Answer:
b. the IP address can be spoofed, so if you want to read response from the deceived party, you can use IP spoofing to hide yourself.
Explanation:
19. When troubleshooting a desktop motherboard, you discover the network port no longer works. What is the best and least expensive solution to this problem? If this solution does not work, which solution should you try next?
Answer:
Disable the network port and install a network card in an expansion slot.
Explanation:
The best and least expensive solution will be to disable the network port, once that is done, you install a network card in the card expansion slot.
This way you can still connect wireless devices to your computer and the network issue will be solved.
Answer: Update the Motherboard Drivers
Explanation: I would try update the motherboard drivers 1st.
Write a function stats that takes an array and the number of elements in the array. Then, it computes and prints the minimum value, maximum value, and the average value of all the values in the array. The output should be formatted with a two-digit precision. The function name: stats The function parameters (in this order): an array, double The number of elements in the array, int The function should not return anything.
Answer:
#include<iostream>
using namespace std;
void stats(double [],int);
int main()
{
int totalElements,i;
cout<<"Enter total elements:"<<endl;
cin>>totalElements;
double array[totalElements];
cout<<"Enter the elements in array:"<<endl;
for(i=0;i<totalElements;i++)
cin>>array[i];
stats(array,totalElements);
}
void stats(double array[],int totalElements)
{
int i;
double minimum,maximum;
double Sum=0.0,average=0.0;
minimum=array[0],maximum=array[0];
for(i=0;i<totalElements;i++)
{
if(array[i]>maximum)
maximum=array[i];
if(array[i]<minimum)
minimum=array[i];
Sum+=array[i];
}
average=Sum/totalElements;
cout<<"Test: ";
for(i=0;i<totalElements;i++)
cout << fixed << setprecision(2) <<array[i]<<" ";
cout<<endl;
cout <<"minimum:"<< fixed << setprecision(2) <<minimum<<endl;
cout <<"maximum:"<< fixed << setprecision(2) <<maximum<<endl;
cout <<"average:"<< fixed << setprecision(2) <<average<<endl;
}
Explanation:
Loop through the total elements to get the input from user and call the stats function.In the stats function check whether a number is maximum, minimum or average.Calculate the average by finding the sum of all the numbers in array and dividing it by total numbers.Finally display the results.
Final answer:
The 'stats' function calculates and prints the minimum, maximum, and average values from a given array with two-digit precision.
Explanation:
The function named stats is designed to compute statistical values such as the minimum value, maximum value, and the average value of elements in an array. The function takes two parameters: an array of type double and the number of elements in the array of type int. It then proceeds to calculate the required statistics and prints them with a precision of two decimal places. This function does not return any value as its primary purpose is to print out the calculated statistics.
Design a 128KB direct-mapped data cache that uses a 32-bit address and 16 bytes per block. The design consists of two components: (1) The 32-bit memory address is subdivided into several sections in bits so that each address can map to its cache location (2) The cache itself including the cache storage and other necessary bits in each cache line. Explain your design.
Answer:
See the attached pictures for detailed answer.
Explanation:
The cache contain 8K number of blocks.each block has 16 Byte worth of data. Each of these bytes are addressable by block offset. Now, each of these blocks are addressable and their address is provided by line offset or set offset. If block number is given in Decimal format then the block to which that block will be mapped is given by expression
mappedToBlock = (bNumber)mod213
TAG and valid but consititues something called cache controller. TAG holds the physical address information because the TAG in physical address get divided into TAG and line offset in cache address. Valid bit tells status of blocks. If it is 1 then data blocks are referred by CPI else not.
Suppose that the UDP receiver computes the Internet checksum for the received UDP segment and finds that it matches the value carried in the checksum field.
Can the receiver be absolutely certain that no bit errors have occurred?
Answer:
No, the receiver cannot be absolutely certain that no bit errors have occurred. This is because of the manner in which the checksum for the packet is calculated. If the corresponding bits (that would be added together) of two 16-bit words in the packet were 0 and 1 then even if these get flipped to 1 and 0 respectively, the sum still remains the same. Hence, the 1s complement the receiver calculates will also be the same. This means the checksum will verify even if there was transmission error
Explanation:
No, the receiver cannot be confident that there haven't been any bit mistakes.
What is UDP?UDP stands for user data protocol. UDP is defined as a messaging protocol that makes it easier for computers to communicate with one another via a network. A communications protocol called User Datagram Protocol (UDP) is largely used to provide low-latency, loss-tolerant connections between internet-based applications. UDP allows data to be transferred before the receiving party provides an agreement, which speeds up transfers.
A checksum field on UDP allows for the detection of IP header errors; if a checksum error is found, the IP header checksum is recalculated before the IP packet is delivered. Therefore, it cannot ensure that your package will reach the receiver. However, UDP does guarantee that if it is delivered, it will be error-free.
Thus, no, the receiver cannot be confident that there haven't been any bit mistakes.
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A variation of client-server computing is ____, in which software that allows for applications are installed on servers and then accessed and executed through desktop clients, instead of installing applications on each individual client computer.
Answer:
Terminal services
Explanation:
Terminal services are multiuser, lean client environment from Microsoft which was built for Windows servers. It was first launched on the Windows NT 4.0 in 1996, as of the 2009 the launch of Windows Server 2008 R2, it became part of Remote Desktop Services (RDS), and exclusively, the "shared sessions" method in RDS.
Just like in the old times of mainframes, Terminal Services supports numerous users connected to a central computer. The user's device can be a compact PC, bare-bones PC or a dedicated terminal, and all of them are to function as input/output (I/O) terminals and use the same OS and applications running in the server. Terminal Services utilizes the Microsoft's Remote Desktop Protocol (RDP) to govern mouse, keyboard and screen transfer.
A case competitions database:You work for a firm that has decided to sponsor case competitions between teams of college business students, and you were put in charge of creating a database to keep the corresponding data. The firm plans to hold about dozens of different regional competitions at various dates that will take place at different branches around the country. For each competition, you need to store a name, date, and the name of the branch where it is going to take place. For each college that has agreed to participate in the competition, you need to store the college name, a contact phone number, and a contact address. Each college can participate in only one competition, and the database should know which competition each college is participating in. On the other hand, each college is allowed to send more than one team to its competition. Each team gives itself a name and a color, and consists of several students of the same college; for each student, you want to store a first name, last name, date of birth, major, and expected graduation date. Question: How many tables do you need
Answer:
We will ned (4) four tables.
Explanation:
For the given scenario we will have to build a relational database. The database will have four tables i.e. Competation, Collage, Team and student.
For each table the database fields are mention below. Note that foreign keys are mentioned in itallic.
Competition:
Competition_ID, Competition_Name, Competition_Data, Competition_Name _of_Branch, College_ID
Collage:
Collage_ID, Collage_Name, Collage_Contact, Collage_address
Team:
Name, Color, Team_ID, College_ID, Competition_ID
College:
Student_ID, First_Name, Last_Name, Date_of_Birth, Major, Expected_Gradiuation_Date, Collage_ID, Team_ID, Competation_ID
To organize the required data for your firm's case competitions, you would need to create four tables: Competitions, Colleges, Teams, and Students. These tables would store competition details, college contact information, team names and colors, and student demographics, respectively.
Explanation:Databases organize information into tables that are composed of rows and columns where each row is a record for an entity and each column is an attribute of the entity. For your case competition database, you will need different tables to store related data without redundancy. The goal here is consistency, efficiency, and comprehensibility. Let's break down the tables you would require:
A Competitions table to store each competition's name, date, and branch location.A Colleges table to store participating college's name, contact phone number, and address, with a reference to the competition they are participating in.A Teams table to record each team's name, color, and associated college.A Students table to keep track of all student participants' first name, last name, date of birth, major, and expected graduation date, with a reference to their team.By designing your database with these four tables, you ensure data is grouped logically and related information is linked appropriately to eliminate the need for repeated data entry and promote efficient data management.
You coded the following class:
try
{
Scanner file = new Scanner( new File("data.txt"));
String s = file.nextLine();
}
catch (ArithmeticException ae)
{
System.out.println(ae.getMessage());
}
Explain what the problem is and how to fix it.
Answer:
The code above tries to read a file using the scanner class in java programming language.
But it was done properly.
First, the scanner class library is not imported to the program.
Second,the syntax used in this program is an invalid way of reading a file in java programming language using the scanner class.
Lastly, "String s = file.nextLine();" is not a proper way of reading each line
To correct this problem; use the following code;
import java.util.Scanner;
import java.io.File;
import java.io.FileNotFoundException;
public class ReadFileWithScanner {
try
{
public static void main(String args[]) throws FileNotFoundException {
File text = new File("data.txt");
Scanner file = new Scanner(text);
int lines = 1;
while(file.hasNextLine()){
String line = filw.nextLine();
System.out.println("line " + lines + " :" + line);
lines++;
}
catch (ArithmeticException ae)
{
System.out.println(ae.getMessage());
}
}
}
Bill is building a project network that involves testing a prototype. he must design the prototype (activity 1), build the prototype (activity 2), and test the prototype (activity 3). activity 1 is the predecessor for activity 2 and activity 2 is the predecessor for activity 3. if the prototype fails testing, bill must redesign the prototype; therefore, activity 3 is a predecessor for activity 1. this is an example of
a. conditional statements.
b. looping.
c. having more than one start node.
d. good network development.
e natural network flow.
Answer:
Bill is building a project network that involves testing a prototype. he must design the prototype (activity 1), build the prototype (activity 2), and test the prototype (activity 3). activity 1 is the predecessor for activity 2 and activity 2 is the predecessor for activity 3. if the prototype fails testing, bill must redesign the prototype; therefore, activity 3 is a predecessor for activity 1. this is an example of
b. looping
Explanation:
The given example is of looping because each activity is leading to another activity on the completion of some conditions. The answer a is not valid as it is not just an example of conditional statements rather it is a loop which will keep moving until unless reached a situation to end it.The option c, d an e are not right options for the given example.Consider a short, 10-meter link, over which a sender can transmit at a rate of 150 bits/sec in both directions. Suppose that packets containing data are 100,000 bits long, and packets containing only control (e.g., ACK or handshaking) are 200 bits long. Assume that N parallel connections each get 1 of the link bandwidth N. Now consider the HTTP protocol, and suppose that each download object is 100 Kbits long and that the initial download object contains 10 referenced objects from the same sender.
Would parallel downloads via parallel instances of non-persistent HTTP make sense in this case? Now consider persistent HTTP. Do you expect significant gains over the non-persistent case? Justify and explain your answer.
Answer:
The Tp value 0.03 micro seconds as calculated in the explanation below is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP.
Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.
Explanation:
Given details are below:
Length of the link = 10 meters
Bandwidth = 150 bits/sec
Size of a data packet = 100,000 bits
Size of a control packet = 200 bits
Size of the downloaded object = 100Kbits
No. of referenced objects = 10
Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.
The formula below is used to calculate the total time delay for sending and receiving packets :
d = dp (propagation delay) + dt (transmission delay)
For Parallel downloads through parallel instances of non-persistent HTTP :
Bandwidth = 150 bits/sec
No. of referenced objects = 10
For each parallel download, the bandwith = 150/10
= 15 bits/sec
10 independent connections are established, during parallel downloads, and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.
Therefore, for parallel downloads, the total time required is calculated as:
(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)
= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)
= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)
= (670 + 4Tp) + (6706 + 4Tp)
= 7377 + 8 Tp seconds
Thus, parallel instances of non-persistent HTTP makes sense in this case.
Let the speed of propogation of the medium be 300*106 m/sec.
Then, Tp = 10/(300*106)
= 0.03 micro seconds
The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.
A nearest neighbor approach is best used: (a) With large-sized data sets. (b) When irrelevant attributes have been removed from the data. (c) When a generalized model of the data is desirable. (d) When an explanation of what has been found is of primary importance. Select only one choice and give additional explanations
Answer:
(b) When irrelevant attributes have been removed from the data.
Explanation:
A nearest neighbor approach -
It refers to the method of searching where the nearest point from the given set can be determined with respect to some specific point , is referred to as the nearest neighbor approach .
The method is used for the purpose of finding soe data in order to replace , edit , remove or add some data .
Hence , from the given question ,
The correct answer about nearest neighbor approach is b.
A Nearest Neighbor approach is best used when irrelevant attributes have been removed from the data. This algorithm works best when the dataset is clean and free of irrelevant, misleading, or noisy data. It is less effective with large data sets, and doesn't provide a generalized model or easy explanation of results.
Explanation:A Nearest Neighbor approach in machine learning is typically best used when irrelevant attributes have been removed from the data. This type of algorithm generally works well when the dataset is clean and free of irrelevant, misleading, or noisy data that could interfere with finding the nearest neighbors. Notably, it does not work as effectively with large data sets due to computational demands. Furthermore, it does not provide a generalized model of the data or an easy explanation of the results, which eliminates options (a), (c), and (d).
To illustrate, consider a dataset used for predicting house prices. Irrelevant attributes including the house color, name of the owner, etc., have to be removed for the Nearest Neighbor algorithm to be effective as they do not have a direct impact on the house price. The algorithm then identifies the 'neighbor' house prices that are most similar to the house price we're trying to predict.
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Consider an array of size nine with the numbers in the following order: 40, 60, 20, 80, 70, 90, 30, 10, 50. (a) Create the heap using the algorithm described in class. Show the heap as a tree. Show the heap as an array. Exactly how many comparisons did heap creation use
Answer:
See attached picture for heap creation steps.
10 comparisons were used.
Explanation:
See attached pictures.
Write a logical expression that is equivalent to the Exclusive OR gate on 2 inputs, called in1, in2: If either one or the other (but not both) input is True, then your expression should evaluate to True. Otherwise (both inputs are True or both inputs are False), then your expression should evaluate to False.
Answer:
F= in1'.in2' + (in1.in2)' + in1'. in2+ in1.in2'
Explanation:
See the truth table in the attachment
I have used apostropy (') to write complement
Write a program to do the following: Load $t0 with 12 and call it x. Load 21 in $t1 and denote it as y. Finally, load 32 in $t2 and refer to it as z. Now, compute 3x2 +10y+5z. Output the result using syscall 1 (remember the result has to be in $a0). Write down the output number on your answer sheet.
Answer:
Complete code is given below:
Explanation:
addi $t0, $zero, 12 # x = 12
addi $t1, $zero, 21 # y = 21
addi $t2, $zero, 32 # z = 32
addi $t3, $zero, 3
mul $t3, $t3, $t0
mul $t3, $t3, $t0 # $t3 = 3x^2
addi $t4, $zero, 10
mul $t4, $t4, $t1 # $t4 = 10y
addi $t5, $zero, 5
mul $t5, $t5, $t2 # $t5 = 5z
add $a0, $t3, $t4
add $a0, $a0, $t5 # $a0 = 3x^2 + 10y + 5z
addi $v0, $zero, 1
syscall
Design a program that gives simple math quizzes. The program should display two random numbers that are to be added, such as
247
+ 129
The program (Java language) should allow the student to enter the answer. If the answer is correct, a message of congratulations should be displayed. If the answer is incorrect, a message showing the correct answer should be displayed
Here is the pseudo-code to follow
//Declare the variables to hold intermediate and final values
Declare Integer rand1, rand2, sum ans
//Get the random numbers generated by the function
//random and assign those values in rand1 and rand2 respectively
Set rand1 = random(1, 999)
Set rand2 = random(1, 999)
//Dsiplay rand1 and rand2 in a formatted way treated for addition
Display
Answer:
import java.util.Scanner; import java.util.Random; public class Main { public static void main(String[] args) { int rand1, rand2, sum, ans; Random rand = new Random(); rand1 = rand.nextInt(1000); rand2 = rand.nextInt(1000); sum = rand1 + rand2; System.out.println(rand1 + " + " + rand2); Scanner input = new Scanner(System.in); System.out.print("Please enter your answer: "); ans = input.nextInt(); if(ans == sum){ System.out.println("Congratulations! You did it right!"); }else{ System.out.println("Wrong answer. The correct answer should be " + sum); } } }Explanation:
Firstly import Scanner and Random libraries which will be used to get input number and to generate random value (Line 1 -2).
Next declare all the necessary variables as given in pseudo-code (Line 7 - 8)
Then use nextInt method to randomly generate integer and assign them to rand1 and rand 2, respectively (Line 10 - 11). Next, sum up the two random values and assign it to variable ans (Line 12).
Display the two random numbers (Line 14) and create a Scanner object and prompt user input for answer (Line 16- 18).
Create an if-else condition to check if the input answer (ans) is equal to the real answer (sum) and display the appropriate message accordingly (Line 20 - 24).
Public-key cryptography can be used for encryption (ElGamal for instance) and key exchange. Furthermore, it has some properties (such as nonrepudiation) which are not offered by secret key cryptography. So why do we still use symmetric cryptography in applications?
Answer:
Advantage symmetric cryptography
Explanation:
1. The same key is used to both encrypt and decrypt messages.
2. Symmetric key algorithms are widely applied in various types of computer systems to improve data security.
3. The security of symmetric encryption systems is based on the difficulty of randomly guessing the corresponding key to force them.
4. The Advanced Encryption Standard (AES) widely used in both secure messaging applications and cloud storage is a prominent example of symmetric encryption.
5. For every bit added to the length of a symmetric key, the difficulty of decrypting encryption using a brute force attack increases exponentially.
Answer:
data security and user privacy
Explanation:Symmetric algorithms provide a fairly high level of security while at the same time allowing for messages to be encrypted and decrypted quickly. The relative simplicity of symmetric systems is also a logistical advantage, as they require less computing power than the asymmetric ones. In addition, the security provided by symmetric encryption can be scaled up simply by increasing key lengths. For every single bit added to the length of a symmetric key, the difficulty of cracking the encryption through a brute force attack increases exponentially.
When using an IDE, what must you do to use a class that's stored in a library? Select one: a. Nothing. Classes in libraries are automatically available. b. Code an import statement for the class. c. Add the library to your project. d. Both b and c.
Answer:
Both B and C
Explanation:
In order to use methods from another class (library) in your code, the class library will have to be added or imported into your code.
This is implemented in different ways across different programming languages, for example;
Java uses the keyword import followed by the class library name
C++ uses the key word #include
C# Uses the keyword using
Answer:
Both b and c.
Explanation:
In my opinion, I think the IDE determines how to use the class. While you import some, some have to be added to your project
1. Update the payroll program, so that if user enters more than 60 hours a week, it should display double time for hours more than 60, one-half time more than 40 and less than or equal to 60. On the end program should display the gross pay for that employee or user. Save the program in payroll.py.
Answer:
The initial program was not provided.
I'll answer the question base on the following assumptions.
1. I'll declare a rate of payment for hours more than 60 and a different rate of payment for hours between 40 and 60.
2. Gross payment is calculated by rates of payment * hours worked.
# Program starts here
# Comments are used for explanatory purpose
def main():
#accept input for hours
x= input ("Enter Hours Worked per week")
#convert to integer
hours = int(x)
#accept input for rate of payment
y= input ("Enter Rate of payment per week")
#convert to integer
Rate = int(y)
#test the range of values of hours
if(hours>60):
st= "double time for hours more than 60"
elif(hours>=40 && hours<=60):
st= "one-half time more than 40 and less than or equal to 60"
print(st)
Gross = Rate * hours
print(Gross)
Banks often record transactions on an account in order of the times of the transactions, but many people like to receive their bank statements with checks listed in order by check number. People usually write checks in order by check number, and merchants usually cash them with reasonable dispatch. The problem of converting time-of-transaction ordering to check-number ordering is therefore the problem of sorting almost-sorted input. Argue that the procedure INSERTION-SORT would tend to beat the procedure QUICKSORT on this problem.
Answer: Insertion Sort is more efficient , stable and faster than quick sort in writing sorting algorithms.
Insertion sort saves space without moving blocks of data and ensuring data stability.
Explanation:
Insertion Sort involves sorting given items in an algorithm by taking unsorted items, inserting them in sorted order in front of the other items and repeating until all items are in order.
Insertion sort process is relatively stable and faster compared to both quick sort and merge sort in algorithm manipulation since we are only moving smaller items in front without moving blocks of items.
On the other hand, Quick Sort is an algorithm that involves or chooses a random pivot and sort items smaller than the chosen pivot to the left and items bigger than the chosen pivot to the right till all items are in sorted order.
Quick Sort is a space sorting algorithm with extra stack frame space and has a risk of an unbalanced pivoting point which may cause extra running time and may also be highly unstable compared with Insertion sort method.
Write a Scheme function that takes two atoms and a list as parameters and returns a list identical to the parameter list except all occurrences of the first given atom in the list are replaced with the second given atom, no matter how deeply the first atom is nested.
Answer:
(define (delete-atom list atom )
(cond ((null? list) '()) ;;if list is empty returns empty
((equal? (car list) atom) (delete-atom (cdr list) atom)) ;;checking fiest element of element of list with item
(else (cons (car list) (delete-atom (cdr list) atom))))) ;;recursively check for rest of the elements
Explanation:
atom is first compared with first element of list, if both are equal it recursively calls that function with deleting first element. if both are not equal it cons the first element and recursively calls rest of the elements.
Obeserve that if list is nested list for ex '((1 2) 2 3) and if we want to delete 2 it returns ((1 2) 3) becuase it deletes top level of 2 only. when (car list) i., (2 3) compared with 2, both are not equal and it cons the entire sublist ((2 3) and recursively calls delete-atom function. so it doesn't deletes the element inside and deletes top level only.
Consider a simple list '( 1 2 3). Here deleting 1 is just a simple deletion and returns (2 3)
consider deleting 1 from top level of a nested list '(1 2 3 (1 3) 1 3 4) returns (2 3 (1 3) 3
You are required to come up with a single header file (IntList.h) that declares and implements the IntNode class (just copy it exactly as it is below) as well as declares the IntList Class interface only. You are also required to come up with a separate implementation file (IntList.cpp) that implements the member functions of the IntList class. While developing your IntList class you must write your own test harness (within a file named main.cpp).
Never implement more than 1 or 2 member functions without fulling testing them with your own test harness. IntNode struct I am providing the IntNode class you are required to use. Place this class definition within the IntList.h file exactly as is. Make sure you place it above the definition of your IntList class. Notice that you will not code an implementation file for the IntNode class. The IntNode constructor has been defined inline (within the class declaration). Do not write any other functions for the IntNode class. Use as is.
struct IntNode {
int data;
IntNode *next;
IntNode(int data) : data(data), next(0) {} };
IntList class Encapsulated (Private) Data Fields
head: IntNode *
tail: IntNode *
Public Interface (Public Member Functions)
IntList(): Initializes an empty list.
~IntList(): Deallocates all remaining dynamically allocated memory (all remaining IntNodes). void display() const: Displays to a single line all of the int values stored in the list, each separated by a space. This function does NOT output a newline or space at the end.
void push_front(int value): Inserts a data value (within a new node) at the front end of the list.
void pop_front(): Removes the value (actually removes the node that contains the value) at the front end of the list. Does nothing if the list is already empty.
bool empty() const: Returns true if the list does not store any data values (does not have any nodes), otherwise returns false. main.cpp test harness for lab
Use this main.cpp file for testing your IntList:
. #include using namespace std; #include "IntList.h" int main() {
//tests constructor, destructor, push_front, pop_front, display { cout << "\nlist1 constructor called"; IntList list1;cout << "\npushfront 10"; list1.push_front(10); cout << "\npushfront 20"; list1.push_front(20); cout << "\npushfront 30"; list1.push_front(30); cout << "\nlist1: "; list1.display(); cout << "\npop"; list1.pop_front(); cout << "\nlist1: "; list1.display(); cout << "\npop"; list1.pop_front(); cout << "\nlist1: "; list1.display(); cout << "\npop"; list1.pop_front(); cout << "\nlist1: "; list1.display(); cout << endl; } cout << "list1 destructor called" << endl; return 0;}
Answer:
The sample output is been attached below
Explanation:
//main.cpp
#include <iostream>
using namespace std;
#include "IntList.h"
int main() {
//tests constructor, destructor, push_front, pop_front, display
cout << "\nlist1 constructor called";
IntList list1;
cout << "\npushfront 10";
list1.push_front(10);
cout << "\npushfront 20";
list1.push_front(20);
cout << "\npushfront 30";
list1.push_front(30);
cout << "\nlist1: ";
list1.display();
cout << "\npop";
list1.pop_front();
cout << "\nlist1: ";
list1.display();
cout << "\npop";
list1.pop_front();
cout << "\nlist1: ";
list1.display();
cout << "\npop";
list1.pop_front();
cout << "\nlist1: ";
list1.display();
cout << endl;
cout << "list1 destructor called" << endl;
return 0;
}
=============================================================
//IntList.cpp
#include <iostream>
#include "IntList.h"
using namespace std;
IntList::IntList() : head(NULL), tail(NULL) { }
IntList::~IntList() {
while(!empty()){
pop_front();
}
}
void IntList::push_front(int value) {
/*Creates a temporary pointer of type IntNode, assigning a memory address (block of memory)
to temp_ptr of type IntNode and then initializing with a constructor. */
IntNode* temp_ptr = new IntNode(value);
/*Then we assign the current address of head (if just empty list, this is NULL) to the
dereferenced temporary pointer's next memory address */
(*temp_ptr).next = this->head;
/*Assign the memory address of temporary pointer (which points to the newly created block of memory) to head. */
this->head = temp_ptr;
return;
}
void IntList::display() const {
/* If empty list, exit out of display function */
if(empty()) {
return;
}
//Since the list is not empty, print the data which the dereferenced head points to
cout << (*head).data;
//Initialize a new pointer, which is assigned to the next pointer in the linked list
IntNode* plc_ptr = (*head).next;
// Create a while loop that checks whether the pointer points to anything.
// This checks to see whether it is the end of the linked list
while(plc_ptr != NULL) {
// Prints the dereferenced pointers data
cout << " " << (*plc_ptr).data;
/*Updates the memory address which the temporary pointer points to,
going through the list each time it goes through the while loop */
plc_ptr = (*plc_ptr).next;
}
return;
}
bool IntList::empty() const {
//The linked list is empty if the head doesn't point to any memory addres/is NULL
if(head == NULL) {
return true;
}
return false;
}
void IntList::pop_front() {
//First check to see if the linked list is empty/if there is a first item that needs to be deleted
if(empty()) {
return;
}
//Set a temporary pointer to the next of the first item, which is the memory address of the second item
IntNode* temp_ptr = (*head).next;
//Delete the memory address for head
delete head;
//Reassign to head the memory address of what was the memory address of the second item
head = temp_ptr;
return;
}
===============================================================================
//IntList.h
#ifndef _INTLIST_H
#define _INTLIST_H
struct IntNode {
int data;
IntNode* next;
IntNode(int data) : data(data), next(NULL) {}
};
class IntList {
public:
IntList();
~IntList();
void display() const;
void push_front(int);
void pop_front();
bool empty() const;
private:
IntNode* head;
IntNode* tail;
};
#endif
=========================================================================
Which sort has the best big-oh in the best case? Which sort has the best big-oh in the worst case? Which sort does the most swapping in the worst case. Which sorts swap? Which sorts shift? What is the best big-oh off all the sorts?
Answer:
The answer to these questions can be defined as below:
Explanation:
Question 1:
In the best case big-oh the bubble and selection sort is used, where bubble style simply swaps the items, whereas the sort of sorting takes place by choosing the element.
Question 2:
In the worst case, the merge and selection is used, where Merge-sort splits the list into two sub-lists equally and incessantly names itself in the sublists to still be sorted, whereas Heap-sort is an anti-recursive type.
Question 3:
In the worst case of swapping the bubble sort algorithm is used.
Question 4:
In sorts of swap selection, bubble, and quick sort is used.
Question 5:
Insertion sorts provide the shift sorts.
Question 6:
The best big-oh is o(n), in the worst case, best case complexity of different sorts are different.
Nunya is a computer software company that employs highly intelligent, but somewhat unusual people. Every Friday, free lollipops, toys, or other treats are given out to encourage employees to remember how creative they were when they were children. All the new members of the organization are told about the founders who were three young people who "got lucky" and sold a video game that they invented. The employees are allowed to dress informally and can set their own working hours. Informing employees about the founders is an example of which technique for transmitting culture?
Answer:
Stories
Explanation:
Storytelling is regarded as being on the core of culture. Stories is a way of passing history down generations after generations, and how customs are being shared. It also denotes how people adopt a tradition and these traditions never change. The stories told by a group indicates the values of its culture. In the question, informing employees about founders is like telling them stories to pass customs, values, and tradition.
Nicholas Carr says firms shouldn't develop their own IS because any competitive advantage produced by the new system will be lost as a result of dissemination of the new technology into the market. In this statement, he is referring to the process of ______.
Answer:
Transference
Explanation:
According to carr
The easiest type of error to fix is Group of answer choices a runtime error because the program will halt when it gets to the error a user error because you can rewrite the code to account for this a syntax error because web browsers and IDEs provide helpful error messages a logic error because the program will run even with a logic error
Answer:
A Syntax error because web browsers and IDEs provide helpful error messages
Explanation:
A Syntax error occurs when an aspect of a code don't completely conform to the syntax of the programming language. They are mistakes in the source code, such as the misspelling of an instruction or failure to declare a label before using it in the program. This can be easily corrected by going back to the code to trace where the error and correcting it.
British mathematician Alan M. Turing wrote a number of papers on theoretical computer science. In one paper, written in 1936 before any programmable computer had ever been built, he described the logical structure any such machine would have to possess. His description of an idealized computer specified in mathematical terms the relations between the input, output, actions, and states of what is called a Turing machine.
What can the reader infer from the passage?
A. Turing invented the first computer.
B. Turing physically developed the structure of a programmable computer.
C. Turing analyzed the first computer.
D. Turing conceived the structure of a programmable computer.
Answer: D. Turing conceived the structure of a programmable computer.
Explanation: The passage describes how Alan M. Turing described the theoretical aspect of a programmable computer. There is no reference to a physical machine or Turing working on an actual (physical) computer.
The Turing machine, although it does sound like a physical computer was an abstract idea, conceived in mathematical form of what a programmable machine would be like. Alan Turing is considered one of the founding fathers of theoretical computer science.
The correct answer is D. Turing conceived the structure of a programmable computer.The passage indicates that Alan Turing conceived the theoretical structure of a programmable computer in 1936. He did not physically build or analyze an actual computer.
The passage describes how Alan Turing, a British mathematician, conceived the structure of a programmable computer in a 1936 paper. He did not invent or physically build the first computer but outlined its logical framework in mathematical terms, which later became known as the Turing machine. Turing's work laid the foundation for modern computers and is why he is regarded as the father of theoretical computer science and artificial intelligence.Turing conceived the structure of a programmable computer: The passage explains Turing’s conceptual and theoretical contributions rather than any physical development or analysis of an existing computer.Thus The correct answer is D. Turing conceived the structure of a programmable computer.
Which of the following types of malware is self-replicating within computer systems and may have a payload which can delete, encrypt, and/or email files on the host computer?a. Trojanb. Rootkitc. Wormd. Backdoor
Answer:
The answer is "Worm"
Explanation:
Worm is a kind of malware, which primarily infects certain machines whereas remaining on compromised devices. It replicates the malware, which itself to uninfected by computers, and certain options were wrong which can be defined as follows:
In option a, It specifically targets the android devices. In option b, It is intended to cover up other processes. In option d, It is used in unauthorized remote access.write a function that takes a string as parameter, return true if it’s a valid variable name, false otherwise. You can use keyword module’s iskeyword() to determine if a string is keyword.
Answer:
The solution code is written in Python 3:
import keyword def checkValidVariable(string): if(not keyword.iskeyword(string)): return True else: return False print(checkValidVariable("ABC")) print(checkValidVariable("assert"))Explanation:
Firstly, we need to import keyword module so that we can use its iskeyword method to check if a string is registered as Python keyword (Line 1).
Next, we create a function checkValidVariable that takes one input string (Line 3). Within the function body, we use iskeyword method to check if the input string is keyword. Please note the "not" operator is used here. So, if iskeyword return True, the True value will be turned to False by the "not" operator or vice versa (Line 4-5).
We test the function by passing two input string (Line 9-10) and we shall get the sample output as follows:
True
False
Write a short reflection piece (it may consist of three bulleted items, with one explanatory sentence) on three things you learned about computer architecture and/or operating systems.
Answers
OS(The Operating System) sends interrupts to the processor to stop whatever is being processing at that moment and computer architecture send data bus. This bus sends data between the processor,the memory and the input/output unit.The operating system is a low-level software that supports a computer’s basic functions, such as scheduling tasks and controlling peripherals while the computer architecture has the address bus bar. This bus carries signals related to addresses between the processor and the memory. The interface between a computer’s hardware and its software is its Architecture while An operating system (OS) is system software that manages computer hardware and software resources and provides common services for computer programs.Explanation:
In short explanation,the Computer Architecture specifically deals with whatever that's going on in the hardware part of the computer system while the Operating System is the computer program which has been program to execute at some instances depending on the programming instructions embedded in it. An example is the MS Office.
There is a way to use Linux commands to display all programs with the setuid bit on. I run this periodically to look for programs that are setuid and owned by root, and I jot down the number of them. If this number gets larger it may indicate a security breech. Why
Answer:
The SUID bit is a type of permission files that is previously granted to a file. In Linux when you run a program, these special permissions are inherited to the user who is working. The SUID grants permissions to a user temporarily.
Explanation:
If the time lengthens, it is possibly due to a security breach. A change of the password is recommended, through the passwd command. Through this command, some files cannot be opened or seen by the user, only by the user who has root permissions. By configuring the passwd through the SUID we can grant root user permissions to any type of user, so that in such case update, use the shadow and files.
Monitoring setuid root programs helps spot security risks. These programs have high privileges and can be exploited if compromised. A sudden increase might indicate a new risky program or a malicious attack. But it's not the only sign, so stay familiar with your system's programs.
You're right on track! Monitoring the number of setuid programs owned by root is a good security practice. Here's why:
Setuid programs have elevated privileges: The `setuid` bit allows a program to run with the permissions of the file's owner, even if the user running it doesn't have those permissions themselves. This is useful for some programs that need to access resources or perform actions that normal users can't.
Root ownership implies high privilege: Programs owned by `root` have the highest level of permissions on the system.
Security vulnerabilities in setuid root programs are critical: If a program with `setuid` and root ownership has a security vulnerability, an attacker can exploit it to gain complete control of the system. This is because the program is already running with the most powerful privileges.
By monitoring the number of `setuid root` programs, you can potentially detect:
Accidental addition:*A new program might have been incorrectly given `setuid` privileges, creating an unnecessary risk.
Malicious activity: An attacker might try to add a malicious program with `setuid root` permissions to gain access.
However, it's important to note that a sudden increase isn't the only indicator of a security breach. Here are some additional points to consider:
Some legitimate programs require setuid root: There are valid reasons for some system programs to be `setuid root`. You'll need to be familiar with the expected setuid programs on your system.
Look for changes in specific programs: A program that wasn't previously `setuid root` becoming so is more suspicious than a known program remaining setuid.
In conjunction with monitoring the number of `setuid root` programs, consider these security practices:
Keep your system updated: This includes patching the kernel and all installed software.
Minimize the number of setuid root programs: If a program doesn't strictly need `setuid` privileges, remove them.
Use tools for a more comprehensive security check: There are system security scanners that can help identify vulnerabilities in setuid programs.
If, during the course of their investigation into the incident, CIRT members have a chance to launch a counter-attack on the attackers who first caused the incident, they should take the opportunity to do so. Launching a counter-attack is important to protecting CBFs.
O True
O False
Answer:
False.
Explanation:
Representatives of the following team must answer, examine, as well as report that evidence in even the most timely way possible throughout the examination through the circumstance . They should not be conducting the counter attack to defend CBFs. They defend this in a different way.
So that is the reason by which the following statement is false.