Answer:
There is a need for some people to see the total transparency of the records that are meant to be shared with a select group of people. There are certain things that can be done for this. The first one is that the sharing rules should be checked. To whom are the details shared with. The next one is the filters that are used.
These can be reported and changed accordingly. Lastly, the whole organization’s defaults can be checked. This is something that can be done when the previous methods that were done did not work that well. Once some changes are done, the people can check if they already work and if the records can be viewed.
Explanation:
Answer:
You can perform the following two steps
Explanation:
- Have the user press the appropriate function key combination to enable the wireless radio and then attempt to connect to the wireless network (since by mistake he could have disabled it).
- Ask the user to turn on the laptop’s airplane mode and attempt to reconnect to the wireless network (this mode basically what it does is disable adapters and activate it will connect the Wi-Fi network).
<u>Client-server</u> implemented a network where hosts are assigned specific roles, such as for file sharing and printing. Other hosts access those resources but do not host services of their own.
<u>Explanation:</u>
The client-server can be utilized on the web just as on a neighborhood (LAN). Instances of customer server frameworks on the web incorporate internet browsers and web servers, FTP customers and servers, and the DNS. Different hosts get to those assets yet don't have administrations of their own. Since it permits arrange permits numerous PCs/gadgets to interface with each other and offer assets.
Answer:
// here is code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int n,no_open=0;
cout<<"enter the number of lockers:";
// read the number of lockers
cin>>n;
// initialize all lockers with 0, 0 for locked and 1 for open
int lock[n]={};
// toggle the locks
// in each pass toggle every ith lock
// if open close it and vice versa
for(int i=1;i<=n;i++)
{
for(int a=0;a<n;a++)
{
if((a+1)%i==0)
{
if(lock[a]==0)
lock[a]=1;
else if(lock[a]==1)
lock[a]=0;
}
}
}
cout<<"After last pass status of all locks:"<<endl;
// print the status of all locks
for(int x=0;x<n;x++)
{
if(lock[x]==0)
{
cout<<"lock "<<x+1<<" is close."<<endl;
}
else if(lock[x]==1)
{
cout<<"lock "<<x+1<<" is open."<<endl;
// count the open locks
no_open++;
}
}
// print the open locks
cout<<"total open locks are :"<<no_open<<endl;
return 0;
}
Explanation:
First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.
Output:
enter the number of lockers:9
After last pass status of all locks:
lock 1 is open.
lock 2 is close.
lock 3 is close.
lock 4 is open.
lock 5 is close.
lock 6 is close.
lock 7 is close.
lock 8 is close.
lock 9 is open.
total open locks are :3
Answer: the answer is c
Explanation:in general, the merges and splits in hierarchical clustering are determined in a greedy manner
