{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mt2-08f-soln

mt2-08f-soln - Answer Key Testname MT2-08F 1 TRUE 2 FALSE 3...

This preview shows pages 1–3. Sign up to view the full content.

Answer Key Testname: MT2-08F 1) TRUE 2) FALSE 3) FALSE 4) TRUE 5) TRUE 6) C 7) A 8) 9) 10) FALSE 11) TRUE 12) FALSE 13) FALSE 14) TRUE 15) B 16) B 17) B 18) A 19) D 20) A 21) B 22) B 23) byte 24) nibble 25) 256 26) Hexadecimal 27) TRUE 28) FALSE 29) TRUE 30) TRUE 31) TRUE 32) TRUE 33) FALSE 34) TRUE 35) FALSE 36) A 37) A 38) B 39) B 40) D 41) Emoticons 42) netiquette 43) shareware 44) fair use 45) Phishing 46) open source 47) TRUE 48) FALSE 49) TRUE 50) TRUE 7

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Answer Key Testname: MT2-08F 51) TRUE 52) FALSE 53) TRUE 54) TRUE 55) D 56) B 57) C 58) D 59) Decryption 60) XOR 61) 62) 8
61) Recall that C in hexadecimal is 12 in decimal. Thus, to convert 2C from hexadecimal to decimal, we multiply 2 × 16 1 and add it to 12 × 16 0 , which gives 32 + 12 = 44. In general, if we have an n + 1 digit hex- adecimal number x n x n - 1 x n - 2 · · · x 1 x 0 , where each x i is a hexadecimal digit, we can convert the number into decimal by computing ( x n × 16 n )+( x n - 1 × 16 n - 1 )+( x n - 2 16 n - 2 )+ · · · +( x 1 × 16 1 )+( x 0 × 16 0 ) . 62) To convert 13 from decimal to binary, we use the following steps: 1. We first determine what is the largest power of 2 that is no more than 13. Note that 2 4 = 16 and 2 3 = 8, so 8 is the largest power
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern