which of the following will you select as x in the series of clicks to clear data validation circles in a worksheet: data tab > data tools group > arrow next to x > clear validation circles?
a.What If Analysis
b.Data Validation
c.Remove Duplicates
d. Consolidate worksheet data.

A pipe cleaner can be recommended to clean in Type III of embrasure space.

What is a pipe cleaner?

An interdental tool called a pipe cleaner is used to clean in between teeth surfaces with large spacings. Pipe cleaners should only be placed partially between the teeth, just far enough to massage the gum tissue and scrape away any plaque. For people with open furcations and Type III embrasures, pipe cleaners are advised.

The pipe cleaner's metal core makes it simpler to use in a wide area, but it should be used carefully as it can harm soft tissues. A section of a pipe cleaner should be inserted into the embrasure space or furcation region, cut into 3-inch lengths. Wrap your fingers around the interior root surface and rock back and forth like you are shining shoes.

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Complete question is:

A pipe cleaner can be recommended to clean in which type of embrasure space?

Question 2 options:

1)

Type I

2)

Type II

3)

Type III

4)

Type IV

Answer:

I dont unverstand your question

The national savings in an open economy is calculated by adding private savings, public savings, and net capital inflows.

What is economy?

The creation, distribution, and exchange of goods and services, as well as their consumption, constitute the economy. It is broadly defined as a "social realm emphasising actions, discourses, and material expressions associated with the development, use, and management of finite resources." A given economy is a set of activities with primary variables such as culture, values, education, technological evolution, history, social organisation, political structure, legal systems, and natural resources. These factors provide context and content, as well as the conditions and parameters that govern an economy. Alternatively, the economic sphere is a social world of interconnected human activities and transactions that does not exist in isolation.

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The cable type that comes in two different modes, single mode and multimode is the fiber optic cable.

Fiber optic cables are made up of thin strands of glass or plastic fibers that transmit data through light waves. The two modes of fiber optic cables are single mode and multimode.

Single mode fiber optic cables have a small core, typically 9 microns, and only allow one mode of light to travel through them.

As a result, these cables are well suited for long-distance transmission because they have low attenuation and a narrow bandwidth. Single mode cables are used in telecommunications and cable TV networks

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❎❎❎❎❎❎❎ sorry but that didn't help me that much

The correct answer is c, Troughed belt conveyor. Among all belt conveyors, the troughed belt conveyor is best for bulk handling. It is made up of a flat belt with forward and backward idlers that form a rectangular cross-section.

This conveyor is the most commonly used conveyor belt for handling large quantities of material because of its adaptability, versatility, and reliability. It is used in mining, power generation, processing, and manufacturing facilities because it is inexpensive to use and maintain.

Flat belt conveyor - A flat belt conveyor is a conveyor that transports objects horizontally or with a slight inclination. The conveyor belt is flattened to create a smooth surface that rotates around pulleys to transport objects from one end to the other. This conveyor type is often used in packaging, food processing, and other industries that require low-cost conveyance solutions.

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The allowable normal strength in the c-section is 48000 pounds per square inch.

The allowable stress depends on both the factor of safety imposed on the object and the yield strength or stress at which an object will be permanently damaged. To calculate the allowable normal strength we can divide the yield strength by the factor of safety.

To determine the allowable normal strength output we can use this equation below:

The allowable normal strength = The number yield strength : A stress-concentration factor

The number yield strength is 84 kpsi.

Then convert kpsi to psi : 84 kpsi = 84000 psi

A stress-concentration factor is 1.75

The formula to find allowable normal strength output is:

84000 psi : 1.75 = 48000

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Some of the craft unions are listed below: ElectriciansPlumbersPipe fitters. Therefore, the correct answer is A, B, and C.

Auto workers are not part of the craft unions. They belong to industrial unions or trade unions that represent a particular industry or profession.The definition of a craft union is one in which the members are skilled workers who perform particular types of work.

These workers are usually required to undergo an apprenticeship program, which trains them in the craft that they are going to work in. Craft unions are typically focused on workers who have similar skills, and they are often restricted to certain professions or trades.

So, correct options are option A, B, and C.

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Answer:plumbers

pipe fitters

electricians

Explanation:

Answer:

See attachment for chart

Explanation:

The IPO chart implements he following algorithm

The expressions in bracket are typical examples

Input

Input Number (5, 4.2 or -1.2) --- This will be passed to the Processing module

Processing

Assign variable to the input number (x)

Calculate the square (x = 5 * 5)

Display the result (25) ----> This will be passed to the output module

Output

Display 25

Answer:

Electrical energy

Answer:A 4-input NAND gate is normally high. It can only go low when all four of the inputs have a logical high on them. Assuming the circuit should see a condition where all four inputs go high, then a constant high output indicates a fault does exist.The first step in troubleshooting this failure is to verify the circuit, under normal operation should be seeing a condition where all four inputs do go high. It is completely possible that the circuit is working properly and the condition you were expecting simply does not occur. Verifying what your inputs should be allow you to confirm whether a failure exists in reality.Second, you should verify that the logic probe, multi-meter, oscilloscope, logic analyzer, or monitoring circuit you are using to test circuit outputs is working properly. This is easiest done by touching the probe of the monitoring device to another output you should see changing states. In one system I once had to trouble-shoot, the device had a built-in monitoring circuit and the failure was actually in the monitoring circuit rather than in the main circuit. Once you have confirmed that your test equipment is working properly, you can begin testing the actual circuit to troubleshoot the failure.Given that the circuit should be seeing periods where the gate should go low and that your test equipment is working properly, then a constant high output on this gate means either the output of the gate is open (assuming their is a tie-up resistor somewhere on the output line) or at least one of the inputs is being held low. A constant low input could be a failed gate driving one of the inputs, a short to ground on the circuit board trace connected to the constant low input, or a tie-up resistor on the input that has opened (Technically, this last cause creates a floating input which could intermittently be seen as either a high or low, but in my experience is invariably seen as a low.) The only way to determine which fault exists is to measure the voltages on each input pin to determine which input is failing and then trace the constant low back to its source.729 viewsView 1 upvote1Add a comment...Sponsored by Purdue University OnlinePurdue's 100% Online Master's Now Accepting Applications.Earn an online MS in Electrical and Computer Engineering degree in 12-24 months. View curriculum guide.Learn MoreLoring Chien, Electrical Engineer for 45 years & IEEE Sr. Life memberAnswered Jul 2, 2020

Explanation:

Answer:

a formula containing arguments

Explanation:

Hopefully that's correct

The f(4) using is found using newton's interpolating polynomials of order 4.

function y=CL10_Exercise(part)

%% Input

% part: string for part a,b,c,d

%

%% Output

% y value of the underlying function at x=4

%

%% Write your code here

X=[1,2,3,5,6];

Y=[15,8,5.5,30,52];

x=4;

y=1;

switch part

case 'a'

%% Newton interpolation (Order 4)

a=X;

b=Y;

%x=input('Enter x: ');

[m,n]=size(a);

fx=0;

for i=1:n

%_____________Calculating Dividing Difference_____________________

s=0;

for j=1:i

p=1;

for k=1:i %Denominator part product

if(k~=j)

p=p*(a(j)-a(k));

end

end

s=s+b(j)/p; %summation f(x)/product

end

%_________________________________________________________________

p=1;

for j=1:i-1 %coefficient part of f[...]

p=p.*(x-a(j));

end

fx=fx+s.*p; %Polynomial!

end

y=fx;

case 'b'

%% not-a-knot spline

case 'c'

%% clamped spline

case 'd'

%% Hermite spline

end

end

Hence, the f(4) using is found using newton's interpolating polynomials of order 4.

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Answer:

This band gap also allows semiconductors to convert light into electricity in photovoltaic cells and to emit light as LEDs when made into certain types of diodes. Both these processes rely on the energy absorbed or released by electrons moving between the conduction and valence bands.

Explanation:

On the internet

The American society of civil engineering give us infrastructure in 2013 A grade of D

The overall score for America's infrastructure in all 16 categories is D+ in the 2013 Report Card, which is slightly higher than the D in ASCE's 2009 Report Card.

Six infrastructure sectors profited through increased private sector investment, focused initiatives in municipalities and states to perform renovations or repairs, or a one-time boost in government money.

Importantly, it is the first time the grades have increased since the American Society of Civil Engineers initially rated the quality of America's network in 1998. Nevertheless, a D+ rating remains unacceptable.

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The engineer's data on the number of failures in each mode for each of the three configurations can be used to determine whether the configuration has an effect on the type of failure.

To analyze this data, statistical methods can be employed to identify any significant differences between the failure modes in each configuration. One approach to this analysis is to use a chi-square test of independence. This statistical test can determine whether there is a significant association between two categorical variables, in this case the configuration and the failure mode. If the p-value of the test is below a predetermined level of significance (usually 0.05), then there is evidence that the configuration has an effect on the type of failure. Another approach to analyzing this data is to use a contingency table to display the frequency of failures in each configuration and failure mode. This can help identify any patterns or trends in the data, which can provide insights into the effects of the different configurations on the type of failure.

Overall, it is possible to determine whether the configuration has an effect on the type of failure by analyzing the data obtained by the engineer. By using statistical methods and contingency tables, it is possible to identify any significant associations or patterns in the data, which can help engineers make informed decisions about the design of the component.

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Answer:

Automotive engineering

Answer:

color

Explanation:

Without more context I cannot be sure that this is correct, but I believe you are looking for the CSS declaration property of "color".

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

Answer:

biodiversity, also called biological diversity, the variety of life found in a place on Earth or, often, the total variety of life on Earth. A common measure of this variety, called species richness, is the count of species in an area.

Explanation:

Determine the operating region of the transistors: This can be done by comparing the values of VGS and VDS for each transistor with their respective threshold voltage (VTH). If VGS < VTH, the transistor is in cutoff region. If VGS > VTH and VDS < VGS - VTH, the transistor is in triode region. If VGS > VTH and VDS > VGS - VTH, the transistor is in saturation region.

Write the expressions for the drain current (ID) of each transistor in the appropriate operating region:

In cutoff region: ID = 0

In triode region: ID = mun Cox [(W/L)(VGS - VTH) VDS - 0.5VDS^2] (where W/L is the width-to-length ratio of the transistor)

In saturation region: ID = 0.5mun Cox (W/L)(VGS - VTH)^2

Apply Kirchhoff's laws to find the voltage and current values indicated in the circuit:

Apply KVL to the loop containing R, Q1, and Q2 to find the voltage drop across R

Apply KCL to the node connecting Q1 and Q2 to find the current flowing through R

Use the equations from steps 2 and 3 to solve for the required values of W and R:

For Q1 and Q2, set their drain currents to the desired values and solve for W using the appropriate ID equation

For R, use the voltage and current values obtained in step 3 to solve for its resistance value

Note that there may be multiple valid solutions to this problem, depending on the desired voltage and current values and the specific circuit configuration.

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The corresponding temperature at the center of the sphere after approximately 0.309 seconds is approximately 198.83°C.

To determine the time required for a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy, we can use the concept of thermal diffusion through a sphere. The time required can be calculated using the equation for the thermal diffusion time constant:

τ = (ρ * c * r^2) / (4 * k)

where:

τ is the thermal diffusion time constant,

ρ is the density of the sphere material (in this case, aluminum) = 2700 kg/m^3,

c is the specific heat capacity of the sphere material (in this case, aluminum) = 950 J/kg⋅K,

r is the radius of the sphere (diameter/2) = 75 mm / 2 = 37.5 mm = 0.0375 m,

k is the thermal conductivity of the sphere material (in this case, aluminum) = 240 W/m⋅K.

Substituting these values into the equation, we can calculate the thermal diffusion time constant:

τ = (2700 kg/m^3 * 950 J/kg⋅K * (0.0375 m)^2) / (4 * 240 W/m⋅K)

Now we can solve for τ:

τ ≈ 0.309 seconds

The thermal diffusion time constant represents the time required for a sphere to reach approximately 63.2% of the maximum possible thermal energy. To calculate the time required to accumulate 90% of the maximum energy, we can use the following relation:

t = τ * ln((90% - 63.2%) / (100% - 63.2%))

Substituting the values into the equation:

t = 0.309 seconds * ln((90% - 63.2%) / (100% - 63.2%))

t ≈ 0.309 seconds * ln(0.271 / 0.368)

t ≈ 0.309 seconds * ln(0.736)

t ≈ 0.309 seconds * (-0.305)

t ≈ -0.094 seconds

The negative value obtained implies that 90% of the maximum possible thermal energy cannot be accumulated within a sphere near the inlet of the system. This suggests that the system may need additional time or adjustments to reach the desired energy level.

To calculate the corresponding temperature at the center of the sphere, we can use the concept of one-dimensional transient heat conduction through a sphere. The equation for this scenario is:

T = Ti + (Tg,i - Ti) * (1 - exp(-t / τ))

where:

T is the temperature at the center of the sphere at time t,

Ti is the initial temperature of the spheres = 25°C,

Tg,i is the gas temperature at the inlet of the system = 300°C,

t is the time,

τ is the thermal diffusion time constant calculated earlier.

Let's calculate the corresponding temperature at the center of the sphere when t = 0.309 seconds

T = 25°C + (300°C - 25°C) * (1 - exp(-0.309 / 0.309))

T ≈ 25°C + 275°C * (1 - exp(-1))

T ≈ 25°C + 275°C * (1 - 0.3679)

T ≈ 25°C + 275°C * 0.6321

T ≈ 25°C + 173.8275°C

T ≈ 198.8275°C

Therefore, the corresponding temperature at the center of the sphere after approximately 0.309 seconds is approximately 198.83°C.

Now, let's consider the advantage of using copper instead

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The dotted decimal representation of the IPv4 address 11001011.00000000.01110001.11010011 is 203.0.113.211.

IPv4 addresses are typically represented in dotted decimal notation. In the dotted decimal notation of IPv4 address, each byte is represented by a decimal number and separated by dots (periods).To convert the binary representation of an IPv4 address to dotted decimal notation, we can separate the binary sequence into four groups of eight bits (one byte) and convert each group to its equivalent decimal number. The resulting four decimal numbers are separated by dots to form the dotted decimal representation of the IPv4 address.In this case, the binary representation 11001011.00000000.01110001.11010011 can be split into four groups of eight bits each as 11001011. 00000000. 01110001. 11010011. Then, each of these groups can be converted to its decimal equivalent, which results in the dotted decimal representation of the IPv4 address 203.0.113.211.

Therefore, the dotted decimal representation of the IPv4 address 11001011.00000000.01110001.11010011 is 203.0.113.211.

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By combining the two plots i.e. magnitude plot and phase plot, we will get Bode plots for H(w).

The magnitude plot starts at 0 dB, goes through a downward slope of -20 dB/decade, a downward slope of -40 dB/decade, and then approaches 0 dB at +20 dB/decade. The phase plot starts at 90 degrees, goes through a downward slope of -45 degrees/decade, and then approaches -90 degrees.

To sketch the Bode plots for the given transfer function H(w), we first need to express it in the standard form:

H(w) = K * (jw + a) / (jw)(1 + jw/b)

where K = 0.2, a = 10, and b = 2. We can then plot the magnitude and phase responses separately on a logarithmic scale for the frequency range of interest.

Magnitude plot:

At very low frequencies (w << b), the denominator dominates, and the magnitude of H(w) is approximately 0 dB. As the frequency increases, the denominator decreases, and the magnitude of H(w) starts to decrease at a slope of -20 dB/decade due to the pole at the origin. At the break frequency (w = b), the slope changes to -40 dB/decade due to the double pole. Finally, at high frequencies (w >> b), the numerator dominates, and the magnitude of H(w) approaches 0 dB at a slope of +20 dB/decade due to the zero at a.

Phase plot:

At very low frequencies (w << b), the phase of H(w) is approximately 90 degrees due to the pole at the origin. As the frequency increases, the phase decreases at a slope of -45 degrees/decade until it reaches -180 degrees at the break frequency (w = b) due to the double pole. Finally, at high frequencies (w >> b), the phase of H(w) approaches -90 degrees due to the zero at a.

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For a wear indicator type ball joint, when the grease fitting boss is flush with the base: B. The ball joint should be replaced.

A wear indicator can be defined as a type of technology that is designed to alert and warn a car owner or driver that the car's brake pad is needs a replacement.

Generally, a wear indicator is designed and fitted with a small boss. Thus, this small boss would continuously recede into its housing as wear occurs. However, when it recedes into the cover plate or flush with the base, it simply means that the ball joint should be replaced.

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Answer:

y'(x) = 7 ;

y'(x) = 15x²

Explanation:

Differentiation from first principle :

y = 7x

Formula to obtain derivative from first principle:

y'(x) = lim(h - - > 0) : [y(x + h) - y(x)] / h

Obtain : y(x + h)

y(x) = 7x

y(x + h) = 7(x + h)

Substitute into formula :

y'(x) = lim(h - - > 0) : [7(x + h) - 7x] / h

Expand :

y'(x) = lim(h - - > 0) : [7x + 7h - 7x] / h

y'(x) = lim(h - - > 0) : 7h/ h

y'(x) = 7

2.)

y = 5x³

y'(x) = lim(h - - > 0) : [y(x + h) - y(x)] / h

Obtain : y(x + h)

y(x) = 5x³

y(x + h) = 5(x + h)³

Substitute into formula :

y'(x) = lim(h - - > 0) : [5(x + h)³ - 5x³] / h

Expand :

y'(x) = lim(h - - > 0) : [5(x³ + 3x²h + 3xh² + h³) - 5x³] / h

y'(x) = lim(h - - > 0) : [5x³ + 15x²h + 15xh² + 5h³ - 5x³] / h

y'(x) = lim(h - - > 0) : [15x²h + 15xh² + 5h³] / h

y'(x) = lim(h - - > 0) : h(15x² + 15xh + 5h²) / h

y'(x) = lim(h - - > 0) : (15x² + 15xh + 5h²)

y'(x) = 15x²

Answer:

One Foot

Explanation:

The general rule is to maintain a distance of one foot per four feet of height.

To formulate the optimization model, we can use the following decision variables:

- Xi: the number of items of type i to buy

- S: the size of the bag to buy

The objective function will be to maximize the expected value of the total value of the items minus the total cost of the items and the bag:

Maximize: E[Σ(Vi * Xi) - Σ(Ci * Xi) - Cb]

Subject to:

- Σ(Xi * Si) <= 36: the size of the bag must be able to hold all the items

- 20 <= S <= 36: the size of the bag must be between 20 and 36

- Xi >= 0: we can't buy negative items

where:

- Vi: the value of item i

- Ci: the cost of item i

- Cb: the cost of the bag

To solve the expected value solution, we can assume that the values of the items and the cost of the bag are uniformly distributed within their respective ranges. We can then use the expected value of these distributions to compute the expected value of the objective function.

To solve the robust solution, we can assume that the values of the items and the cost of the bag are uncertain and can take any value within their respective ranges. We can then use a worst-case analysis to compute the robust objective function.

To formulate the two-stage program, we can assume that the cost of the bag is uncertain and can take any value within its range. We can then generate different scenarios for the cost of the bag and solve the problem for each scenario.

Thus, we can then use the solutions to these sub-problems to compute the expected value of the objective function.

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To group the rocks into different categories based on their physical attributes. She conducted research on texture and composition.

Two criteria—TEXTURE and COMPOSITION—are used to categorize rocks. The sizes, shapes, and relationships between the many mineral grains and other components that make up a rock are what determine its texture. The process that created the rock controls these elements. Color, hardness, luster, crystal formations, density, and cleavage are characteristics that aid in the identification of minerals in rocks by geologists. Crystal structure at the atomic level principally determines crystal shape, cleavage, and hardness. The chemical makeup mostly determines color and density. The most prevalent physical characteristics are those related to crystal structure, color, hardness, cleavage, and specific gravity. Examining a mineral's crystal shape is one of the finest ways to identify it.

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