<h2>
</h2>
<u>The computer is the combination of hardware and software. Hardware is the physical component of a computer like motherboard, memory devices, monitor, keyboard etc., while software is the set of programs or instructions. Both hardware and software together make the computer system function.</u>
<h2><u>E</u><u>x</u><u>a</u><u>m</u><u>p</u><u>l</u><u>e</u><u>s</u><u> </u><u>:</u><u>-</u><u> </u></h2>
<h3><u>Mainframe Computer</u></h3>
- Mainframe Computer. 20 in Kickin Technology Series: MainFrame Computers. MORE POWEeeer! Movie 1. ...
- Desktop Computer. Movie: 2. Desktop Computers. ...
- Laptop or Notebook Computer. Movie: 3. Laptop Computers. ...
- Palmtop Computer or Personal Digital Assistant (PDA)
Answer:
C++ code explained below
Explanation:
#include<bits/stdc++.h>
#include <iostream>
using namespace std;
int FiboNR(int n)
{
int max=n+1;
int F[max];
F[0]=0;F[1]=1;
for(int i=2;i<=n;i++)
{
F[i]=F[i-1]+F[i-2];
}
return (F[n]);
}
int FiboR(int n)
{
if(n==0||n==1)
return n;
else
return (FiboR(n-1)+FiboR(n-2));
}
int main()
{
long long int i,f;
double t1,t2;
int n[]={1,5,10,15,20,25,30,35,40,45,50,55,60,65,70,75};
cout<<"Fibonacci time analysis ( recursive vs. non-recursive "<<endl;
cout<<"Integer FiboR(seconds) FiboNR(seconds) Fibo-value"<<endl;
for(i=0;i<16;i++)
{
clock_t begin = clock();
f=FiboR(n[i]);
clock_t end = clock();
t1=double(end-begin); // elapsed time in milli secons
begin = clock();
f=FiboNR(n[i]);
end = clock();
t2=double(end-begin);
cout<<n[i]<<" "<<t1*1.0/CLOCKS_PER_SEC <<" "<<t2*1.0/CLOCKS_PER_SEC <<" "<<f<<endl; //elapsed time in seconds
}
return 0;
}
I wore haute you using Brainly.
False. Nothing is lacking in movement.
That was was pretty difficult, had to do some research for that one. So it may not be to correct one.
Answer:
A ) 200 Zombie systems
B ) 200 Zombie systems
Explanation:
maximum uplink data rate : 512 kbps = 512 * 1000 * 8 = 4096000 bits/sec
Determine the number of 512 byte ICMP echo packets = 512 * 8 = 4096 bits
hence the maximum number of 512-byte ICMP echo request (ping) packets a single zombie computer can send per second = 400 packets/sec
i.e. for 512kbps = 400 packets/sec
Determine the number of Zombie systems that the attacker will need to flood a target server
A) For a fast ethernet
A fast ethernet = 100 mbps
The number of Zombie systems needed = 1 zombie system * 200
and 1 zombie system = 512 kbps
therefore for a Fast ethernet the number of zombies systems needed = 200
B) For A Gigabit Ethernet
same as a fast ethernet system i.e. = 200
