Challenge 2: Variadic Average — Possible Solution
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def average(*args):
    return sum(args) / len(args)

print(average(4, 8))
print(average(1, 2, 3, 4, 5))

Output:
6.0
3.0

WHY THIS WORKS AS AN ANSWER
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*args in the function signature reuses this chapter's own *args
example directly — it collects however many positional arguments the
function is called with into a single tuple named args, regardless of
whether that's 2 numbers or 5.

sum(args) adds up every number in that tuple, exactly like the
chapter's total(*args) example. len(args) then counts how many
numbers were passed in — reusing len() from earlier chapters, but
applied to the args tuple itself rather than a list or string.

Dividing sum(args) by len(args) computes the average generically for
any number of arguments, which is the entire reason *args is useful
here — a fixed-parameter function like average(a, b) could only ever
average exactly two numbers, but average(*args) works whether it's
called with 2 numbers, 5 numbers, or any other count.