Answer:
Method:- This is related to hackers technique and way of accessing to copy other data. It also includes the skill, knowledge, tools and other things with which to be able to pull off the attack.
Opportunity:- this is related to how a user gives way to access to the hackers. It includes the time, the chance, and access to accomplish the attack. 
Motive:- This may relate to the hacker to destroy the reputation of another or for money. It is reason to want to perform this attack against this system
 
        
             
        
        
        
Answer:
D. All of these choices are correct
Explanation:
Electronic invoice presentment and payment, otherwise known as EIPP are business-to-business systems that combine e-invoicing and e-payment processes for the purpose of sending invoices to customers via Web portal or via a secure network by the use of a third party service provider. They are also used for receiving payer-initiated, third-party-processed, and ACH-network-settled wire transfers.
 
        
                    
             
        
        
        
So that they can lower the I2R losses
        
             
        
        
        
Answer:
C. can boost server utilization rates to 70% or higher.
Explanation:
Virtualization - 
It is the process of running virtual instance of the computer in the abstracted layer from the very actual hardware , is known as virtualization . 
In this process multiple operating system can be operated on the single computer setup , it is useful for increasing the speed of the program.
Hence, from the question, 
The correct statement with respect to virtualization is C. can boost server utilization rates to 70% or higher . 
 
        
             
        
        
        
Answer:
void printC()  
{  
    int i, j;  
    for (i = 0; i < 4; i++) //i indicate row number. Here we have 5 rows
        {  
          printf("C"); //print C for every row  
          for (j = 0; j < 6; j++) //j indicate column number. Here we have 7 Rows
          {  
            if (i == 0 || i == 4) //For first and last row  
                printf("C"); //print 'CCCCCCC'
           else if (i = 1|| i= 3) //for Second forth row  
                 printf("C        +      +"); //print 'C    +    +'
           else if (i = 2) For second row  
                 printf("C       +++++"); //print 'C +++++'
            else
                continue; //to jump to next iteration
          }  
          printf("\n"); // print in next line
 }  
}