Answer:
A) True
Explanation:
Java provides collections architecture or framework used to store and manipulate a group of objects or collections.
The collection framework has interfaces which include; Set, Queue, Deque, List, as well as classes which include; Hashset, ArrayList, LinkedList, LinkedHashset, PriorityQueue, Vector and TreeSet.
There are also many methods declared in the collection interface which include; add(), addAll(), remove(), removeAll(),retainAll(), clear(), size(), iterator(), toArray() etc
Answer:
cout << setprecision(2)<< fixed << number;
Explanation:
The above statement returns 12.35 as output
Though, the statement can be split to multiple statements; but the question requires the use of a cout statement.
The statement starts by setting precision to 2 using setprecision(2)
This is immediately followed by the fixed manipulator;
The essence of the fixed manipulator is to ensure that the number returns 2 digits after the decimal point;
Using only setprecision(2) in the cout statement will on return the 2 digits (12) before the decimal point.
The fixed manipulator is then followed by the variable to be printed.
See code snippet below
<em>#include <iostream> </em>
<em>#include <iomanip>
</em>
<em>using namespace std; </em>
<em>int main() </em>
<em>{ </em>
<em> // Initializing the double value</em>
<em> double number = 12.3456; </em>
<em> //Print result</em>
<em> cout << setprecision(2)<< fixed << number; </em>
<em> return 0; </em>
<em>} </em>
<em />
Answer:
See explaination
Explanation:
class Taxicab():
def __init__(self, x, y):
self.x_coordinate = x
self.y_coordinate = y
self.odometer = 0
def get_x_coord(self):
return self.x_coordinate
def get_y_coord(self):
return self.y_coordinate
def get_odometer(self):
return self.odometer
def move_x(self, distance):
self.x_coordinate += distance
# add the absolute distance to odometer
self.odometer += abs(distance)
def move_y(self, distance):
self.y_coordinate += distance
# add the absolute distance to odometer
self.odometer += abs(distance)
cab = Taxicab(5,-8)
cab.move_x(3)
cab.move_y(-4)
cab.move_x(-1)
print(cab.odometer) # will print 8 3+4+1 = 8
Answer:
Option d is the correct answer for the above question.
Explanation:
- The first loop of the program has a second loop and then the statement. In this scenario, the second loop executes for the value of the first loop and the statement executes for the value of the second loop.
- The first loop executes 4 times, Then the second loop or inner loop executes n times for the n iteration of the first loop, for example, 1 time for the first iteration of the first loop, 2 times for the second iteration of the first loop and so on.
- Then the inner loop executes (1+2+3+4) iteration which gives the result 10 iterations.
- The sum initial value is 0 and the "sum++", increase the value of the sum by 1.
- So the value of the sum becomes 10 after completing 10 iterations of the inner for loop.
- Hence the 10 will be the output. So the Option d is the correct answer while the other is not.
Answer:
Following are the solution to the given point:
Explanation:
The solution to this question is defined in the attached in the attached file.
