Answer:
INSERT INTO Movie(Title,Rating,ReleaseDate)
VALUES("Raiders of the Lost ArkPG",'PG',DATE '1981-06-15'),
("The Godfaher",'R',DATE '1972-03-24'),
("The Pursuit of Happyness",'PG-13',DATE '2006-12-15');
Explanation:
The SQL statement uses the "INSERT" clause to added data to the movie table. It uses the single insert statement to add multiple movies by separating the movies in a comma and their details in parenthesis.
Answer: B) Shape is a base class, and circle and square are derived classes of Shape.
Explanation:
Shape is a base class because circle and squares are the shapes so these are the derived class of the shape, which is inherited by the shape like circle and square. As, the base class (shape) is the class which are derived from the other classes like circle and square and it facilitates other class which can simplified the code re-usability that is inherited from the base class. Base class is also known as parent class and the super class.
Answer:
Option (c) is the correct answer of this question.
Explanation:
RAM (Random Access Memory) It is a type of computer memory which can also be retrieved and adjusted in either sequence, usually still had to preserve operating memory and bytecode.
It is a type storage device, and maintains the data used when the device is operating.RAM makes file access more frequently compared with certain digital storage types.
<u>For Example</u>:- PCs, tablets, smartphones ,printers etc.
Other options are not related to the given scenario.
Answer:
Big Oh notation is used to asymptotically bound the growth of running time above and below the constant factor.
Big Oh notation is used to describe time complexity, execution time of an algorithm.
Big Oh describes the worst case to describe time complexity.
For the equation; T(N) = 10000*N + 0.00001*N^3.
To calculate first of all discard all th constants.
And therefore; worst case is the O(N^3).
Answer:
C++.
Explanation:
<em>Code snippet.</em>
#include <map>
#include <iterator>
cin<<N;
cout<<endl;
/////////////////////////////////////////////////
map<string, string> contacts;
string name, number;
for (int i = 0; i < N; i++) {
cin<<name;
cin<<number;
cout<<endl;
contacts.insert(pair<string, string> (name, number));
}
/////////////////////////////////////////////////////////////////////
map<string, string>::iterator it = contacts.begin();
while (it != contacts.end()) {
name= it->first;
number = it->second;
cout<<word<<" : "<< count<<endl;
it++;
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////
I have used a C++ data structure or collection called Maps for the solution to the question.
Maps is part of STL in C++. It stores key value pairs as an element. And is perfect for the task at hand.
