A product has demand of 4,000 units per year. Ordering cost is $20 and holding cost is $4 per unit per year. The cost-minimizing solution for this product is to order all 4,000 units at one time 200 units per order 1000 units per order O400 units per order A PERT activity has an optimistic time of 3 days, pessimistic time of 15 days and an expected time of 7 days. The most likely time of the activity is 9 days 6 days 7 days 8 days

A pie chart is a circular graphical representation used to display data as slices of a whole. It is commonly used to illustrate proportions or percentages. The three main chart elements included in a pie chart are Slices, Labels, Legend.

1. Slices: The pie chart is divided into slices, also known as sectors or wedges, which represent different categories or data points. The size of each slice is proportional to the value it represents, reflecting the relative importance or quantity of the data.

2. Labels: Each slice is typically accompanied by a label that identifies the category or data point it represents. The labels are placed either inside or outside the corresponding slice, allowing viewers to easily understand the meaning of each segment.

3. Legend: The legend is a key component of the pie chart that provides additional information about the categories or data points represented by the slices. It displays the corresponding labels and their respective colors or patterns, helping viewers to interpret the chart accurately.

Together, these three elements work harmoniously to present the data in a visually appealing and informative manner, enabling viewers to grasp the distribution or composition of the whole dataset at a glance.

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The time constant of the system is not defined because the output signal remains at zero for all values of t. The sensitivity of the system is a = 0.

The time constant and sensitivity of a measurement system can be determined from the equation y = a (1 - e ^ -t/τ). Here, τ is the time constant and a is the sensitivity. The output signal is initially steady at 60 units and the input signal is then suddenly increased to 90 units.

The equation of the system is given by:

y = a (1 - e ^ -t/τ)

Given that the initial output signal is steady at 60 units, it implies that when t=0, y=60. Thus the equation becomes:

60 = a (1 - e ^ 0/τ)

60 = a (1 - 1)

60 = 0

This equation is not possible. It implies that the value of a is zero. Thus, the equation of the system is simplified to:

y = 0

Therefore, the answer  is a = 0.

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Angular velocity: The rate of change of angular displacement of a body with respect to time is angular velocity. It is de Angular velocity:

Angular velocity is a physical quantity used to describe the rate of change of angular displacement of an object or particle with respect to time. It is measured in radians per second (rad/s) or degrees per second (°/s). The angular velocity is a vector quantity and its direction is perpendicular to the plane of rotation. The magnitude of the angular velocity depends on the rate of change of the angle swept by an object as it rotates about a fixed point. It is commonly used in fields such as physics, engineering, and astronomy to describe the motion of rotating objects or particles. It is also an important parameter for understanding the dynamics of rotating systems and can be used to calculate other physical quantities such as angular acceleration, torque, and moment of inertia.

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Answer:

Alvaro designs machinery and equipment for ships. His job title is best described as Marine Engineer.

Jeanette draws maps and plans for freeway projects. Her job title is best described as Civil Drafter.

Joseph designs a new airport. His job title is best described as Transportation Engineer.

Geraldine designs imaging systems that use light. Her job title is best described as Photonics Engineer.

Hope this helps sorry I am late to respond, but anyone else who needs the answer it is here.

Based on the above, Alvaro is a Marine Engineer;

Machinery and Equipment are known to be a type of industrial tools, devices, and items that aids facilities.

Conclusively, They are said to be tangible personal property that are used as a tool for repair parts, etc. Professions in ship jobs includes, Marine Engineer, Photonics Engineer, etc.

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Structural Safety Assessment and Traffic Control Strategies of Widened Highway Bridges under Maintenance Works Requiring Partial Lane Closure journal is based on the  way and techniques that can be used in maintaining  the Safety on the road as well as the Assessment of Traffic Control.

There are Devices that are typically used in the traffic control, and this can  include speed bumps as well as the barricades to block streets.

Others techniques are:

Traffic restraint can also entails the includes programs  that can foster bicycle as well as the pedestrian travel.

The rapid structural safety assessment help in inspecting and evaluating the concrete structure as well in determining  the damaged structure if it is  unsafe for personnel within the building .

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The error message suggests a type mismatch between the parameter and the value assigned, specifically between 'std string (*)[*]' and 'std string [()][()]'.

The type mismatch error occurs because the code is attempting to initialize a parameter of one type with a value of a different type. The parameter is expected to be a pointer to a string array, but the value being passed is a two-dimensional string array. This mismatch in types results in the compilation error.

To resolve the error, the code needs to be modified to ensure that the parameter and the value being passed have compatible types. This could involve changing the parameter type or modifying the value being passed to match the expected type.

Type mismatches and how to resolve them in programming to ensure correct parameter initialization and value assignment.

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Answer:

Increase in stroke volume brings about increase in pump rate .the relationship between them is linearly proportional.

Explanation:

Increase in stroke volume brings about increase in pump rate .the relationship between them is linearly proportional.

The rate at which blood is been pumped by the heart depends on the Stroke volume as well as heart rate. The stroke volume gives the quantity of blood that is been pumped by the heart Everytime it beats.

One of the important factor that determine the Cardiac output is Stroke volume. And injection fraction is calculated as (stroke volume/

end-diastolic volume)

or...................

Answer:

i believe it is true

Explanation:

Similarities between magnetic field and electric field:Both magnetic fields and electric fields are fundamental concepts in electromagnetism.They are both vector fields, meaning they have both magnitude and direction.

Both fields are produced by a source. Electric fields are produced by electric charges, while magnetic fields are produced by moving charges (currents) or changing electric fieldsBoth fields exhibit the property of superposition, meaning that the total field at a point is the vector sum of the individual fields produced by each source.Both fields obey the principle of inverse square law, which states that the strength of the field decreases with the square of the distance from the source.Differences between magnetic field and electric field:Electric fields are produced by stationary electric charges, while magnetic fields are produced by moving electric charges or changing electric fields.

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Answer:

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

Explanation:

We must estimate the monthly storage change of the lake by considering inflows, outflows, seepage and evaporation losses and precipitation. That is:

\(\Delta V_{storage} = V_{inflow} -V_{outflow}-V_{seepage}-V_{evaporation}+V_{precipitation}\)

Where \(\Delta V_{storage}\) is the monthly storage change of the lake, measured in cubic feet.

Monthly inflow

\(V_{inflow} = \left(30\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{inflow} = 77.76\times 10^{6}\,ft^{3}\)

Monthly outflow

\(V_{outflow} = \left(27\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{outflow} = 66.98\times 10^{6}\,ft^{3}\)

Seepage losses

\(V_{seepage} = s_{seepage}\cdot A_{lake}\)

Where:

\(s_{seepage}\) - Seepage length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{seepage} = 1.5\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{seepage} = (1.5\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{seepage} = 2.86\times 10^{6}\,ft^{3}\)

Evaporation losses

\(V_{evaporation} = s_{evaporation}\cdot A_{lake}\)

Where:

\(s_{evaporation}\) - Evaporation length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{evaporation} = 6\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{evaporation} = (6\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{evaporation} = 11.44\times 10^{6}\,ft^{3}\)

Precipitation

\(V_{precipitation} = s_{precipitation}\cdot A_{lake}\)

Where:

\(s_{precipitation}\) - Precipitation length gain, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{precipitation} = 4.25\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{precipitation} = (4.25\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{precipitation} = 8.10\times 10^{6}\,ft^{3}\)

Finally, we estimate the storage change of the lake during the month:

\(\Delta V_{storage} = 77.76\times 10^{6}\,ft^{3}-66.98\times 10^{6}\,ft^{3}-2.86\times 10^{6}\,ft^{3}-11.44\times 10^{6}\,ft^{3}+8.10\times 10^{6}\,ft^{3}\)

\(\Delta V_{storage} = 4.58\times 10^{6}\,ft^{3}\)

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

The volume of water gained and the loss of water through flow,

seepage, precipitation and evaporation gives the storage change.

Response:

Given parameters:

Area of the lake = 525 acres

Inflow = 30 ft.³/s

Outflow = 27 ft.³/s

Seepage loss = 1.5 in. = 0.125 ft.

Total precipitation = 4.25 inches

Evaporator loss = 6 inches

Number of seconds in a month is found as follows;

\(30 \ days/month \times \dfrac{24 \ hours }{day} \times \dfrac{60 \, minutes}{Hour} \times \dfrac{60 \, seconds}{Minute} = 2592000 \, seconds\)

Number of seconds in a month = 2592000 s.

Volume change due to flow, \(V_{fl}\) = (30 ft.³/s - 27 ft.³/s) × 2592000 s = 7776000 ft.³

1 acre = 43560 ft.²

Therefore;

525 acres = 525 × 43560 ft.² =  2.2869 × 10⁷ ft.²

Volume of water in seepage loss, \(V_s\) = 0.125 ft. × 2.2869 × 10⁷ ft.² = 2,858,625 ft.³

Volume gained due to precipitation, \(V_p\) = 0.354167 ft. × 2.2869 × 10⁷ ft.² = 8,099,445.123 ft.³

Volume evaporation loss, \(V_e\) = 0.5 ft. × 2.2869 × 10⁷ ft.² = 11,434,500 ft.³

Which gives;

ΔV = 7776000 - 2858625 + 8099445.123 - 11434500 = 1582823.123

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Answer:

lol what

smmsdnndnsnsnnd

Answer:

d) "Napoleon Declares Himself Holy Roman Emperor"

Explanation:

The French Revolution is defined as a period of the major social upheaval which began in the year 1787 and lasted till year 1799. This revolution completely redefined the the very nature of the political power in France. and also the relationship between the rulers of France and the people they governed.

The 1789 Estates-General was the 1st meeting since year 1614 of the French Estates-General. It is a general assembly which represents the French estates of  realm.

During the French revolution was at peak, the National Assembly issued the Declaration of the rights of the man to the public.

Maximilien Robespierre was considered to be one of the most influential figure and most important statesman during the French Revolution.

Thus all the options (a),(b) and (c) are headlines about the French Revolution, except option (d).

Answer:

In air-breathing propulsion engines, the most significant source of excess enthalpy available to drive a propeller, fan or produce jet momentum downstream of the high pressure turbine is the compression of the air. When the air is compressed, its temperature and pressure increase. This increase in temperature and pressure results in an increase in the enthalpy of the air. The high enthalpy air can then be used to drive a propeller, fan, or produce jet momentum.

The other sources of excess enthalpy in air-breathing propulsion engines are the combustion of the fuel and the expansion of the exhaust gases. However, the compression of the air is the most significant source of excess enthalpy because it occurs at a much higher pressure than the combustion of the fuel or the expansion of the exhaust gases.

Explanation:

Answer:

it is false john deere invented the self-scouring steel plow

Explanation:

Answer:

Sorry whats the question

Explanation:

To design a critically damped parallel RLC circuit with a resistor R = 1Ω and an inductor L = 2H, you need to select the value of the capacitor C according to the following formula: C = 1 / (4 * R * L) Plug in the values for R and L: C = 1 / (4 * 1 * 2) C = 1 / 8 So, you need to select a capacitor with a value of 1/8 F (0.125 F) for the circuit to be critically damped.

To calculate the value of the capacitor required to make the parallel RLC circuit critically damped, we need to use the formula for the damping ratio, which is given by: ζ = R / (2√(L/C)) where R is the resistance, L is the inductance, C is the capacitance, and ζ is the damping ratio. For critically damped behavior, ζ = 1, which means: 1 = R / (2√(L/C)) Substituting the given values of R = 1 Ω and L = 2 H, we get: 1 = 1 / (2√(2/C)) Squaring both sides and rearranging, we get: C = 8/9 F Therefore, the value of the capacitor required to make the parallel RLC circuit critically damped is 8/9 F.

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To make a parallel RLC circuit critically damped, the value of the capacitor should be chosen so that the damping factor is equal to 1. In a parallel RLC circuit, the damping factor can be calculated using the formula:

damping factor = R / (2 * √(L * C))

Given that R = 1 Ω and L = 2 H, we can rearrange the formula to find the value of the capacitor (C):

C = (R^2) / (4 * L)

Plugging in the values, we get:

C = (1^2) / (4 * 2) = 1 / 8

Therefore, the value of the capacitor needed for the circuit to be critically damped is C = 1/8 F (farads).

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The diameter, bolt M27 is known to be 3.0mm in pitch and 41mm across flats.

This is a term that is often seen as the Major diameter. The diameter of a bolt is known to be a kind of a Shank diameter which is said to be often shown or expressed in the unit called millimeters in regards to Metric bolts.

This is so due to the  fact that it is almost the same as the Major or Thread diameter.

Therefore, The diameter, bolt M27 is known to be 3.0mm in pitch and 41mm across flats.

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Answer:

The major financial flop that led to the end of the Sega Dreamcast and ultimately caused Sega to stop making game consoles altogether is:

3: A game called Shenmue, that cost more than $70 million to make, meant that everyone who owned a Dreamcast needed to buy two copies of the game just for Sega to make back the money they had spent to develop it-which didn't happen.

Explanation:

Shenmue, released on December 29, 1999, was created for Dreamcast by Suzuki.  It was widely described as the most expensive video game ever produced.  It had an estimated production and marketing cost of between US$47 and $70 million, according to the latest available data.

Answer:

yes maybe

Explanation:

I have a project and everyone in the world will know about it.

Answer:

both

Explanation:

Answer:

the class is different and the topic treated in class 1 is different from class 3

Pressure in compressed air tank reads 43.2 pounds per square inch (lb/in)

Now, convert pressure to atmosphere (atm)

Recall that 1 atm = 14.695964 lb/in

so, let Z = 43.2 lb/in

To get the value of Z, cross multiply

43.2 lb/in x 1 atm = 14.695964 lb/in x Z

43.2 lb/in atm = 14.695964 lb/in x Z

Z = 43.2 lb/in atm / 14.695964 lb/in

Z = 2.94 atm

The pressure of the compressed air tank reads 2.94 atmosphere.The relation between the volume, the pressure, and the temperature is PV = mRT. Then the amount of air added to the tank is 1.2062 kg.

It is a branch of science that deals with heat and work transfer. A rigid tank contains 20 lbm of air at 20 psi and 70°F. More air is added to the tank until the pressure and temperature rise to 23.5 psi and 90°F, respectively.

Therefore, Pressure in compressed air tank reads 43.2 pounds per square inch (lb/in)

Now, convert pressure to atmosphere (atm)

Recall that 1 atm = 14.695964 lb/in

so, let Z = 43.2 lb/in

To get the value of Z, cross multiply

43.2 lb/in x 1 atm = 14.695964 lb/in x Z

43.2 lb/in atm = 14.695964 lb/in x Z

Z = 43.2 lb/in atm / 14.695964 lb/in

Z = 2.94 atm

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Is an instrument used in boilers to indicate the water level in the boiler. It consists of three valves, which are arranged vertically on the boiler, with the top valve indicating the steam level, the middle valve indicating the water level, and the bottom valve indicating the level of sediment or sludge in the boiler.

The water level in a boiler is critical to maintain safe and efficient operation, as both low and high water levels can result in damage to the boiler and safety hazards. The tri indicator allows the boiler operator to monitor the water level and take corrective action as needed, such as adding water to the boiler or shutting down the system if the water level falls below a safe level.

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Solution :

A wide band pass filter has (Q < 10)

Gain in dB of overall filter = 17 dB

\($F_L = 20 \ Hz $\)

\($F_H = 20 \ kHz$\)

\($17 = 10 \log A_0$\)

\($1.7 = \log A_0 $\)

\($A_0 = 5.47 $\)

\($F_L = 20 \ Hz $\)

\($F_L = \frac{1}{2 \pi R_1C_1}$\)

Let \($C_1 = 0.1 \mu F$\)

\(20 = \frac{1}{2 \pi \times R_1 \times 0.1 \times 10^{-6}}$\)

\($R_1 = \frac{10^6}{4 \pi \times 0.1} $\)

\($R_1 = 79.6 \ k \Omega$\)

\($F_H = 20 \ kHz$\)

\($F_H = \frac{1}{2 \pi R_2C_2}$\)

\($20 \times 10^3 = \frac{1}{2 \pi R_2 \times 0.1 \times 10^{-6}}$\)

\($R_2 = \frac{10^3}{4 \pi \times 10^3} $\)

\($R_2 = 79.6 \ \Omega$\)

Since the overall gain off filter is 5.47

\($A_0 = 5.47$\)

\($A_0 = A_{01} \times A_{02}$\)

\($A_0 = 3.4 \times 1.6$\)

\($A_0 \sim 5.47$\)

\($A_{01} = 1+ \frac{R_F_1}{R_i_1}$\)

\($3.4 = 1+ \frac{R_F_1}{R_i_1}$\)

\($2.4 = \frac{10 k}{R_i_1}$\)

\($R_i_1 = 4166 \ \Omega$\)

\($R_i_1 = 4.1 \ k \Omega $\)

\($A_{02} = 1+ \frac{R_F_2}{R_i_2} $\)

\($1.6 = 1+ \frac{R_F_2}{R_i_2}$\)

\($0.6 = \frac{R_F_2}{R_i_1}$\)

\($R_i_2 = \frac{0.1 k}{0.6}$\)

\($R_i_2 = 166 \ \Omega $\)

Answer:

scope statement and deliverables

Explanation:

Francisco Franco Bahamonde was a Spanish general who ruled over Spain from 1939 to 1975 as a dictator., assuming the title Caudillo.

Franco's ideology included the following consistent points:

Authoritarianism

Nationalism

National Catholicism

Militarism, conservatism

Franco is responsible for decomposing the work for a project to host a major conference.

Scope statement and deliverables should be used to get started.

Answer:

technician b?

Explanation:

The goal of Boeing, Inc is to determine whether an existing machine can mill an engine part that has a design specification of 5.0mm +- 0.10 mm. The process has a standard deviation of σ = 0.02 mm.

Given that the standard deviation of the process is σ = 0.02 mm. Now, we need to determine if the existing machine can produce a part that meets the design specification using the given standard deviation (σ).The tolerance limit (T.L) can be calculated as;

T.L = (Upper Limit – Lower Limit) / 2 where;

Upper Limit = 5.0 + 0.10 = 5.10 mm

Lower Limit = 5.0 – 0.10 = 4.90 mm

T.L = (5.10 – 4.90) / 2 = 0.10 / 2 = 0.05 mm

The value of T.L = 0.05 mm

The process standard deviation (σ) = 0.02 mm

Using the formula;

P = 0.6827, approximately 68.27% of the parts produced by the existing machine fall within the tolerance limit (5.0 ± 0.10)

With a process standard deviation of σ = 0.02 mm and a tolerance limit of 0.05 mm, approximately 68.27% of the parts produced by the existing machine fall within the tolerance limit (5.0 ± 0.10). It can, therefore, be concluded that the existing machine can produce parts that meet the design specification.

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The reason for low power factor in no load test in induction motor is high rotor reactance at lower loading since the rotor reactance value gets decreased with increased loading on induction motor

Usually at no load, the Cosine of the phase angle between the stator current(I1) and the stator applied voltage(V1) is the power factor of an induction motor.

In conclusion, the reason for low power factor in no load test in induction motor is high rotor reactance at lower loading since the rotor reactance value gets decreased with increased loading on induction motor

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Answer: Why is even here then.

Explanation: