Problem 1. Byte Stuffing Decode the following "raw" sequence of bits into bytes using HDLC decoding. 111110111001111101

31 Convert the hexadecimal IEEE format floating point single precision number 0x40200000 to decimal: First convert hex to binary  (Provide a space between four bit groups.  Example 111 1010 not 1111010)  Sign bit Exponent in decimal after removing excess 127.   Just the exponent.  Example 3 not 23 Mantissa or Significand  (Provide a space between four bit groups.  Example 111 1010 not 1111010)  Put it together and convert to decimal

a. Represent the number 1.01011010101 x 2-19 in IEEE Standard 754 single precision floating point binary. b. Convert each of the following decimal values to 8-bit two’s complement binary. i) -3810 ii) 5510 iii) -6510 iv) 7210

H - For the IEEE 754 single-precision floating point, write the hexadecimal representation for the following decimal values: (i)–1.0 (ii)– 0.0 (iii)256.015625

Show Steps Please 8. Decode the following ASCII message, assuming 7-bit ASCII characters and no parity: 1001010 1001111 1001000 1001110 0100000 1000100 1001111 1000101

4) Write the binary representation of the decimal number 63.25 assuming the IEEE 754 double precision format. (32-bit double precision: 11 bits for exponent (1023 bias) + 52 bits for base) Show your work leading to your answer (sign bit, exponent bits, base bits) as to how you derived your final 64-bit answer.

1. unsigned binary and decimal representations.a. Give the 16-bit unsigned binary and hexadecimal representations of thedecimal number (2040).b. Give the 16-bit unsigned binary and decimal representations of the hexnumber (B2CF).

6. Convert the following binary numbers into Hexadecimal (base 15) format. a. 001010101010101010101010₂ b. 010101010101010001110₂

Floating-Point Representation. What is the floating-point value for binary 4. a) 11000010100000 using the simple model (14-bit format, 5 bits for the exponent with a bias 15, a normalized significand of 8 bits, and a single sign bit for the number)?

Explain what the following MIPS code does: (do not explain each line but explain what these two lines of codes do together) li $v0, 5syscall