Reflection
This time that I took the test, I seemed to perform slighty worse than the last time that I took the test. I think this may be because of the time constraint that we were given this time compared to the much larger amount of time we had to take the last mcq. I think that in this test I was able to go at a better pace because we had less time. And I feel like I improved at reading the questions but there is still much room for improvement.
Corrections
Q3: The mistake here is that I did not follow the order of the code correctly which made me convert the final code to 200 for first and second rather than converting first to 100 and the moving on to make second 200.
Q9: Some programming languages can be used to manipulate audio data, this is not how audio data are represented digitally. At the lowest level, all digital data are represented as sequences of bits.
Q10: I read the binary code wrong because I forgot that the first number in binary is 1 or 0 when I thought it was 2. This messed up the whole sum of values.
Q12: In the answer I chose, the robot moves forward and rotates to the left only is it can move forward. Once the robot goes through one iteration, it would end up facing a wall which makes the function loop without end because it will no longer be able to move forward.
Q14: The procedure would return true any time it encounters a value that is greater than or equal to the preceding value. It won’t check any subsequent values in the list.
Q17: I did not see the extra path that B can travel to C through and realize that there are 4 connections that need to be broken (all connections to B) in order for B and C to not connect.
Q24: Lossy compression cannot be used when you want to restore a picture to its original size because the point of lossy compression is to reduce the size to be able to run without much problems.
Q30: Data packets can be sent to recipients in different and varied orders.
Q31: The binary scale increases by the power of 2. So, when you add three zeroes, the number will be 2^3 times bigger which is 8 times bigger, not 1000. I thought the scale was increased by 10 instead of 2.
Q33: I did not see that the problem was not undecidable because algorithm 1 gives a possible solution to the problem.
Q42: I did not know what item mod 2 meant in the function but later saw that 4 was the wrong answer because the function counted the number of odd numbers which was 3. Item mod 2 is the remainder of a number when divided by two. So, when plugging in 4, the remainder would be 0 because 4/2=2 and has no remainder.
Q52: A binary search require less list elements because when examining it divides the list number by 2. and 2^7 is 128 so it fits in a binary search.
Q55: The function that I chose was wrong because it doesn’t insert the last element into the first spot rather it returns it to the original place which leaves the list unchanged.
Q59: The information of others in her contacts list would be used by Adrianna’s phone while the correct answer was asking for data that would not be from her phone which is the location of other runners.
Q64: Line 3 is incorrect because it should be there as the variable result is assigned the value “adult” by default and then assigned the value “senior citizen” or “minor” only when appropriate.
Q65: The incorrect option had hot AND humid evaluate to false, the body of the IF statement is not executed, and nothing is displayed.
Q67: The chances of each color were out of order in the option I chose so the probabilities would be off.
Q70: This code segment assigns the characters to the right of position n to newStr. The code segment then concatenates newStr with the substring consisting of the characters to the left of position n. For example, if oldStr is “best” and n is 3, the code segment assigns “t” to newStr, then concatenates “t” and the substring “be” and assigns the result “tbe” to newStr.