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1. What is the second line printed? ``

python
for row in range(1, 4):
    cells = []
    for col in range(1, 4):
        cells.append(str(row * col + 1))
    print(",".join(cells))

``

3 5 72,4,63,5,73,6,9

2. What is the third line printed? ``

python
n = 4
for i in range(n):
    for _ in range(n - 1 - i):
        print(" ", end="")
    for j in range(1, i + 2):
        print(j, end="")
    for j in range(i, 0, -1):
        print(j, end="")
    print()

``

123321 12321 1232112321

3. This should print # on three lines, but it prints #, ## and ###. ``

python
line = []
for row in range(1, 4):
    line.append("#")
    print("".join(line))

`` What is the fix?

Change range(1, 4) to range(1, 3)Add line.clear() at the end of each rowMove print("".join(line)) after the loopChange append to insert(0, "#")

4. This code should print a top row of 5 characters for n = 5, but the top row has 6. ``

python
n = 5
for i in range(n):
    for _ in range(n - i):
        print(" ", end="")
    for j in range(1, i + 2):
        print(j, end="")
    for j in range(i, 0, -1):
        print(j, end="")
    print()

`` Which change fixes it?

Change the space loop to range(n - 1 - i)Add one more space in front of each rowChange range(n - i) to range(n - i + 1)Change range(i, 0, -1) to range(i, 0)

5. A learner writes row = "* *" and then calls row.replace(" ", "."), expecting row to hold *..*. Is that expectation right?

True, because replace changes the string that row refers toFalse, because replace only works on lists, not on stringsFalse, because replace returns a new string, so row still holds the spaces

Rows and mirrors