# 1 Analyze the following code: int number; Keyboard k; k = new Keyboard( ); Sys
ID: 3569272 • Letter: #
Question
# 1 Analyze the following code:
int number;
Keyboard k;
k = new Keyboard( );
System.out.println("Enter a number:");
number = k.readInt( );
if (number <= 0)
{
System.out.println(number);
}
A. The if statement is wrong because it does not have the else clause;
or D. number entered from the keyboard cannot be negative.
# 2 Analyze the following code:
int number;
Keyboard k;
k = new Keyboard( );
System.out.println("Enter a number:");
number = k.readInt( );
if (number <= 0)
{
System.out.println(number);
}
System.out.println(number);
number is printed out twice if number is positive;
# 3 Which of the following is equivalent to (x > 1).
Consider the following code fragment when answering questions 4
B. If number is zero, number is displayed;Explanation / Answer
1. B. If number is zero, number is displayed; because condition is number is less than or equal to 0.
2. C. number is printed out twice if number is negative; Once inside the loop condition is number is less than or equal to 0
and once outside the loop.
3. !(x <= 1) because boundary condition x = 1 also gives the correct result.
4. C. an object of class String.
5. a variable because it's of primitive data type.
6. B. a class. Keyboard is a class and k is the object.
7. C.a String value
8. String
4 becasue it has 4 parameters.
9. 11 300
10. a -> String
b -> Boolean
c -> int
z -> String
11. done -> String
n -> int
y -> String
k -> Keyboard
z -> String
12.
if( a > b){
System.out.println(
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