What does 32mb SVGA do

Solution :

ENTITY 1               CARDINALITY RATIO                                     ENTITY 2

1. Student                        1 to many                                     Social security card

                      A student may have more than one          

                     social security card (legally with the

                      same unique social security number),

                     and every social security number belongs

                   to a unique.

2. Student                   Many to Many                                    Teacher

                    Generally students are taught by many

                    teachers and a teacher teaches many students.

3.ClassRoom               Many to Many                                   Wall

                      Do not forget that the wall is usually

                      shared by adjacent rooms.

4. Country                 1 to 1                                             Current President

                       Assuming a normal country under normal

                       circumstances having one president at a

                      time.

5. Course               Many to Many                               TextBook

                        A course may have many textbooks and

                       text book may be prescribed for different

                      courses.

6. Item                             Many to Many                           Order

                        Assuming the same item can appear

                         in different orders.

7. Student                       Many to Many                          Class

                        One student may take several classes.

                        Every class usually has several students.

8. Class                                 Many to 1                             Instructor

                       Assuming that every class has a unique

                        instructor. In case instructors were allowed

                        to team teach, this will be many-many.

9. Instructor                  1 to 1                                             Office

                       Assuming every instructor has only one

                      office and it is not shared. In case of offices

                     shared by 2 instructors, the relationship

                    will be 2-1. Conversely, if any instructor has a joint

                    appointment and offices in both departments,

                    then the relationship will be 1-2. In a very general

                  case, it may be many-many.

10. E-bay Auction item             1-Many                           E-bay bid

                        1 item has many bids and a bid is unique

                        to an item.

Complete Question:

A technician needs to manage a Linux-based system from the GUI remotely.  Which of the following technologies should the technician deploy?

A. RDP  

B. SSH  

C. VNC  

D. Telnet

Answer:

Managing a Linux-based System from the GUI Remotely

The technology that the technician should deploy to manage the Graphic User Interface (GUI) remotely is:

C. VNC

Explanation:

VNC stands for Virtual Network Computing. This computing technology enables a remote user to access and control another computer to execute commands. It acts as a cross-platform screen-sharing system and is used to remotely access and control another computer. VNC makes a computer screen, keyboard, and mouse available to be seen and used from a remote distance.  The remote user does everything from a secondary device without being in front of the primary computer or device.  An example of a good VNC is TeamViewer.

Answer:

Explanation:

The following Java program has the method calcAverage(). It creates a sum variable which holds the sum of all the values in the array and the count variable which holds the number of elements in the array. Then it uses a For Each loop to loop through the array and go adding each element to the sum variable. Finally it calculates the average by dividing the sum by the variable count and returns the average to the user. A test case has been created in the main method using the example in the question. The output can be seen in the image attached below.

class Brainly {

   public static void main(String[] args) {

       int[] myArr = {1, 2, 3, 4, 5};

       System.out.println(calcAverage(myArr));

   }

   public static double calcAverage(int[] myArr) {

       double sum = 0;

       double count = myArr.length;

       for (int x: myArr) {

           sum += x;

       }

       double average = sum / count;

       return average;

   }

}

Answer and Explanation:

public class Main {

int Carmodel;

int Purchaseprice;

int Currentyear;

Public void setPurchasePrice(int Purchaseprice){

this.Purchaseprice=Purchaseprice;

}

Public void getPurchasePrice(){

return Purchaseprice;

}

static void printInfo(int Carmodel int Currentyear ){

this.Carmodel=Carmodel;

this.Currentyear=Currentyear;

System.out.println(getPurchasePrice() Carmodel Currentyear);

}

}

The above java program defines a class that has three methods, a get method that returns purchase price of the object, a set method that sets purchase price of the object, and a print method that print out the information about the car object(model, year, price). The print method also takes arguments and assigns values of the arguments/parameters to the object, then prints all.

(c) 4

The function strpos() is a function in php that returns the position of the first occurrence of a particular substring in a string. Positions are counted from 0. The function receives two arguments. The first argument is the string from which the occurrence is to be checked. The second argument is the substring to be checked.

In this case, the string is "Get Well Soon!" while the substring is "Well"

Starting from 0, the substring "Well" can be found to begin at position 4 of the string "Get Well Soon!".

Therefore, the function strpos("Get Well Soon!", "Well") will return 4.

Then, with the echo statement, which is used for printing, value 4 will be printed as output of the code.

Answer:

Phishing Victim

Explanation:

Replying to this email could make you a victim of a phishing scam. Phishing attacks are common challenges on security that internet users are faced with. It could lead to others getting hold of sensitive information like your password or bank details. email is one way the hackers used to get this information. replying to such an email could lead one to an unsecure website where information could be extracted

Answer:

You typically create a database during Oracle Database software installation. However, you can also create a database after installation.

Reasons to create a database after installation are as follows:

Base to be converted to was not included but we would assume conversion to base 10(decimal)

Answer and Explanation:

The 3DF in base 16 is a hex number so we need to convert to its equivalent binary/base 2 form:

We therefore each digit of given hex number 3DF in base 16 to equivalent binary, 4 digits each.

3 = 0011

D = 1101

F = 1111

We then arrange the binary numbers in order

3DF base 16 = 1111011111 in base 2

We then convert to base 10 =

= 1x2^9+1x2^8+1x2^7+1x2^6+0x2^5+1x2^4+1x2^3+1x2^2+1×2^1+1×2^0

= 991 in base 10

Answer:

import java.util.Scanner;

class Main {

 public static int calcSeries(int n) {

   int sum = 0;

   for(int i=10; i>=n; i--) {

     sum += i;

   }

   return sum;

 }

 public static void main(String[] args) {

   Scanner reader = new Scanner(System.in);

   int n = 0;

   do {

     System.out.print("Enter n: ");

     n = reader.nextInt();

     if (n >= 10) {

       System.out.println("Please enter a value lower than 10.");

     }

   } while (n >= 10);

   reader.close();

   System.out.printf("sum: %d\n", calcSeries(n));

 }

}

Answer:

Speak to a U-Phone representative.

Explanation:

A representative will tell you the necessary details about the U-phone without any biases whatsoever, as they´re the closest to the company when it comes to detail about their products.

Answer:

Here the code is given as follows,

Explanation:

Answer:

Replace /* Your solution goes here */

with the following:

courseStudent.SetName("Smith");

courseStudent.SetAge(20);

courseStudent.SetID(9999);

courseStudent.PrintAll();

cout <<courseStudent.GetID();

Explanation:

From the given code segment, we have the following methods defined under the StudentData class;

SetName -> It receives name from the main

SetAge -> It receives age from the main

SetID --> It receives ID from the main and passes it to GetID method

printID --> Prints all the required output.

So, we have:

Pass name to setName

courseStudent.SetName("Smith");

Pass age to setAge

courseStudent.SetAge(20);

Pass ID to setID

courseStudent.SetID(9999);

Print all necessary outputs

courseStudent.PrintAll();

Manually print the student ID

cout <<courseStudent.GetID();

Answer:

Header and footer

Explanation:

The page layout tab usually found in document programs such as Microsoft Word hold options used to configure several page options which is to be applied to a document. The page layout option gives users options to set page margins, orientation, page numbering, indentation and the other page layout options which is primarily how we would like our document to look like when printed. The page layout option also allows the insertion of a header and footer which is a note which can be written at the topmost and bottomost part of the document. Once, a header and footer option is selected, it is applied to every page of the document being worked on at the moment.

Answer:

"if" is the right response.

Explanation:

Answer:

a. Youthful

Explanation:

The river gets its water from youth stage fluvial water. The river meanders from side to side. East Brownsville will be less stable in international boundary in terms of physical position on the landscape because its base level drops.

Answer:

import math

n = int(input("Enter n: "))

for number in range(1, n+1):

   if number % 2 == 1:

       print("{:.2f}".format(math.sqrt(number)), end=" ")

Explanation:

In order to calculate the square roots of the numbers, import math. Note that math.sqrt() calculates and gives the square root of the given number as a parameter

Ask the user to enter the n

Create a for loop that iterates from 1 to n+1 (Since n is included, you need to write n+1)

Inside the loop, check whether the number is odd or not using the if structure and modulo. If the number is odd, calculate its square root using the math.sqrt() and print the result with two digits after the decimal

Answer:

removeDuplicates() function:-

//removeDuplicates() function removes duplicate elements form linked list.

   void removeDuplicates() {

       //declare 3 ListNode pointers ptr1,ptr2 and duplicate.

       //initially, all points to null.

       ListNode ptr1 = null, ptr2 = null, duplicate = null;

       //make ptr1 equals to head.

       ptr1 = head;

        //run while loop till ptr1 points to second last node.

       //pick elements one by one..

       while (ptr1 != null && ptr1.next != null)

       {

               // make ptr2 equals to ptr1.

               //or make ptr2 points to same node as ptr1.

           ptr2 = ptr1;

           //run second while loop to compare all elements with above selected element(ptr1->val).

           while (ptr2.next != null)

           {

              //if element pointed by ptr1 is same as element pointed by ptr2.next.

               //Then, we have found duplicate element.

               //Now , we have to remove this duplicate element.

               if (ptr1.val == ptr2.next.val)

               {

                  //make duplicate pointer points to node where ptr2.next points(duplicate node).

                       duplicate = ptr2.next;

                       //change links to remove duplicate node from linked list.

                       //make ptr2.next points to duplicate.next.

                   ptr2.next = duplicate.next;

               }

             //if element pointed by ptr1 is different from element pointed by ptr2.next.

               //then it is not duplicate element.

               //So, move ptr2 = ptr2.next.

               else

               {

                   ptr2 = ptr2.next;

               }

           }

           //move ptr1 = ptr1.next, after check duplicate elements for first node.

           //Now, we check duplicacy for second node and so on.

           //so, move ptr1  by one node.

           ptr1 = ptr1.next;

       }

   }

Explanation:

Complete Code:-

//Create Linked List Class.

class LinkedList {

       //Create head pointer.

       static ListNode head;

       //define structure of ListNode.

       //it has int val(data) and pointer to ListNode i.e, next.

   static class ListNode {

       int val;

       ListNode next;

       //constructor to  create and initialize a node.

       ListNode(int d) {

               val = d;

           next = null;

       }

   }

//removeDuplicates() function removes duplicate elements form linked list.

   void removeDuplicates() {

       //declare 3 ListNode pointers ptr1,ptr2 and duplicate.

       //initially, all points to null.

       ListNode ptr1 = null, ptr2 = null, duplicate = null;

       //make ptr1 equals to head.

       ptr1 = head;

       //run while loop till ptr1 points to second last node.

       //pick elements one by one..

       while (ptr1 != null && ptr1.next != null)

       {

               // make ptr2 equals to ptr1.

               //or make ptr2 points to same node as ptr1.

           ptr2 = ptr1;

           //run second while loop to compare all elements with above selected element(ptr1->val).

           while (ptr2.next != null)

           {

              //if element pointed by ptr1 is same as element pointed by ptr2.next.

               //Then, we have found duplicate element.

               //Now , we have to remove this duplicate element.

               if (ptr1.val == ptr2.next.val)

               {

                  //make duplicate pointer points to node where ptr2.next points(duplicate node).

                       duplicate = ptr2.next;

                       //change links to remove duplicate node from linked list.

                       //make ptr2.next points to duplicate.next.

                   ptr2.next = duplicate.next;

               }

             //if element pointed by ptr1 is different from element pointed by ptr2.next.

               //then it is not duplicate element.

               //So, move ptr2 = ptr2.next.

               else

               {

                   ptr2 = ptr2.next;

               }

           }

           //move ptr1 = ptr1.next, after check duplicate elements for first node.

           //Now, we check duplicacy for second node and so on.

           //so, move ptr1  by one node.

           ptr1 = ptr1.next;

       }

   }

   //display() function prints linked list.

   void display(ListNode node)

   {

       //run while loop till last node.

       while (node != null)

       {

               //print node value of current node.

           System.out.print(node.val + " ");

           //move node pointer by one node.

           node = node.next;

       }

   }

   public static void main(String[] args) {

       //Create object of Linked List class.

       LinkedList list = new LinkedList();

       //first we create nodes and connect them to form a linked list.

       //Create Linked List 1-> 2-> 3-> 2-> 4-> 2-> 5-> 2.

       //Create a Node having node data = 1 and assign head pointer to it.

       //As head is listNode of static type. so, we call head pointer using class Name instead of object name.

       LinkedList.head = new ListNode(1);

       //Create a Node having node data = 2 and assign head.next to it.

       LinkedList.head.next = new ListNode(2);

       LinkedList.head.next.next = new ListNode(3);

       LinkedList.head.next.next.next = new ListNode(2);

       LinkedList.head.next.next.next.next = new ListNode(4);

       LinkedList.head.next.next.next.next.next = new ListNode(2);

       LinkedList.head.next.next.next.next.next.next = new ListNode(5);

       LinkedList.head.next.next.next.next.next.next.next = new ListNode(2);

       //display linked list before Removing duplicates.

       System.out.println("Linked List before removing duplicates : ");

       list.display(head);

       //call removeDuplicates() function to remove duplicates from linked list.

       list.removeDuplicates();

       System.out.println("")

       //display linked list after Removing duplicates.

       System.out.println("Linked List after removing duplicates :  ");

       list.display(head);

   }

}

Output:-

From method names, I am compelled to believe you are creating some sort of a Lexer object. Generally you implement Lexer with stratified design. First consumption of characters, then tokens (made out of characters), then optionally constructs made out of tokens.

Hope this helps.

Answer:

1. Identification of a vacancy and development of the job description.

2. Recruitment planning

3. Advertising

4. Assessment and Interview of applicants

5. Selection and Appointment of candidates

6. Onboarding

Explanation:

The Recruitment process refers to all the stages in the planning, assessment, and absorption of candidates in an organization. The stages involved include;

1. Identification of a vacancy and development of the job description: This is the stage where an obvious need in the organization is identified and the duties of the job requirement are stipulated.

2. Recruitment planning: This is the stage where the HR team comes together to discuss the specific ways they can attract qualified candidates for the job.

3. Advertising: The HR team at this point seeks out job sites, newspapers, and other platforms that will make the opportunities accessible to applicants.

4. Assessment and Interview of applicants: Assessments are conducted to gauge the candidates' knowledge and thinking abilities. This will provide insight into their suitability for the job.

5. Selection and Appointment of candidates: Successful candidates are appointed to their respective positions. A letter of appointment is given.

6. Onboarding: Candidates are trained and guided as to the best ways of discharging their duties.

Answer:

Explanation:

The answer should be 1.16 or 61.1

Answer:

Here the answer is given as follows,

Explanation:

import seaborn as sns  

import pandas as pd  

df = {'country': ['US', 'US', 'US', 'US', 'UK', 'UK', 'UK'],  

  'year': [2008, 2009, 2010, 2011, 2008, 2009, 2010],  

  'Happiness': [4.64, 4.42, 3.25, 3.08, 3.66, 4.08, 4.09],  

  'Positive': [0.85, 0.7, 0.54, 0.07, 0.1, 0.92, 0.94],  

  'Negative': [0.49, 0.09, 0.12, 0.32, 0.43, 0.21, 0.31],  

  'LogGDP': [8.66, 8.23, 7.29, 8.3, 8.27, 6.38, 6.09],  

  'Support': [0.24, 0.92, 0.54, 0.55, 0.6, 0.38, 0.63],  

  'Life': [51.95, 55.54, 52.48, 53.71, 50.18, 49.12, 55.84],  

  'Freedom': [0.65, 0.44, 0.06, 0.5, 0.52, 0.79, 0.63, ],  

  'Generosity': [0.07, 0.01, 0.06, 0.28, 0.36, 0.33, 0.26],  

  'Corruption': [0.97, 0.23, 0.66, 0.12, 0.06, 0.87, 0.53]}  

dataFrame = pd.DataFrame.from_dict(df)  

explanatory_vars = ['LogGDP', 'Support', 'Life', 'Freedom', 'Generosity', 'Corruption']  

plot_vars = ['Happiness'] + explanatory_vars  

sns.pairplot(dataFrame,  

            x_vars = explanatory_vars,  

            dropna=True,  

            palette="Blues")    

Answer:

Explanation:

The code provided had many errors. I fixed the errors and changed the userAgeCheck function as requested. Checking the age of that the user has passed as an input and changing the variables as needed depending on the age. The background color that was changed was for the form field as the question was not very specific on which field needed to be changed. The piece of the code that was created can be seen in the attached image below.

var validColor = "LightGreen";

var invalidColor = "Orange";

var userAgeInput = document.getElementById("userAge");

var formWidget = document.getElementById("userForm");

var userAgeValid = false;

function userAgeCheck(event) {

   if ((userAgeInput >= 25) && (userAgeInput <= 35)) {

       document.getElementById("userForm").style.backgroundColor = validColor;

       userAgeValid = true;

   } else {

       document.getElementById("userForm").style.backgroundColor = invalidColor;

       userAgeValid = false;

   }

};

function formCheck(event) {};

if (!userAgeValid) {

   userAgeInput.addEventListener("input", userAgeCheck);

   formWidget.addEventListener('submit', formCheck);  

   event.preventDefault();

}

Answer:

Explanation:

The following code is written in Java. It creates the raiseToPower method that takes in two int parameters. It then uses recursion to calculate the value of the first parameter raised to the power of the second parameter. Three test cases have been provided in the main method of the program and the output can be seen in the attached image below.

class Brainly {

   public static void main(String[] args) {

       System.out.println("Base 5, Exponent 3: " + raiseToPower(5,3));

       System.out.println("Base 2, Exponent 7: " + raiseToPower(2,7));

       System.out.println("Base 5, Exponent 9: " + raiseToPower(5,9));

   }

   public static int raiseToPower(int base, int exponent) {

       if (exponent == 0) {

           return 1;

       } else if (exponent == 1) {

           return base;

       } else {

           return (base * raiseToPower(base, exponent-1));

       }

   }

}

Answer:

Memory-mapped I/O and port-mapped I/O are two complementary methods of performing input/output between the central processing unit and peripheral devices in a computer. An alternative approach is using dedicated I/O processors, commonly known as channels on mainframe computers, which execute their own instructions.

Answer:

The given statement is "True".

Explanation:

Answer:

The answer is "False"

Explanation:

It is put to use Six Sigma had 3.4 defects per million opportunities (DPMO) from the start, allowing for a 1.5-sigma process shift. However, the definition of zero faults is a little hazy. Perhaps the area beyond 3.4 DPMO is referred to by the term "zero faults.", that's why Before being released, large computer programs, such as operating systems, must have no faults the wrong choice.

Answer:

3) If number is zero, number is displayed;

Explanation:

The code snippet created is supposed to take any input number from the user (positive or negative) and print it to the console if it is less than 0. In this code the IF statement does not need an else clause and will work regardless. The System.out.println() statement does not need to be inside braces since it is a simple one line statement. Therefore, the only statement in the question that is actually true would be ...

3) If number is zero, number is displayed;

Answer:

If number is zero, number is displayed

Explanation:

Given

The above code segment

Required

What is true about the code segment

Analyzing the options, we have:

(1) Wrong statement because there is no else clause

The above statement is not a requirement for an if statement to be valid; i.e. an if statement can stand alone in a program

Hence, (1) is false

(2) The print statement must be in { }

There is only one statement (i.e. the print statement) after the if clause. Since the statement is just one, then the print statement does not have to be in {} to be valid.

Hence, (2) is false

(3) number is displayed, if input is 0

In the analysis of the program, we have: number <=0

i.e. the if statement is true for only inputs less than 0 or 0.

Hence, number will be printed and (3) is false

(4) number is displayed for positive inputs

The valid inputs have been explained in (3) above;

Hence, (4) is false.

(5) is also false

Answer:

The function is as follows:

int returnsFixed(int arr [],int n){

   int retVal = -1;

   for(int i = 0; i<n;i++){

       if(i == arr[i]){

           retVal = i;

           break;

       }

   }

   return retVal;

}

Explanation:

This defines the functionl it receives the array and the length of the array

int returnsFixed(int arr [],int n){

This initializes the return value to -1

   int retVal = -1;

This iterates through the array

   for(int i = 0; i<n;i++){

This checks if i equals arr[i]

       if(i == arr[i]){

If yes, the return value (i.e. the fixed point) is set to i

           retVal = i;

And the code is exited

           break;

       } This ends the if condition

   } This ends the iteration

This returns the calculated fixed point

   return retVal;

}

Answer:

Denial of service attack.

Explanation:

In Cybersecurity, vulnerability can be defined as any weakness, flaw or defect found in a software application or network and are exploitable by an attacker or hacker to gain an unauthorized access or privileges to sensitive data in a computer system.

This ultimately implies that, vulnerability in a network avail attackers or any threat agent the opportunity to leverage on the flaws, errors, weaknesses or defects found in order to compromise the security of the network.

Data theft can be defined as a cyber attack which typically involves an unauthorized access to a user's data with the sole intention to use for fraudulent purposes or illegal operations. There are several methods used by cyber criminals or hackers to obtain user data and these includes DDOS attack, SQL injection, man in the middle, phishing, sniffing, DOS attack, etc.

A denial of service (DOS) attack is a form of cyber attack that involves either the exploitation of a software flaw or overwhelmingly floods a computer system in a network with traffic in order to prevent authorized users from performing legitimate activities or transactions.

Some of the ways to prevent vulnerability or cyber attacks in a network are;

1. Ensure you use a very strong password with complexity through the use of alphanumerics.

2. You should use a two-way authentication service.

3. You should use encrypting software applications or services.

Answer:

it is the process that binds together the data and code into a single unit and keeps both from being safe from outside interference and misuse. In this process, the data is hidden from other classes and can be accessed only through the current class's methods