Reflection
Going into this test I was very confident, I predicted a score aorund 95% and that is exactly what I recieved. My scores between the past two tests and our most recent stayed consistant, both being in the mid 90% range. The total score I achieved on this test was a 67/70. I managed this score by participating in lessons to understand the topics, helping construct and personally reviewing the frequently missed questions for he past MCQ, and reviewing personal mistakes and missed topics on my past MCQ’s. I took my time on this test, as last time I made much more mistakes by rushing, I utilized the entire period to test and even went into lunch. Ultimately I made 3 mistake that I think will be easy to correct.
Q16
I misread the problem and thought I was bringing down devices, not removing connections. My thought process was that if I took down devices A, D, and F, the connection would be severed. However, we are removing connections, not devices. Removing any three connections will not prevent devices B and C from communicating. For example, if connections A-C, C-D, and B-D are removed, then devices B and C can still communicate along the path B-E-F-C. The correct answer would be to remove the four first connections branching off of B to isolate it.
Q55
This is incorrect because this code segment assigns the value of the last element of the list to the variable temp, then removes the last element of the list, then appends temp to the end of the list. The resulting list is the same as the original list. Answer C assigns the value of the last element of the list to the variable temp, then removes the last element of the list, then inserts temp as the first element of the list.
Q70
I answered this question partially correct. Option A is correct because it separates the characters to the left of position n into the variable ‘left’ and the characters to the right of position n into the variable ‘right’. Then, it combines ‘left’ and ‘right’ and assigns the result to ‘newStr’. However, my other answer, Option B, was incorrect because it assigns the characters from the beginning of the string up to one position past n to ‘left’. Then, it assigns the characters from one position before n to the end of the string to ‘right’. After that, it concatenates ‘left’ and ‘right’ and assigns the result to ‘newStr’. For example, if oldStr is “best” and n is 3, Option B assigns “best” to ‘left’, “est” to ‘right’, and “bestest” to ‘newStr’. Option C is another correct answer because it assigns the characters to the left of position n to ‘newStr’. Then, it concatenates ‘newStr’ with the substring consisting of the characters to the right of position n. For instance, if oldStr is “best” and n is 3, Option C assigns “be” to ‘newStr’, then concatenates “be” with the substring “t”, and assigns the result “bet” to ‘newStr’.