1. The bits corresponding to a three digit BCD number are 011000100101 what is t
ID: 2291931 • Letter: 1
Question
1. The bits corresponding to a three digit BCD number are 011000100101 what is the number in base 10? Write the decimal number 123 as an 8-bit binary number. What is the 6-digit two's-complement for 010101? Write the base-two binary number 01011010 as a decimal number. Convert the binary number 11010111 to hexadecimal. Convert the hexadecimal number 8C5 to decimal (base 10) Write e 2.718281828... as a binary number using base 2 scientific notation (that is, a base 2 number between 0.5 and 1 -the mantissa times 2" where n is the exponent). Compute the mantissa to at least 8 binary digits. What are the MSB and LSB for the binary number 01101011? Multiply the binary number 01100110 by 2 and write the result as a binary number. If you divide the binary number 01111011 by 4, what is the remainder?Explanation / Answer
Answer :- The BCD number 011 000 100 101 has equivalent number in base 10 as 3045.
Answer :- Decimal 123 = 64 + 32 + 16 + 8 + 2 + 1. So binary form is 01111011 = 0111 1011.
Answer :- Inverting the bits we get, 101010, adding one at LSBwe get 6-didgit 2's complement number as 101011.
Answer :- 01011010 has decimal equivalent as 26 + 24 + 23 + 21 = 64 + 16 + 8 + 2 = 90. The positions having 1 in binary is replaced by 2position and then added. Position starts from LSB with LSB position as zero.
Answer :- 1101_0111 has hexadecimal equivalent as 0xD7. We take combinations of 4 digits from LSB and then we write the equivalent values for each combinations.
Answer :- 0x8C5 has decimal value as 8*162 + 12*16 + 5*16 = 2245. Note C has value of 12 in decimal.
Answer :- MSB is 0 and LSB is 1, 01101011 . MSB and LSB both has been highlighted.
Answer :- Multiplying a binary number by 2 is same as shifting the binary number bits left by one bit. So after multiplying by 2, the binary digits would be 011001100.
Answer :- Dividing a binarynumber by 2 causes right shift by one bit. Since we are dividing by 4, the quotient value will be 011110. The reaminder would be 11.
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