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katovenus [111]
3 years ago
9

Select the correct answer. Which is the bottom-most layer in the OSI model?

Computers and Technology
1 answer:
tatiyna3 years ago
3 0

Answer:

B.Physical layer

Explanation:

I am sure.Hope it helps

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Rainbow [258]
我沒有看到問題?如果您可以在此問題上發布問題,那將非常有幫助!
6 0
2 years ago
Input 10 integers and display the following:
LekaFEV [45]

Answer:

// code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;

   int largest=INT_MIN;

   int smallest=INT_MAX;

   int n;

   cout<<"Enter 10 Integers:";

   // read 10 Integers

   for(int a=0;a<10;a++)

   {

       cin>>n;

       // find largest

       if(n>largest)

       largest=n;

       // find smallest

       if(n<smallest)

       smallest=n;

       // if input is even

       if(n%2==0)

       {  

           // sum of even

           sum_even+=n;

           // even count

           eve_count++;

       }

       else

       {

           // sum of odd    

          sum_odd+=n;

          // odd count

          odd_count++;

       }

   }

   

   // print sum of even

   cout<<"Sum of all even numbers is: "<<sum_even<<endl;

   // print sum of odd

   cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;

   // print largest

   cout<<"largest Integer is: "<<largest<<endl;

   // print smallest

   cout<<"smallest Integer is: "<<smallest<<endl;

   // print even count

   cout<<"count of even number is: "<<eve_count<<endl;

   // print odd cout

   cout<<"count of odd number is: "<<odd_count<<endl;

return 0;

}

Explanation:

Read an integer from user.If the input is greater that largest then update the  largest.If the input is smaller than smallest then update the smallest.Then check  if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.

Output:

Enter 10 Integers:1 3 4  2 10 11 12 44 5 20                                                                                

Sum of all even numbers is: 92                                                                                            

Sum of all odd numbers is: 20                                                                                              

largest Integer is: 44                                                                                                    

smallest Integer is: 1                                                                                                    

count of even number is: 6                                                                                                

count of odd number is: 4

3 0
3 years ago
Write a program that dynamically allocates an array large enough to hold a user-defined number of test scores. Once all the scor
Anastasy [175]

Answer:

// Program written in C++

// Comments are used to explain some lines

#include <iostream>

#include <iomanip>

using namespace std;

// Functions

void getData(double *, int);

void selectionSort(double *, int);

double getAverage(double *, int);

void displayData(double *, int, double);

int main() //Main Method

{

double *ToTest, // To dynamically allocate an array

Average; // To hold the average of the scores

int Scores; // To hold number of scores

// Get number of scores

cout << "Number of average to find? ";

cin >> Scores;

// Allocate an array to number of scores

ToTest = new double[Scores];

getData(ToTest, Scores);

selectionSort(ToTest, Scores);

Average = getAverage(ToTest, Scores);

printData(ToTest, Scores, Average);

return 0;

}

//Get Data

void getData(double *ToTest, int Scores)

{

cout << "Enter each scores.\n";

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

{

do

{

cout << "Score #" << (i + 1) << ": ";

cin >> *(ToTest + i);

if (*(ToTest + i) < 0)

{

cout << "Scores must be greater than 0.\n"

<< "Re-enter ";

}

} while (*(Test + i) < 0);

}

}

// Selection Sort

void selectionSort(double *ToTest, int Scores)

{

int startscan, minIndex;

double minValue;

for (startscan = 0; startscan < (Scores - 1); startscan++)

{

minIndex = startscan;

minValue = *(ToTest + startscan);

for (int i = startscan + 1; i < Scores; i++)

{

if (*(ToTest + i) < minValue)

{

minValue = *(Test + i);

minIndex = i;

}

}

*(ToTest + minIndex) = *(ToTest + startscan);

*(ToTest + startscan) = minValue;

}

}

// Calculate Average

double getAverage(double *ToTest, int Scores)

{

double Total;

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

{

Total += *(ToTest + i);

}

return Total / Scores;

}

// Print Data

void printData(double *ToTest, int Scores, double Avg)

{

cout << "\n Test scores\n";

cout << "Number of scores: " << Scores << endl;

cout << "Scores in ascending-order:\n";

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

{

cout << "#" << (i + 1) << ": " << *(ToTest + i) << endl;

}

cout << fixed << showpoint << setprecision(2);

cout << "Average score: " << Avg << endl;

}

6 0
3 years ago
What are some ways to accomplish full-duplex (FDX) digital communications on guided medium at the physical layer (OSI Layer 1)?
labwork [276]

Answer:

d. All of the above

Explanation:

Physical layer is the lowest layer of the OSI reference model that deals with the setup of physical connection to the network and with transmission and reception of signals.

Full duplex (FDx) is a bidirectional type of communication system where two end nodes send and receive data signals at the same time, and a single carrier is simultaneously used for dual communication.

To send or receive data, the participating nodes in a Full Duplex system do not have to wait for a free carrier/medium. It employs a medium that has at least two internal channels - one for sending and one for receiving.

Based on the above explanations, options A, B and C are valid and are correct ways to accomplish full-duplex (FDX) digital communications on guided medium at the physical layer (OSI Layer 1). There all of them are correct.

6 0
2 years ago
How many times do you need to click the Format Painter button to apply copied formats to multiple paragraphs one right after the
const2013 [10]

The answer is (A) Twice

Format painter is that icon tool that looks like a paintbrush that is used to pick up formatting of an existing text and then paints it on the next text you select. If you wish to apply the formatting to more than one element, you should double-click the format painter and press esc key to deactivate.


5 0
3 years ago
Read 2 more answers
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