Answer:
Explanation:
Depends on the configuration of the email because there are two protocols POP and IMAP, the most recent protocol is IMAP, we can delete an email and this It moves to a To be Deleted folder, this happens because the email is stored in the server, but with the protocol POP the email is stored in the server and downloaded to the application, if you delete an email, this is deleted in all devices.
Answer: Please find below the answer along with explanation.
Explanation:
For a given communication channel (for instance, a LAN segment using Ethernet ) the Bandwidth refers to the theoretical maximum data rate that the channel can support, for instance, 100 Mbps in a 100Base T network.
The throughput, instead, refers to the actual data rate achieved in a given communications channel, taking into account the different channel impairments.
For instance, in a LAN segment that uses the original Ethernet 802.3 standard (CSMA/CD), a frequent occurrence of collisions can take down the actual data rate from the theoretical 100 Mbps to a very lower figure, i.e., 5 Mbps.
Answer:
The answer is "14".
Explanation:
Let the IP address 
When it borrowed 4 bits
If the borrowed bits are left out then:
The Number of useable host addresses:
Answer:
Explanation:
#include<iostream>
#include<ctime>
#include<bits/stdc++.h>
using namespace std;
double calculate(double arr[], int l)
{
double avg=0.0;
int x;
for(x=0;x<l;x++)
{
avg+=arr[x];
}
avg/=l;
return avg;
}
int biggest(int arr[], int n)
{
int x,idx,big=-1;
for(x=0;x<n;x++)
{
if(arr[x]>big)
{
big=arr[x];
idx=x;
}
}
return idx;
}
int main()
{
vector<pair<int,double> >result;
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
int choice;
cin>>choice;
while(choice!=2)
{
int n,m;
cout<<"Enter N"<<endl;
cin>>n;
cout<<"Enter M"<<endl;
cin>>m;
int c=m;
double running_time[c];
while(c>0)
{
int arr[n];
int x;
for(x=0;x<n;x++)
{
arr[x] = rand();
}
clock_t start = clock();
int pos = biggest(arr,n);
clock_t t_end = clock();
c--;
running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;
}
double avg_running_time = calculate(running_time,m);
result.push_back(make_pair(n,avg_running_time));
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
cin>>choice;
}
for(int x=0;x<result.size();x++)
{
cout<<result[x].first<<" "<<result[x].second<<endl;
}
}
