Answer:
Drill and Practice
Explanation:
<em>Drill and Practice</em> is a type of Computer-based Instruction (CBI) program. In <em>Drill and Practice</em>, answered questions are given immediate feedback. These problems or exercises are structured and answered on the program to provide instant feedback to the person taking the test.
<em>For instance</em>, when going through an evaluation test but you are required to provide the correct answer before moving on. Once a question is answered, the program will indicate whether correct or wrong, if the question is correct, move to the next question but if wrong you start again. This is a typical example of Drill and practice.
This is exactly what is seen in our scenario in the question.
Answer:
Preventive control is not a classification that can be applied to security controls.
Explanation:
At the most basic level, we should protect resources and assets that are of value to an organization by mitigating the risk to those assets and resources. Security controls include any type of policy, technique, procedure, or solution that help mitigate risks. There are models that define security control objectives and are classified by control type and by function
By Control Type
- Physical control: Anything that can be touched and can be used to detect and prevent unauthorized access from adversaries and threat actors. Examples include CCTV, electric fences, Biometrics
- Technical controls: Examples include firewalls, AMSI solutions, IDSs and IPSs that help protect an organization’s resources and assets. They can be both hardware and software solutions.
- Administrative controls: These are the overall design of the protocols or guidelines that define business needs based upon the organization’s security goals that help implement a secure environment. Things like staff awareness and training are among the examples of administrative controls.
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Answer:
#include <iostream>
using namespace std;
class ProblemSolution {
private:
int num1, num2;
public:
ProblemSolution(int n1, int n2) {
num1 = n1;
num2 = n2;
}
int calculateSum() {
int sum = 0;
sum = num1 + num2;
return sum;
}
void printSum() {
// calculateSum will return sum value that will be printed here
cout <<"Sum = "<< calculateSum();
}
~ProblemSolution() {
cout << "\nDestructor is called " << endl;
};
};
int main() {
int a, b;
cout << "Enter a: ";
cin >> a;
cout << "Enter b: ";
cin >> b;
// Initiallizing object pointer of type ProblemSolution
ProblemSolution *objPtr = new ProblemSolution(a,b);
// printing Sum
objPtr->printSum();
// delete objPtr to relaease heap memory :important
delete objPtr;
return 0;
}
Explanation:
we will initialize a pointer "objPtr" and initallize the constructor by passing 2 values a and b followed by the keyword "new". the keyword "new" allocates memory in the heap. we can access class member functions using arrow "->". it is important to delete objPtr at the end of the program so that the heap memory can be freed to avoid memory leakage problems.
Answer:
RAM Data can be changed while ROM its content can be read but cannot be changed during normal computer operations