Which types of scenarios would the NETWORKDAYS function help calculate? Check all that apply.
1 answer:
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Answer:
Explanation:
Since the array is not provided, I created a Python function that takes in the array and loops through it counting all of the words that are longer than 5. Then it returns the variable longer_than_five. To test this function I created an array of words based on the synapse of Pride and Prejudice. The output can be seen in the attached picture below.
def countWords(p_and_p_words):
longer_than_five = 0
for word in p_and_p_words:
if len(word) > 5:
longer_than_five += 1
return longer_than_five
The recursive function would work like this: the n-th odd number is 2n-1. With each iteration, we return the sum of 2n-1 and the sum of the first n-1 odd numbers. The break case is when we have the sum of the first odd number, which is 1, and we return 1.
int recursiveOddSum(int n) {
if(2n-1==1) return 1;
return (2n-1) + recursiveOddSum(n-1);
}
To prove the correctness of this algorithm by induction, we start from the base case as usual:
by definition of the break case, and 1 is indeed the sum of the first odd number (it is a degenerate sum of only one term).
Now we can assume that returns indeed the sum of the first n-1 odd numbers, and we have to proof that
returns the sum of the first n odd numbers. By the recursive logic, we have
and by induction, is the sum of the first n-1 odd numbers, and 2n-1 is the n-th odd number. So,
is the sum of the first n odd numbers, as required: