If you mean 10 + 6 that's 16 which is 10000 in binary
To represent the days of the week, you would need at least 3 bits. With 3 bits, you can represent up to 8 different values (2^3 = 8), which is sufficient to cover all 7 days of the week (Monday to Sunday). Each additional bit would double the number of possible values, but 3 bits are the minimum required to uniquely represent all 7 days.
To consider the difference between straight binary and BCD, the binary numbers need to be split up into 4 binary digits (bits) starting from the units. In 4 bits there are 16 possible values from 0000 to 1111 (0 to 15). In straight binary all of these possible combinations are used, thus: 4 bits can represent the decimal numbers 0-15 8 bits can represent the decimal numbers 0-255 12 bits can represent the decimal numbers 0-4095 16 bits can represent the decimal numbers 0-65535 etc In arithmetic, all combinations of bits are used, thus: 0000 1001 + 0001 = 0000 1010 In BCD or Binary Coded Decimal, only the representations of the decimal numbers 0-9 are used (that is 0000 to 1001 in binary), and the 4-bits (nybbles) are read as decimal digits, thus: 4 bits can represent the decimal digits 0-9 8 bits can represent the decimal digits 0-99 12 bits can represent the decimal digits 0-999 16 bits can represent the decimal digits 0-9999 In arithmetic, only the representations of decimal numbers are used, thus: 0000 1001 + 0001 = 0001 0000 When BCD is used each half of a byte is read directly as a decimal digit. BCD is obviously inefficient as storage (for large numbers) as each nybble is only holding 3/8 of the possible numbers, however, it is sometimes easier and quicker to work with decimal digits (for example when there is lots of display of counting numbers to do there is less binary to decimal conversion needing to be done).
18427(10) = 1000111111111011(2)So, it will need 16 bits (16 digits from the binary value) for 18427 itself. For the complement (the sign), add 1 more bit: the answer is 17.
log2 200 = ln 200 ÷ ln 2 = 7.6... → need 8 bits. If a signed number is being stored, then 9 bits would be needed as one would be needed to indicate the sign of the number.
8
Most modern digital cameras use 24 bits (8 bits per primary) to represent a color. But more or less can be used, depending on the quality desired. Many early computer graphics cards used only 4 bits to represent a color.
45 in binary is 101101, so you need at least 6 bits to represent 45 characters.
4.1 bit for 2,2 bits for 4,3 bits for 8,4 bits for 16.
If you mean 10 + 6 that's 16 which is 10000 in binary
To represent the days of the week, you would need at least 3 bits. With 3 bits, you can represent up to 8 different values (2^3 = 8), which is sufficient to cover all 7 days of the week (Monday to Sunday). Each additional bit would double the number of possible values, but 3 bits are the minimum required to uniquely represent all 7 days.
"recommended setting" There are 19 characters including the space between the two words. If the old convention of using 1 byte to represent a character, then we would need (19 x 8) which is 152 bits. If we use unicode as most modern computers use (to accommodate all the languages in the world) then 2 bytes will represent each character and so the number of bits would be 304.
To determine the number of bits in three dollars, we need to first convert the dollar amount to cents, as there are 100 cents in a dollar. Three dollars is equal to 300 cents. Next, we need to calculate the number of bits in 300 cents. Since 2^8 (256) is the closest power of 2 to 300, we would need at least 8 bits to represent 300 cents accurately.
You need four bits for each hex digit since 4 bits can support a value from 0-15. Binary_____Hexidecimal 0_________0 1_________1 10________2 11________3 100_______4 101_______5 110_______6 111_______7 1000______8 1001______9 1010______A 1011______B 1100______C 1101______D 1111______F
Bit -- Value 1 -- 1 11 -- 3 111 -- 7 1111 -- 15 11111 -- 31 111111 -- 63 1111111 -- 127 11111111 -- 255 111111111-- 511 1111111111 -- 1023 Therefore - 10 bits would be more than adequate (unless you had to represent capitals and lower case - in which case you would need one more bit).
You need 20 bits of address bus to address 1 Mb of memory.
11 bits (which actually allows -1024 to 1023)