What is the output of the following variable assignment?x = 75def myfunc():    x = x + 1    print(x) myfunc()print(x)

The code you provided will result in an error. This is because the variable x is defined in the global scope and then again in the local scope of the function myfunc().

When you try to modify x in myfunc() by doing x = x + 1, Python gets confused because it thinks x is a local variable (since it's being assigned a new value in the function), but it hasn't been defined in the function yet.

Here's a step-by-step breakdown:

1. x = 75 assigns the value 75 to x in the global scope.
2. def myfunc(): begins the definition of a function called myfunc.
3. x = x + 1 attempts to assign a new value to x in the local scope of myfunc. However, x hasn't been defined in this scope yet, so Python raises an error.
4. print(x) and myfunc() are never executed because of the error.
5. print(x) in the global scope is also never executed because the program has already stopped due to the error.

To fix this, you can use the global keyword to tell Python that you want to use the global x in myfunc(), like this:

x = 75

def myfunc():
    global x
    x = x + 1
    print(x)

myfunc()
print(x)

This will output:

76
76

This is because myfunc() modifies the global x and then prints it, and then the global x is printed again outside the function.