Choose the Correct Answer(s)What will be the output after the following statements?def xyz(): a = 56 xyz() print(a)OptionsNameErrora = 56xyz56

Break Down the Problem

1. We need to analyze the code provided and determine what happens when it is executed.
2. The function xyz is defined, and within it, a local variable a is assigned the value 56.
3. After calling the function xyz, there is an attempt to print the value of a.

Relevant Concepts

1. Understand the scope of variables in Python: variables defined inside a function are local to that function and cannot be accessed outside of it.
2. Recognize that an attempt to access a variable that is not in the global scope will raise a NameError.

Analysis and Detail

1. The function xyz() sets a local variable a = 56.
2. After calling xyz(), the variable a defined inside xyz does not exist in the global scope.
3. The line print(a) is outside of the function and thus attempts to access the global variable a, which has not been defined.

Verify and Summarize

Based on the understanding of variable scope in Python, the attempt to print a will result in a NameError since a is not available in the global scope.

Final Answer

NameError