My Score on my MCQ

My full score on my MCQ test, some of the questions were difficult and I had to take a more increased time on them, some I thought they were easy to I didn’t take as much for them.

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Revising the questions

Question 1

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The answer (A) I chose was wrong because, the code segment assigns the initial value of first to temp, then assigns the initial value of second to first. The initial value of second, which has been stored in first, is then assigned to second. Therefore, the value of second has not changed from its initial value.
looking back answer (B) would have been the correct option, the code segment assigns the initial value of first to temp, then assigns the initial value of second to first. The initial value of first, which has been stored in temp, is then assigned to second. Therefore, the initial values of first and second have been interchanged.

Question 7

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The answer I chose (A) was wrong because the values 1 and 3 are the original values of start and current, not the final values.
Revising I didnt read the question through, the code segment initially sets start to 1, end to 20, and current to 3. Next, the code segment sets start to the value of current, which is 3. Then current is increased from 3 to 4. The final values of start and current are 3 and 4, respectively. Only when you use 3 and 4

Question 9

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The answer I chose was (A) saying, it allows programmers to implement algorithmic solutions to otherwise unsolvable problems. This would be incorrect, unsolvable problems cannot be solved with an algorithm. because I thought that the new
Reading the question further I shouldve chosen answer (D) , it incrementally adding code segments to correct, working programs can help programmers identify new errors that are introduced instead of fixing unsolvable problems.

Question 13

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The answer I chose was (A), I was thinking while looking at the daily messages sent and the app relase date and seeing there was little to no affect
Answer (D) Wouldve been the better answer, the average number of characters per message appears to decrease after the mobile app was released.

Question 14

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Answer (C) Would’ve been the correct answer, The programs each display ten values, but each value displayed by program B is one greater than the corresponding value from program A. Program A displays 1 2 3 4 5 6 7 8 9 10 and program B displays 2 3 4 5 6 7 8 9 10 11. Program A was greater than Program B by one value.

Question 19

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I tried using basic knowledge, I was just unlucky with this question. I thought that data e-books would still have data from previous versions.

Question 23

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I chose answer (D), There are multiple paths from P to S (for example, P to R to S, P to Q to S, P to R to Q to S, and P to Q to R to S) so this is why it would be wrong cant connect multiple times.
The correct answer would be (B), Redundant routing is impossible if there is only one possible path from one device to another. There is only one possible path from P to S (P to R to Q to S) it would go in one straight direction not multiple.

Question 24

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I chose answer (D) because, I didn’t have much knowledge in this area so I had to guess with what I already knew
The answer I should’ve chosen was (C) because, The transformation is lossless because an encoded string can be restored to its original version. For example, Open quotation, percent, hash, underscore, hash, underscore, percent E, underscore, BEST, underscore, W, hash, H, close quotation can be restored to Open quotation, THIS, underscore, IS, underscore, THE, underscore, BEST, underscore, WISH, close quotation by replacing all instances of Open quotation, percent, close quotationwith Open quotation, T H, close quotation and by replacing all instances of Open quotation, hash, close quotation with Open quotation, I S, close quotation. I had to research this as well as examples to understand it

Question 25

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I chose answer (A) same as the last question I didn’t understand it so I tried to guess based on what I saw from the first question breaking banana into b -an -an -a and i realized that it would still be possible when we are looking for not possible.
The answer I would choose know is (B) LEVEL_UP, It is not possible to use byte pair encoding in the string Open quotation, LEVEL, underscore, UP, close quotationbecause no pair of characters appears in the string more than once thus being not possible

Question 26

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I chose answer (C), I realized that the code I put for my answer would only move the robot up and down from its inital position
The answer I instantly realized knowing my mistakes is (A), this would move the robot forward and only move right when its possible, I swapped the roles thus making the mistake

Question 27

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The answer I chose (A), I was thinking that you would need to know if win is true before to see if you would display the win screen
Answer (B) would do what I thought A would do. Inserting Win, left arrow, true between line 9 and line 10 will cause the loop to terminate when the guess is correct.

Question 28

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The answer I chose was (B), I should’ve chosen (D), my mistake was I thought bits would not be multipled by 6 but to the power of 6.
Using 6 bits will only allow for up to 64 characters open parenthesis, 2 raised to the sixth power equals 64, close parenthesis.

Question 29

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With these inputs, the AND gate will produce an output of false and the first OR gate will produce an output of false. Since both inputs to the second OR gate will be false, the circuit will have an output of false.
I chose answer (B) and I realized, with these inputs, the AND gate will produce an output of true and the first OR gate will produce an output of false. The second OR gate will have one true input and one false input, causing the circuit to have an output of true.

Question 30

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I realized that when I chose answer (B), this approximation assumes that the Analysis procedure is called only twice. and we can’t assume so knowing this I knew it was answer (D). Each call to the Analysis procedure requires one hour of program execution time and (D) answers the requirments.

Question 31

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I thought that program one only moved the robot up left then it would create a circle, but eventually it would stay in the same position because it would move forward then right in the grey circle forcing it to stay there,
Knowing this I would chose answer (C)

Question 33

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Answer (A) would have been the correct answer because the flowchart sets include to true whenever floor is greater than 1 0 or bedrooms equal 3, and sets include to false otherwise. Therefore, the algorithm is equivalent to Include, left arrow, open parenthesis, floor greater than 10, close parenthesis, OR, open parenthesis, bedrooms equal 3, close parenthesis.
My answer was Question (B), this expression would be used for a flowchart to set include to true whenever floor is greater than 1 0 and bedrooms equal 3. This does not correctly set include to true in cases where only one of the two conditions is true.

Quesiton 38

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The answer I accidentally chose was (B), I clicked wrong and I didn’t notice, I thought I got this right fast, the expression count, MOD, 10 equals 0 evaluates to true when count is a multiple of 1 0, and so this code segment sets cost to 0 under the appropriate conditions meeting the requirments

Question 40

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The answer I chose was (B), I should’ve looked at it a second time, The average total points earned per student can be determined using the result of the total points calculation for each student. The average increase in total points per student as a result of the score replacement policy can be determined by calculating the differences between each student score before and after the replacement policy was applied. The proportion of students who improved their total points as a result of the score replacement policy can be determined by comparing the midterm and final scores for each student with the result of the total points calculation.

Question 42

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For this qustion I was stumped, It didn’t make sense and I tried to use calulcaltions or math to see If I was able to find the pattern or anything but I still ended up getting it wrong
The right answer was (D), With 32-bit addressing, IPv4 has 2 raised to the thirty second power possible addresses. With 128-bit addressing, IPv6 has 2 raised to the one hundred twenty eighth power possible addresses. Since 2 raised to the thirty second power times 2 raised to the ninety sixth power equals 2 raised to the one hundred twenty eighth power, IPv6 has 2 raised to the ninety sixth power times as many possible addresses as IPv4.

Quesiton 46

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An undecidable problem is one in which no algorithm can be constructed that always leads to a correct yes-or-no answer.

Question 47

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In public cryptography, a message is encrypted with a recipient’s public key and decrypted with the recipient’s private key.

Question 48

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This option causes the experiment to be successful when RANDOM, open parenthesis 1 comma 100, close parenthesis produces a result from 1 to 7 5, or 75% of the time.

Question 50

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The procedure implements a linear search, which sequentially compares each element of the list with the target value. The list does not need to be sorted because the procedure checks list elements until either the target is found or it reaches the end of the list.

Question 52

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Options II and III perform the steps in a correct order. In order to generate the desired list, the algorithm must perform the “shorten” step after the “keep palindromes” step, otherwise the “keep palindromes” step would not be able to determine whether the original word was a palindrome.

Question 55

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The expression open parenthesis, open parenthesis, response equals, open quotation, y, close quotation, close parenthesis, AND, open parenthesis, response equals, open quotation, yes, close quotation, close parenthesis, close parenthesisalways evaluates to false because it is not possible for the variable response to be equal to both Open quotation, y, close quotation and Open quotation, yes, close quotation. Therefore, the procedure will always return false.

Question 56

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The value of count starts at 1 and increases by twos, so it counts odd integers. The loop iterates 10 times, adding each intermediate value of count each time. Therefore, the program displays the sum of the odd integers starting at 1 and ending at 19.

Question 60

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The value of count starts at 1 and increases by twos, so it counts odd integers. The loop iterates 10 times, adding each intermediate value of count each time. Therefore, the program displays the sum of the odd integers starting at 1 and ending at 19.