A big/potent virus could eliminate economic statuses/money, maybe bankrupting the company.
Answer: Reusing software device has some economical challenges such as:
- Investment cost in reusing the device is considered as the extra cost.It might require some extra components for the working to become usable
- Requires skilled person who can develop and design the software to be used again.The creating and designing of new software design is comparatively easy but making the system reusable requires someone who has more designing skill who will be highly paid for the work
- Cost of writing and reading of the software can also be considered as the economical challenge as the reused system created is to be studied by some other organization members or sources not familiar with the functioning. .
Answer:
0.01 second ; 0.008 seconds; 800 seconds
Explanation:
Given that:
Sending rate = 1000 bps
Rate of 1000 bps means that data is sent at a rate of 1000 bits per second
Hence, to send out 10 bits
1000 bits = 1 second
10 bits = x
1000x = 10
x = 10 / 1000
x = 0.01 second
2.)
A single character 8 - bits
1000 bits = 1 second
8 bits = (8 / 1000) seconds
= 0.008 seconds
3.)
100,000 characters = (8 * 100,000) = 800,000
1000 bits = 1 second
800,000 bits = (800,000 / 1000)
= 800 seconds
Answer:
Option A. is correct
Explanation:
It now not possible for a country to be "occupied" by an invisible invader that arrives through airwaves and wireless networks. It is almost impossible to block foreign countries’ satellite broadcasts and Internet transmissions inspite of Spy satellites and other communications technology and Global positioning systems.
Option A. is correct
Answer:
// Program is written in C++ Programming Language
// Comments are used for explanatory purpose
// Program starts here
#include<iostream>
using namespace std;
int main()
{
// Declare integer variable n which serves as the quotient.
int n;
// Prompt to enter any number
cout<<"Enter any integer number: ";
cin>>n;
// Check for divisors using the iteration below
for(int I = 1; I<= n; I++)
{
// Check if current digit is a valid divisor
if(n%I == 0)
{
// Print all divisors
cout<<I<<" ";
}
}
return 0;
}
