Apply Four (4) R Functions to explore an R Built-in Data Set R is a widely popular programming language for data analytics. Being skilled in using R to solve problems for decision makers is a highly marketable skill.In this discussion, you will install R, install the R Integrated Development Environment (IDE) known as RStudio, explore one of R's built-in data sets by applying four (4) R functions of your choice. You will document your work with screenshots. You will report on and discuss with your peers about this learning experience.To prepare for this discussion:Review the module’s interactive lecture and its references.Download and install R. Detailed instructions are provided in the Course Resources > R Installation Instructions.Download and install R Studio. Detailed instructions are provided in the Course Resources > RStudio Installation Instructions.View the videos in the following sections of this LinkedIn Learning Course: Learning RLinks to an external site.:Introduction What is R?Getting Started To complete this discussion:Select and introduce to the class one of the R built-in data sets.Using R and RStudio, apply four (4) R functions of your choice to explore you selected built-in data set and document your exploration with screenshots.Explain your work, along with your screenshot, and continue to discuss your R experiment throughout the week with your peers.

Answer:

primary consumer because YES

Answer:predator and secondary consumer

Explanation:

Answer:

approaching a lady or a man

ᴄᴏᴍᴘʟᴇᴛᴇɴᴇꜱꜱ ᴍᴇᴀɴꜱ ᴛʜᴇ ꜱᴛᴀᴛᴇ ᴏꜰ ʙᴇɪɴɢ ᴄᴏᴍᴘʟᴇᴛᴇ ᴀɴᴅ ᴇɴᴛɪʀᴇ; ʜᴀᴠɪɴɢ ᴇᴠᴇʀʏᴛʜɪɴɢ ᴛʜᴀᴛ ɪꜱ ɴᴇᴇᴅᴇᴅ.

ʙᴇᴀᴜᴛʏ ᴍᴇᴀɴꜱ combination of qualities, such as shape, colour, or form, that pleases the aesthetic senses, especially the sight.

Deep foundations are used when the surface soils are not capable of supporting the loads of a structure. Two common types of deep foundations are replacement and displacement foundations.

1. Replacement foundations:

Replacement foundations are created by removing the soil beneath a structure and replacing it with concrete or other materials. This creates a solid, stable foundation that can support the weight of the structure.

Illustration: The process of creating a replacement foundation begins by excavating the soil beneath the structure, leaving a large hole. The hole is then filled with concrete, which is allowed to cure and harden. The concrete provides a solid base for the structure to rest on, ensuring that it is stable and secure.

2. Displacement foundations:

Displacement foundations work by pushing soil aside as they are installed, creating a cavity in the ground that is then filled with concrete. These foundations are often used when soil conditions are too difficult to excavate or when a more efficient installation method is desired.

Illustration: The process of creating a displacement foundation involves driving a steel pile or tube into the ground. The pile is typically filled with concrete as it is being driven, creating a cavity in the soil. Once the pile has reached the desired depth, it is left in place, and the cavity is filled with concrete. This creates a solid, stable foundation that can support the weight of the structure.

In summary, both replacement and displacement deep foundations are effective methods for creating stable foundations for structures when the soil conditions are not suitable for shallow foundations. The choice of foundation type depends on the specific site conditions, load requirements, and project budget.

The typical journal-to-bearing clearance for engines is,

B) 0.0005 to 0.0025 inch

Given that,

To find the typical journal-to-bearing clearance for engines.

Now, The range 0.0005 to 0.0025 inch represents the recommended clearance between the rotating journal of the crankshaft and the bearing surface to ensure proper lubrication and minimal friction.

Hence, The correct option for typical journal-to-bearing clearance for engines is,

B) 0.0005 to 0.0025 inch.

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The statement that assigns the value of the first element of a tuple to variable k can be written as:

k = t[0]

Here, t refers to the tuple and the [0] index represents the first element of the tuple. The value of this first element is then assigned to the variable k.

In Python, tuples are ordered and immutable sequences of elements, which can be of different types. Tuples are accessed using index numbers starting from 0 for the first element, similar to accessing elements in lists. However, unlike lists, tuples cannot be modified once they are created.

In the statement k = t[0], we are assigning the value of the first element of the tuple t to the variable k. This is a common operation when working with tuples, especially when we want to extract specific values from a tuple and use them in our code. The statement works by using the indexing operator [0] to access the first element of the tuple, and then assigning its value to the variable k.

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To define a public static method named getAllodd that takes an ArrayList with all of the odd values in the argument ArrayList.

To start with, let's look at the method signature we need to define:

```
public static ArrayList getAllodd(ArrayList myList) {
   // Your code here
}
```

As you can see, the method takes an ArrayList of Doubles called `myList` as its argument, and returns an ArrayList of Doubles that contains all of the odd values in `myList`.

Now, let's take a look at how we can implement this method. Here's some code that should do the trick:

```
public static ArrayList getAllodd(ArrayList myList) {
   ArrayList oddList = new ArrayList();
   for (Double num : myList) {
       if (num % 2 != 0) {
           oddList.add(num);
       }
   }
   return oddList;
}
```

Here's how the code works:

1. First, we create a new ArrayList called `oddList` to hold the odd values we find.

2. Next, we use a for loop to iterate over each element in `myList`. Inside the loop, we check if the current element is odd by using the modulus operator (`%`) to check if it has a remainder when divided by 2. If it does have a remainder, we know it's odd, so we add it to `oddList`.

3. Finally, we return `oddList` once we've checked all the elements in `myList`.

To test our method, we can create an ArrayList called `myList` that contains the values (3, 2, 7.5, 8, 6), and then call `getAllodd` with `myList` as its argument:

```
ArrayList myList = new ArrayList(Arrays.asList(3.0, 2.0, 7.5, 8.0, 6.0));
ArrayList oddList = getAllodd(myList);
System.out.println(oddList);
```

This should output the ArrayList `[3.0, 7.5]`, which contains all the odd values from `myList`.

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Answer:

Engineer A results will be more accurate

Explanation:

Analytical method is better than numerical method. Engineer A has used analytical method and therefore his results will be more accurate because he used simplified method. Engineer B has used software to solve the problem related to heat transfer his results will be approximate.

Answer:

Explanation:

Considering the flow of mercury in a tube:

When it comes to laminar flow of mercury, the thermal entry length is quite smaller than the hydrodynamic entry length.

Also, the hydrodynamic and thermal entry lengths which is given as DLhRe05.0= for the case of laminar flow. It should be noted however, that Pr << 1 for liquid metals, and thus making the thermal entry length is smaller than the hydrodynamic entry length in laminar flow, like I'd stated in the previous paragraph

Light olefin is produced by cracking propane in a steady flow reactor at 1.2 bar and 750 K under ideal gas conditions.

Cracking is a chemical process in which larger hydrocarbon molecules are broken down into smaller molecules by the application of heat. In this case, propane is being cracked to produce light olefins, such as ethylene and propylene. The cracking reactions occur in a steady flow reactor, which maintains a constant flow of reactants and products. The reactor operates at a pressure of 1.2 bar and a temperature of 750 K, which are ideal gas conditions. These conditions promote the breaking of propane molecules into smaller olefin molecules. The resulting light olefins can be used as raw materials in the production of various chemical products, such as plastics, rubber, and synthetic fibers.

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Steel-framed homes have insufficient insulation and poor energy efficiency. This is due to thermal bridging, which reduces the insulating characteristics of steel by 60%. Steel conducts heat more efficiently than wood.

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The address of the element stored at arr[20][30] in the 2-D array declaration char Arr[100][100] would be 30 + 20 * 100 = 2030.

The declaration of the 2-D array is shown below:

char Arr[100][100]

Here, Arr is a 2-D array consisting of 100 rows and 100 columns. This means that there are a total of 10,000 elements in this array. Each element of this array is of type char. Therefore, each element will occupy a single byte of memory.

The array is byte-addressable. This means that each element of the array is accessible using its byte address. Since each element occupies a single byte of memory, the byte address of an element is the same as its memory address.

To calculate the address of the element stored at arr[20][30], we first need to understand how the elements are stored in the array.

The elements of a 2-D array are stored in row-major order. This means that the elements of the first row are stored first, followed by the elements of the second row, and so on. Within a row, the elements are stored from left to right.Now, to calculate the address of the element stored at arr[20][30], we need to calculate the byte address of this element. Since the array is byte-addressable, we can calculate the byte address of an element by multiplying its row number by the number of columns in the array and adding its column number. This gives us the following formula:

Byte Address of Element = Base Address + (Row Number * Number of Columns + Column Number)

Since the base address of the array is 0, we can simplify this formula to:

Byte Address of Element = Row Number * Number of Columns + Column Number

Using this formula, we can calculate the byte address of the element stored at arr[20][30] as follows:

Byte Address of Element = 20 * 100 + 30 = 2030

Therefore, the address of the element stored at arr[20][30] is 2030.

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I guess you got the one

hope it helps you better

Answer:

fan

Explanation:

Windowing is a technique used in networking to control the flow of data between two devices. It is a process by which the sending device sends a specified number of data packets and then waits for an acknowledgement from the receiving device before sending any further packets. This technique helps to avoid overloading the receiving device and ensures that the data is transmitted efficiently.

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Answer:

15625 moles of methane is present in this gas  deposit

Explanation:

As we know,

PV = nRT

P = Pressure = 230 psia = 1585.79 kPA

V = Volume = 980 cuft = 27750.5 Liters

n = number of moles

R = ideal gas constant = 8.315

T = Temperature = 150°F = 338.706 Kelvin

Substituting the given values, we get -

1585.79 kPA * 27750.5 Liters = n * 8.315 * 338.706 Kelvin

n = (1585.79*27750.5)/(8.315 * 338.706) = 15625

Explanation:

D the answer is D and anyone can answer this Question

Based on the provided options, the scientific expression that is the same as 1545.5347 is 154553.4e3. Therefore, the correct option is option D.

A scientific expression, often known as scientific notation, is a method of representing extremely big or extremely small integers. It is often used to simplify the representation of such numbers in scientific and mathematical operations. A number is expressed in scientific notation as a product of two parts: a coefficient and a power of ten.

The coefficient is a value between 1 and 10, and the power of 10 determines how far the decimal point should be moved. The "e3" signifies multiplying the integer by 10 raised to the power of 3, which is similar to moving the decimal point three places to the right in this formula. As a result, 154553.4e3 equals 154553.4103, which simplifies to 154553400.

Therefore, the correct option is option D.

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Answer:

Emails and phone calls are both common forms of communication that are used in professional and personal settings. There are several similarities between email and phone calls:

1. Both are asynchronous forms of communication: Unlike instant messaging or face-to-face conversations, both emails and phone calls allow the sender or recipient to respond at their convenience. They don't require immediate attention or an instant response.

2. Both are written forms of communication: While phone calls rely on spoken words, emails are written. As a result, both can be used to convey detailed information and allow the sender to carefully consider their words before sending.

3. Both are forms of direct communication: Emails and phone calls both allow for direct communication between two parties. This can be beneficial for discussing sensitive information or resolving issues quickly.

4. Both can be used for formal and informal communication: Emails and phone calls can be used in both personal and professional contexts. They are both flexible forms of communication that can be adapted to fit different situations.

5. Both require attention to tone and etiquette: Just like with phone calls, emails require attention to tone and proper etiquette. Both forms of communication should be approached professionally and respectfully to ensure effective communication.

In conclusion, while there are differences between emails and phone calls, there are also similarities that make them useful communication tools. Both allow for direct, asynchronous communication and can be adapted to fit different situations.

Explanation:

B: Contaminated groundwater is the potential geologic hazard that is NOT represented by a feature on the figure given.

A geologic hazard refers to an extreme natural event in the crust of the earth that poses a threat to life and property. For example, volcanic eruptions, tsunamis (tidal waves),  earthquakes,  groundwater contamination, and landslides. As per the context of the given picture, groundwater contamination is a potential geologic hazard that is not represented by a feature in the image attached.

Groundwater contamination is a phenomenon that occurs when man-made products such as road salts gasoline, oil, and chemicals get into the groundwater and make it to become unsafe and unfit for human use.

Image is attached showing potential geologic hazards such as earthquake, volcano, flood-prone areas, and landslide.

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Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature \(T_1 = -16^0\ C\)

Quality \(x_1 = 0.2\)

Outlet:

Temperature \(T_2 = -16^0 C\)

Quality  \(x_2 = 1\)

The following data were obtained at saturation properties of R134a at the temperature of -16° C

\(v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg\)

\(v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg\)

\(m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}\)

\(\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}\)

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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The ultimate BOD (BODu) of the biodegradable industrial wastewater is approximately 949.6 mg/L.

To determine the ultimate BOD (BODu) of a biodegradable industrial wastewater with a BOD5 of 600 mg/L and a first-order rate constant of 0.20 per day, follow these steps:

1. Identify the given values: BOD5 = 600 mg/L, rate constant (k) = 0.20 per day.
2. Recall the first-order reaction formula: BODu = BOD5 / (1 - e^(-kt)), where BODu is the ultimate BOD, BOD5 is the 5-day BOD, k is the rate constant, and t is the time in days.
3. Since we're determining the BODu, plug in the given values: BODu = 600 / (1 - e^(-0.20 * 5)).
4. Calculate the exponent part: e^(-0.20 * 5) = e^(-1) = 0.3679 (approx).
5. Calculate the denominator: 1 - 0.3679 = 0.6321.
6. Divide BOD5 by the denominator: BODu = 600 / 0.6321 = 949.6 mg/L (approx).

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The solution to the nonlinear system by Newton’s method with the given initial estimates and the first five iterations is x = 1.7220 and y = 1.5305.

Newton’s Method is a method that is commonly used to find the root of a nonlinear system. It involves using an initial estimate of the root and calculating a new estimate of the root at each iteration until the desired accuracy is achieved. Below is the solution to the given nonlinear system by Newton’s method.

The given nonlinear system is as follows.

\(f1(x,y) = - 2x³ + 3y² + 42 = 0 f2(x,y) = 5x² + 3y³ - 69 = 0\)

The Jacobian matrix of the system is given as:

\(J = ∂f1/∂x ∂f1/∂y ∂f2/∂x ∂f2/∂y = [-6x 6y] [10x 9y²]\)

We will use the initial estimates x = 1 and y = 1.

The first five iterations are shown below. \(Iteration 1:  x1 = x0 - [J⁻¹(x0,y0) . f(x0,y0)] = [1,1] - [(J⁻¹(1,1)) . f(1,1)]\)

Where J⁻¹ is the inverse of the Jacobian matrix and f(x0,y0) is the value of the system at the point x0 and y0.

Substituting values into the above formula, we have:

\(x1 = [1,1] - [(J⁻¹(1,1)) . f(1,1)] = [1,1] - [(1/54) . (-43,-24)] = [1.7963, 1.4444]\)

Iteration 2:\(x2 = x1 - [J⁻¹(x1,y1) . f(x1,y1)]\)

Substituting values into the above formula, we have:

\(x2 = [1.7963, 1.4444] - [(J⁻¹(1.7963,1.4444)) . f(1.7963,1.4444)] = [1.7283, 1.5206]\)

Iteration 3: \(x3 = x2 - [J⁻¹(x2,y2) . f(x2,y2)]\)

Substituting values into the above formula, we have:

\(x3 = [1.7283, 1.5206] - [(J⁻¹(1.7283,1.5206)) . f(1.7283,1.5206)] = [1.7222, 1.5304] Iteration 4: x4 = x3 - [J⁻¹(x3,y3) . f(x3,y3)]\)

Substituting values into the above formula, we have:

\(x4 = [1.7222, 1.5304] - [(J⁻¹(1.7222,1.5304)) . f(1.7222,1.5304)] = [1.7220, 1.5305] Iteration 5: x5 = x4 - [J⁻¹(x4,y4) . f(x4,y4)]\)

Substituting values into the above formula, we have:

\(x5 = [1.7220, 1.5305] - [(J⁻¹(1.7220,1.5305)) . f(1.7220,1.5305)] = [1.7220, 1.5305]\)

Therefore, the solution to the nonlinear system by Newton’s method with the given initial estimates and the first five iterations is x = 1.7220 and y = 1.5305.

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The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.

For a layer of 3.8 m thickness, we were given the following parameters:

For Sf, we have:

Sf = Sc/U

Sf = 7.3/0.5

Sf = 14.6

Therefore, Sf for a layer of 38 m thickness is given by:

Sf = 14.6 × 38/3.8

Sf = 146 cm.

At 50%, the time for a layer of 3.8 m thickness is:

\(t_{50}\) = 1.5 year.

At 50%, the time for a layer of 38 m thickness is:

\(t_{50}\) = 1.5 × (38/3.8)²

\(t_{50}\) = 150 years.

For a thickness of 38 m, U₂ is given by:

\(\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25 \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05\)

The new settlement after 1.5 year is:

Sc = U₂Sf

Sc = 0.05 × 146

Sc = 7.3 cm.

For time, t₂ = 5 year:

\(U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25 \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09\)

The new settlement after 5 year is:

Sc = U₂Sf

Sc = 0.09 × 146

Sc = 13.14 cm.

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Answer:

Efficiency is defined as any performance that uses the fewest number of inputs to produce the greatest number of outputs. Simply put, you're efficient if you get more out of less.

Explanation:

Potential difference need to be apply to plates of capacitors is 5.63.

The capacitor is a two-terminal electrical device that stores energy in the form of electric charges.

C=55.0μf

E=155j

E=1/2cv^2

E=1/2*55.0*v^2=155

v=155*2/55.0=5.63v

potential difference is 5.63v.

The energy stored in a capacitor is nothing but the electric potential energy and is related to the voltage and charge on the capacitor. If the capacitance of a conductor is C, then it is initially uncharged and it acquires a potential difference V when connected to a battery.

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Answer:

faster

Explanation:

because in a big grinder you ca only grind bigger things but not Small things , some parts of your things that will

remain as they were in the beginning, so it will take more time to grind Small things .

If I had to guess, I'd say the answer is

Drier conditions will likely result in more wildfires

Studies show that warmer and drier areas will double in wildfires by 2050.