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antoniya [11.8K]
3 years ago
5

Which term refers to the capability of a switch to copy data from any or all physical ports on a switch to a single physical por

t?
Computers and Technology
1 answer:
pantera1 [17]3 years ago
3 0

Answer:

The correct answer to the following question will be "Port mirroring".

Explanation:

A network traffic analysis tool, recognized as the Port Mirroring.

  • Also known as the Switch port analyzer.
  • The switch transfers a replica among all networking transmissions viewed on one port (or even a whole VLAN) to the next port, wherein the packet could be evaluated, through port mirroring disabled.

Therefore, Port mirroring is the right answer.

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__________ Refers to the difference between the original analog data and the approximation of that data using the techniques for
otez555 [7]

Answer:

Quantizing error

Explanation:

In digitisation of analog signal refers to the rounding off of values which are approximately equal to the analog signals. The method of sampling chooses a few point on the analog signal and then these points are joined to round off the value to a near stabilised value. such process is known as Quantization. For any system, during its functioning, there is always a difference in the values of its input and output. The processing of the system results in an error, which is the difference of those values.

The difference between an input value and its quantized value is called a Quantization Error. A Quantizer is a logarithmic function that performs Quantization rounding off the value. An analog-to-digital converter (ADC) works as a quantizer.

8 0
3 years ago
PLEASE HELP, NEED THIS TURNED IN IN 20 MINUTES!!!!!! WILL GIVE BRAINLIEST!!!
ratelena [41]

Answer:

We cant help with that. Maybe give me more information?

5 0
3 years ago
Write a C program that stores N random numbers in an array named rA[ ] and determine the largest, smallest, and average values o
Aliun [14]

Answer:

// program in C.

// headers

#include <stdio.h>

// headers

#include <stdlib.h>

// headers

#include <limits.h>

// main function

int main(int argc, char** argv)

{

   // read value of N from command line arguments

   int N=atoi(argv[1]);

   // variables

   int max=INT_MIN;

   // variable

   int min=INT_MAX;

int rA[N],i;

// fill the array with random number

for(i=0;i<N;i++)

{

    // generate random number from -50 to +50

    rA[i]=rand()%101-50;

    // find the Maximum

    if(rA[i]>max)

    max=rA[i];

    // find the Minimum

    if(rA[i]<min)

    min=rA[i];

}

// print values of array

printf("values of the array are:");

for(i=0;i<N;i++)

{

    printf("%d ",rA[i]);

}

// print Maximum

printf("\nMaximum value is:%d",max);

// print Minimum

printf("\nMinimum value is:%d",min);

return 0;

}

Explanation:

Read value of N from command line.Then create an array of size N.Fill the array with random number from -50 to +50.Then find the Maximum of all the elements and  assign it to variable "max" and find Minimum from all and assign it to variable "min". then print all the elements of the array, Minimum and Maximum.

Output:

command line argument=10

values of the array are:-18 -18 4 -38 2 6 -42 -20 -6 44 -6 -11 15 -31 1                                                    

Maximum value is:44                                                                                                        

Minimum value is:-42

4 0
3 years ago
One of the most notable impacts of IT on business is improved
Katena32 [7]

One of the most notable impacts of IT on business is improved  a lot of companies as it helps to speed up the way that business is been conducted.

<h3>What are the impacts of IT in business?</h3>

The Impact of IT Industry on Business is that businesses are said to be able to have a lot of access to a lot of tools that aid them solve and handle complex issues.

Hence, One of the most notable impacts of IT on business is improved  a lot of companies as it helps to speed up the way that business is been conducted.

Learn more about information technology from

brainly.com/question/1162014

#SPJ1

4 0
2 years ago
Read 2 more answers
Consider the problem of making change for n cents using the fewest number of coins. Assume that each coins value is an integer.
Oksana_A [137]

Answer:

There are two algorithms in which apply different optimal solutions.

They are: A Dynamic and Naive recursive programs

Explanation:

// A Naive recursive C++ program to find minimum of coins  

// to make a given change V  

#include<bits/stdc++.h>  

using namespace std;  

// m is size of coins array (number of different coins)

int minCoins(int coins[], int m, int V)  

{  

// base case  

if (V == 0) return 0;  

// Initialize result

int res = INT_MAX;  

// Try every coin that has smaller value than V  

for (int i=0; i<m; i++)  

{  

if (coins[i] <= V)  

{  

 int sub_res = minCoins(coins, m, V-coins[i]);  

 // Check for INT_MAX to avoid overflow and see if  

 // result can minimized

 if (sub_res != INT_MAX && sub_res + 1 < res)  

  res = sub_res + 1;  

}  

}  

return res;  

}  

// Driver program to test above function  

int main()  

{  

int coins[] = {9, 6, 5, 1};  

int m = sizeof(coins)/sizeof(coins[0]);  

int V = 11;  

cout << "Minimum coins required is "

 << minCoins(coins, m, V);  

return 0;  

}  

.........................................

// A Dynamic Programming based C++ program to find minimum of coins  

// to make a given change V  

#include<bits/stdc++.h>  

using namespace std;  

// m is size of coins array (number of different coins)  

int minCoins(int coins[], int m, int V)  

{  

// table[i] will be storing the minimum number of coins  

// required for i value. So table[V] will have result  

int table[V+1];  

// Base case (If given value V is 0)  

table[0] = 0;  

// Initialize all table values as Infinite  

for (int i=1; i<=V; i++)  

 table[i] = INT_MAX;  

// Compute minimum coins required for all  

// values from 1 to V  

for (int i=1; i<=V; i++)  

{  

 // Go through all coins smaller than i  

 for (int j=0; j<m; j++)  

 if (coins[j] <= i)  

 {  

  int sub_res = table[i-coins[j]];  

  if (sub_res != INT_MAX && sub_res + 1 < table[i])  

   table[i] = sub_res + 1;  

 }  

}  

return table[V];  

}  

// Driver program to test above function  

int main()  

{  

int coins[] = {9, 6, 5, 1};  

int m = sizeof(coins)/sizeof(coins[0]);  

int V = 11;  

cout << "Minimum coins required is "

 << minCoins(coins, m, V);  

return 0;  

}  

5 0
3 years ago
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