Using an online AES calculator, show the output of two blocks that differ by exactly one bit. a. Create the equivalent of 1 block of data (128-bits), set a key, use ECB mode and run the calculator. Show the resulting screen. b. Modify the plaintext data to differ by exactly one bit and run the calculator with the same key and mode as in part a. Show the new resulting output values. c. List two observations about the results of parts a and b above.

4, in the fixed-point machine, the following statement is wrong (   ).A. In addition to the complement, the original code and the inverted code cannot indicate that the original code of −1 B. +0 is not equal to −0C. the inverse of +0 is not equal to the negative of −0 D. For the same machine word length, the complement code can represent a negative number more than the original code and the inverted code.

In one of the labs performed you were to examine a .jpg and write the first 2 bytes of Hexadecimal information representing that .jpg. Which of the following is the Hexadecimal representation of that .jpg?     FF E8     FF D8     E8 FF     D8 FF

Example: Assuming m-7 bits and using three of those bits for representing the fractional part, ( 1100.000 )2 ( 0110.000 )2 ( 0011.000 )2 ( 0001.100 )2 ( 0000.110 )2 ( 0000.011 )2 (12)10 %3D (12)10 / 2 (6)10 (6)10 /2 (3)10 (3)10 /2 (1.5)10 (1.5)10 / 2 (0.75)10 (0.75)10 / 2 (0.375)10 In this case, each successive division by two has shifted the binary sequence to the right by one bit.

7. Add the following two 8-bit binary numbers, producing an 8-bit result. Indicate the status of the carry and overflow bits at the end of the addition by circling the appropriate choices. The space in between is only for readability.    1101 0110 + 1001 0111 --------------   Overflow?       Yes    or    No Carry?       Yes   or     No

Refer to figure 8 and enter the decimal value for the Sequence number starting at offset 0x26 (length 4- bytes). .......! pMo...E. .(..@... .f...1.. %M.X.$P. 0000 do do fd c4 09 94 00 21 70 4d 4f ae 08 00 45 00 00 28 19 8е 40 00 80 06 fa 66 се a8 1е бс cd bc За ба 06 9а өө 8f f6 a5 0010 0020 25 4d c6 58 ob 24 50 11 0030 fb 8e d8 dø 00 o0 00 00 00 00 00 00 Figure 8 Byte Details Question 37 Refer to figure 6 and enter the binary bits which represents "only" the IP header length. dø do fd c4 09 94 00 21 00 28 19 8e 40 00 80 06 .......! pMO...E. .(..@... .f...1.. %M.X.$P. 0000 70 4d 4f ae 08 00 45 00 0010 fa 66 сө а8 1е бс cd bc 0020 За 0а 06 9а Ө0 8f f6 a5 25 4d c6 58 ob 24 50 11 0030 fb 8e d8 dø ø0 ø0 00 00 00 00 00 00 Figure 6 Byte Details Question 41 Complete the command to scan the subnet 130.101.108.0:

Using GDB online Debugger | Compiler . Thank you

Fill in the table for the transition function for the FSA with the alphabet E = {0,1,2} that recognizes the language of ternary numbers that are divisible by 4. Please use the names of the states listed on the left as your answers in the table. The start state is zero. state 1 2 zero one two three The accept states are |A. zero В. one C. three D. two

Please answer the question in image. With simple and easy explanation. Thank you