Question

suponiendo que la siguiente asignación se haya ejecutado con éxito

the_list = \1\, 1, 2

¿cuál de las siguientes expresiones se evalúa como true ?
(seleccione dos expresiones.)

the_list.index(1) == 0

1.1 in the_list1:2

len(the_list0:2) < 2

the_list.index(\1\) in the_list

Explanation:

To solve this, we analyze each expression with the list the_list = ["1", 1, 1]:

1. Analyze the_list.index('1') == 0

The index() method returns the first occurrence of the element. For the_list, the first "1" is at index 0. So the_list.index('1') is 0, and 0 == 0 is True.

2. Analyze 1.1 in the_list[1:2]

Slicing the_list[1:2] gives [1] (since slicing is up to but not including the end index). The element 1.1 is not in [1], so this is False.

3. Analyze len(the_list[0:2]) < 3

Slicing the_list[0:2] gives ["1", 1]. The length of this slice is 2. Since 2 < 3 is true, this expression is True.

4. Analyze the_list.index("1") in the_list

the_list.index("1") is 0. Now, check if 0 is in the_list (which is ["1", 1, 1]). The elements are "1", 1, 1—0 is not in the list, so this is False.

Final Answer (Two True Expressions):

• the_list.index('1') == 0
• len(the_list[0:2]) < 3

So the correct options are:

• the_list.index('1') == 0
• len(the_list[0:2]) < 3

Answer:

To solve this, we analyze each expression with the list the_list = ["1", 1, 1]:

1. Analyze the_list.index('1') == 0

The index() method returns the first occurrence of the element. For the_list, the first "1" is at index 0. So the_list.index('1') is 0, and 0 == 0 is True.

2. Analyze 1.1 in the_list[1:2]

Slicing the_list[1:2] gives [1] (since slicing is up to but not including the end index). The element 1.1 is not in [1], so this is False.

3. Analyze len(the_list[0:2]) < 3

Slicing the_list[0:2] gives ["1", 1]. The length of this slice is 2. Since 2 < 3 is true, this expression is True.

4. Analyze the_list.index("1") in the_list

the_list.index("1") is 0. Now, check if 0 is in the_list (which is ["1", 1, 1]). The elements are "1", 1, 1—0 is not in the list, so this is False.

Final Answer (Two True Expressions):

• the_list.index('1') == 0
• len(the_list[0:2]) < 3

So the correct options are:

• the_list.index('1') == 0
• len(the_list[0:2]) < 3