To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) 0.0 - This option is incorrect because the variable f is declared and initialized inside the try block and is not accessible outside the try block. Therefore, it cannot be printed in the finally block.

Option B) Compilation fails - This option is correct. The code will not compile because the variable f is declared inside the try block and is not accessible outside the try block. Therefore, when trying to print f in the finally block, a compilation error will occur.

Option C) A ParseException is thrown by the parse method at runtime - This option is incorrect because the code does not handle a ParseException. Instead, it catches a NumberFormatException in the catch block.

Option D) A NumberFormatException is thrown by the parse method at runtime - This option is incorrect because the code does not throw a NumberFormatException. Instead, it catches a NumberFormatException in the catch block.

The correct answer is B. Compilation fails because the variable f is declared inside the try block and is not accessible outside the try block.

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• Wrong explanation

To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) Line 35 will not compile. In Java, when declaring a variable, the variable name must start with a letter, underscore, or a dollar sign. In this case, the variable name starts with a pound sign (#), which is not a valid character. Therefore, Line 35 will not compile.

Option B) Line 36 will not compile. In Java, the variable type should come before the variable name. In this case, the variable type is specified as "int$", which is not a valid type declaration. Therefore, Line 36 will not compile.

Option C) Line 37 will not compile. Line 37 declares a variable named "Double_height" of type Double. The variable name starts with an uppercase letter, which is allowed in Java, but it is considered a bad practice. However, the code will compile and run without any issues. Therefore, Line 37 will compile.

Option D) Line 38 will not compile. In Java, the variable name cannot contain special characters like "~". Therefore, Line 38 will not compile.

The correct answer is A and D. Line 35 will not compile because the variable name starts with a pound sign (#), and Line 38 will not compile because the variable name contains a special character (~).

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• Wrong explanation

To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) Foo.beta() is a valid invocation of beta() - This option is incorrect because the beta() method is an instance method and cannot be invoked using the class name. It should be invoked on an instance of the Foo class.

Option B) Foo.alpha() is a valid invocation of alpha() - This option is correct. The alpha() method is a static method, and static methods can be invoked using the class name.

Option C) Method beta() can directly call method alpha() - This option is correct. Since both beta() and alpha() are methods of the same class, the beta() method can directly call the alpha() method without any issues.

Option D) Method alpha() can directly call method beta() - This option is incorrect. The alpha() method is a static method, while the beta() method is an instance method. Static methods cannot directly call instance methods without creating an instance of the class.

The correct answers are B and C.

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• Wrong explanation

To answer this question, we need to understand the concept of method overriding and access modifiers in Java.

In Java, when a class extends another class, it inherits all the methods from the superclass. However, if the subclass needs to provide a different implementation of a method inherited from the superclass, it can override the method.

In this scenario, class Two extends class One, which means that it inherits the method foo() from class One. To correctly complete class Two, we need to override the foo() method with the correct access modifier.

Let's go through each option to understand why it is correct or incorrect:

Option A) int foo() { /* more code here */ } This option is incorrect because it changes the return type of the foo() method, which is not allowed when overriding a method.

Option B) void foo() { /* more code here */ } This option is correct because it overrides the foo() method with the same return type and access modifier.

Option C) public void foo() { /* more code here */ } This option is correct because it overrides the foo() method with the same return type and a broader access modifier (public is broader than the default access modifier).

Option D) protected void foo() { /* more code here */ } This option is correct because it overrides the foo() method with the same return type and a broader access modifier (protected is broader than the default access modifier).

The correct answers are B), C), and D). These options correctly complete class Two by overriding the foo() method with the same return type and appropriate access modifiers.

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• Wrong explanation