Question: How many hashes (#) will the following code display in the output? 1st = list(range(1, 31, 2)) for r in range(len(1st)): for c in range(len(1st)):

 How many hashes (#) will the following code display in the
output? 1st = list(range(1, 31, 2)) for r in range(len(1st)): for c
in range(len(1st)): if Ist[c] % 2 l= 0: print("#") O a. 45
O b. None of the given options, O c. 15 O d.
225 O e. 105 What is the output of the following code

How many hashes (#) will the following code display in the output? 1st = list(range(1, 31, 2)) for r in range(len(1st)): for c in range(len(1st)): if Ist[c] % 2 l= 0: print("#") O a. 45 O b. None of the given options, O c. 15 O d. 225 O e. 105 What is the output of the following code if the user enters 15 and 42 in two separate lines? x = int(input()) y = int(input()) print(x, y, X, sep="0") O a. No output. The program will produce an error O b. 150428150 O c. 15842015 O d. 0156420150 0 15:42 15 Assuming that data is a correctly created list, then the fact that lists are mutable means that the following statement: data[5] data[-2] + sum(data) O a. Maybe illegal if the list contains integers. O b. Can be executed if and only if the list contains at least six elements. O c is illegal. O d. Is fully correct. O e. Can be executed if and only if the list contains at least five elements. Which of the following is false about the print() function? O a. In the print() function, the default for the sep parameter is a single whitespace. O b. The print() function can take multiple arguments to be printed. The arguments should be separated by commas. O c. By default, the print function ends output with a newline character. O d. The brand function can only take string literals as arguments. The following code snippet def func1(a): return None def func2(a): if a > 2: return func1(a) * func1(a) else: return func1(a * 4) print(func2(4)) O a. Will outputke. b. Will output oc Will cause a runtime error. Od. Will output 16

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