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

Answers

Answer 1

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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Soru The approximate solution time is 5 minutes. (10 points) NOTE: • If you need to use √3 , you have to use it as 1.7. Don't use any unit when you fill the blanks. • You have to write your answers by using three decimal digits considering the rounding. A 380 kV three-phase transmission line is 300 km long. The series impedance is 0.065+j0.3 ohm per phase per km, and the shunt admittance is j0.00012 Siemens per phase per km. You assume that the line is lossless. Find the propagation constant. Propagation constant equals

Answers

The propagation constant (γ) can be calculated using the following formula:

γ = √(Z * Y)

where Z is the series impedance and Y is the shunt admittance. In this case, Z = 0.065 + j0.3 ohm/km and Y = j0.00012 S/km.

Substituting the given values into the formula:

γ = √((0.065 + j0.3) * (j0.00012))

Let's calculate the propagation constant step by step:

First, let's multiply the series impedance and the shunt admittance:

(0.065 + j0.3) * (j0.00012) = (0.0000078 + j0.000036) + j(0.000036 - j0.00000156)

= 0.0000078 + j0.000036 + j0.000036 + j^2(0.00000156)

= 0.0000078 + j0.000072 + j^2(0.00000156)

Next, let's simplify the expression:

j^2 = -1

Therefore, the expression becomes:

0.0000078 + j0.000072 - 0.00000156 = 0.0000078 + j0.00007044

Now, let's calculate the square root of the expression:

√(0.0000078 + j0.00007044) = 0.002790 + j0.026565

Rounding to three decimal places:

γ ≈ 0.002 + j0.027

Hence, the propagation constant is approximately 0.002 + j0.027 (per km).

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hat are the default file & directory permissions if the umask is o022? (Explain your answer and show the calculations. (b1) [4 marks] 25) To display current working directory in Linux, we use the command (b) [1 mark] a. Is b. pwd d. passwd e. cal 26) Explain what the following command will do if you use it in Linux? (b) mount -t ext3 /dev/sdcl /var/FirstPar [4 marks] 27) To display a calendar in Linux, we use the command (b) a. date b. time c. cal d. man 28) What is the difference between Linux work station and Linux server?

Answers

25) The default file and directory permissions, if the umask is o022, are as follows:

Umask defines default file permissions in Linux, therefore, a new file is created with the permission 0666 - 022 = 0644. Here, umask value is subtracted from the maximum permission value that can be set. A directory is created with permission 0777 - 022 = 0755.

UMASK is used to control the permissions of newly created files. It subtracts the mask value from the permissions set by the user. In the case of default file and directory permissions, if the umask is o022, then, it means that the default permission for the file is 644 and the directory is 755.

For the default permission of the file:

Maximum permission = 666Mask = 02

2Default permission = 666 - 022 = 644

For the default permission of the directory:

Maximum permission = 777

Mask = 022

Default permission = 777 - 022 = 755

26) The given command is as follows:

The given command will mount the partition sdcl of type ext3 on the /var/FirstPar directory.

In Linux, the mount command is used to mount a device or a file system. It attaches a file system to the root file system, making it accessible at a certain mount point

To mount the file system, you need to use the command "mount". Here, the "-t" flag specifies the file system type, "ext3" is the type of the file system, "/dev/sdcl" is the device name or block device name for the file system that you want to mount, and "/var/FirstPar" is the mount point where you want to mount the file system.

27) The command to display a calendar in Linux is:

In Linux, the cal command is used to display a calendar on the terminal. This command takes no arguments and displays a calendar of the current month. It also highlights the current date.

28) The difference between Linux workstation and Linux server is as follows:

A Linux workstation is designed for use by a single user or a group of users working together in a small office or home office (SOHO) environment. It is usually a high-end desktop or laptop computer running a Linux-based operating system that is optimized for desktop use.

Linux server, on the other hand, is designed for use in a network environment where many users need to access shared resources. It is a dedicated computer that provides services to other computers or devices on the network. It runs specialized server software and can be accessed remotely from other computers or devices on the network.

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Implement method. public void printByLevel (Queue,Node){}
Using Queue to print levels of a binary tree
public interface Queue { // interface is a blueprint of class contains methods without implemenation
//A Blueprint Interface is a collection of one or more functions
/** Returns the number of elements in the queue. */
int size( );
/** Tests whether the queue is empty. */
boolean isEmpty( );
/** Inserts an element at the rear of the queue. */
void enqueue(E e);
/** Returns, but does not remove, the first element of the queue (null if empty). */
E first( );
/** Removes and returns the first element of the queue (null if empty). */
E dequeue( );
}
public class Tree {
private Node root;
}
public class Node {
E data;
Node leftChild;
Node rightChild;
public E getData() {
return data;
}
public Node(int k,E e)
{
key=k;
data=e;
leftChild=null;
rightChild=null;
}
public void display() {
System.out.print(key+":");
System.out.println(data);
}
}

Answers

The provided code snippet includes a method called "printByLevel" in the Tree class, which takes a Queue and Node as parameters. The method is intended to print the levels of a binary tree using the provided Queue.

The Tree class represents a binary tree structure and has a private instance variable "root" of type Node. The Node class represents a node in the binary tree and contains data, leftChild, and rightChild attributes.

The printByLevel method aims to traverse the binary tree in a level-by-level manner and print the elements using the provided Queue interface. However, the implementation details of the printByLevel method are not provided in the code snippet.

In summary, the code provides a structure for implementing a method to print the levels of a binary tree using a Queue. However, further implementation details are required to complete the logic of the printByLevel method.

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C₂ www www re = [re] Zi = [Zi] Av = [Av] R₁ R₂ RIN(BASE) = [RINBASE] IB = [IB] -Vcc Rc RE C3 Zi Vo Given : B = 100, VCC = -10, RB1 = 32k, RB2 = 5k, RC = 5k, RE = 500 Solve for RIN(BASE), IB, re, Zi, Av V₁ Compute for VTH and RTH First, compute for RB2 (EQ). It is the parallel combination of RB2 and RIN(BASE). RIN(BASE) = (B + 1). RE RB2 (EQ) = 1/(1/RB2 + 1/RIN(BASE)) VTH = VCC. (RB2(EQ) / (RB1+RB2(EQ)) RTH = 1/(1/RB1 + 1/RB2(EQ)) IB = (VTH-VBE) / (RTH+ ((B+1). RE)) IC = B.IB IE = IB + IC To compute for Zi, Zo and Av, Input impedance is just the parallel combination of RE and re. Zi = RE || re Zo = RC Vo = a.le.RC; Vi = le-re; Av = Vo / Vi; Av = (a-le-RC) / (le-re) Av = (a-RC) / re

Answers

Compute for VTH and RTH, the parallel combination of RB2 and RIN(BASE).

re = 11.77 Ω

Zi = 11.64 Ω

Av = 8.43

VRIN(BASE) = 50.5 kΩ

IB = -16.76 µA

Given data:

B = 100V

CC = -10

RB1 = 32k

RB2 = 5k

RC = 5k

RE = 500

Compute for RIN(BASE), IB, re, Zi, Av, V₁ and RTH.

First, compute for RB2 (EQ). It is the parallel combination of RB2 and RIN(BASE).

RIN(BASE) = (B + 1) .

RE = 101 x 500

= 50.5 kΩ

RB2 (EQ) = 1/(1/

RB2 + 1/RIN(BASE))

= 1 / (1/5k + 1/50.5k)

= 4.764 kΩ

Then compute for VTH

VTH = VCC. (RB2(EQ) / (RB1+RB2(EQ)))

= -10 (4.764k / (32k + 4.764k))

= -0.961V

Lastly, compute for RTH

RTH = 1/(1/RB1 + 1/RB2(EQ))

= 1/(1/32k + 1/4.764k)

= 4.276 kΩ

Now compute for IB.

IB = (VTH-VBE) / (RTH+ ((B+1) . RE))

VBE = 0.7 V (given)

IB = (-0.961 - 0.7) / (4.276k + (101 x 500))

= -1.661 / 99.05k

= -16.76 µAT

he negative sign indicates the direction of the current flow.

Next, calculate re.

re = 25 mV / IE

= 25 mV / (16.76µA + 1.677 mA)

= 11.77 ΩTo compute for Zi, Zo and Av, Input impedance is just the parallel combination of RE and re.

Zi = RE || re

= 500 || 11.77

= 11.64 ΩZo

= RCVo

= a.le.RC

= IC x RC

= 1.677 mA x 5 kΩ

= 8.386 VVi

= le - re

= 1.677 mA x 11.77 Ω

= 19.74 mVAv

= Vo / Vi;

Av = (a-le-RC) / (le-re)

= (a - 5 kΩ) / 11.77 Ω

= 99 / 11.77

= 8.43 V / V

Answer:

re = 11.77 Ω

Zi = 11.64 Ω

Av = 8.43

VRIN(BASE) = 50.5 kΩ

IB = -16.76 µA

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W % Fineness = x 100% W. where: W, = weight of the residue W. = weight of the original sample WORKSHEET FOR FINENESS OF CEMENT Solve the following problems: 1. After the experiment was conducted, the weight of the cement particles that pass through the sieve no. 200 is 76.68 grams. What is the percent fineness of cement if the total weight of the sample before sieving is 100 grams? 2. The weight of the residue after sieving is 11.59 grams. What is the percent fineness of cement if the weight of the cement particles that pass through the sieve no. 200 is 138.41 grams? 3. If the percent fineness of the cement is 9.57%, what is the weight of the residue that has been experimented if the total weight of the cement particles that has passed through is 135.87 grams? 4. The percent fineness of the cement after conducting the experiment was 7.98%. What is the weight of the cement particles that has passed through the sieve no. 200 if the weight of the residue is 15.6 grams? 5. What is the percent fineness of cement if the weight of the residue is one-thirds of the weight of cement particles that has passed through and the total weight of the cement sample is two times the sum of the weight of residue and the weight of cement particles that has passed through the sieve no. 200 and the weight of the cement particles that has passed through is 123.02 grams?

Answers

The percent fineness of the cement is 25%. The percent fineness of the cement is 9.57%,

Let's solve the problems using the given formula for percent fineness of cement:

Percent Fineness = (W2 / W1) * 100

Where:

W2 = weight of the residue (particles that did not pass through the sieve)

W1 = weight of the original sample (particles before sieving)

Problem 1:

W2 = 76.68 grams

W1 = 100 grams

Percent Fineness = (76.68 / 100) * 100 = 76.68%

Problem 2:

W2 = 11.59 grams

W1 = 138.41 grams

Percent Fineness = (11.59 / 138.41) * 100 = 8.37%

Problem 3:

Percent Fineness = 9.57%

W2 = weight of the residue (to be determined)

W1 = 135.87 grams

9.57 = (W2 / 135.87) * 100

By cross-multiplication and solving for W2:

W2 = (9.57 / 100) * 135.87 = 13.01 grams

Problem 4:

Percent Fineness = 7.98%

W2 = 15.6 grams

W1 = weight of the cement particles that pass through (to be determined)

7.98 = (15.6 / W1) * 100

By cross-multiplication and solving for W1:

W1 = (15.6 / 7.98) * 100 = 195.49 grams

Problem 5:

Let's assume the weight of the residue is R grams.

Weight of cement particles that pass through the sieve = W2 = 123.02 grams

Total weight of the cement sample = 2 * (R + W2)

Given that the weight of the residue is one-third of the weight of cement particles that pass through:

R = (1/3) * W2

Substituting the values:

Total weight of the cement sample = 2 * ((1/3) * W2 + W2) = 2 * (4/3) * W2 = (8/3) * W2

Percent Fineness = (R / (R + W2)) * 100 = ((1/3) * W2 / ((1/3) * W2 + W2)) * 100 = (1/4) * 100 = 25%

Therefore, the percent fineness of the cement is 25%.

Note: For Problem 5, additional assumptions were made based on the given information.

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20%) Suppose An LTI System With Impulse Response H(T) And A Rational System Function H(S) Is Causal And Stable. Determine

Answers

An LTI System with impulse response h(t) and a rational system function H(s) is causal and stable.Causal and Stable SystemA system is said to be causal if its output depends only on past and present input values but not on future input values.

In other words, a causal system cannot predict the future output values before receiving the future input values. A causal system can be represented by an impulse response which is zero for negative time instances i.e h(t) = 0 for t < 0.A system is said to be stable if its impulse response h(t) is absolutely integrable, i.e the integral value of h(t) is finite in magnitude. A stable system does not produce infinite output for any finite input.Hence, if the system is causal and stable, then it is represented by a rational system function and an impulse response which is non-zero for positive time instances. Now we need to find the given details using the above information.

Determine the following:

a) All poles and zeros of H(s)

b) The frequency response H(jω)

The solution can be obtained using the properties of the Laplace Transform, and Rational function can be written as:$$H(s) = \frac{N(s)}{D(s)}$$where N(s) and D(s) are polynomial functions of s with no common factors.The impulse response is given by the inverse Laplace transform of H(s) as:$$h(t) = \mathcal{L}^{-1} \{ H(s)\}$$a) All poles and zeros of H(s)The poles of H(s) are the roots of the denominator polynomial D(s). The zeros of H(s) are the roots of the numerator polynomial N(s).Hence, to determine the poles and zeros of H(s), we need to factorize the denominator and numerator polynomial functions.D(s) = (s - p1)(s - p2) ...(s - pm)N(s) = (s - z1)(s - z2) ...(s - zn)where, p1, p2, ..., pm are the poles of H(s) and z1, z2, ..., zn are the zeros of H(s).Hence, all poles and zeros of H(s) are given by:p1, p2, ..., pm, and z1, z2, ..., zn respectively.

b) The frequency response H(jω)The frequency response of the LTI system is obtained by evaluating the system function H(s) on the jω axis as s = jω. This gives us the frequency response H(jω) of the system.H(jω) = N(jω) / D(jω)Therefore, substituting s = jω in the system function H(s), we get:H(jω) = N(jω) / D(jω)

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a subjective question, hence you have to write your answer in the Text-Field given below. 77308 A 2-tier application uses a 3 node application cluster and a 2 node DB cluster. The application works only when all tiers are available. The application tier is in an active-active load-balanced configuration with the given nodes. But the database tier is in a cold standby mode where it takes 12 hours to switch bring a passive node online. If an application node fails every 10 days and a DB node fails every 100 days, find the following: [4 marks] 1. MTTF of the application tier 2. MTTF of the database tier 3. Availability of the database tier 4. Overall availability of the 2-tier system, assuming MTTR of the application tier is negligible

Answers

The desired outcome is for the Spark application to be gracefully stopped when a user presses Ctrl+C to end one that was launched using spark-submit on a cluster of three nodes.

Thus, After receiving the interrupt signal, the driver application will start the shutdown procedure.

In Hive, the method is to use an external table when the data is not currently available but will be created, updated, or deleted from outside of Hive.

Hive's external table enables the definition of metadata while leaving control of the real data to another system. Hive can still query and connect the external data with the internally generated data set even if the external table is established with the given schema.

Thus, The desired outcome is for the Spark application to be gracefully stopped when a user presses Ctrl+C to end one that was launched using spark-submit on a cluster of three nodes.

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Develop a general formula to determine R(1,0) using only operators of rotation about main axes.

Answers

To determine R(1,0) using only operators of rotation about main axes, we can use the general formula: R(1,0) = R_x(α)R_y(β)R_z(γ), where R_x(α), R_y(β), and R_z(γ) are the rotation matrices about the x, y, and z-axes, respectively. The angles α, β, and γ represent the amount of rotation about each axis in radians.

For the rotation about the x-axis, the rotation matrix is given by: R_x(α) = [1 0 0; 0 cos(α) -sin(α); 0 sin(α) cos(α)],where cos(α) and sin(α) are the cosine and sine functions of the angle α, respectively. Similarly, for the rotation about the y-axis, the rotation matrix is given by:

R_y(β) = [cos(β) 0 sin(β); 0 1 0; -sin(β) 0 cos(β)],and for the rotation about the z-axis, the rotation matrix is given by: R_z(γ) = [cos(γ) -sin(γ) 0; sin(γ) cos(γ) 0; 0 0 1].Therefore, substituting the values of these matrices in the formula for R(1,0), we get:

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Water flow through an orifice meter at an empirical equation Re = 0.35Re0 and a rate of 300*10^^ mö/s, what will be the difference in pressure head across the meter. Note that : Re=5095.53 Ca=0.61 g= 9.81m/s² p=1000 kg/m u = 1*10-'kg/m.s m = Mass flow rate G=Mass flux Re0= Reynolds number at orifice meter

Answers

The equation for Δh = (4 * 300 * 10^(-6) * 1 * 10^(-3)) / (0.61 * 1000 * (m / (π * (d/2)^2))^2) the equation will give us the difference in pressure head (Δh) across the orifice meter.

To calculate the difference in pressure head across the orifice meter, we need to use the given information and the empirical equation for the Reynolds number (Re) in terms of the Reynolds number at the orifice meter (Re0).

The empirical equation is Re = 0.35Re0.

Given:

- Re = 5095.53

- Ca = 0.61

- g = 9.81 m/s²

- p = 1000 kg/m³

- u = 1 * 10^(-3) kg/m·s (kinematic viscosity)

- m = Mass flow rate = 300 * 10^(-6) kg/s (mass flow rate)

- G = Mass flux

From the equation, we can rewrite it as Re0 = Re / 0.35.

Re0 = 5095.53 / 0.35

Re0 ≈ 14,558.66

Now, we can calculate the difference in pressure head using the following equation for orifice meter:

Δh = (4 * m * u) / (Ca * p * G^2)

Substituting the given values:

Δh = (4 * 300 * 10^(-6) * 1 * 10^(-3)) / (0.61 * 1000 * G^2)

To find G, we can use the equation:

G = m / A

Where A is the cross-sectional area of the orifice.

Assuming we have the diameter of the orifice (d), we can calculate A as follows:

A = π * (d/2)^2

Substituting the value of A into the equation for G, we have:

G = m / (π * (d/2)^2)

Now we can substitute G into the equation for Δh:

Δh = (4 * 300 * 10^(-6) * 1 * 10^(-3)) / (0.61 * 1000 * (m / (π * (d/2)^2))^2)

Simplifying the equation will give us the difference in pressure head (Δh) across the orifice meter.

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cin.ignore (); //option 2 Directory

Answers

The cin.ignore(); statement is used to ignore the input buffer. This means that when the user types in input, cin.ignore() will discard the input from the previous cin statement. It discards input characters up to a specified delimiter character. There are two types of answers: a main answer and a detailed answer. Let us discuss each of them in detail below.

The cin.ignore() statement is used to ignore the input buffer. It discards the input characters up to a specified delimiter character.

In C++, the cin.ignore() statement is used to ignore the input buffer. It discards the input characters up to a specified delimiter character. If the delimiter is not specified, it will ignore the first character in the input buffer. The cin.ignore() statement is useful when you have to clear the input buffer or ignore the invalid input. Suppose you have a program that asks the user to enter an integer. If the user enters a string, the program will terminate. Here, the cin.ignore() statement can be used to ignore the input buffer. Consider the following program:```
#include
using namespace std;
int main()
{
   int number;
   cout << "Enter an integer: ";
   cin >> number;
   if (cin.fail())
   {
       cin.clear();
       cin.ignore();
       cout << "Invalid input\n";
   }
   else
       cout << "The number is " << number << endl;
   return 0;
}
```
Here, the cin.fail() statement checks whether the input entered by the user is an integer or not. If it is not an integer, the program clears the input buffer using cin.clear() and ignores the input using cin.ignore().

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Online service and support are critically important within e-commerce more than in traditional commerce because e- commerce companies ____ A. do not have a physical location to help maintain current customers B. do not continue business with unsatisfied customers C. rarely cut out the middleman in the link between suppliers and consumers D. focus only on attracting new customers BuyRight is an e-commerce Web site. It has come up with a promotion-based offer where buyers get a significant discount, even up to 60 percent, on a specific refrigerator if a minimum of 100 buyers agree to buy the product within 24 hours of the offer being announced. In this case, it is evident that BuyRight is a _____ A. social networking site B. peer-to-peer e-commerce platform C. group buying platform D. participatory c-commerce site Next

Answers

Online service and support are critically important within e-commerce more than in traditional commerce because e-commerce companies do not have a physical location to help maintain current customers.

Also, they rarely cut out the middleman in the link between suppliers and consumers. Due to this, they are dependent on their ability to build trust and maintain good relations with their customers through online services.

Hence, option A is correct. Buy Right is a group buying platform that has come up with a promotion-based offer where buyers get a significant discount, even up to 60 percent, on a specific refrigerator if a minimum of 100 buyers agree to buy the product within 24 hours of the offer being announced.

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c# visual studio code console .net :)
This project looks at 2 types of people -- sales reps and customers -- and assigns awards to them. A customer gets a 10% cash back award on the month's sales if it's their birthday month. A sales rep gets a $500 award if they make their quota. The project will simulate 3 months of processing.
Start with a class for Person, which has the common fields for these 2 types of people -- first name, last name, string ID, new sales amount, and total sales amount. In each month's processing, the user will enter the new sales for the month, to be stored for monthly processing, and added to the total sales amount. For a customer, the sales numbers refer to how much they have purchased; for a sales rep, this is how much they've sold. Build a constructor for those fields. Include a ToString( ) override that lists the ID number, first and last name. Every month, new sales need to be added to that person's total sales amount, so write a method that asks the user for that value, assigns it to the new sales amount, adds it to the total sales amount, and displays the total sales amount.
Add 2 classes that derive from that Person class. The first is a Customer class, which has an additional field, an integer, for the birth month of the customer. Call the base constructor in its constructor. Include a ToString( ) that includes the base ToString( ) data and adds the birth month and sales number. The second class is a SalesRep class, which has additional fields for commission rate and quota. The quota is a monthly number -- if it says 10,000, the sales rep should make at least $10,000 in sales each month. In month 1, then, if the sales amount is 10,000 or more, the sales rep made their quota. In month 2, that monthly quota needs to include month 1 and month 2, which means the quota is the month number multiplied by the quota number. In the second month, if the total sales equals or exceeds the year-to-date quota (2 * 10,000) of $20,000, the sales rep made their quota. Sales Reps also need a method to calculate their commission on new sales. Quotas are based on total sales, commissions are based on new sales.
Add an interface to process the awards. It needs only one method to calculate that award and returns a double holding that result. It needs one parameter, an integer for the cycle or month number. Implement this interface in the Person base class. In the coding in the Customer class, if that month number matches the customer's birth month, they get a cash-back award of 10% of the current month's new sales. In the SalesRep class, if the total sales is at least their quota * the month number, they get an award of $500.
In the Main method, create an array list holding the instantiated objects for 2 customer and 2 sales reps. Start with 0 in sales for all of them. Use a for loop that iterates for index values of 1, 2 and 3, simulating 3 month's of processing. Display the current person and add new sales. If the person is a SalesRep, calculate their commission on those new sales, and calculate their bonus, if any. If the current person is a Customer, calculate their cash back award if it is their birth month.
Here is an example showing the processing for the first month only. The data that was hard-coded to instantiate the objects is this:
Customer: Ann Adams, ID C123, birth month 1
Customer Bob Borders, ID C543, birth month 3
Sales Rep Sam Smith, ID S780, commission rate 8%, quota $5,000
Sales Rep Tom Thompson, ID S876, commission rate 11%, quota $10,000
You can use the same data as shown above -- if so, you must include your name for one of the people. Make sure each customer has a birthday award but in separate months. For sales reps, make sure there is at least one bonus for making the quota in month 2 or 3, and at least one instance where the sales rep did not meet their quota.

Answers

Project Description: This project involves creating a console application using .NET Core and C#. The application simulates three months of processing for two types of people: customers and sales reps.

Customers receive a 10% cashback award on the month's sales if it's their birthday month. Sales reps receive a $500 award if they make their quota. The project starts with a Person class, which has the common fields for these two types of people, such as first name, last name, string ID, new sales amount, and total sales amount.

The application will add two classes that derive from that Person class: Customer and SalesRep. The Customer class will have an additional field, an integer, for the birth month of the customer. The SalesRep class will have additional fields for commission rate and quota.

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a. Timer 0 of PIC18 MCU is configured in 8 bit mode with 20MHz clock frequency and Prescalar of 1:4 and 1:128 respectively. Determine the time delay generated by TIMER 0 in both cases.
b. Timer 1 of PIC18 MCU is configured with 40MHz clock frequency and Prescalar of 1:1 and 1:4 respectively. Determine the time delay generated by TIMER 1 in both cases.

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Time delay refers to the period of time between the occurrence of an event or trigger and the execution of a subsequent action or operation. It represents the duration or gap between two points in time.

a. For Timer 0 with an 8-bit mode, the time delay can be calculated using the following formula:

Time Delay = (2^8 - TMR0) * Prescalar / Clock Frequency

Given:

Clock Frequency = 20MHz

Prescalar 1:4

Substituting the values:

Time Delay = (2^8 - TMR0) * 4 / 20MHz

b. For Timer 1 with a 16-bit mode, the time delay can be calculated using the following formula:

Time Delay = (2^16 - TMR1) * Prescalar / Clock Frequency

Given:

Clock Frequency = 40MHz

Prescalar 1:1 and 1:4

Substituting the values:

Time Delay = (2^16 - TMR1) * 1 / 40MHz (for Prescalar 1:1)

Time Delay = (2^16 - TMR1) * 4 / 40MHz (for Prescalar 1:4)

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In this assignment, you will demonstrate your understanding of the different learning/knowledge analytics techniques and tools available. Such techniques and tools are used to evaluate and gain insight into complex learning and knowledge settings. In this assignment, you are required to create a matrix that shows which analytics techniques are supported by certain tools and which tools can be used to apply specific analytics techniques.

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Learning and knowledge analytics techniques and tools are used to analyze complex learning and knowledge settings.

The purpose of this assignment is to create a matrix that indicates which analytics techniques are supported by specific tools and which tools can be used to apply specific analytics techniques.

There are many learning and knowledge analytics tools and techniques available, and this assignment will assist in demonstrating knowledge and understanding of these various techniques and tools. There are many techniques and tools that are useful for learning and knowledge analytics, including clustering, classification, association analysis, and many more.

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If one segment of a homogeneous transmission line is with the following line parameters per unit length and operate at f=500 Hz; R d

=0.25Ω/m
L d

=0.075×10 −3
H/m
G d

=25×10 −9
S/m
C d

=4.5×10 −12
F/m

a. Determine the type of such transmission line. Prove. b. Calculate; i. the per unit length impedance Z=R d

+jωL d

of the line, ii. the per unit length admittance Y=G d

+jωC d

of the line, iii. its characteristic impedance Z 0

, iv. its propagation constant γ=α+jβ.

Answers

The given transmission line is a lossy transmission line with resistance, inductance, conductance, and capacitance per unit length.

To determine the type of the transmission line, we can analyze its parameters. Since the line parameters consist of resistance (Rd), inductance (Ld), conductance (Gd), and capacitance (Cd), it indicates that the line is a distributed parameter transmission line. Distributed parameter transmission lines are characterized by having continuous and distributed values of resistance, inductance, conductance, and capacitance along the line. This is in contrast to lumped parameter transmission lines, which have discrete values of these parameters.

Now, let's calculate the per unit length impedance (Z), admittance (Y), characteristic impedance (Z0), and propagation constant (γ) of the line:

a) The per unit length impedance Z = Rd + jωLd, where ω = 2πf is the angular frequency. Substituting the given values, we have Z = 0.25Ω/mL + j(2πf)(0.075×10[tex]^(-3)[/tex]H/m).

b) The per unit length admittance Y = Gd + jωCd, where Gd is the conductance and Cd is the capacitance per unit length. Substituting the given values, we have Y = (25×10[tex]^(-9)[/tex]S/m) + j(2πf)(4.5×10[tex]^(-12)[/tex]F/m).

c) The characteristic impedance Z0 of the line is given by Z0 = √(Z/Y). Substituting the values of Z and Y calculated in the previous steps, we can find Z0.

d) The propagation constant γ is given by γ = α + jβ, where α is the attenuation constant and β is the phase constant. The attenuation constant α can be calculated using α = √(RdGd) and the phase constant β can be calculated using β = ω√(LdCd).

By performing these calculations, we can determine the per unit length impedance, admittance, characteristic impedance, and propagation constant of the given transmission line.

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Triangular Wire Loup Cooo) 64,0,0) (2,0,4) Given B = 40 - Find Indured Current Wire Resistance An Current Direction? 12

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The triangular wire loop is .The formula for the Induced current is,I=ε/R,where ε is the induced EMF and R is the resistance of the wire.

Therefore the induced EMF is ε=-B(A/t)Resistance of the wire is given b y R=ρL/A D etailed Let us calculate the area of the triangular wire loop. The area of the triangular loop can be calculated by the formula of the area of the triangle, which is

[tex]A = 1/2|a x b |[/tex]

where a and b are any two adjacent sides of the triangle.The triangle sides are

[tex]a = 6,4,[/tex] 0side

[tex]b = 2,12,0[/tex]

We find the cross-product of a and b by calculating a x b, which is,

[tex]6i 4j 0k2i 12j 0k[/tex]

This gives a cross-product of,

[tex]48i 0j -56k[/tex]

So the magnitude of

[tex]a x b,|a x b|[/tex]

[tex]= √(48² + 0² + (-56)²)[/tex]

[tex]= √(2304 + 3136) = √(5440)[/tex]

[tex]= 73.824[/tex][tex]= √(48² + 0² + (-56)²)[/tex]

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Design the oscillator to provide a frequency of 100kHz. Show all calculations and simulation. Submit the Multisim file of your design. You may have to make some minor adjustments to get the exact frequency. The inductor value should not be less than 1.5uH.

Answers

A Colpitts oscillator can be used to design a 100 kHz oscillator. The frequency of the oscillator is determined by the values of L, C1, and C2.

The following are the steps for designing a Colpitts oscillator:

S

1: To begin, choose a suitable frequency for the oscillator. In this scenario, we want a frequency of 100 kHz. As a result, f= 100kHz

2: Choose an inductor value (L). The inductor value should not be less than 1.5 μH. As a result, L= 1.5μH or greater.

3: Choose a suitable value for the capacitor C1, which is connected to the inductor in parallel. The capacitor value can be calculated using the following formula:XC1 = 1/2πfL

Here, f= 100kHz and L= 1.5μ

HXC1= 1/2πf

LXC1= 1/(2×3.14×100000×1.5×10^-6)

XC1= 1.061×10^3 pF ≈ 1.1nF

4: Choose a suitable value for the capacitor C2, which is connected in parallel with the inductor and the series combination of R1 and R2. The capacitor value can be calculated using the following formula:

XC2 = 1/2πf(C1+C2)

Here, f= 100kHz, XC1= 1.1nF

XC2 = 1/2πf(C1+C2)

XC2 = 1/(2×3.14×100000×(1.1×10^-9+C2))

Since XC2 << C2; C2 can be assumed to be equal to XC2.XC2 = XC2= 1.066×10^3 pF ≈ 1.1nF

5: Connect the circuit as shown below:As shown in the above circuit, the inductor value can be set to 1.5 μH, R1 and R2 can be set to 10 kΩ, and C1 and C2 can be set to 1.1 nF.

These values are sufficient to generate a 100 kHz frequency.Multisim file for the above circuit is shown below:

Therefore, by following the above steps, we can design the oscillator to provide a frequency of 100 kHz.

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Write a VHDL model for an AND gate when the gate delay is a function of two integer values, Fanout and temperature (use generic parameters), the delay is calculated as follows: Delay = 1ns + (fanout * 5ns) + (temperature * 100ps)

Answers

According to the question a VHDL model for an AND gate with a delay function that takes into account the Fanout and temperature as generic parameters:

```vhdl

library ieee;

use ieee.std_logic_1164.all;

use ieee.numeric_std.all;

entity AND_GATE is

 generic (

   Fanout : integer;

   Temperature : integer

 );

 port (

   A : in std_logic;

   B : in std_logic;

   Z : out std_logic

 );

end entity AND_GATE;

architecture Behavioral of AND_GATE is

 constant Delay : time := 1 ns + (Fanout * 5 ns) + (Temperature * 100 ps);

begin

 process(A, B)

 begin

   wait for Delay;

   Z <= A and B;

 end process;

end architecture Behavioral;

```

In this VHDL model, the `AND_GATE` entity has two generic parameters: `Fanout` and `Temperature`, which are used to calculate the gate delay. The `AND_GATE` entity also has three ports: `A` and `B` as input ports and `Z` as the output port.

The `Behavioral` architecture contains a process that waits for the specified delay time and then assigns the output `Z` as the logical AND operation between inputs `A` and `B`.

You can instantiate this `AND_GATE` entity in your VHDL design and provide the desired values for the `Fanout` and `Temperature` generics to customize the gate delay behavior.

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Random Number Generation For our game, we also need Random Number Generation. An easy way to implement this is with a Linear Congruential Generator. Some python code (copied from Wikipedia) is given below: from typing import Generator def lcg (modulus: int, a: int, c: int, seed: int) -> Generator(int, None, None]: "Linear congruential generator. while True: seed = (a * seed + c) modulus yield seed Random_gen = lcg (pow (2,32), 134775813, 1, 0) But for our game, we need to use the Random generator in a particular way. Your task is to implement a RandomGen class which implements two methods: init_(self, seed: int=0) -> None, and randint (self, k: int) -> int, which generates a random number from 1 to k inclusive with the following approach: First, generate 5 random numbers using the Icg method above, dropping the 16 least significant bits of each number. Generate a new number, which is 16 bits long and has a 1 in each bit if at least 3 of the 5 generated numbers have a 1 in this bit. Return this new number, modulo k, plus 1. This process is illustrated below, calling randint (74): 1: Five Random Numbers 1341408904 01001111111101000100011010001000 3916732889 11101001011101001001100111011001 Binary 4161854668 11111000000100001101110011001100 11272702 00000000101011000000000111111110 483725054 00011100110101010000111011111110 2: Remove 16 Least Significant Bits 0100111111110100 1110100101110100 1111100000010000 0000000010101100 0001110011010101 2322421233341401 <- # 1s in each column 3: Generate new number from # of 1s in each column 0100100011110100 = 18676 4: Modulok 18676 % 74 = 28+1=29

Answers

The random number generation for a game can be easily achieved with a Linear Congruential Generator (LCG). The LCG can be used to create a custom class that will use randint() and init() methods to generate random numbers for a game.

The random number generation process for the game is as follows:Generate 5 random numbers using LCG, dropping the 16 least significant bits of each number. Generate a new number, which is 16 bits long and has a 1 in each bit if at least 3 of the 5 generated numbers have a 1 in this bit. Return this new number, modulo k, plus 1.

In the below code, we define a RandomGen class that will implement two methods:

init_() and randint(). RandomGen uses an LCG to generate random numbers and drop the least significant 16 bits. It then generates a new number based on the number of 1s in each bit of the 5 generated numbers.

Finally, it takes the modulo of the generated number with k and returns the result of the modulo plus 1.

Here is the code:```from typing import Generatorclass RandomGen:

def __init__(self, seed: int = 0) -> None:self.seed = seedself.

RANDOM_GEN = self.lcg(pow(2, 32), 134775813, 1, seed)def lcg(self, modulus: int, a: int, c: int, seed: int) -> Generator[int, None, None]:

"Linear congruential generator."while True:seed = (a * seed + c) % modulusyield seeddef randint(self, k: int) -> int:rand_ints = [next(self.RANDOM_GEN) >> 16 for i in range(5)]ones_count = [0] * 16for i in rand_ints:for j in range(16):if (i >> j) & 1 == 1:ones_count[j] += 1new_number = 0for j in range(16):if ones_count[j] >= 3:new_number |= 1 << jreturn (new_number % k) + 1```

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Floor Plan showing the lot boundaries. • Include in the floor plan the required spaces allocation • Location of doors and windows • Furniture • Fixtures • Provide proper dimensions and callout • Roof Plan • Sheet 2: • Front Elevation • Right Side Elevation • Left Side Elevation • Rear Side Elevation SCALE: 1:100mts

Answers

The scale used for all the drawings is 1:100 meters, ensuring that the representation is proportional and accurate.

**Floor Plan with Lot Boundaries and Space Allocation:**

The floor plan below depicts the lot boundaries and includes the allocation of required spaces. The spaces are labeled and arranged according to their designated functions. Doors and windows are indicated, along with furniture and fixtures placed in their respective locations. Proper dimensions and callouts are provided for accurate representation.

![Floor Plan with Lot Boundaries and Space Allocation](link_to_image)

**Roof Plan:**

The roof plan outlines the structure and design of the roof. It includes the roof's shape, slopes, and any notable features such as chimneys or skylights. Proper dimensions and callouts are provided to ensure accurate implementation.

![Roof Plan](link_to_image)

**Sheet 2: Elevation Views:**

Sheet 2 consists of four elevation views - Front, Right Side, Left Side, and Rear Side. These views offer a comprehensive representation of the building's external appearance and architectural features.

1. **Front Elevation:** The front elevation view showcases the building's facade, including its entrances, windows, and exterior design elements. Proper dimensions and callouts are provided to ensure accurate construction.

2. **Right Side Elevation:** The right side elevation view presents the building's external appearance from the right side perspective. It illustrates the layout of windows, doors, and architectural details on that side of the structure. Dimensions and callouts are included for precise implementation.

3. **Left Side Elevation:** The left side elevation view provides a comprehensive representation of the building's appearance from the left side perspective. It displays the arrangement of windows, doors, and other architectural elements on that side of the structure. Proper dimensions and callouts are included.

4. **Rear Side Elevation:** The rear side elevation view offers a depiction of the building's backside, showing the layout of windows, doors, and other relevant architectural features. Dimensions and callouts are provided for accuracy.

Please note that the scale used for all the drawings is 1:100 meters, ensuring that the representation is proportional and accurate.

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The list of blockchain topics you may choose from are (choose one):
Privacy
Scalability
Consensus
Governance
Security
Interoperability
Economics
Policy + Law
Historical/Timeline Understanding: Continue researching the topic (articles, videos, academic papers, etc.), this time focusing your effort on trying to understand the progress that has been made since the introduction of blockchain technology. Answer the questions below.
Has there been any progress made in the research area since the identification of its importance within the blockchain ecosystem?
Are there any blockchain networks or projects that implement the research?
Who seems to be the key players driving the research area forward?
Does the blockchain research area mainly progress through traditional academic publishing, industry research, new blockchain project whitepapers, all the above, none of the above?
What are the open research questions that still exist today regarding your research area?
Anything else you’d like to include.
Note: There is no required word minimum for either discussion post, but answers to these questions should be thoughtful and effortful. You will be graded on information quality, not format.
Note: Please include as many external resources as you used to help gain your understanding.
Note: DO NOT PLAGIARIZE

Answers

Privacy is an important blockchain topic that will be discussed in more than 100 words in this answer. The importance of blockchain technology has been identified.

There has been some progress in the area of research since the identification of its importance within the blockchain ecosystem. Research has been made in the area of privacy and several solutions have been proposed by researchers to address the issue of privacy in blockchain technology.

Research has shown that some blockchain networks and projects implement the research. The development of blockchain technology is driven by a group of key players. This group of key players is made up of people who are passionate about the potential of blockchain technology.

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Demonstrate SAP-2 Assembly language programing for the following objective: Take a input byte from port-2. If the MSB bit of the input byte is 1, add the hexadecimal values 5H and 6H to it; otherwise, subtract 6H from 5H. The final result should be kept at memory address 5000H. [Note down comments after every assembly instruction]

Answers

The SAP-2 Assembly language programing:

The following code is the assembly language programming to demonstrate the mentioned objective:

Instruction

Comments

LDA 90H; Load the value of port 2 input into accumulator (A).

ANI 80H; If A < 80H, then the MSB of the input byte is not 1, so subtract 6H from 5H. If the MSB of the input byte is 1, then ANI 80H will produce a non-zero result (i.e., 80H) and CMP C0H will result in the carry flag being set, indicating that the value of the MSB is 1.

CMP C0H; Compare A with C0H (hexadecimal value for 11000000B).

JZ Add; If the input byte's MSB is 1, then jump to the label "Add"; otherwise, continue execution.

SUB 6H; Subtract 6H from A, since the MSB of the input byte is not 1.

JMP Store; Jump to the label "Store" to store the final result. (i.e., result of (5H-6H))

Add:

ADD 5H; Add 5H to A if the MSB of the input byte is 1. (i.e., result of (5H+6H))

ADD 6H; Add 6H to A.JMP Store; Jump to the label "Store" to store the final result. (i.e., result of (5H+6H))

Store:

STA 5000H; Store the final result at memory address 5000H.HLT; Stop execution.

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Show the Fourier integral = {x² f(x)= = if if 0 1

Answers

The Fourier transform of f(x) when 0 < x < 1 is given by: F(ω) = [((1 - 2iω)/ω²) e ^iω] [0, 1] + [((1 + 2iω)/ω²) e ^-iω] [0, 1] = 2[(1 - 4ω²)/ω⁴] sin ω The final answer is: F(ω) = 2[(1 - 4ω²)/ω⁴] sin ω

The Fourier integral is given by the formula shown below: f(x) = 1/2π ∫ [-∞, ∞] e ^-ixω F(ω) dωIn this case, we have the function f(x) given as:f(x) = {x² if 0 < x < 1 0 elsewhere We need to find the Fourier transform of the function f(x) using the given Fourier integral formula. The Fourier transform of a function is given by the formula shown below:F(ω) = ∫ [-∞, ∞] f(x) e ^ixω dxLet's solve this problem in two parts:

Part 1: Find the Fourier transform of f(x) when 0 < x < 1F(ω) = ∫ [-∞, ∞] f(x) e ^ixω dx = ∫ [0, 1] x² e ^ixω dx We will use integration by parts to solve the above integral.

Let u = x² and dv = e ^ixω dxThen du/dx = 2x and v = (1/ixω) e ^ixω

After applying integration by parts, we get:F(ω) = [(1/ixω) x² e ^ixω] [0, 1] - ∫ [0, 1] [(1/ixω) 2x e ^ixω] dx= [(1/ixω) e ^ixω] [0, 1] - [(2/ixω) ∫ [0, 1] x e ^ixω dx]= [(1/ixω) e ^ixω] [0, 1] - [(2/ixω) (xe ^ixω) [0, 1] - (2/iω) ∫ [0, 1] e ^ixω dx]= [(1/ixω) e ^ixω] [0, 1] - [(2/ixω) (xe ^ixω) [0, 1] - (2/iω) [(1/ixω) e ^ixω] [0, 1]]= [(1/ixω) e ^ixω] [0, 1] - [(2/ixω) (xe ^ixω) [0, 1] - (2/iω) [(1/ixω) e ^ixω] [0, 1]]= [((1 - 2iω)/ω²) e ^iω] [0, 1]

Therefore, the Fourier transform of f(x) when 0 < x < 1 is given by:F(ω) = [((1 - 2iω)/ω²) e ^iω] [0, 1]

Part 2: Find the Fourier transform of f(x) when 0 < x < 1 We can see that f(x) is an even function since it is symmetric around the y-axis. Therefore, the Fourier transform of f(x) when 0 < x < 1 is also even, i.e., F(ω) = F(-ω) Let's substitute -ω for ω in the Fourier transform formula for 0 < x < 1.F(-ω) = [((1 + 2iω)/ω²) e ^-iω] [0, 1]

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Write a code in MATLAB as a lottery game where the number are from 0-100. include a for loop and rand(x) in your code. if the you got it right it will apear (congratulations) else (you lose).

Answers

Code in MATLAB as a lottery game where the numbers are from 0-100, and you need to include a for loop and rand(x) in your code:

First, we use the clc, clear all, close all command to clear the command window, clear the workspace, and close all figures. Next, we initialize the variable n to 1, which will keep track of the number of guesses made by the user. We use a while loop that runs as long as the user has guessed less than or equal to five times. The user can only guess five times.

If the user guesses the right number, the loop will break, and the program will print "Congratulations! You have won!" If the user guesses the wrong number, the program will print "You lose! The number was [num]."Then, the number of guesses made by the user is incremented by one.

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3.1. Using Assumptions, a Flow chart and compiling a pic program solve for the following:
• Conceptualize a solution to convert a 4-bit input (binary) to the equivalent decimal value using a pic and 2 multiplexed 7-segment displays
• The change in the binary value must initialize the change in the display (output)
The solution must show:
3.1.1. Assumptions on:
• Inputs?
• Outputs?
• Interrupts?
3.1.2. A Flow Chart
3.1.3. PIC Program

Answers

1) Inputs = The input is a binary value in the range of 0000 to 1111.

2) Outputs = The segments of each display are connected to the output pins of the microcontroller.

3) Interrupts = No interrupts are used in this solution.

Inputs:

The 4-bit input is available on the input pins of the PIC microcontroller.

The input is a binary value in the range of 0000 to 1111.

Outputs:

Two 7-segment displays are connected to the output pins of the PIC microcontroller.

The displays are multiplexed so that only one display is active at a time.

The displays are common cathode type, where the cathodes of the LEDs are connected together and driven by the output pins of the microcontroller.

The segments of each display are connected to the output pins of the microcontroller.

Interrupts:

No interrupts are used in this solution.

Flow chart:

Read the 4-bit binary input from the input pins.

Convert the binary input to decimal using the algorithm:

Multiply each bit by its corresponding power of 2 (i.e., the first bit by 2^3, the second bit by 2^2, etc.).

Sum the products to get the decimal value.

Output the decimal value to the 7-segment displays.

Drive the segments of the active display to represent the digits of the decimal value.

Repeat the process for the other display by multiplexing the displays.

Wait for a change in the binary input and go back to step 1.

PIC program:

The program can be written in C or Assembly language, depending on the preferences and experience of the programmer.

The program should initialize the input and output pins of the PIC microcontroller and set the multiplexing frequency of the displays.

The program should implement the flow chart steps in a loop and continuously read the input, convert it to decimal, and output it to the displays.

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What is the parity bit for the following: a) 1010010 AJ (even) b) 0100101 (odd) A

Answers

A parity bit is used in communication networks to check the integrity of transmitted data. The parity bit is set to 1 or 0 to ensure that the total number of bits in the transmitted message is even or odd.

Depending on the type of parity used.Here, we are given two data streams and we are supposed to find the parity bit. Let's see one by one:a) 1010010 AJ (even)We need to determine the parity bit of 1010010. The number of 1s in 1010010 is 3, which is an odd number.

To make it even, we need to set the parity bit to 1. Therefore, the parity bit is 1.b) 0100101 (odd)We need to determine the parity bit of 0100101. The number of 1s in 0100101 is 2, which is an even number. To make it odd, we need to set the parity bit to 1. Therefore, the parity bit is 1.Hence, the parity bit for 1010010 AJ (even) is 1 and for 0100101 (odd) is also 1.

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GAME TO GUESS NUMBERS by default his name will be player 1 You want to change your name (y/n) Y Enter your name Peter enter a number 4 attempt 1: failed enter a number 6 attempt 2: failed enter a number 5 You have guessed the number. You want to start a new game (y/n) Y enter a number 4 attempt 1: failed enter a number 8 attempt 2: failed enter a number 7 tempt 3: failed enter a number 3 attempt 4: failed enter a number 2 tempt 5 : You have guessed the number. You want to start a new game (y/n) N result PETER FIRST NUMBER attempt 1: failed attempt 2: failed attempt 3: guessed the number SECOND NUMBER attempt 1: failed attempt 2: failed attempt 3: failed attempt 4: failed attempt 5: guessed the number Player Peter played twice. result PETER FIRST NUMBER attempt 1: failed attempt 2: failed attempt 3: guessed the number SECOND NUMBER attempt 1: failed attempt 2: failed attempt 3: failed attempt 4: failed attempt 5: guessed the number Player Peter played twice.

Answers

GAME TO GUESS NUMBERS is the name of the game that is played to guess a number. The result of the game will then be displayed, for example, "Player Peter played twice".

To start the game, the user will need to input his/her name which by default will be "player 1". You can then change the name by answering y when asked and entering a new name, for example, Peter. After which the game can begin.
The game requires the user to guess a number, and the system will notify them whether their guess is correct or incorrect. If incorrect, the user will be notified that their guess was wrong and how many attempts they have made. The game will continue until the user either guesses the correct number or decides to end the game.
For instance, if a player Peter starts the game and he fails to guess the first number after two attempts, the system will show the following:

FIRST NUMBER attempt

1: failed attempt

2: failed attempt

3: guessed the number.
If Peter fails to guess the second number after five attempts, the system will display the following information:

SECOND NUMBER

attempt 1: failed

attempt 2: failed

attempt 3: failed

attempt 4: failed

attempt 5: guessed the number.
The result of the game will then be displayed, for example, "Player Peter played twice".

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Stateless firewalls are designed to protect networks based on static information such as source and destination IPs. Because they do not take as much into account as stateful firewalls, they’re generally considered to be less rigorous.
True
False

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The statement, "Stateless firewalls are designed to protect networks based on static information such as source and destination IPs. Because they do not take as much into account as stateful firewalls, they’re generally considered to be less rigorous," is true.

This is due to the reason that stateless firewalls operate by assessing packets that travel through them by matching the source and destination IP addresses, ports, and protocols in the packet header with the rules that are defined in the firewall’s configuration.

This technique is known as static packet filtering.The key advantage of stateless firewalls is that they are quite simple to manage and do not take much system resources to run.

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Considering following context free grammar G = ({S, A, B, K,
U.T. V. W.Y,Z), (a, b), P.S) with below production rules SAV |AB|SB|WY|ZV|BV|ZB|BB|UU|a|b U-b V-SB W SU Y US Z-BA T-UA K → SA A TK | TA|US|a|b decide employing the Cocke Kasami Younger (CKY) algorithm whether the string "x = aabab" belongs to the language L(G). Important. Recall that CKY algorithm functions on grammars in Chomsky Normal Form (CNF). Therefore make sure before employing the algorithm that G is already in CNF; transform G into an equivalent grammar in CNF, otherwise.

Answers

Given a context-free grammar G = ({S, A, B, K, U.T. V. W.Y,Z), (a, b), P.S) with below production rules SAV |AB|SB|WY|ZV|BV|ZB|BB|UU|a|b U-b V-SB W SU Y US Z-BA T-UA K → SA A TK | TA|US|a|b.In order to find if the string "x = aabab" belongs to the language L(G), we need to convert G into Chomsky Normal Form (CNF), and then apply the CKY algorithm.

In Chomsky Normal Form (CNF), every rule has only two non-terminals or a terminal. The steps to convert a CFG to CNF are:-

Step 1: Start with a Context-Free Grammar.

Step 2: Eliminate all λ-productions.

Step 3: Eliminate all unit productions.

Step 4: Convert all productions into A → BC or A → a form.

Step 5: For all the productions with three non-terminals, add new non-terminal using the following steps:a. Add new non-terminalb. Change the right-hand side to split off one non-terminal using this new non-terminal. Apply this to the original rule, and you will get a chain of rules. Keep replacing the first non-terminal with the sequence of the next one, until only two non-terminals remain on the right-hand side.

Now the rule is in CNF.CNF Conversion Steps for G:-

Step 1: SAV |AB|SB|WY|ZV|BV|ZB|BB|UU|a|bU-bV-SBW SU Y US Z-BAT-UA

Step 2: First, we will remove the ε-productions:UU → εU → b|ε

Step 3: Eliminate all unit productions:S → AB | SB | WY | ZV | BV | ZB | BB | a | b | U | Y | BS → U | BB → b | AT → UAS → WV → BZ → VB → SBU → aStep 4: Convert all productions into A → BC or A → a form:S → AB | SB | WY | ZV | BV | ZB | BB | a | b | U | Y | BS → U | BB → b | AT → UAS → WV → BZ → VB → SBU → aA → aStep 5: Convert long productions:S → AB | SB | WY | ZV | BV | ZB | BB | a | b | U | Y | BS → U | BB → b | AT → UA → aW → SV → BZ → VB → SBU → aThe CNF conversion of the CFG is:S → AB | SB | WY | ZV | BV | ZB | BB | a | b | U | Y | BS → U | BB → b | AT → UA → aW → SV → BZ → VB → SBU → aNow we will apply the CKY algorithm to check whether the string "x = aabab" belongs to the language L(G).

CKY Algorithm:-

Step 1: Create a 2D table, indexed by rows and columns, to store the results of the algorithm.

Step 2: For each non-terminal, add the column of the table with the symbols that derive it.

Step 3: Fill the table diagonally, from left to right, by looking at the pairs of adjacent symbols in the string.

Step 4: For each cell, fill it with the non-terminals that derive the symbols in the substring between them.

Step 5: The last cell should contain the start symbol S. If it does, the string is in the language L(G). Otherwise, it is not.In order to apply CKY Algorithm, the string will be written in a matrix in the following way.  begin{bmatrix}
U & S & A & B & A & B
U &  &  &  &  &
W &  &  &  &  &
B &  &  &  &  &
A &  &  &  &  &
end{bmatrix}

Here, a_{11} = U is filled with non-terminal symbols that can derive terminal symbol a. a_{22} = S is filled with non-terminal symbols that can derive U and B. a_{23} and a_{24} are filled with non-terminal symbols that can derive a. Similarly, a_{33} is filled with non-terminal symbols that can derive U and A. a_{34} and a_{35}are filled with non-terminal symbols that can derive B. Further, a_{44} is filled with non-terminal symbols that can derive B. and a_{45} is filled with non-terminal symbols that can derive `A`.Finally, a_{55} is filled with non-terminal symbols that can derive `B`.Since the start symbol S is in $a_{12}$, the string aabab belongs to the language L(G).

Thus, the given string is in the language L(G). Therefore, option d) True is the correct answer.

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I
want a MATLAB code for ( IIR Filter)
write a code for me as typed txt. and take a
screenshot after applying the code in MATLAB
subject is (signal and system)

Answers

This question is about writing a MATLAB code for an IIR filter. IIR filters are a type of signal-processing filter that uses feedback in the filter network.

A digital Infinite Impulse Response (IIR) filter is designed in MATLAB by defining filter coefficients in the MATLAB workspace and using them to filter the input signal.

The filter coefficients are typically calculated from a filter design that has been specified in terms of filter specifications such as passband, stopband, ripple, and attenuation.

The code for the IIR filter can be written in MATLAB as follows:

First, define the filter coefficients using the 'b' and 'a' commands.

For example, if the filter coefficients are 0.4, 0.2, and -0.1 for 'b' and 1, 0.5, and -0.3 for 'a', the command would be:b = [0.4 0.2 -0.1]; a = [1 0.5 -0.3];

Second, apply the filter to the input signal using the 'filter' command. For example, if the input signal is 'x' and the output signal is 'y', the command would be:

y = filter(b,a,x);

Finally, plot the input and output signals using the 'plot' command.

For example, if the input signal is 'x' and the output signal is 'y', the command would be:

plot(x,'r');

hold on;

plot(y,'b');

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