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Ilya [14]
2 years ago
15

A(n) ________ is published by the receiver and used by the sender to encrypt messages.

Computers and Technology
1 answer:
____ [38]2 years ago
7 0
The answer is "Key".  A public key and sometimes a private key are used by the sender and receiver to encrypt and decrypt messages.
You might be interested in
Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To
SVETLANKA909090 [29]

Answer:

<h2>Function 1:</h2>

#include <stdio.h> //for using input output functions

// start of the function PrintPopcornTime body having integer variable //bagOunces as parameter

void PrintPopcornTime(int bagOunces){

if (bagOunces < 3){ //if value of bagOunces is less than 3

 printf("Too small"); //displays Too small message in output

 printf("\n"); } //prints a new line

//the following else if part will execute when the above IF condition evaluates to //false and the value of bagOunces is greater than 10

else if (bagOunces > 10){

    printf("Too large"); //displays the message:  Too large in output

    printf("\n"); //prints a new line }

/*the following else  part will execute when the above If and else if conditions evaluate to false and the value of bagOunces is neither less than 3 nor greater than 10 */

else {

/* The following three commented statements can be used to store the value of bagOunces * 6 into result variable and then print statement to print the value of result. The other option is to use one print statement printf("%d",bagOunces * 6) instead */

    //int result;

    //result = bagOunces * 6;

    //printf("%d",result);

 printf("%d",bagOunces * 6);  /multiplies value of bagOunces  to 6

 printf(" seconds");

// seconds is followed with the value of bagOunces * 6

 printf("\n"); }} //prints a new line

int main(){ //start of main() function body

int userOunces; //declares integer variable userOunces

scanf("%d", &userOunces); //reads input value of userOunces

PrintPopcornTime(userOunces);

//calls PrintPopcornTime function passing the value in userOunces

return 0; }

Explanation:

<h2>Function 2:  </h2>

#include <stdio.h> //header file to use input output functions

// start of the function PrintShampooInstructions body having integer variable numCycles as parameter

void PrintShampooInstructions(int numCycles){

if(numCycles < 1){

//if conditions checks value of numCycles is less than 1 or not

printf("Too few."); //prints Too few in output if the above condition is true

printf("\n"); } //prints a new line

//else if part is executed when the if condition is false and else if  checks //value of numCycles is greater than 4 or not

else if(numCycles > 4){

//prints Too many in output if the above condition is true

printf("Too many.");

printf("\n"); } //prints a new line

//else part is executed when the if and else if conditions are false

else{

//prints "N: Lather and rinse." numCycles times, where N is the cycle //number, followed by Done

for(int N = 1; N <= numCycles; N++){

printf("%d",N);

printf(": Lather and rinse. \n");}

printf("Done.");

printf("\n");} }

int main() //start of the main() function body

{    int userCycles; //declares integer variable userCycles

   scanf("%d", &userCycles); //reads the input value into userCycles

   PrintShampooInstructions(userCycles);

//calls PrintShampooInstructions function passing the value in userCycles

   return 0;}

I will explain the for loop used in PrintShampooInstructions() function. The loop has a variableN  which is initialized to 1. The loop checks if the value of N is less than or equal to the value of numCycles. Lets say the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 1<2. So the program control enters the body of loop. The loop body has following statements. printf("%d",N); prints the value of N followed by

printf(": Lather and rinse. \n"); which is followed by printf("Done.");

So at first iteration:

printf("%d",N); prints 1 as the value of N is 1

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

1: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 2.

Now at second iteration:

The loop checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 2=2. So the program control enters the body of loop.

printf("Done."); prints Done after the above two lines.

printf("%d",N); prints 2 as the value of N is 2

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

2: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 3.

The loop again checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to false as N<numCycles  which means 3>2. So the loop breaks.

Now the next statement is:

printf("Done."); which prints Done on the screen.

So as a whole the following output is displayed on the screen:

1: Lather and rinse.

2: Lather and rinse.

Done.

The programs along with their outputs are attached.

6 0
2 years ago
You cannot change data directly in the PivotTable. Instead, you must edit the Excel table, and then ____, or update, the PivotTa
Fudgin [204]

Answer:

d. refresh

Explanation:

The Excel software does not allow a user to change values/data directly in the Pivottable. You will get an error message like "Cannot change this part of a PivotTable report" when you try to type data directly.  You will have to edit the Excel Table and then refresh to reflect the updated data.

5 0
2 years ago
Read 2 more answers
Write a while loop that prints userNum divided by 2 (integer division) until reaching 1. Follow each number by a space. Example
jasenka [17]

Answer:

public class Main

{

public static void main(String[] args) {

    int userNum = 40;

    while(userNum > 1){

        userNum /= 2;

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

    }

}

}

Explanation:

*The code is in Java.

Initialize the userNum

Create a while loop that iterates while userNum is greater than 1. Inside the loop, divide the userNum by 2 and set it as userNum (same as typing userNum = userNum / 2;). Print the userNum

Basically, this loop will iterate until userNum becomes 1. It will keep dividing the userNum by 2 and print this value.

For the values that are smaller than 1 or even for 1, the program outputs nothing (Since the value is not greater than 1, the loop will not be executed).

4 0
2 years ago
_____ is the physical link between a network and a workstation. a. The session layer b. An html address c. An adapter card d. A
AveGali [126]

Answer:

(d) Network topology

Explanation:

The physical or logical layout of all connecting devices including their connecting materials (such as cables) of a network is called the topology of the network. It is also the systematic and schematic arrangement of the devices (such as printers, scanners, computers, routers, bridges) that make up a network and their communication media. Examples of these communication media are twisted pair cable, optical fiber cable e.t.c

Types of topology (especially in a LAN - Local Network Area) are;

i. bus topology

ii. ring topology

iii. mesh topology

iv. star topology

4 0
2 years ago
Read 2 more answers
int) You are the head of a division of a big Silicon Valley company and have assigned one of your engineers, Jim, the job of dev
sergeinik [125]

Answer:

Correct option is E

Explanation:

a) 2n^2+2^n operations are required for a text with n words

Thus, number of operations for a text with n=10 words is 2\cdot 10^2+2^{10}=1224 operation

Each operation takes one nanosecond, so we need 1224 nanoseconds for Jim's algorithm

b) If n=50, number of operations required is 2\cdot 50^2+2^{50}\approx 1.12589990681\times 10^{15}

To amount of times required is 1.12589990681\times 10^{15} nanoseconds which is

1125899.90685 seconds (we divided by 10^{9}

As 1$day$=24$hours$=24\times 60$minutes$=24\times 60\times 60$seconds$

The time in seconds, our algortihm runs is \frac{1125899.90685}{24\cdot 60\cdot 60}=13.0312 days

Number of days is {\color{Red} 13.0312}

c) In this case, computing order of number of years is more important than number of years itself

We note that n=100 so that 2(100)^2+2^{100}\approx 1.267650600210\times 10^{30} operation (=time in nanosecond)

Which is 1.267650600210\times 10^{21} seconds

So that the time required is 1.4671881947\times 10^{16} days

Each year comprises of 365 days so the number of years it takes is

\frac{1.4671881947\times 10^{16}}{365}=4.0197\times 10^{13} years

That is, 40.197\times 10^{12}=$Slightly more than $40$ trillion years$

4 0
3 years ago
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