To answer this question, you need to understand the relationship between JSP (JavaServer Pages) and servlets.

Option A) Servlets are built on JSP semantics and all servlets are compiled to JSP pages for runtime usage. This option is incorrect because servlets are not built on JSP semantics. Servlets and JSPs are separate technologies and have different semantics.

Option B) JSPs are built on servlet semantics and all JSPs are compiled to servlets for runtime usage. This option is correct. JSPs are built on servlet semantics, meaning that JSPs are translated into servlets before being executed. During runtime, the JSP pages are compiled into servlets by the JSP container, which is then used to process the requests and generate the responses.

Option C) JSP and servlets are unrelated technologies. This option is incorrect. JSP and servlets are related technologies and are often used together in Java web applications. JSPs are a higher-level abstraction that allows developers to embed Java code within HTML-like pages, while servlets are Java classes that handle the processing of HTTP requests and responses.

Option D) None of the Above. This option is incorrect because Option B is the correct answer.

The correct answer is B. JSPs are built on servlet semantics, and all JSPs are compiled to servlets for runtime usage.

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

Option A) The running application crashes - This option is incorrect because if a Runtime Exception is thrown in the finalize method, it does not necessarily mean that the entire application will crash. The finalize method is called by the garbage collector before an object is garbage collected, and any exception thrown in this method does not directly affect the running application.

Option B) The exception is simply ignored and the object is garbage collected - This option is correct. When a Runtime Exception is thrown in the finalize method, it is simply ignored by the garbage collector. The object will still be eligible for garbage collection and will be collected at some point in the future.

Option C) The exception is simply ignored, but the object is not garbage collected - This option is incorrect. Regardless of whether an exception is thrown in the finalize method or not, the object will still be eligible for garbage collection. The exception does not prevent the object from being garbage collected.

Option D) If a Runtime Exception is thrown in the finalize method - This option is incomplete and does not provide a complete statement.

The correct answer is B) The exception is simply ignored and the object is garbage collected. This option is correct because any exception thrown in the finalize method is ignored and does not prevent the object from being garbage collected.

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

Option A) Arrays in Java are essentially objects - This option is correct because arrays in Java are indeed objects. In Java, arrays are a special type of object that can hold a fixed number of elements of the same type.

Option B) It is not possible to assign one array to another. Individual elements of the array can, however, be assigned - This option is incorrect. In Java, it is possible to assign one array to another using the assignment operator. For example: int[] array1 = {1, 2, 3}; int[] array2 = array1;. This assigns array1 to array2, meaning both arrays now refer to the same memory location.

Option C) Array elements are indexed from 1 to the size of the array - This option is incorrect. In Java, array elements are indexed from 0 to size-1. The first element of an array is accessed using the index 0, not 1.

Option D) If a method tries to access an array element beyond its range, a compile warning is generated - This option is incorrect. In Java, if a method tries to access an array element beyond its range, it will not generate a compile warning. Instead, it will throw an ArrayIndexOutOfBoundsException at runtime.

The correct answer is A) Arrays in Java are essentially objects. This option is correct because arrays in Java are objects.

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To answer this question, we need to understand how Oracle manages the storage of segments.

When a user creates an object (such as a table or index) without specifying a TABLESPACE clause, Oracle will store the segment in the default tablespace for the user.

Option A) System tablespace - This option is incorrect because the system tablespace is used to store the data dictionary and other system-related objects, not user-created segments.

Option B) Users tablespace - This option is incorrect because there is no specific "users" tablespace in Oracle. Each user has their own default tablespace, which is where the segment would be stored.

Option C) Default tablespace for the user - This option is correct because Oracle will store the segment in the default tablespace for the user who created the object.

Option D) Default tablespace for system - This option is incorrect because there is no specific "default tablespace for system" in Oracle. The default tablespace is specific to each user.

The correct answer is Option C) Default tablespace for the user. This option is correct because Oracle will store the segment in the default tablespace for the user who created the object.

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