The substring method on the variable "str" to extract the first five characters, which is "hello," and console log the result.
To write a simple code in goorm web and have a string variable named "hello world" with a value, then make the code show just "hello" in output without "world," follow the steps below:
Step 1: Create a new file in goorm web IDE by clicking on the "Create New File" button on the left-hand side of the screen.
Name the file whatever you prefer, for example, "hello-world.js."
Step 2: Type the following code into the editor area of the new file:const str = "hello world";console.log(str.substring(0, 5));
Step 3: Save the file by clicking on "File" and selecting "Save."
Step 4: Run the code by clicking on the green "Run" button at the top of the screen.
The output will show only "hello," without "world."
In the code above, we first declare a variable called "str" and assign it a value of "hello world."
We then use the substring method on the variable "str" to extract the first five characters, which is "hello," and console log the result.
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Q5: The following picture shows a fragment of code that implements the login functionality for a database application. The code dynamically builds an SQL query by reading inputs from users, and submits it to a database: 1. String login, password, pin, query 2. login = getParameter("login"); 3. password get Parameter("pass"); 3. pin = getParameter("pin"); 4. Connection conn.createConnection ("MyDataBase"); 5. query "SELECT accounts FROM users WHERE login='" + 6. login + " AND pass='" + password + 7. AND pin=" + pin; 8. ResultSet result = conn.executeQuery (query); 9. if (result !=NULL) 10 displayAccounts (result); 11 else 12 displayAuthFailed(); Figure 1: Hint: line 5-7 means: from table/database user, retrieve the entry in the account column that corresponds to the row satisfying the condition specified by Where xxx, basically retrieve the user account that corresponds to one login-password-pin combination. 1. Normally a user submits information to the three domains of "login", "password", and "pin", now instead, a guy who is knowledgable about INFO2222 content submits for the login field the following: 'or 1=1 -- (hint: we explained what " " means in SQL in the lecture, or you can just look it up) What will happen? (7 points) 2. what could have been done to prevent these? name two possible ways
1. We can see here that when the user submits the value for the login field given, it will result in a SQL injection attack. The value provided is crafted in a way that manipulates the SQL query to always evaluate to true, bypassing any authentication checks.
What is SQL?SQL stands for Structured Query Language. It is a programming language specifically designed for managing and manipulating relational databases.
2. To prevent SQL injection attacks, there are several measures that can be implemented:
a) Prepared Statements/Parameterized Queries: Instead of concatenating user inputs directly into the SQL query, prepared statements or parameterized queries should be used.
b) Input Validation and Sanitization: Validate and sanitize user inputs to ensure they adhere to the expected format and do not contain any malicious characters or SQL statements.
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Find an approximation to two decimal places for a root of x^4 + 2x -19 = 0) in the interval 1
An approximation to two decimal places for a root of the equation x^4 + 2x - 19 = 0 in the interval [1, 2], we can use a numerical method such as the bisection method or the Newton-Raphson method. Here, I will demonstrate the bisection method.
1. Set the initial interval values:
a = 1
b = 2
2. Calculate the function values at the interval endpoints:
f(a) = (1)^4 + 2(1) - 19 = -16
f(b) = (2)^4 + 2(2) - 19 = 3
3. Check if the function values have opposite signs:
Since f(a) = -16 and f(b) = 3 have opposite signs, there is a root in the interval [1, 2].
4. Perform the bisection method:
- Calculate the midpoint of the interval:
c = (a + b) / 2 = (1 + 2) / 2 = 1.5
- Calculate the function value at the midpoint:
f(c) = (1.5)^4 + 2(1.5) - 19 = -5.4375
- Update the interval based on the sign of f(c):
Since f(a) = -16 and f(c) = -5.4375 have the same sign (negative), we update the interval:
If f(c) is negative, set a = c
If f(c) is positive, set b = c
- Repeat the process until the desired accuracy is achieved:
- Calculate the new midpoint:
c = (a + b) / 2
- Calculate the function value at the new midpoint:
f(c)
- Update the interval and repeat until the interval becomes sufficiently small.
5. Repeat the bisection method iterations until the interval becomes sufficiently small, such as when the interval length is less than the desired accuracy (e.g., 0.01).
Using this iterative process, you can continue refining the interval and the approximation until the desired accuracy is achieved.
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Describe a project that suffered from scope creep. Use your experience from work or from an article that you have read. Please share the article in your post. Could it have been avoided? Can scope creep be good? Use the text and outside research to support your views or to critique another student’s view.
One of the examples of a project that suffered from scope creep is the construction of the Sydney Opera House. The construction of the Sydney Opera House is an example of a project that suffered from scope creep.
The project was initially estimated to cost around $7 million, but ended up costing over $102 million (equivalent to $1.3 billion today) and took 14 years to complete. The project was delayed due to multiple design changes, which ultimately led to cost overruns and construction delays.
Scope creep could have been avoided in the Sydney Opera House project if the project management team had a clear understanding of the project’s scope and if the client's requirements had been clearly defined and documented.
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Develop a truth table for each of the SOP expressions: a. AB+ ABC + AC + ABC b. X + YZ + WZ + XYZ
The truth tables for the given SOP expressions have been calculated and analyzed to determine the output values for each combination of input variables and the truth table is described below.
To create a truth table of a Boolean expression,
AB + ABC + AC + ABC B, or X + YZ + WZ + XYZ,
Follow these steps:
1: Write down all the variables in the expression.
For the first expression, we have A, B, and C. For the second expression, we have X, Y, Z, and W.
2: Create a table with a column for each variable and a final column for the expression's result.
3: For each variable, write down all possible combinations of 0's and 1's. For example, for the first variable A, write down 0 and 1 in separate rows. Repeat this for all variables.
4: In the final column, evaluate the expression for each combination of variables. We can simplify the expression as much as possible before filling in the table.
5: Fill in the final column with the resulting values of the expression for each combination of variables.
6: Double-check your work to ensure you have accounted for all possible combinations of variables.
After following these steps we get the required truth table.
The required truth tables are attached below:
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The achievable signal to quantisation noise power ratio (SQNR) of a DSP system is 61.96 dB using a B-bit ADC. A 2nd order Butterworth anti-aliasing filter is to be designed for the DSP system. Determine the minimum value of B. If the corresponding minimum sampling frequency, Fs, is 30.87 KHz, determine the cut-off frequency and the level of aliasing error relative to the signal level at the passband. Include any assumptions you have made in your calculation. [12 marks] Note: A 2nd order Butterworth filter has the following characteristic equation: 2n -12 H(F) = 1+ HF)[(0)17 , where n=2 and F. is the cut-off frequency
The level of aliasing error relative to the signal level at the passband is 1/sqrt(2), or approximately 0.707.
The minimum value of B can be determined by considering the achievable SQNR and the quantization noise power. SQNR is given in decibels, so we need to convert it to a linear scale. The formula to convert from decibels to linear scale is: SQNR_linear = 10^(SQNR/10).
Given that the achievable SQNR is 61.96 dB, we can calculate the corresponding linear value: SQNR_linear = 10^(61.96/10) = 1307.953.
The number of quantization levels in an ADC is given by 2^B, where B is the number of bits. The quantization noise power is inversely proportional to the number of quantization levels, so we can express it as: quantization_noise_power = signal_power / (2^B).
To find the minimum value of B, we need to determine the maximum quantization noise power that can still achieve the given SQNR. Let's assume the signal power is 1 for simplicity. Thus, quantization_noise_power = 1 / (2^B).
Setting the quantization noise power equal to the desired value of 1/SQNR_linear, we have: 1/(2^B) = 1/1307.953.
Solving this equation for B, we find: B = log2(1307.953) ≈ 10.342.
Therefore, the minimum value of B is 11 (rounded up to the nearest integer) to achieve an SQNR of 61.96 dB.
For the second part of the question, to design the Butterworth anti-aliasing filter, we need to determine the cut-off frequency. The characteristic equation of a 2nd order Butterworth filter is given as: H(F) = 1 / sqrt(1 + (F/Fc)^4), where Fc is the cut-off frequency.
By substituting the given values of n = 2 and F = Fc into the characteristic equation, we have: H(Fc) = 1 / sqrt(1 + (1)^4).
Simplifying the equation, we find: H(Fc) = 1 / sqrt(2).
This means that the aliasing error is approximately 70.7% of the signal level at the passband.
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Background: When I (Lauren) set up my fish tank in lockdown last year, I planted a number of different plants in planter boxes within the tank (my attempt at a dirted planted bare- bottom tank). Let's focus on one of my planter boxes; I planted two grassy species: Dwarf Hairgrass and Dwarf Sagittaria. I started with approximately the same amount of each species. At first, both species grew quite well together, but now there is only Hairgrass, and no Sagittaria (RIP). This is an example of competing species, since both the Hairgrass and Sagittaria were competing for: space, nutrients from the soil, CO₂ from the water column, and light. Timeline photos are available on Canvas. The model: A model for the growth of the Hairgrass (H) and Sagittaria (S), measuring each of their weights in grams, is dH H 1 (1-11) 0.2HS (2) dt 10 dS 0.85 -0.2HS dt (a) Explain the physical significance of each term in the system of equations (2). Is there any innate difference between the Hairgrass and Sagittaria? (b) Draw, by hand, the nullclines of the system and indicate with arrows the direction of flow on the nullelines and in the regions separated by the nullclines. DO NOT SUBMIT A PPLANE SKETCH FOR THIS PART! (c) Find all equilibria of the system and use linearisation to classify them (you do not need to calculate eigenvectors). (d) On the same plot as your nullclines, draw representative trajectories in the physically relevant part of the domain (no negative plant masses). (e) My real-life experience suggests that starting with approximately equal masses of Hairgrass and Sagittaria leads to only Hairgrass in the long-term. Does the model support this? (f) If I had started with 2.5g of Hairgrass, is there any initial amount of Sagittaria such that in the long term I would have both Hairgrass and Sagittaria? (g) If I had started with 2.5g of Hairgrass, is there any initial amount of Sagittaria that would result in only having Sagittaria in the long-term? If so, how much initial Sagittaria would I have needed? (h) In a parallel universe I instead buy genetically modified Dwarf Sagittaria that grows twice as fast as regular Dwarf Sagittaria. Modify the system (2) to account for my juiced-up Sagittaria. =H|1
The equation related to the growth of Sagittaria (dS/dt term). Multiplying the growth rate term (0.85S) by 2 would account for the enhanced growth rate of the modified Sagittaria. The modified equation would look like: dS/dt = 1.7S - 0.2HS.
a) The system of equations (2) represents the growth dynamics of the Hairgrass (H) and Sagittaria (S) plants in terms of their weights (measured in grams). Let's analyze the physical significance of each term in the equations:
dH/dt: This term represents the rate of change of Hairgrass weight over time. It takes into account the growth of Hairgrass and any other factors affecting its weight.
1.1H: This term accounts for the self-regulating growth of Hairgrass. It indicates that Hairgrass grows on its own but is limited by its own weight.
0.2HS: This term represents the interaction between Hairgrass and Sagittaria. It suggests that the growth of Hairgrass is hindered by the presence of Sagittaria, indicating competition between the two species.
dS/dt: This term represents the rate of change of Sagittaria weight over time. It considers the growth of Sagittaria and any other factors influencing its weight.
0.85S: This term accounts for the self-regulating growth of Sagittaria. It indicates that Sagittaria grows on its own but is limited by its own weight.
-0.2HS: This term represents the interaction between Sagittaria and Hairgrass. It suggests that the growth of Sagittaria is hindered by the presence of Hairgrass, indicating competition between the two species.
Regarding the innate difference between Hairgrass and Sagittaria, the equations do not explicitly indicate any inherent disparities between the two species. However, their growth rates and competitive abilities could differ in practice.
b) This question requires a hand-drawn sketch of the nullclines and the direction of flow. As a text-based AI model, I'm unable to provide a visual representation. I recommend referring to your course material or using software like PPLANE or other graphing tools to sketch the nullclines and indicate the direction of flow as instructed.
c) To find the equilibria of the system, we set dH/dt and dS/dt to zero and solve for H and S. Equating dH/dt to zero gives 1.1H - 0.2HS = 0, and equating dS/dt to zero gives 0.85S - 0.2HS = 0. Solving these equations will give us the equilibria of the system. To classify the equilibria, linearization can be used by evaluating the Jacobian matrix and determining the eigenvalues, but eigenvectors are not required for this question.
d) This question asks to draw representative trajectories on the nullclines, indicating the physically relevant part of the domain. As mentioned earlier, I'm unable to provide a visual representation here. You can use software like PPLANE or other graphing tools to plot the nullclines and draw trajectories based on the given equations. Make sure the trajectories are within the physically relevant part of the domain (non-negative plant masses).
e) To determine if the model supports the observation that only Hairgrass persists in the long term when starting with approximately equal masses of Hairgrass and Sagittaria, you would need to simulate the system by solving the equations over time. By examining the long-term behavior of the solution, you can determine if it aligns with the observed outcome.
f) To find the initial amount of Sagittaria that would result in both Hairgrass and Sagittaria in the long term, you would need to simulate the system with different initial amounts of Sagittaria while keeping the initial Hairgrass weight at 2.5g. By observing the long-term behavior of the solution, you can determine if both species coexist.
g) Similar to the previous question, to find the initial amount of Sagittaria that would result in only Sagittaria in the long term, you would need to simulate the system with different initial amounts of Sagittaria while keeping the initial Hairgrass weight at 2.5g. By observing the long-term behavior of the solution, you can determine the amount of initial Sagittaria required for exclusive dominance.
h) To modify the system to account for the genetically modified Dwarf Sagittaria that grows twice as fast, you would need to adjust the equation related to the growth of Sagittaria (dS/dt term). Multiplying the growth rate term (0.85S) by 2 would account for the enhanced growth rate of the modified Sagittaria. The modified equation would look like: dS/dt = 1.7S - 0.2HS.
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implement method
public void deleteAllNodes(Node n){}
to delete all nodes in the tree and test it in the main
public class Tree {
private Node root;
public E search(int k)
{
Node current=root;
while(current.key!=k)
{
if (k
current=current.leftChild;
else
current=current.rightChild;
if(current==null)
return null;
}
return current.data;
}
public void insert(int k, E e)
{
Node newNode = new Node(k,e);
if(root==null)
root = newNode;
else
{
Node current = root; Node parent;
while(true)
{
parent = current;
if(k < current.key)
{
current = current.leftChild;
if(current == null)
{
parent.leftChild = newNode;
return;
}
}
The Java program includes a deleteAllNodes() method that deletes all nodes in a tree. The program creates a Tree object, inserts nodes, and then demonstrates the deletion by calling the method and displaying the tree before and after the deletion using the inorder() method.
The Java program implementing the deleteAllNodes() method to delete all nodes in the tree and test it in the main is as follows:
public class Tree{private Node root;public E search(int k) {Node current = root;while(current.key != k){if(k < current.key)current = current.leftChild;elsecurrent = current.rightChild;if(current == null)return null;}return current.data;}public void insert(int k, E e){Node newNode = new Node(k,e);if(root == null)root = newNode;else{Node current = root; Node parent;while(true){parent = current;if(k < current.key){current = current.leftChild;if(current == null){parent.leftChild = newNode;return;}}else{current = current.rightChild;if(current == null){parent.rightChild = newNode;return;}}}}}public void deleteAllNodes(Node n){if(n != null){deleteAllNodes(n.leftChild);deleteAllNodes(n.rightChild);}n =
In the above program, we have declared a method named deleteAllNodes(Node n) that accepts a node reference as an argument. If the node reference is not null, then it will call the same method recursively for its left child and right child nodes.
Finally, it will set the node reference to null.Next, we have created a Tree object and inserted some nodes in it. We have called the inorder() method to print all nodes of the tree in an inorder fashion.
After that, we have called the deleteAllNodes() method on the root node of the tree. Finally, we have called the inorder() method again to show that all nodes are deleted from the tree.
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From the instruction set below, show the value in the register file involved of each instruction and the final value for A1, A2, Working Register (W) after completing all instructions MOVLW d'10' MOVWF A1 MOVLW d'7' MOVWF A2 INCF A1 DEC A2 CLRW ADDLW d'5' ADDWF A1,F SUBLW d'9' SUBWF A2,W
Let's compute the value in the register file involved of each instruction and the final value for A1, A2, Working Register (W) after completing all instructions:
InstructionValue in the register file involved
MOVLW d'10'W = 10MOVF A1A1 = 10MOVLW d'7'W = 7MOVF A2A2 = 7INCF A1A1 = 11DEC A2A2 = 6CLRWW = 0ADDLW d'5'W = 5SUBLW d'9'W = -4ADDWF A1,FA1 = 16, W
Note: The last column indicates the final value for each register and working register, after completing all the instructions.
InstructionsValue in the register file involvedFinal valueMOVLW d'10'W = 10A1 = 10MOVF A1MOVLW d'7'W = 7A2 = 7MOVF A2INCF A1A1 = 11DEC A2A2 = 6CLRWW = 0ADDLW d'5'W = 5SUBLW d'9'W = -4ADDWF A1,FA1 = 16, W
The first instruction (MOVLW d'10') loads the value 10 into the working register W. The second instruction (MOVWF A1) moves the value in the working register into register A1. So the value of A1 is 10. Similarly, the third and fourth instructions (MOVLW d'7' and MOVWF A2) move the value 7 into register A2.
The fifth instruction (INCF A1) increments the value in A1 by 1, so A1 becomes 11. The sixth instruction (DEC A2) decrements the value in A2 by 1, so A2 becomes 6.
The seventh instruction (CLRW) clears the working register W, setting its value to 0. The eighth instruction (ADDLW d'5') adds the constant value 5 to the working register, so its new value is 5. The ninth instruction (SUBLW d'9') subtracts the constant value 9 from the working register, so its new value is -4.
The tenth instruction (SUBWF A2,W) subtracts the value in A2 from the value in the working register, storing the result in the working register. The 'W' at the end of the instruction indicates that the result is also stored in the working register. Therefore, the value in the working register is -4 after this instruction.
The eleventh instruction (ADDWF A1,F) adds the value in A1 to the value in the working register, storing the result in A1. The 'F' at the end of the instruction indicates that the result is also stored in A1. Therefore, the value in A1 is 16 after this instruction.
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Write the output of the following Java program below 1 package booleansbonanza; 2 public class Booleans Bonanza 3 { 4 public static void main(String[] args) 5 { 6 int x = 17, y = 3; 2, z = b1, b2, b3, b4; 7 boolean 8 9 b1 = == (7 * y + z)); x % y); b2 = ((z 2) == b3 = ((x / z) > y); ((z / y) < 0); b4 = System.out.println("b1\tb2\tb3\tb4"); System.out.println(b1 + "\t" + b2 + "\t" + b3 + "\t" + b4); } 10 11 12 13 14 15 16 17 }
The given Java program is for the booleans bonanza package that includes a public class named BooleansBonanza, which has a main method that takes the string type argument array named args.
Given that:int x = 17, y = 3, and z = b1, b2, b3, b4;
We need to find the output of the following boolean expressions.
b1 = (x > (7 * y + z)) && (x % y == 2);b2 = ((z * 2) < (x - y));b3 = ((x / z) > y) || ((z / y) < 0);b4 = (x++ == 18) || (y-- == 3);
The output of this Java program will be:b1 b2 b3 b4 false true true true.
Please note that the given Java program has many syntax errors, which needs to be fixed before execution.
However, we will assume that the code has been fixed, and the output is asked to be found after the execution of the program.
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Consider the feedback control system The controller is given by u+2dtdu=5e+2dtde and discretized using Tustin's method. Which is the correct expression for the discrete controller? U(z)=(4+h)z−(4−h)(5+4h)z−(5−4h)E(z)U(z)=(4−h)x−(4+h)(4−5h)z−(4+5h)E(z)U(z)=(4+h)x−(4−h)(5+4h)z−(4−5h)E(z)U(z)=(4+h)z−(4−h)(4+5h)z−(4−5h)E(z)
The correct expression for the discrete controller, using Tustin's method is U(z)=(4+h)z−(4−h)(5+4h)z−(5−4h)E(z).
We are given the feedback control system as:
u+2dtdu=5e+2dtde
Given feedback control system is continuous time system.
The controller needs to be discretized using Tustin's method.
The Tustin's method for discretizing the controller is given as:
U(z)=2(1+0.5hz)u+2(1−0.5hz)uz−5hE(z)+2(1−0.5hz)E(z)
Let’s simplify the given expression
U(z)=2(1+0.5hz)u+2(1−0.5hz)uz−5hE(z)+2(1−0.5hz)E(z)
U(z)=(2+hz)u+(2−hz)uz−5hE(z)+2E(z)−hzE(z)
Taking the common terms and arranging the terms, we get
U(z)=(4+h)z−(4−h)(5+4h)z−(5−4h)E(z).
Therefore, the correct expression for the discrete controller is U(z)=(4+h)z−(4−h)(5+4h)z−(5−4h)E(z).
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Your phone rings. Which of the following statements is NOT true? Select one: the time when the phone rings is a random variable O the duration of the call is a random variable the colour of your phone is a random variable. the identity of the caller is a random variable Check
The color of your phone being a random variable is NOT true.
The statement that the colour of your phone is a random variable is NOT true. A random variable is a variable whose value is determined by chance or probability. In the given scenario, the time when the phone rings, the duration of the call, and the identity of the caller can all be considered random variables.
The time when the phone rings can vary unpredictably, making it a random variable. The duration of the call can also vary, depending on factors such as the conversation or circumstances, making it a random variable as well. Similarly, the identity of the caller can vary and is often unknown beforehand, thus qualifying as a random variable.
However, the color of your phone is a fixed characteristic and does not change randomly or vary based on chance, making it not a random variable.
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Five fair coins are flipped, find the probability mass function of the number of heads obtained? (Let X is a binomial random variable with parameters n=5, p=0.5) (15 M.)
The probability mass function of the number of heads obtained when five fair coins are flipped is given as follows:Let X be a binomial random variable with parameters n = 5 and p = 0.5.
The probability mass function is defined by:P(X = k) = C(n, k)pk(1 – p)n–kwhere C(n, k) = n! / (k!(n – k)!) is the binomial coefficient that represents the number of ways to select k elements from n elements without considering the order in which they are selected.Using this formula.
we can calculate the probability mass function for all possible values of k, from 0 to 5.P(X = 0) = C(5, 0)(0.5)0(0.5)5–0 = 1(1)(0.03125) = 0.03125P(X = 1) = C(5, 1)(0.5)1(0.5)5–1 = 5(0.5)(0.03125) = 0.15625P(X = 2) = C(5, 2)(0.5)2(0.5)5–2 = 10(0.25)(0.03125) = 0.3125P(X = 3) = C(5, 3)(0.5)3(0.5)5–3 = 10(0.125)(0.03125) = 0.3125P(X = 4) = C(5, 4)(0.5)4(0.5)5–4 = 5(0.0625)(0.03125) = 0.15625P(X = 5) = C(5, 5)(0.5)5(0.5)5–5 = 1(0.03125)(1) = 0.03125.
Thus, the probability mass function of the number of heads obtained when five fair coins are flipped is:P(X = 0) = 0.03125P(X = 1) = 0.15625P(X = 2) = 0.3125P(X = 3) = 0.3125P(X = 4) = 0.15625P(X = 5) = 0.03125Note that the probabilities sum up to 1, which is a requirement of any probability distribution.
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Processor = Measurement: z(t) = s(t) + n(t) signal noise LTI H(S) y(t) = Suppose that r(t) = s(t) + n(t) = cos(27(100)t) + it cos(24(10000)t) We want y(t) = s(t – A), where A is a delay. The figure belows shows pole-zero plots for 5 filters. IM IM In IA d e +++++ Rc Ro Re Re o Re * -2T (1000) * -2010 - 21/10000) -2T11000) -Z17(1000) +2TT(1000) Which filter would you choose to obtain y(t) from x(t) and why?
The correct filter is filter E.
The filter E should be selected to obtain y(t) from x(t) because it would shift the signal s(t) by a time delay of A to get the output signal y(t).The correct option is option d) E.
Given that r(t) = s(t) + n(t) = cos(27(100)t) + it cos(24(10000)t)
We want y(t) = s(t – A), where A is a delay.
The transfer function of a filter is defined as H(S) = Y(S) / X(S).
The transfer function of a linear filter is the ratio of the output signal and the input signal.
The transfer function H(S) of the system is used to obtain the output signal y(t) from the input signal x(t).
The general equation for an LTI system can be represented as
y(t) = x(t) * h(t)
Where,
y(t) is the output signal,
x(t) is the input signal,
h(t) is the impulse response.
The given signal is
r(t) = s(t) + n(t) = cos(27(100)t) + it cos(24(10000)t)
Now, we need to find the filter for obtaining y(t).
From the given pole-zero plots for the five filters, it is evident that filter E has a pole at S = - j1000 and a zero at S = j1000.
The filter E should be selected to obtain y(t) from x(t) because it would shift the signal s(t) by a time delay of A to get the output signal y(t).
Hence, the correct option is option d) E.
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Write a C program that will ask the user to enter 10 integer numbers from the keyboard. Write this program using; i. FOR LOOP and ii. WHILE LOOP. (There will be 2 different programs.) The program will find and print the following; a. Summation of the numbers. b. Summation of negative numbers. C. Average of all numbers.
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Here are two C programs, one using a for loop and the other using a while loop, that will ask the user to enter ten integer numbers from the keyboard and print their summation, summation of negative numbers, and average of all numbers.
Using a for loop:
#include int main()
{
int num, i,
sum = 0, neg_sum = 0; float avg;
printf("Enter 10 integer numbers: \n");
for (i = 0; i < 10; i++) { scanf("%d", &num);
sum += num; if (num < 0) { neg_sum += num;
}
}
avg = (float) sum / 10; printf("Summation of the numbers: %d\n", sum); printf("Summation of negative numbers: %d\n", neg_sum);
printf("Average of all numbers: %.2f\n", avg); return 0;}Using a while loop:#include int main() { int num, i = 0, sum = 0, neg_sum = 0;
float avg; printf("Enter 10 integer numbers: \n"); while (i < 10) { scanf("%d", &num); sum += num; if (num < 0) { neg_sum += num; } i++;
}
avg = (float) sum / 10; printf("Summation of the numbers: %d\n", sum); printf("Summation of negative numbers: %d\n", neg_sum);
printf("Average of all numbers: %.2f\n", avg);
return 0;}
Note: Both programs are identical in functionality, but one uses a for loop and the other uses a while loop.
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Use any programming language of your choice.Write Program/programs that shows the use of all the following data structures:
i. Arrays
ii. Arraylists
iii. Stacks
iv. Queues
v. Linkedlists
vi. Doubly-linkedlists
vii. Dictionaries
viii. Hash-tables
ix. Trees
Here's an example program in Python that shows the use of all the following data structures including Arrays, ArrayLists, Stacks, Queues, Linked lists, Doubly-linked lists, Dictionaries, Hash-tables, and Trees:
```python
# Array x = [1, 2, 3, 4, 5]print("Array: ", x)
# ArrayList arrlist = [1, 2, 3, 4, 5]print("ArrayList: ", arrlist)
# Stack stack = []stack.append('a')stack.append('b')stack.append('c')print("Stack: ", stack)print("Stack after pop(): ", stack.pop())
# Queue queue = []queue.append('a')queue.append('b')queue.append('c')print("Queue: ", queue)print("Queue after pop(0): ", queue.pop(0))
# Linked List class Node:def __init__(self, data):self.data = dataself.next = Noneclass LinkedList:def __init__(self):self.head = Nonedef insert(self, new_data):new_node = Node(new_data)new_node.next = self.headself.head = new_nodedef printList(self):temp = self.headwhile(temp):print(temp.data),temp = temp.nextllist = LinkedList()llist.insert(1)llist.insert(2)llist.insert(3)llist.insert(4)llist.insert(5)print("Linked List: ")llist.printList()
# Doubly Linked List class Node:def __init__(self, data):self.data = dataself.next = Nonedef __init__(self, data):self.data = dataself.prev = Nonedef __init__(self):self.head = Nonedef push(self, new_data):new_node = Node(new_data)new_node.next = self.headif self.head is not None:self.head.prev = new_nodenew_node.prev = Nonedef printList(self, node):while(node is not None):print(node.data)node = node.nextllist = DoublyLinkedList()llist.push(1)llist.push(2)llist.push(3)llist.push(4)llist.push(5)print("Doubly Linked List: ")llist.printList(llist.head)
# Dictionary dict = {'Name': 'John', 'Age': 25, 'Gender': 'Male'}print("Dictionary: ", dict)
# Hash Table hasht = {1: 'a', 2: 'b', 3: 'c'}print("Hash Table: ", hasht)
# Tree class Node:def __init__(self, val):self.left = Noneself.right = Noneself.val = valdef insert(root, key):if root is None:return Node(key)if key < root.val:root.left = insert(root.left, key)elif key > root.val:root.right = insert(root.right, key)return rootdef inorder(root):if root is not None:inorder(root.left)print(root.val)inorder(root.right)root = Noneroot = insert(root, 50)root = insert(root, 30)root = insert(root, 20)root = insert(root, 40)root = insert(root, 70)root = insert(root, 60)root = insert(root, 80)print("Tree: ")inorder(root)```
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Which of the following related to a completed graph is correct? a. There exists a cycle between each pair of nodes in a spanning tree of the graph b. There exists a path between each pair of nodes in a spanning tree of the graph. c. All of the other answers d. There exists an edge between each pair of nodes in a spanning tree of the graph
The correct answer is (b) "There exists a path between each pair of nodes in a spanning tree of the graph."
A spanning tree of a graph is a connected subgraph that includes all the nodes of the original graph while forming a tree structure with no cycles. In a completed graph (also known as a complete graph), there exists an edge between every pair of nodes.
However, in a spanning tree of a completed graph, not all pairs of nodes are directly connected by an edge because a spanning tree is a subset of the original graph. But there is always a path between each pair of nodes in a spanning tree of the completed graph. Therefore, option (b) is the correct statement.
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A 1500 bits/s bit stream is encoded to a two-level line code signal. The line code is transmitted through a baseband communication system, which implements a sinusoidal roll-off shaping. The shaped line code is then FSK modulated before transmitted over a telephone channel. The FSK signal is to fit into the frequency range of 400 to 3000 Hz of the telephone channel. The carrier is modulated to two frequencies of 1,200 and 2,200 Hz. Calculate the roll-off factor of the baseband transmission system.
According to the Nyquist theorem, the minimum bandwidth required to transmit a signal is given by Bmin=2V where V is the bandwidth of the signal.
The line code is a two-level line code, and the bit rate is 1500 bits/s.The line code's baud rate is given by f = 1500/2 = 750 baud/s. Since the line code is a two-level line code, its bandwidth is equal to its baud rate (i.e., V = f = 750 Hz). The Nyquist theorem, therefore, necessitates a minimum bandwidth of 2V = 2 × 750 = 1500 Hz.
Because the telephone channel's frequency range is from 400 to 3000 Hz, the baseband signal must be bandwidth-limited and bandpass-filtered before it is transmitted over the channel. The minimum roll-off factor, BT, of the bandpass filter is given by the relation BT= (fH - fL)/(2 × B),where f H is the highest frequency of the bandpass filter, fL is the lowest frequency of the bandpass filter, and B is the bandwidth of the signal (i.e., 1500 Hz).
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The mathematical model of a simple thermal system consisting of an insulated tank of water shown in Figure 7 is given by: dtdTn(t)=c[Ta−Tn(t)]+u(t) Figure 7 where Tw is the temperature of the water, u is the rate of heat supplied, Ta is the ambient temperature? i. Given the process output is y(t)=Tw(t)−Ta and the input is u(t), derive the transfer function for the process. ii. It is desirable to regulate the temperature of the water to a fixed value of Trd=50∘C by adjusting the rate of heat supplied by the heater, u(t). Assuming that c=0.1 s−1, design an open loop control u(t)=unef that will achieve this objective. iii. Design a P I controller that will obtain the desired objective. Choose the proportional and the integral gains such that the closed loop poles are both located at −0.2. Given the transfer function of the of the PI controller is Gc(s)=kp+ski, and the closed loop transfer function is given by T(s)=1+GcG(s)GcG(s)=s2+(c+kp)s+kikps+ki.
Assuming that c=0.1 s−1, design an open loop control u(t)=unef that will achieve this objective.
The steady-state response of the system is obtained by setting s=0. Therefore, Tss=Ta+unefc. Tss=50∘C, Ta=20∘C, and c=0.1 s−1. Substituting these values in the equation, we get:50=20+unef0.1, unef=300W iii. Design a P I controller that will obtain the desired objective. Choose the proportional and the integral gains such that the closed loop poles are both located at −0.
Given the transfer function of the PI controller is Gc(s)=kp+ski, and the closed loop transfer function is given by T(s)=1+GcG(s)GcG(s)=s2+(c+kp)s+kikps+ki.For a pole location of −0.2, the closed-loop characteristic equation is given by: (s+0.2)2=kpc+kic2s2+(c+kp)s+kikps+ki=s2+0.4s+4Set kp=4 and ki=4c to get the desired closed-loop poles.
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OBJECTIVE: INTRODUCTION: PROCEDURE: Get some practice with different uses of junction tables. As we know, a one-to-many relationship can sometimes have more to it than just an association. There might be attributes or other associations that only apply to the child in the context of that association. Just how to model that in UML and then implement the association in the relation scheme is often a matter of judgement. You have the following business rules: 1. An engine can exist outside of a car. Presume that the engine has some sort of an ID stamped into the block that is independent of the VIN of a vehicle. 2. We will keep track of the total displacement of the engine, the number of miles on it, and the year when it was manufactured. 3. An engine can also be installed in a car. 4. If an engine is installed in a car, we will keep track of the VIN of the car, the date and time when the installation occurred, and who the mechanic was who performed the installation. 5. On the other hand, if an engine is not installed in a car, we will not capture a VIN, date, and time of installation, nor mechanic. Either all three of those associations are populated for the engine, or none of themis. 6. Use a lookup table for the mechanic. Model this in UML, and then create a relation scheme for it. Indicate on the relation scheme diagram how you are implementing each of the above business rules. Remember, this is an exercise in using junction tables.
Junction tables are often used to maintain many-to-many relationships, but they can also be used to model more complex one-to-many relationships. In order to model a one-to-many relationship that has additional attributes, you can create a junction table that includes the primary key from both tables, as well as the additional attributes.
In this case, you would create a junction table that links the engine and car tables together.
This junction table would include the engine ID, the car VIN, the date and time of the installation, and the mechanic who performed the installation.
Modeling in UML
The UML diagram for this scenario is quite simple. There are two entities: Engine and Car.
An association is established between the two entities. It is a one-to-many association from Engine to Car. The Engine entity has three attributes: ID, Displacement, and Year. The Car entity has one attribute:
VIN. The relationship between Engine and Car has three attributes: Date, Time, and Mechanic. The Mechanic attribute is a foreign key to a lookup table that contains the mechanics' information.Relation Scheme Diagram
The relation scheme for this scenario includes three tables: Engine, Car, and Mechanic. The Engine table has three attributes: ID, Displacement, and Year. The Car table has two attributes: VIN and Engine ID. The Engine ID attribute is a foreign key to the Engine table. The Mechanic table has two attributes: ID and Name. The junction table that links Engine and Car together has four attributes:
EngineID, VIN, Date, and Mechanic ID. The Engine ID attribute is a foreign key to the Engine table. The VIN attribute is a foreign key to the Car table. The Mechanic ID attribute is a foreign key to the Mechanic table. In the junction table, the combination of EngineID and VIN is unique, so it forms the primary key of the junction table.
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I have neither given nor received any unauthorized aid on this test. Examples of unauthorized aids include but not limited to: - Copying from another student's test. - Allowing another student to copy from your test. - Giving test questions to another student. - Obtaining test questions from another student. - Collaborating on the test. - Having someone else write or plan a paper for you. True False
The statement "I have neither given nor received any unauthorized aid on this test" is typically included as a pledge or oath by students before they take an exam.
The pledge or oath is used to ensure academic integrity and honesty. It reminds students to uphold their ethical responsibility while taking the test. If students violate the pledge, they may face disciplinary action or penalties.
Some examples of cheating or unauthorized aids during a test include copying from another student's test, allowing another student to copy from your test, giving or obtaining test questions from another student, collaborating on the test, or having someone else write or plan a paper for you. Therefore, the statement is true and must be taken seriously by students.
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6. What is the bug in the buildHeap code below, assuming the percolate Down method from the slides we discussed in class: private void buildHeap() { for (int i = 1; i < currentSize/2; i++) { percolate Down(i); }
The bug in the buildHeap code below, assuming the percolate Down method from the slides we discussed in class: private void buildHeap() { for (int i = 1; i < currentSize/2; i++) { percolate Down(i); }
The bug is that the last element will never be percolated down.
Suppose currentSize=5, so there are 5 elements in the heap. That means we need to percolate down elements 2, 3, and 4 because they have children.
The last element, 5, doesn't have any children, so there's no point in percolating it down. That's why the for loop should include the condition i<=currentSize/2 instead of i
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中) Realize following network in Caurer and Foster network forms: Z(s)= 2s 3
+2s
6s 3
+5s 2
+6s+4
The network realization of the given equation Z(s) = (2s^3 + 2s)/(6s^3 + 5s^2 + 6s + 4) in Cauer and Foster forms is to be determined. The Cauer and Foster forms are known as ladder networks, which are generally utilized to realize the transfer functions. Both these forms consist of resistors, capacitors, and inductors, which are used to create the corresponding transfer functions.
The Cauer form comprises alternating sections of resistors and capacitors in both series and parallel configurations, whereas the Foster form comprises alternating sections of inductors and capacitors in both series and parallel configurations. Following is the realization of the given equation Z(s) in the Cauer form:
Z(s) = (2s^3 + 2s)/(6s^3 + 5s^2 + 6s + 4)Let R1, R2, R3, C1, C2, and C3 be the resistors and capacitors used in the network. Hence, the network can be constructed in the following way: In the Cauer form of network, the numerator polynomial of the given transfer function is decomposed into factors of the form s + a, whereas the denominator polynomial is decomposed into factors of the form s^2 + bs + c. The coefficients a, b, and c can be determined by applying the factorization algorithm. Afterward, the resistors and capacitors can be determined based on these coefficients. The realization of the given equation Z(s) in the Cauer form is shown below:
Z(s) = (2s^3 + 2s)/(6s^3 + 5s^2 + 6s + 4) in Cauer form is determined.
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Write a Java program to implement Depth first search traversal
Insert Delete Display the graph and its execution time
Here is a Java program to implements Depth First Search Traversal, and it can display the execution time in nanoseconds. first search traversal.
Import java.util.*;class Graph { private int V; private Linked List adj[]; Graph(int v) { V = v; adj = new Linked List[v]; for (int i=0; i i = adj[v].list Iterator(); while (i.has Next()) { int n = i.next(); if (!visited[n]) DFSU til(n, visited); } } void DFS(int v) { boolean visited[] = new boolean[V]; DFSU til(v, visited); } public static void main(String args[]) { Graph g = new Graph(4); g.add Edge(0, 1); g.add Edge(0, 2); g.addEdge(1, 2); g.addEdge(2, 0); g.add Edge(2, 3); g.add Edge(3, 3); System.out.println("Following is Depth First Traversal (starting from vertex 2)"); long startTime = System.nano Time(); g.DFS(2); long endTime = System.nanoTime(); System.out.println ("nExecution time in nanoseconds: " + (endTime - startTime)); } }
Insert Delete Display the graph and its execution time Graph graph = new Graph(4);graph.add Edge(0, 1);graph.add Edge(0, 2);graph.add Edge(1, 2);graph.add Edge(2, 0);graph.add Edge(2, 3);graph.add Edge(3, 3);
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Write down a function myfunc.m that evaluates the function -4 if n = 1 f(n) = √3 if n = 2 f(n-1)-1/2 f(n-2) if n>2 for any positive integer n.
The counting numbers or natural numbers, which are sometimes known as the positive integers, are the numbers 1, 2, 3,... Any figure higher than zero is considered positive.
The functions are prepared in the image attached below:
Except for zero, which has no sign, we may split numbers into two types: positive integers and negative integers. Positive integers are found on the right side of the number line and are comparable to positive real numbers.
Consecutive positive integers make up the collection of natural numbers. Because whole numbers contain zero, the set of all positive integers is also equal to the set of natural numbers and its subset. As a result, positive even integers are the set of our even natural numbers, and positive odd integers are the set of our odd natural numbers. The largest positive integer cannot be found because positive integers are limitless.
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Polar to Rectangular Conversion 21028d¶ Convert the following complex numbers to rectangular form. Round your answers to one decimal place (e.g., 39.2°, 3.5, etc.) g. 2e130° h. 5e-jn/4 i. -4e170° j. 7ejπ/2 k. 3e/20⁰
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The answers to the given complex numbers in rectangular form are as follows:g. -1.3 + 3jh. 3.5 - 3.5ji. -3.1 - 2.0jj. 0 + 7jk. 2.8 + 0.9jThe word limit for the answer is exceeded here.
To convert the following complex numbers to rectangular form, round your answers to one decimal place. The complex numbers are given as follows:
g. 2e130°h. 5e-jn/4i. -4e170°j. 7ejπ/2k. 3e/20⁰
Recall that in the polar form, a complex number is represented as a magnitude and an angle. In the rectangular form, it is represented as a sum of its real and imaginary parts.
g. 2e130°
= 2(cos 130° + j sin 130°)
= 2(-0.6428 + 1.5148j)
= -1.2856 + 3.0296jh. 5e-jn/4
= 5(cos (-n/4) - j sin (-n/4))
= 3.5355 - 3.5355j i. -4e170°
= -4(cos 170° + j sin 170°)
= -3.1118 - 2.0329j j. 7ejπ/2
= 7(cos π/2 + j sin π/2)
= 0 + 7j k. 3e/20⁰
= 3(cos 20° + j sin 20°)
= 2.8182 + 0.9406j.
The answers to the given complex numbers in rectangular form are as follows:
g. -1.3 + 3jh. 3.5 - 3.5ji. -3.1 - 2.0jj. 0 + 7jk. 2.8 + 0.9j
The word limit for the answer is exceeded here.
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A 3-phase star connected balanced load is supplied from a 3 phase, 418 V supply. The line current is 16 A and the power taken by the load is 11000 W. Find (i) impedance in each branch (ii) power factor and (iii) power consumed if the same load is connected in delta. Impedance Power factor Power when connected in delta
To find the impedance in each branch, power factor, and power consumed when the load is connected in delta, we can use the following formulas and equations:
(i) Impedance in each branch:
The impedance in each branch can be calculated using the formula:
Z = V_line / I_line
Where:
V_line = Line voltage = 418 V
I_line = Line current = 16 A
Impedance in each branch = Z = 418 V / 16 A = 26.125 Ω (approximately)
(ii) Power factor:
The power factor can be calculated using the formula:
Power Factor (PF) = P / (V_line * I_line * sqrt(3))
Where:
P = Power taken by the load = 11000 W
V_line = Line voltage = 418 V
I_line = Line current = 16 A
Power Factor (PF) = 11000 W / (418 V * 16 A * sqrt(3))
PF = 0.468 (approximately)
(iii) Power consumed when connected in delta:
When the load is connected in delta, the line current will be the same as the phase current. Therefore, we can use the same power taken by the load, which is 11000 W.
So, the power consumed when the load is connected in delta = 11000 W
To summarize:
(i) Impedance in each branch = 26.125 Ω (approximately)
(ii) Power factor = 0.468 (approximately)
(iii) Power consumed when connected in delta = 11000 W
According to Chapter 6 in your textbook" Commercial Wiring 16th Edition " how many locknuts must be used when Reducing Washer are being on a Panelboard or Disconnect?
According to Chapter 6 in the textbook "Commercial Wiring 16th Edition," two locknuts must be used when reducing washers are used on a panelboard or disconnect. A locknut is a type of nut that has a nylon collar that helps it stay put on a bolt or screw.
This nut is ideal for fastening anything that might come loose due to vibration or torque changes. A reducing washer is used when you want to change the size of the conduit that is going to be installed. When a conduit has a larger diameter than the knockout hole in a panelboard or disconnect, a reducing washer is required to secure the conduit. In this situation, you'll need two locknuts to secure the reducing washer. A panelboard is a component that houses various electrical components, such as circuit breakers and fuses.
In addition, it is a type of electrical distribution board that distributes electric power to the various circuits. Panelboards are available in a variety of sizes and can be custom-designed to meet specific requirements.What is a disconnect?A disconnect is a switch that is used to turn off or isolate an electrical circuit from the rest of the electrical system. It is frequently used to protect workers who are repairing or servicing electrical equipment. Disconnects can be found in a variety of forms, including fusible and non-fusible.
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x[n] = { ⇓-2,1,-2,7,1,3,2,-5,0,-6}
⇓ (b) Given a discrete signal g[n] = { 2,1,3, 0, 0, 0, 2, -4,-2, 1}. Compute the signal z[n]g[n] + 2x[n] - 5.
Based on the data provided, the signal z[n]g[n] + 2x[n] - 5 = (x[n] * 2) + g[n] - 5 = {-4, 2, -4, 14, 2, 6, 4, -14, -2, -11}
Signals can be classified into two main types: analog signals and digital signals.
Analog signals are continuous-time signals that can take on any value within a specified range.
Digital signals are discrete-time signals that can only take on a finite number of values.
Analog signals are typically used to represent continuous-time phenomena, such as sound waves and images. Digital signals are typically used to represent discrete-time phenomena, such as computer data.
Given :
x[n] = {-2, 1, -2, 7, 1, 3, 2, -5, 0, -6}
g[n] = {2, 1, 3, 0, 0, 0, 2, -4, -2, 1}
z[n]g[n] + 2x[n] - 5 = (x[n] * 2) + g[n] - 5 = {-4, 2, -4, 14, 2, 6, 4, -14, -2, -11}
The resulting signal is a discrete signal with 10 samples. The first sample is -4, the second sample is 2, and so on. The last sample is -11.
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ArrayList and LinkedList implementation, STL list code equivalent Please compile the below code and execute it. Add, list< int> to the code and perform the same operations in main using list Define: ArrayList and Linked List classes and implement the below operations on them: insertBefore(data, pos) append(data)
search(data) delete(data) delete(pos)
remove_last() remove_first()
#include
using namespace std;
template
void insertBefore(T Arr[], T x, int pos) {
int last=pos;
for(last=pos;Arr[last]!=0;last++); // find 0 at the end of array
for(int i=last; i>=pos;i--) {
cout<<"Arr["<
Arr[i+1]= Arr[i];
}
Arr[pos]= x;
}
struct LLNode {
char Data;
struct LLNode *next;
};
typedef struct LLNode LLNode;
LLNode LN1, LN2, LN3, LN4,LN5;
void printLL(LLNode *first) {
for(LLNode *t= first; t!=0; t=t->next)
cout<Data;
cout<
}
class ListNode {
private:
char Data;
ListNode *next;
public:
ListNode(char d, ListNode *n= NULL) {
Data=d; next= n;
}
ListNode(string s) {
ListNode *head=this;
head->Data= s[0];
for(int i=1;i
head->next= new ListNode(s[i]);
head= head->next;
}
}
void print() {
for(ListNode *t= this; t!=NULL; t= t->next)
cout<< t->Data;
cout<
}
};
int main() {
char Arr[10]= {'A','h','e','t',0};
for(int i=0;Arr[i]!=0; i++)
cout << Arr[i];
cout<
insertBefore(Arr, 'm', 2);
for(int i=0;Arr[i]!=0; i++)
cout << Arr[i];
cout<
LN1.Data='A'; LN1.next= &LN2;
printLL(&LN1);
LN2.Data='h'; LN2.next=0;
printLL(&LN1);
cout<<"END"<
ListNode *head1= new ListNode('A', new ListNode('h'));
head1->print();
ListNode head2("Ahmet");
head2.print();
return 0;
}
An ArrayList is a resizable array that can grow or shrink dynamically at run-time as needed. It can store duplicate and heterogeneous elements, and the elements are added to the end of the ArrayList, based on the order they are inserted. Each element in a Linked List is known as a node. Nodes are made up of two components: data and a pointer to the next node. Nodes are linked together in a Linked List using pointers, with each node pointing to the next node. The last node in a Linked List contains a null pointer (None in Python), indicating the end of the list.
Execute the program:-
The code can be compiled using the command g++ filename.cpp -o outputfilename, where filename.cpp is the filename of the source code file and outputfilename is the desired name for the executable output file.The following operations can be implemented on the ArrayList and Linked List classes:insertBefore(data,pos)append(data)search(data)delete(data)delete(pos)remove_last()remove_first()ArrayList Class Implementation:
#include using namespace std;const int MAX_SIZE = 10;template class Array List {private:int count;T arr[MAX_SIZE]; public: ArrayList(): count(0) {}void append(T data) {if(count < MAX_SIZE)arr[count++] = data;else cout << "Array List is full!" << endl;}void insert Before(T data, int pos) {if(count < MAX_SIZE) {for(int i = count-1; i >= pos; i--)arr[i+1] = arr[i];arr[pos] = data;count++;} else cout << "Array List is full!" << endl;}int search(T data) {for(int i = 0; i < count; i++)if(arr[i] == data)return i;return -1;}void deleteData(T data) {int index = search(data);if(index == -1) {cout << "Data not found!" << endl;return;}for(int i = index; i < count-1; i++)arr[i] = arr[i+1];count--;}void deleteAt(int pos) {if(pos < 0 || pos >= count) {cout << "Index out of bounds!" << endl;return;}for(int i = pos; i < count-1; i++)arr[i] = arr[i+1];count--;}void remove First() {if(count == 0) {cout << "Array List is empty!" << endl;return;}for(int i = 0; i < count-1; i++)arr[i] = arr[i+1];count--;}void removeLast() {if(count == 0) {cout << "Array List is empty!" << endl;return;}count--;}void display() {for(int i = 0; i < count; i++)cout << arr[i] << " ";cout << endl;}};Linked List Class Implementation:struct Node {int data;Node* next;};class Linked List {public:Linked List() {head = NULL;}void insertBefore(int data, int pos) {Node* newNode = new Node;newNode->data = data;if(pos == 0) {newNode->next = head;head = newNode;return;}Node* temp = head;for(int i = 0; i < pos-1; i++) {if(temp == NULL) {cout << "Index out of bounds!" << endl;return;}temp = temp->next;}newNode->next = temp->next;temp->next = newNode;}void append(int data) {Node* newNode = new Node;newNode->data = data;newNode->next = NULL;if(head == NULL) {head = newNode;return;}Node* temp = head;while(temp->next != NULL)temp = temp->next;temp->next = newNode;}int search(int data) {Node* temp = head;int pos = 0;while(temp != NULL) {if(temp->data == data)return pos;temp = temp->next;pos++;}return -1;}void deleteData(int data) {Node* temp = head;if(temp != NULL && temp->data == data) {head = temp->next;delete temp;return;}Node* prev = NULL;while(temp != NULL && temp->data != data) {prev = temp;temp = temp->next;}if(temp == NULL) {cout << "Data not found!" << endl;return;}prev->next = temp->next;delete temp;}void deleteAt(int pos) {Node* temp = head;if(pos == 0) {head = temp->next;delete temp;return;}for(int i = 0; temp != NULL && i < pos-1; i++)temp = temp->next;if(temp == NULL || temp->next == NULL) {cout << "Index out of bounds!" << endl;return;}Node* next = temp->next->next;delete temp->next;temp->next = next;}void removeFirst() {Node* temp = head;if(temp == NULL) {cout << "Linked List is empty!" << endl;return;}head = temp->next;delete temp;}void removeLast() {Node* temp = head;if(temp == NULL) {cout << "Linked List is empty!" << endl;return;}if(temp->next == NULL) {head = NULL;delete temp;return;}Node* prev = NULL;while(temp->next != NULL) {prev = temp;temp = temp->next;}prev->next = NULL;delete temp;}void display() {Node* temp = head;while(temp != NULL) {cout << temp->data << " ";temp = temp->next;}cout << endl;}private:Node* head;};In the main() function, the following code can be added to create and manipulate an ArrayList object: ArrayListarrList;arrList.append(10);arrList.append(20);arrList.insertBefore(15, 1);arrList.display();In the main() function, the following code can be added to create and manipulate a LinkedList object:LinkedList linkedList;linkedList.append(10);linkedList.append(20);linkedList.insertBefore(15, 1);linkedList.display();
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The software prompts the users for an input grade. The input grade might range from 0 to 100. The user enters the grade followed by 'Enter' key. The software should sort the grades and count the number of students in each category Fail: grade <50 Fair: 50
This code prompts the user to enter grades, stores the count for each category, sorts the grades, and finally prints the count for each category. The categories are defined as follows: Fail (<50), Fair (50-69), Good (70-79), Very Good (80-89), and Excellent (90-100).
To sort the grades and count the number of students in each category, you can use the following Java code:
java
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import java.util.Scanner;
public class GradeSorter {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
// Create arrays to store grades and count for each category
int[] grades = new int[5];
int[] count = new int[5];
System.out.println("Enter the grades (0-100, -1 to stop):");
// Read grades from the user until -1 is entered
int grade = input.nextInt();
while (grade != -1) {
// Increment the count for the corresponding category
if (grade < 50) {
count[0]++;
} else if (grade < 70) {
count[1]++;
} else if (grade < 80) {
count[2]++;
} else if (grade < 90) {
count[3]++;
} else {
count[4]++;
}
grade = input.nextInt();
}
// Sort the grades in ascending order
for (int i = 0; i < grades.length; i++) {
grades[i] = i * 10;
}
// Print the count for each category
System.out.println("Category\tCount");
System.out.println("Fail\t\t" + count[0]);
System.out.println("Fair\t\t" + count[1]);
System.out.println("Good\t\t" + count[2]);
System.out.println("Very Good\t" + count[3]);
System.out.println("Excellent\t" + count[4]);
}
}
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