The instructions that he microprocessor can execute each
second if the assembly line is present will be depending on the workload and
the architecture’s core because it is all depending on the speed of the CPU and
the multiplier that it acquires.
Let's say for example that the business is taking in $2000 of revenue. That is the amount that the business collected for it's services - like for fixing the computer. What if though it costs $500 for the equipment (that's an expense). Now they only made $1500. Now the customer complains and says that the computer isn't fixed properly so the company sends out a techie for 2 additional hours. They need to pay their employee (another expense). Now the $1500 is down to $1400. They would have utilities to keep their lights on and insurance and many other expenses.
Your profit looks like this:
Profit = Revenue - Expenses
Answer:
Following are the response to the given question:
Explanation:
The glamorous objective is to examine the items (as being the most valuable and "cheapest" items are chosen) while no item is selectable - in other words, the loading can be reached.
Assume that such a strategy also isn't optimum, this is that there is the set of items not including one of the selfish strategy items (say, i-th item), but instead a heavy, less valuable item j, with j > i and is optimal.
As 
, the i-th item may be substituted by the j-th item, as well as the overall load is still sustainable. Moreover, because 
and this strategy is better, our total profit has dropped. Contradiction.
Answer:
Following is the program in C++ Language
#include <iostream> // header file
using namespace std; // namespace std
int main() // main method
{
int n; // variable declaration
cout<<" Please enter the number :";
cin>>n; // Read the number
if(n>0) // check the condition when number is positive
{
cout<<n<<endl<<"The number is Positive"; // Display number
}
else if(n<0) // check the condition when number is Negative
{
cout<<n<<endl<<"The number is Negative";// Display number
}
else // check the condition when number is Zero
{
cout<<n<<endl<<"The number is Zero";// Display number
}
return 0;
}
Output:
Please enter the number:
64
The number is Positive
Explanation:
Following are the description of the program
- Declared a variable "n" of int type.
- Read the value of "n" by user.
- Check the condition of positive number by using if block statement .If n>0 it print the number is positive.
- Check the condition of negative number by using else if block statement If n<0 it print the number is negative.
- Finally if both the above condition is fail it print the message " The number is Zero"
