2

ΔPl = {0.02492 × 8 × 0.9994 × (3.75)²} / 0.2 × 2

ΔPl = 2.8018 / 0.4

ΔPl = 7.0045 N/m²

Therefore pressure deference is 7.0045 N/m²

Answer:

[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy at constant pressure:

[tex]ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}[/tex]

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

[tex]\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,[/tex]

We obtain the expression to compute the specific entropy change:

[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]

Best regards.

Answer: a = change in w =0.00656

b = q = 5.1kj/kg

Explanation:

Find explanation in the attached file

The amount of steam added to the air  a = change in w =0.00656 b = q = 5.1kj/kg

The digital game retail and distribution service Steam is provided by Valve. In order to allow Valve to automatically update its games, it was first released as a software client in September 2003. In late 2005, it was expanded to include the distribution and sale of games from other publishers.

a) We can use the absolute humidity we and wg to determine the amount

of moisture added Aw.

Aww3-W2

Aw= 0.01291 -0.00635

Aw= 0.00656

b) To determine the heat transfer q we will need the enthalpies h and h2.

kJ

kg 9 = 36.2

kJ  kg

31.1

q=5.1

kJ

kg

RESULT

Do = 0.00656

kJ

kg

9 = 5.1

Therefore, The amount of steam added to the air  a = change in w =0.00656 b = q = 5.1kj/kg

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Answer:

  leakage

Explanation:

That current is "leakage current."

Answer:

1. Yes, they are all necessary.

2. Both written and verbal communication skills are of the utmost importance in business, especially in engineering. Communication skills boost you or your teams' performance because they provide clear information and expectations to help manage and deliver excellent work.

Answer:

True

Explanation:

Bona Fide is a Latin term which means in good faith or without any intention to deceive. The business established on a bona fide basis means that there is an absence of fraud. The marketing agency has devoted its services to public relations, advertising and promoting the services of clients. There is no intention of fraud in the business.

Answer:

A) flow depth after jump = 2.46 m,  Froude number after jump = 0.464

B) head loss = 0.572 m

C) dissipation ratio = 0.173

Explanation:

Given data :

width of channel = 10-m

velocity of before jump (V1) = 7 m/ s

flow depth before jump (y1) = 0.8 m

A) determining the flow depth and the Froude's number after the jump

attached below is the solution

B) head loss

HL = Y1 -Y2 + [tex]\frac{V_{1} ^2 - V_{2} ^2}{2g}[/tex] = 0.8 - 2.46 + [tex]\frac{49 - 5.1984}{19.62}[/tex] = 0.572

C) dissipation ratio

HL / Es1 = 0.572 / 3.3 = 0.173

Es1 = 0.8 + [tex]\frac{7^2}{2*9.81}[/tex] = 0.173

Answer:

Creating an app is both an expression of our self and a reflection of what we see is missing in the world. We find ourselves digging deep into who we are, what we would enjoy working on, and what needs still need to be fulfilled. Generating an app idea for the first time can be extremely daunting. Especially with an endless amount of possibilities such as building a church app.

The uncertainty has always spawned a certain fear inside creators. The fear of creating something no one will enjoy. Spending hundreds of dollars and hours building something which might not bring back any real tangible results. The fear of losing our investment to a poor concept is daunting but not random. But simple app ideas are actually pretty easy to come by.

Great app idea generation is not a gift given to a selected few, instead, it is a process by which any of us are able to carefully explore step by step methods to find our own solution to any problem. Whether you are a seasoned creator or a novice, we have provided a few recommendations to challenge and aid you as you create your next masterpiece.

if I am right then make me brainliest

Answer:

A) The relative humidity : 0.818 (81.8%), Temperature at C = 14.4⁰c

B) The rate of energy destruction = 0.0477 kw  

Explanation:

Given data :

at point 1 : m1 = 3 kg/min , T1 = 30⁰c, p1 = 1 bar,  ∅ = 0.50 ( 50%)

at point 2 : T2 = 5⁰c,  P2 = 1 bar, m2 = 5 kg/min

at point 3 : p3 = 1 bar

Answer:

moment of inertia = 4.662 * 10^6 [tex]mm^4[/tex]

Explanation:

Given data :

Mass of machine = 400 kg = 400 * 9.81 = 3924 N

length of span = 3.2 m

E = 200 * 10^9 N/m^2

frequency = 9.3 Hz

Wm ( angular frequency ) = 2 [tex]\pi f[/tex] = 58.434 rad/secs

also Wm = [tex]\sqrt{\frac{g}{t} }[/tex]  ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t = [tex]\frac{wl^3}{48EI}[/tex]

insert the value of t into equation 1

W[tex]m^2[/tex] = [tex]\frac{g*48*E*I}{WL^3}[/tex]   make I the subject of the equation

I ( Moment of inertia about the neutral axis ) = [tex]\frac{WL^3* Wn^2}{48*g*E}[/tex]

I = [tex]\frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}[/tex]  = 4.662 * 10^6 [tex]mm^4[/tex]

Answer: answer provided in the explanation section.

Explanation:

Weather phenomenons that would impart Aviation Operations in Santa Barbara -

1. Although winters are cold, wet, and partly cloudy here. It is in general favorable for flying. But sometimes strong winds damage this pleasant weather.

2.  The Sundowner winds cause rapid warming and a decrease in relative humidity. The wind speed is very high surrounding this area for this type of wind.  

3. Cloud is an important factor that affects aviation operations. Starting from April, here the sky is clouded up to November. The sky is overcast (80 to 100 percent cloud cover) or mostly cloudy (60 to 80 percent) 44% on a yearly basis. Thus extra cloud cover can trouble aviation operations.

4. The average hourly wind speed can also be a factor. This also experiences seasonal variations, these variations are studied carefully in the aviation industry. The windier part of the year starts in January and ends in June. In April, the wind speed can reach 9.5 miles per hour.

This and more are some factors to look into when considering wheather conditions that would affect aviation operations.

I hope this was a bit helpful. cheers

Answer: hello attached below is the diagram which is part of your question

Total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k  it violates Clausius increase of entropy which is Sgen > 0

Explanation:

Clausius statement states that it is impossible to transfer heat energy from a cooler body to a hotter body in a cycle or region without any other external factors affecting it .  

applying the increase in entropy principle to prove this

temp of cold reservoir (t hot)= 600 k

temp of hot reservoir(t cold) = 1220 k

energy (q) = 100 kj

total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k

entropy change in cold reservoir = Q/t cold = 100 / 600 = -0.166 kj/k

entropy change in hot reservoir = Q / t hot = 100 / 1220 = 0.083 kj/k

hence it violates  Clausius inequality of increase of entropy principle which is states that generated entropy has to be > 0

Answer:

Explanation:

In a particular application involving airflow over a heated surface, the boundary layer temperature distribution, T(y), may be approximated as:

[ T(y) - Ts / T∞ - Ts ] = 1 - e^( -Pr (U∞y / v) )

where y is the distance normal to the surface and the Prandtl number, Pr = Cpu/k = 0.7, is a dimensionless fluid property. a.) If T∞ = 380 K, Ts = 320 K, and U∞/v = 3600 m-1, what is the surface heat flux? Is this into or out of the wall? (~-5000 W/m2 , ?). b.) Plot the temperature distribution for y = 0 to y = 0.002 m. Set the axes ranges from 380 to 320 for temperature and from 0 to 0.002 m for y. Be sure to evaluate properties at the film temperature.

Answer:

Average heat transfer coefficient =  31 kw/m^2 k

Heat transfer rate per meter length of pipe =  116.808 KW

Explanation:

water temperature = 20⁰c,  

free-stream velocity = 1.5 m/s

circular pipe diameter = 2.0 cm = 0.02 m

surface temperature = 80⁰c

A) calculate average heat transfer coefficient

we apply the formula below :

m = αAv

A (area) = [tex]\pi /4 (d)^2[/tex]

m = 10^3 * [tex]\pi / 4 ( 0.02)^2[/tex] * 1.5

   = 10^3 * 0.7857( 0.0004) * 1.5

   = 0.4714 kg/s

Average heat transfer coefficient  

h = [tex]\frac{m(cp)}{A}[/tex]  ,  A = [tex]\pi DL[/tex]

L = 1 m , m = 0.4714 kgs , cp = 4.18

back to equation

h = [tex]\frac{0.4714*4.18}{\pi * 0.02 }[/tex]   = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k

B) Heat transfer rate per meter length of pipe

Q = ha( ΔT ),  a = [tex]\pi DL[/tex]

   = 31 * 0.0628 * ( 80 - 20 )

  = 31 * 0.0628 * 60 = 116.808 KW

Answer:

A) 13.80 ft/s^2

B) 1.714 Ib

Explanation:

Magnitude of acceleration center G

mass = W / g = 8 / 32.2 = 0.2484 Ib.s^2/ft

calculate the acceleration along x direction

A = ra

r = radius

a = angular acceleration

A = 6 in [tex]\frac{1 ft}{12 in}[/tex] * a

a= 2A

equation of the plane along the x-direction

w sin∅ - F = ma

8* sin40 - F = 0.2484 * a

hence F = 5.1423 - 0.2484 a

next find the moment of inertia along the z axis

I = 1/2 mr^2

  = 1/2 * 0.2484 * (6/12)^2  = 0.03105 Ib.ft.s^2

Applying moment balance equation

F * r = inertia * a

(5.1423 - 0.2484 a)*0.5 = 0.03105 * 2A

2.57115 = 0.1863 A      hence

A = 13.80 ft/s^2  ( acceleration of the cylinder )

B) Calculate the friction force exerted by the ramp on the cylinder

F = 5.1423 - 0.2484 A

  = 5.1423 - 0.2484 ( 13.80 )

  = 1.714 Ib

The magnitudes of the acceleration and the friction force are;

Acceleration = 13.8 ft/s²

Friction Force = 1.714 lb

The image of the solid homogeneous cylinder is missing and so i have attached it.

From the image we see that;

We are given;

      We know that formula for weight is; W = mg

Thus; m = W/g

where g is acceleration due to gravity = 32.2 ft/s²

m = 8/32.2

mass; m = 0.2484 lb.s²/ft

     Now, to get the acceleration along the x-axis, we will use the formula;

a = rα

where α is angular acceleration. Thus;

a = 0.5α

α = 2a   ----- (eq 1)

    Now, resolving forces along the x-direction gives;

W*sinθ - F = ma

Plugging in the relevant values;

8*sin 40 - F = 0.2484a

F = 8*sin 40 - 0.2484a    -----(eq 2)

     Now, moment of inertia of the cylinder along the z-axis is gotten from;

I = ¹/₂mr²

I = ¹/₂ × 0.2484 × 0.5²

I = 0.03105 lb.ft/s²

     Taking equilibrium of moments we have;

F*r = I*α

Thus;

(8*sin 40 - 0.2484a)0.5 = 0.03105α

⇒ 2.57115 - 0.1242a = 0.03105α

⇒ 0.03105α + 0.1242a = 2.57115

From eq 1, α = 2a. Thus;

0.03105(2a) + 0.1242a = 2.57115

0.1863a = 2.57115

a = 2.57115/0.1863

a = 13.8 ft/s²

F = 8*sin 40 - 0.2484a

F = 5.1423 - 0.2484(13.8)

F = 1.714 lb

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Answer:

D) is due to an increase of the armature current.

Explanation:

Option D is correct because on the DC motor, when the load increases, it leads to an increase in the armature current.

The armature current then sets up a magnetic flux which opposes the main field flux. The net field flux gets reduced. It is at this point, the armature reaction occurs.

Armature reaction is seen as the effect of magnetic flux which is usually set up by an armature current. This occurs when there is the distribution of flux under the main poles.

There are two effects the armature flux causes on the main field flux.

1. The main field flux is distorted by the armature reaction.

2. The magnitude of the main field flux is reduced by the armature flux.

Answer:

B) gate-source junction is reverse-biased

Explanation:

FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow"   in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".

In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

[tex]n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm[/tex]

2) The speed of the rotor is the motor speed. The slip is given by:

[tex]Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm[/tex]

3) The frequency of the rotor is given as:

[tex]f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz[/tex]

4) At standstill, the speed of the motor is 0, therefore the slip is 1.

The frequency of the rotor is given as:

[tex]f_r=slip*f_s\\f_r=1*60=60\ Hz[/tex]

Answer:

a) Mt = 0.0023229

b) = U1 = 214.07

c) = V₁  = 0.861 m³/kg

d) = Vr1 = 621.2

Explanation:

Given that

R = 0.287 KJ/kg.K, T1 = 300 K , P1 = 100 kPa , V1 = 500 cm³, r = 10 , Q = 60 kW , Speed N = 5600 RPM, Number of cylinders K = 4

specific heat at constant volume Cv = 0.7174 kJ/kg.K

Specific heat at constant pressure is 1.0045 Kj/kg.K

a)  To determine the total mass (kg) of air in the engine.

we say

P1V1 = mRT1

we the figures substitute

(100 x 10³) ( 500 x  10⁻⁶) = m ( 0.287 x  10³) ( 300 )

50 = m x 86100

m = 0.00005 / 86100 = 0.0005807 ( mass of one cylinder)

Total mass of 4 cylinder

Mt = m x k

Mt = 0.0005807 x 4

Mt = 0.0023229

b) To determine the specific internal energy (kJ/kg) at state 1

i.e at T1 = 300

we obtain the value of specific internal energy U1 at 300 K ( state 1) from the table ideal gas properties of air.

U1 = 214.07

c) To determine the specific volume (m³/kg) at state 1.

we say

V₁ = V1/m

V₁ = (500 x  10⁻⁶) / 0.0005807

V₁  = 0.861 m³/kg

d) To determine the relative specific volume at state 1.

To obtain the value of relative specific volume at 300 K ( i.e state 1) from the table ideal gas properties of air.

At T1 = 300 k

Vr1 = 621.2

Answer:

[tex]w_{out}=319.1\frac{BTU}{lbm}[/tex]

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

[tex]h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}[/tex]

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

[tex]s_2=s_1[/tex]

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

[tex]x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965[/tex]

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

[tex]h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}[/tex]

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

[tex]h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}[/tex]

Best regards.

Answer: A prototype

Explanation:

The scale model that proves the initial concept is called a domain model.

A copy or depiction of something where all parts have the same dimensions as the original. A scale model is an image or copy of an object that is either larger or smaller than the object being represented's actual size.

A domain model is a type of conceptual model that is used to depict the structural elements and conceptual constraints within a domain of interest.

A domain model will include all of the entities, their attributes, and relationships, as well as the constraints that govern the conceptual integrity of the structural model elements that comprise that problem domain.

Therefore, a domain model is the scale model that proves the initial concept.

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Answer:

efficiency

Explanation:

Answer:

The correct answer is Efficiency.

Explanation:

I got it right on the plato test.

Answer:

a. 115

b. 135

c. 220

d. 185

Explanation:

Spheriodite is microscopic constituents in some steels which is composed of spherically shaped cementide particle. It is most ductile and softest type of steel. Pearlite is two phased lamellar compose of alternating layer of ferrite and cementite. It is hard and strong but not tough. It is applied on cutting tools like chopper, blades and knives.

Answer:

As the asteroid falls closer to the Earth's surface its Gravitational Potential energy decreases and its Kinetic energy increases.

Answer:

The answers to your questions are given below.

Explanation:

The following data were obtained from the question:

Red (R) = 4

White (W) = 7

1. Determination of the sample space, S.

The box contains 4 red balls and 7 white balls. Therefore, the sample space (S) can be written as follow:

S = {R, R, R, R, W, W, W, W, W, W, W}

nS = 11

2. Determination of the probability of all results that appeared in the sample space.

From the question, we were told that the two balls was drawn with return. There, the probability of all results that appeared in the sample space can be given as follow:

i. Probability that the first draw is red and the second is also red.

P(R1) = nR/nS

Red (R) = 4

Space space (S) = 11

P(R1) = nR/nS

P(R1) = 4/11

P(R2) = nR/nS

P(R2) = 4/11

P(R1R2) = P(R1) x P(R2)

P(R1R2) = 4/11 x 4/11

P(R1R2) = 16/121

Therefore, the Probability that the first draw is red and the second is also red is 16/121.

ii. Probability that the first draw is red and the second is white.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(R) = nR/nS

P(R) = 4/11

P(W) = nW/nS

P(W) = 7/11

P(RW) = P(R) x P(W)

P(RW) = 4/11 x 7/11

P(RW) = 28/121

Therefore, the probability that the first draw is red and the second is white is 28/121.

iii. Probability that the first draw is white and the second is also white.

White (W) = 7

Space space (S) = 11

P(W1) = nW/nS

P(W1) = 7/11

P(W2) = nW/n/S

P(W2) = 7/11

P(W1W2) = P(W1) x P(W2)

P(W1W2) = 7/11 x 7/11

P(W1W2) = 49/121

Therefore, the probability that the first draw is white and the second is also white is 49/121.

iv. Probability that the first draw is white and the second is red.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(W) = nW/nS

P(W) = 7/11

P(R) = nR/nS

P(R) = 4/11

P(WR) = P(W) x P(R)

P(WR) = 7/11 x 4/11

P(WR) = 28/121

Therefore, the probability that the first draw is white and the second is red is 28/121.

Answer:

b. What is the operator's efficiency for the day?

                                                      AND

e. What would be the proper piece rate (rate expressed in money) for this job, assuming that the above time standard is correct?

Explanation: