In the context of 1860, "6 bits" typically referred to a currency value rather than a digital measurement, as the term "bit" was used to denote an eighth of a dollar or 12.5 cents in U.S. currency. Therefore, 6 bits would equal 75 cents. This was a common way of expressing monetary values in the 19th century.
Tuesday, November 6, 1860
November 6, 1860
November 6, 1860
2 bits equal 25 cents. So 6 bits would be 75 cents.
Well, honey, with 6 bits, you can represent numbers from 0 to 63. So, technically speaking, the largest number you can make with 6 bits is 63. Don't go expecting any bigger miracles with just 6 bits, darling.
6$ a year 6$ a year
Since 8 bits = 1 byte, yes.
6 bits
A standard die has 6 faces, so to represent the faces on a single die, you need 3 bits (since 2^3 = 8, which can cover the 6 faces). For a pair of dice, you have 2 dice, which means you need 3 bits for each die, resulting in a total of 6 bits (3 bits for the first die and 3 bits for the second die). Therefore, 6 bits are required to represent the faces on a pair of dice.
In asynchronous transmission using a 6-bit code with two parity bits (one for each nibble), one start bit, and one stop bit, the total number of bits transmitted per codeword would be 10 bits (6 data bits + 2 parity bits + 1 start bit + 1 stop bit). This results in a data efficiency of 60% (6 bits of actual data out of 10 total bits). This means that for every 10 bits transmitted, only 6 bits are useful data, making it less efficient compared to systems with fewer overhead bits.
41 in decimal is 0100 0001 in BCD (this is 8 bits not 6 bits)41 in decimal is 101001 in binary (this is 6 bits, but binary not BCD)There is no 6 bit BCD representation of the decimal number 41!
A bit is 1/8 of a dollar or 12.5 cents so if you have six bits it is 75¢