You are responsible for managing Windows updates on 80 Windows 10 Enterprise computers at the office. You currently receive alert notifications when a new critical or security update is available for approval. Once you approve the updates, they are eventually installed. You would like these updates to apply automatically for the Windows 10 computers, and ensure that they are applied within 1 day of approval. How can this be accomplished with Intune

Answer:

The reason is due to proprietary design of the Operating System (OS) which require a virtualization software to blanket or "disguise" the hardware (processor) borderlines of the computer onto which it is to be installed.

Explanation:

An Apple system that has the RISC processor and system architecture which has an operating system made entirely for the Apple system architecture

If the above Apple OS is to be installed on a windows computer, then the procedure to setup up the OS has to be the same as that used when on an Apple system, hence, due to the different processors and components of both systems, a virtualization will be be needed to be provided by a Virtual box, Parallels desktop or other virtualization software.

Answer: Font

Explanation:

Answer:

Check the explanation

Explanation:

INCLUDE Irvine32.inc

TRUE = 1

FALSE = 0

.data

gradeAverage WORD ?

credits WORD ?

oKToRegister BYTE ?

str1 BYTE "Error: Credits must be between 1 and 30" , 0dh,0ah,0

main PROC

call CheckRegs

exit

main ENDP

CheckRegs PROC

push edx

mov OkToRegister,FALSE

; Check credits for valid range 1-30

cmp credits,1 ; credits < 1?

jb E1

cmp credits,30 ; credits > 30?

ja E1

jmp L1 ; credits are ok

; Display error message: credits out of range

E1:

mov edx,OFFSET str1

call WriteString

jmp L4

L1:

cmp gradeAverage,350 ; if gradeAverage > 350

jna L2

mov OkToRegister,TRUE ; OkToRegister = TRUE

jmp L4

L2:

cmp gradeAverage,250 ; elseif gradeAverage > 250

jna L3

cmp credits,16 ; && credits <= 16

jnbe L3

mov OkToRegister,TRUE ; OKToRegister = TRUE

jmp L4

L3:

cmp credits,12 ; elseif credits <= 12

ja L4

mov OkToRegister,TRUE ; OKToRegister = TRUE

L4:

pop edx ; endif

ret

CheckRegs ENDP

END main

Answer:

This program has been written using JAVA. The code is written beloe:

// import library

import java.util.Scanner;

public category Main operate

public static void main(String[] args)

{

// scanner category

Scanner scan=new Scanner(System.in);

System.out.println("Enter Number:");

int n=scan.nextInt();

//All squares but n

System.out.println("All squares but n are: ");

int temp=0,square=0,i=0;

while(square

range

square=i*i;

if(square>=n)break;

System.out.print(square+" ");

i++;

}

//All positive numbers that are divisible by 10 but less than n

System.out.println(" ");

System.out.println(" ");

System.out.println("All positive numbers are divisible by 10 and less than n are: ");

i=1;

while(i

{

if(i%10==0)

System.out.print(i+" ");

if(i>=n)break;

i++;

}

System.out.println(" ");

System.out.println(" ");

System.out.println("All powers of 2 that are less than n are: ");

//All powers of 2 less than n

i=0;

while(i

mathematics.pow(2, i);

if(num_power

System.out.print(num_power+" ");

i++;

}

}

}

Explanation:

With SimpleWriter out and int n already declared in each case this program makes use of a while loop condition to satisfy the following;

1. All squares less than n

2. positive numbers that are divisible by 10 and less than n.

3. All powers of two less than n.

Answer:

Information security policy are used for the prevention of intruders hacking a network when an organization start getting IT related attacks.

Explanation:

Information security policy are used for the prevention of intruders hacking a network when an organization start getting IT related attacks.

An information security policy are set of rules/policies designed to guide employees for the protection of the security of company information and IT systems. The reasons for these policies are:

Answer:

The answer to this question can be described as follows:

Explanation:

Given data:

Performance of the CPU:  

The Fastest Factor Fraction of Work:

[tex]f_1=65 \% \\\\=\frac{65}{100} \\\\ =0.65[/tex]

Current Feature Speedup:

[tex]K_1=[/tex] 1.5

CPU upgrade=6000

Disk activity:

The quickest part is the proportion of the work performed:

Current Feature Speedup:

[tex]k_2=3[/tex]

Disk upgrade=8000

System speedup formula:

[tex]s=\frac{1}{(1-f)+(\frac{f}{k})}[/tex]

Finding the CPU activity and disk activity by above formula:

CPU activity:

[tex]S_{CPU}=\frac{1}{(1-f_1)+(\frac{f_1}{k_1})} \\\\=\frac{1}{(1-0.65)+(\frac{0.65}{1.5})} \\\\=1.276 \% ...\rightarrow (1) \\[/tex]

Disk activity:

[tex]S_{DISK} = (\frac{1}{(1-f_2)+\frac{f_2}{k_2}}) \\\\ S_{DISK} = (\frac{1}{(1-0.35)+\frac{0.35}{3}}) \\\\ = -0.5\% .... \rightarrow (2)[/tex]

CPU:

Formula for CPU upgrade:

[tex]= \frac{CPU \ upgrade}{S_{CPU}}\\\\= \frac{\$ 6,000}{1.276}\\\\= 4702.19....(3)[/tex]

DISK:

Formula for DISK upgrade:

[tex]=\frac{Disk upgrade} {S_{DISK}}\\\\= \frac{\$ 8000}{-0.5 \% }\\\\= - 16000....(4)[/tex]

equation (3) and (4),

Thus, for the least money the CPU alternative is the best performance upgrade.

b)

From (3) and (4) result,

The disc choice is therefore the best choice for a quicker system if you ever don't care about the cost.

c)

The break-event point for the upgrades:

=4702.19 x-0.5

= -2351.095

From (2) and (3))

Therefore, when you pay the sum for disc upgrades, all is equal $ -2351.095

Answer:

The final tables in 3NF are as follows.

Project( ProjectNbr, ProjectName)

Employee( EmployeeNbr, EmployeeName, JobClass)

Job( JobClass, HourlyRate)

ProjectBilling( ProjectNbr, EmployeeNbr, HoursBilled)

Explanation:

The given table is given below.

ProjectBilling( ProjectNbr, ProjectName, EmployeeNbr, EmployeeName, JobClass, HourlyRate, HoursBilled)  

ProjectNbr -> ProjectName

EmployeeNbr -> EmployeeName  

JobClass -> HourlyRate

ProjectNbr, EmployeeNbr -> HoursBilled

1NF

1. All the fields in the given table contain only a single value. The table is in 1NF.

2NF

2. New tables are formed based on the given functional dependencies.

Project( ProjectNbr, ProjectName)

Employee( EmployeeNbr, EmployeeName)

Job( JobClass, HourlyRate)

ProjectBilling( ProjectNbr, EmployeeNbr, HoursBilled)

3. Every table is assigned a primary key which are as follows.

ProjectNbr is the primary key for Project table.

EmployeeNbr is the primary key for Project table.

JobClass is the primary key for Project table.

(ProjectNbr, EmployeeNbr) is the composite primary key for the ProjectBilling table.

4. The tables which are related to each other are linked via primary key and foreign key.

5. In the ProjectBilling table, the composite primary key, ProjectNbr, EmployeeNbr is composed of the primary keys of the Project and Employee tables, i.e., ProjectNbr and EmployeeNbr respectively.

6. Job table is related to Employee table. Hence, primary key of Job table, JobClass, is introduced as foreign key in Employee table.

Employee( EmployeeNbr, EmployeeName, JobClass)

7. Partial dependency arises when composite primary key exists and non-prime attributes (columns other than the primary key) depend on a part of the primary key, i.e., partial primary key.

In the ProjectBilling table, no partial dependency exists.

8. All the tables are in 2NF as given below.

Project( ProjectNbr, ProjectName)

Employee( EmployeeNbr, EmployeeName, JobClass)

Job( JobClass, HourlyRate)

ProjectBilling( ProjectNbr, EmployeeNbr, HoursBilled)

3NF

9. All the tables are in 2NF.

10. In every table, all non-prime attribute depend only on the primary key.

11. No transitive dependency exists, i.e., all non-prime attributes do not depend on other non-prime attributes.

12. Hence, all the conditions are satisfied and the tables are in 3NF.

Answer: True

Explanation:

Answer:

D

Explanation:

If your options are

a) to demonstrate a sophisticated programming style

b) to demonstrate a flexible programming style

c)to ensure that programs will work on any computer platform

d) to ensure that programs run properly and return expected results

Then yes this is very much correct!

Answer:Backspace

Explanation:

If you want to delete words you use the backspace button.

Answer:

Backspace

Explanation:

Answer:

import java.util.Scanner;

public class StudentScores

{

public static void main(String[] args) {

 Scanner scnr = new Scanner(System.in);

 final int NUM_POINTS = 4;

 int[] dataPoints = new int[NUM_POINTS];

 int controlValue;

 int i;

 controlValue = scnr.nextInt();

 for (i = 0; i < dataPoints.length; ++i) {

     dataPoints[i] = scnr.nextInt();

 }

 for (i = 0; i < dataPoints.length; ++i) {

     System.out.print(dataPoints[i] + " ");

 }

 System.out.println();

 for (i = 0; i < dataPoints.length; ++i) {

     if(dataPoints[i] < controlValue){

         dataPoints[i] = dataPoints[i] * 2;          

     }

     System.out.print(dataPoints[i] + " ");

 }

}

}

Explanation:

*Added parts highligted.

After getting the control value and values for the array, you printed them.

Create a for loop that iterates through the dataPoints. Inside the loop, check if a value in dataPoints is smaller than the contorolValue. If it is, multiply that value with 2 and assign it to the dataPoints array. Print the elements of the dataPoints

Answer:

import java.util.Scanner;

public class numm3 {

   public static void main (String [] args) {

       Scanner scnr = new Scanner(System.in);

       final int NUM_POINTS = 4;

       int[] dataPoints = new int[NUM_POINTS];

       int controlValue;

       int i;

       System.out.println("Enter the control Variable");

       controlValue = scnr.nextInt();

       System.out.println("enter elements for the array");

       for (i = 0; i < dataPoints.length; ++i) {

           dataPoints[i] = scnr.nextInt();

       }

       for (i = 0; i < dataPoints.length; ++i) {

           System.out.print(dataPoints[i] + " ");

       }

       System.out.println();

       //Doubling elements Less than Control Variable

       for (i = 0; i < dataPoints.length; ++i) {

           if (dataPoints[i]<controlValue){

               dataPoints[i] = dataPoints[i]*2;

           }

           System.out.print(dataPoints[i] + " ");

       }

       System.out.println(); } }

Explanation:

See the additional code to accomplish the task in bold

The trick is using an if statement inside of a for loop that checks the condition (dataPoints[i]<controlValue) If true, it multiplies the element by 2

Answer:

I believe it is science, but before you do anything with this answer, research more on it, just in case I'm wrong! :)

Explanation:

Answer:

see explaination

Explanation:

The data is collected in a comma-separated values (CSV) format that always includes the following fields:

? date: string

? time: string

? client_ip: string

? server_ip: string

? url_stem: string

? url_query: string

? client_bytes: integer

? server_bytes: integer

What should you do?

a. Load the data as lines of text into an RDD, then split the text based on a comma-delimiter and load the RDD into DataFrame.

# import the module csv

import csv

import pandas as pd

# open the csv file

with open(r"C:\Users\uname\Downloads\abc.csv") as csv_file:

# read the csv file

csv_reader = csv.reader(csv_file, delimiter=',')

# now we can use this csv files into the pandas

df = pd.DataFrame([csv_reader], index=None)

df.head()

b. Define a schema for the data, then read the data from the CSV file into a DataFrame using the schema.

from pyspark.sql.types import *

from pyspark.sql import SparkSession

newschema = StructType([

StructField("date", DateType(),true),

StructField("time", DateType(),true),

StructField("client_ip", StringType(),true),

StructField("server_ip", StringType(),true),

StructField("url_stem", StringType(),true),

StructField("url_query", StringType(),true),

StructField("client_bytes", IntegerType(),true),

StructField("server_bytes", IntegerType(),true])

c. Read the data from the CSV file into a DataFrame, infering the schema.

abc_DF = spark.read.load('C:\Users\uname\Downloads\new_abc.csv', format="csv", header="true", sep=' ', schema=newSchema)

d. Convert the data to tab-delimited format, then read the data from the text file into a DataFrame, infering the schema.

Import pandas as pd

Df2 = pd.read_csv(‘new_abc.csv’,delimiter="\t")

print('Contents of Dataframe : ')

print(Df2)

Answer:thats is so easy obviously 69

Explanation:

Answer: Provided in the explanation section

Explanation:

Code provided below in a well arranged format

inputNos.txt

70

95

62

88

90

85

75

79

50

80

82

88

81

93

75

78

62

55

89

94

73

82

___________________

StandardDev.java

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class StandardDev {

public static void main(String[] args) {

//Declaring variables

String filename;

int count = 0, i = 0;

Scanner sc1 = null;

int min, max;

double mean, stdDev;

int nos[] = null;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

//Getting the input entered by the user

System.out.print("Enter the name of the filename :");

filename = sc.next();

try {

//Opening the file

sc1 = new Scanner(new File(filename));

//counting no of numbers in the file

while (sc1.hasNext()) {

sc1.nextInt();

count++;

}

sc1.close();

sc1 = new Scanner(new File(filename));

//Creating an Integer based on the Count

nos = new int[count];

//Populating the values into an array

while (sc1.hasNext()) {

nos[i] = sc1.nextInt();

i++;

}

sc1.close();

//calling the methods

min = findMinimum(nos);

max = findMaximum(nos);

mean = calMean(nos);

stdDev = calStandardDev(nos, mean);

//Displaying the output

System.out.println("Read from file :" + count + " values");

System.out.println("The Minimum Number is :" + min);

System.out.println("The Maximum Number is :" + max);

System.out.printf("The Mean is :%.2f\n", mean);

System.out.printf("The Standard Deviation is :%.2f\n", stdDev);

} catch (FileNotFoundException e) {

e.printStackTrace();

}

}

//This method will calculate the standard deviation

private static double calStandardDev(int[] nos, double mean) {

//Declaring local variables

double standard_deviation = 0.0, variance = 0.0, sum_of_squares = 0.0;

/* This loop Calculating the sum of

* square of eeach element in the array

*/

for (int i = 0; i < nos.length; i++) {

/* Calculating the sum of square of

* each element in the array    

*/

sum_of_squares += Math.pow((nos[i] - mean), 2);

}

//calculating the variance of an array

variance = ((double) sum_of_squares / (nos.length - 1));

//calculating the standard deviation of an array

standard_deviation = Math.sqrt(variance);

return standard_deviation;

}

//This method will calculate the mean

private static double calMean(int[] nos) {

double mean = 0.0, tot = 0.0;

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Calculating the sum of all the elements in the array

tot += nos[i];

}

mean = tot / nos.length;

return mean;

}

//This method will find the Minimum element in the array

private static int findMinimum(int[] nos) {

int min = nos[0];

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Finding minimum element

if (nos[i] < min)

min = nos[i];

}

return min;

}

//This method will find the Maximum element in the array

private static int findMaximum(int[] nos) {

int max = nos[0];

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Finding minimum element

if (nos[i] > max)

max = nos[i];

}

return max;

}

}

_____________________

Output:

Enter the name of the filename :inputNos.txt

Read from file :22 values

The Minimum Number is :50

The Maximum Number is :95

The Mean is :78.45

The Standard Deviation is :12.45

cheers i hope this helped !!!

Answer:

surveys and questionnaires

Answer:

Survey

Explanation:

Answer:

the person drinks water if the answer to “thirsty?” is yes

Answer:

C

Explanation:

Answer:

  hardware-initiated reset

Explanation:

Once in protected mode, the 80286 is designed to remain there until it is reset by hardware.

__

External hardware can be designed so that software can cause such a reset. In the 1984 PC/AT, such hardware combined with code in the BIOS allowed real mode re-entry and returned execution control to the program that caused the reset.

Answer:

Big data is a field that treats ways to analyze, systematically extract information from, or otherwise deal with data sets that are too large or complex to be dealt with by traditional data-processing application software. ... Big data was originally associated with three key concepts: volume, variety, and velocity.

SPANISH TRANSLATION

Big data es un campo que trata formas de analizar, extraer sistemáticamente información de, o de otra manera tratar con conjuntos de datos que son demasiado grandes o complejos para ser manejados por el software tradicional de aplicación de procesamiento de datos. ... Big data se asoció originalmente con tres conceptos clave: volumen, variedad y velocidad.

Answer:

Flaws and limitations identified in this program includes;

1.There was a not necessary usage of variable retrieval. Would have made use of canConvert.

2. Looking at the program, one will notice numerous typos. One of which is the fact that in JAVA we make use of Boolean instead of bool.

3.We rather use Integer.parseInt in JAVA and not Int16, cant make use of Int16.

4. The exception cant be printed

5. JAVA makes use of checkConversion instead of convertNumber as used in the program.

6. It cant work for decimal numbers, 0 and big integers.

Explanation:

See Answer for the detailed explaination of the flaws and limitations identified in the program.

Answer:

The program code is written in the explanation.

Explanation:

def turn_clockwise(d):

if d=="N":

return "E"

elif d=="E":

return "S"

elif d=="S":

return "W"

elif d=="W":

return "N"

else:

return "None"

Output:

Answer:

vi [ -| -s ] [-l] [-L] [-R] [ -r [ filename ] ] [-S] [-t tag] [-v] [-V]

  [-x] [-w] [-n ] [-C] [+command | -c command ] filename

Explanation:

See attached image file

Answer:

at this day and age, any. My questions would focus around other applications at this point.  250 gb or larger.

Explanation:

Office 365 only uses 4 GB "Microsoft's store page", the smallest hard drives commonly available are 250 GB and larger.  

Answer:

Explanation:

The objective here is to write a collection of MySetOperations in Python.

During the process of  submitting this answer after computing the program ; I keep getting a response message that goes thus "It appears that your answer contains either a link or inappropriate words. Please correct and submit again!"  . In the bid to curb such dilemma , i have created a word document for the collection of MySetOperations in Python.

The document can be seen in the attached file below.

Answer:

The program goes as follows  

Comments are used to explain difficult lines

#include<iostream>

#include<fstream>

#include<sstream>

#include<string>

using namespace std;

int main()

{

//Declare String to accept file name

string nm;

//Prompt user for file name

cout<<"Enter File Name: ";

getline(cin,nm);

//Create File

ofstream Myfile(nm.c_str());

//Prompt user for 4 names of movies

string movie;

for(int i = 1;i<5;i++)

{

 cout<<"Please enter a movie title #"<<i<<": ";

 cin>>movie;

 //Write to file, the names of each movies

 Myfile<<movie<<endl;

}

Myfile.close(); // Close file

//Create an Array for four elements

string myArr[4];

//Prepare to read from file to array

ifstream file(nm.c_str());

//Open File

if(file.is_open())

{

 for(int i =0;i<4;i++)

 {

  //Read each line of the file to array (line by line)

  file>>myArr[i];

 }

}

file.close(); // Close file

//Create a reverseOrder.txt file

nm = "reverseOrder.txt";

//Prepare to read into file

ofstream Myfile2(nm.c_str());

for(int j = 3;j>=0;j--)

{

 //Read each line of the file to reverseOrder.txt (line by line)

 Myfile2<<myArr[j]<<endl;

}

Myfile2.close(); //Close File

return 0;

}

See attached for .cpp program file

Answer:

see explaination

Explanation:

#include <fstream>

#include <iostream>

#include <string>

//to set precision

#include <iomanip>

#include <bits/stdc++.h>

//structure of student is created.

struct Student

{

std::string name;

double grade1;

double grade2;

double grade3;

double Average;

// Constructor initializes semester scores to 0.

Student()

{

grade1 = 0;

grade2 = 0;

grade3 = 0;

Average = 0;

}

//Read from the file and calculate the average, min and max grade with name.

void read(std::fstream& fil)

{

std::string temp;

fil >> name;

fil >> temp;

grade1 = std::stod(temp);

fil >> temp;

grade2 = std::stod(temp);

fil >> temp;

grade3 = std::stod(temp);

Average = ((grade1 + grade2 + grade3) /3.0);

}

//write to file- name and average with precision 2.

void write(std::fstream& fil)

{

fil << name << " " << std::setprecision(2) << std::fixed << Average << std::endl;

}

//Logic to calculate the minimum grade of the class.

double min()

{

double min = grade1;

if(min > grade2)

{

min = grade2;

}

if(min > grade3)

{

min = grade3;

}

return min;

}

//Logic to calculate the maximum grade of the class.

double max()

{

double max = grade1;

if(max < grade2)

{

max = grade2;

}

if(max < grade3)

{

max = grade3;

}

return max;

}

//Print the name and average with precision 2 in the console.

void print()

{

std::cout << name << " " ;

std::cout << std::setprecision(2) << std::fixed << Average << std::endl;

}

};

//As per this question,students are set to 3.

Student students[3];

//main method.

int main()

{

// Read from the file question2.txt

std::fstream fil;

fil.open("question2.txt", std::ios::in);

for(int i = 0; i < 3; i++)

students[i].read(fil);

fil.close();

// finding the max grade of the class.

int max = students[0].max();

int temp = students[1].max();

if(max <temp)

max = temp;

temp = students[2].max();

if(max < temp)

max = temp;

// finding the min grade of the class.

int min = students[0].min();

temp = students[1].min();

if(min > temp)

min = temp;

temp = students[2].min();

if(min > temp)

min = temp;

// print the output in console.

std::cout << "Max Grade " << max << std::endl;

std::cout << "Min Grade " << min << std::endl;

for(int i = 0; i < 3; i++)

students[i].print();

// write to output file output2.txt .

fil.open("output2.txt", std::ios::out);

fil << "Max Grade " << max << std::endl;

fil << "Min Grade " << min << std::endl;

for(int i = 0; i < 3; i++)

students[i].write(fil);

fil.close();

return 0;

}

Answer:

Check the explanation

Explanation:

The Python program that frequently asks the user what pets the user has,

until the user enters "rock", in which case the loop ends can be analysed in the written codes below.

'''

# count of pets

count = 0

# read the user input

pet = input()

# loop that continues till the user enters rock

# strip is used to remove the whitespace

while pet.strip() != 'rock':

  # increment the count of pets

  count += 1

  # output the pet name and number of pets read till now

  print('You have a %s with a total of %d pet(s)' %(pet.strip(),count))

  pet = input() # input the next pet

#end of program

Following are the python program to use the loop to count the number of pets until the user enters the "rock":

c = 0#defining a variable c that initilzes with 0

p = input()#defining a variable p that input value

while p != 'rock':#defining a loop that check p value not equal to rock

  c += 1#using c variable for counts total number of pets

  print('You have a %s with a total of %d pet(s)' %(p,c))#print total number of pet

  p = input() #input value

Output:

Please find the attached file.

Program Explanation:

Find out more about the loop here:

brainly.com/question/17067964

Answer:

ierror(VLOOKUP(Q2,CBFStaff[[Staff ID]:[Name]],2,FALSE), "Invalid Staff ID")

Explanation:

Let me try as much as I can to explain the concept or idea of iferror in vlookup.

iferror have a typically function and result like an if else statement, its syntax is IFERROR(value,value _ if _ error), this simply means that if the the error is equal to value, value is returned if not, the next argument is returned.

Having said that, feom the question we are given,

let's substitute the value with vlookup function and add an else argument, it will look exactly this way;

IFERROR(VLOOKUP(),"Invalid Staff ID")// now this will set the message if vlookup cannot find the.

On the other hand using the values given, we will have;

ierror(VLOOKUP(Q2,CBFStaff[[Staff ID]:[Name]],2,FALSE), "Invalid Staff ID")

Answer:

A  C++ program was used in writing a function in merging ( vector a, vector b)   or that will merge the two vectors into one new one and return the merged vector.

Explanation:

Solution

THE CODE:

#include <iostream>

#include <vector>

using namespace std;

vector<int> merge(vector<int> a, vector<int> b) {

   vector<int> vec;

   int i = 0, j = 0;

   while(i < a.size() || j < b.size()) {

       if(i >= a.size() || (j < b.size() && b[j] < a[i])) {

           if(vec.empty() || vec[vec.size()-1] != b[j])

               vec.push_back(b[j]);

           j++;

       } else {

           if(vec.empty() || vec[vec.size()-1] != a[i])

               vec.push_back(a[i]);

           i++;

       }

   }

   return vec;

}

int main() {

   vector<int> v1 = {2, 4, 6, 8};

   vector<int> v2 = {1, 3, 7, 10, 13};

   vector<int> vec = merge(v1, v2);

   for(int i = 0; i < vec.size(); ++i) {

       cout << vec[i] << " ";

   }

   cout << endl;

   return 0;

}

Answer:

I believe it is just a task. Since there exists(on windows) the Task Manager application, where you can stop any running task, I think that they are called tasks

Explanation:

In most operating systems, a running application is typically referred to as a process. A process is an instance of a program that is being executed by the operating system. It represents the execution of a set of instructions and includes the program code, data, and resources required for its execution.

Each process has its own virtual address space, which contains the program's code, variables, and dynamically allocated memory. The operating system manages and schedules these processes, allocating system resources such as CPU time, memory, and input/output devices to ensure their proper execution.

The operating system provides various mechanisms to manage processes, such as process creation, termination, scheduling, and inter-process communication.

Learn more about operating systems here:

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Answer:

t

Explanation: