If 8 bits make a byte what do 4 bits constitute?
397669
8 bits is a byte... 4 bits is a byte... whether your binary looks like 10011000, 11111111, or 00000001 or 00000000, it's all a byte. If you have 1001, the machine will automatically make it 00001001. It will just add zero's in front of it to make it a byte.
977730
bit (b)\n1 byte (B) = 8 bits \n1 Kilobyte (K / KB) = 10^3 bytes = 1,000 bytes \n1 Megabyte (M / MB) = 10^6 bytes = 1,000,000 bytes \n1 Gigabyte (G / GB) = 10^9 bytes = 1,000,000,000 bytes \n1 Terabyte (T / TB) = 10^12 bytes = 1,000,000,000,000 byte\n\n 1 Megabyte = 1,024,000 bytes.\n\n\nso\nThere are 1,000 KB (Kilo Bytes) in on MB (MegaByte)\nThere are 1,000 MB in one GB (GigaByte)
290645
1GB=1,000,000,000 (a billion bytes)\n\n1MB = 1,000,000 (a million bytes)\n\n0.4 GB = 0,400,000,000 bytes\n\nTherefore, 0.4GB ==> 400 MB
227540
001100100001\n\nSorry braxton_paul, you're second answer is incorrect, as it is binary, BCD, or BINARY CODED DECIMAL, is different. as stated by wd5gnr, BCD wastes space, this is why the BCD number is 12 digits, and your binary is only 9. This is due to the fact that BCD has no base (at least in the normal sense), wheras binary is base two. Just because BCD uses only two characters, doesn't mean its base is two. This is due to the fact that BCD has no "place weight", for example, if you have 95473 (for a random example), in a base 10 (decimal) system, the 4 would have a value of 400. Say you have a binary system, and lets just use the BINARY of 321, or 101000001, (from RIGHT to LEFT) the ones place has a "weight of 1, the two's has a weight of 2, the three's has a wieght of 4, position 4 is 8, 5 = 16, so forth (each time you multiply the previous space by two)... So to convert BIN to DEC, just multiply the digit ocupying the space by the spaces value, so from right to lef, you would get 1x1=1, (the zeros do not matter since anything times zero is zero) 1x64=64, and 1x256=256... then simply add the values together, and you get the DEC equiv. of your BIN, so 1+64+256=321... Just try that with BCD... and you do indeed get 801... that is why it is called binary CODED DECIMAL. the lack of a base is why BCD can only easily be converted to and from Decimal and Binary, just try converting it to octal, or hex... it is a pain in the a$$ unless you first convert it to DEC...\n\nso (as if anybody cares):\n\n321 (DEC, or base ten) = 141 (HEX, or base sixteen) = 501 (OCT, or base eight) = 101000001 (BIN, or base two) = 001100100001 (BCD, or base ten incoded in base two)\n\nwheras:\n\n801 (DEC) = 321 (HEX [coinciednce]) = 1441 (OCT) = 1100100001 (BIN) = 100000000001 (BCD)\n\nBinary, and (B)inary (C)oded (D)ecimal are not the same!\n(and wd5gnr's BCD to DEC conversion table is correct)\n\nUsualy a dead give away when it comes to BCD vs. BIN, is that BCD usaly has preceeding zeros listed, but BIN does not (though preceeding 0's could also mean ASCII, BCH, or BCO, or prettymuch binary coded anything)
709064
A byte is commonly used as a unit of storage measurement in computers, regardless of the type of data being stored. It is also one of the basic integral data types in many programming languages.
705667
This number (512 or 256 etc etc) does means that, that many bits can be accessed simultaneously. That is more the number more the data that you can access and process.
349881
In 8 bit Intel assembly:\n\nYou have 3 integers defined as:\n\nNumber1High   ;high byte of number1\nNumber1Low    ;low byte of nuber1\n\nNumber2High   ;high byte of number2\nNumber2Low    ;low byte of number2\n\nResultHigh    ;high byte of result\nResultLow     ;low byte of result\n\nto add:\nCLR  C;\nMOV  A,Number1Low;\nADD  A,Number2Low;\nMOV  ResultLow,A;\nMOV  A,Number1High;\nADDC A,Number2High;\nMOV  ResultHigh,A;
924671
According to the site below, there are 6, but I suppose you could combine bits and Bytes into one, since Bytes are made of bits.
407425
8 bits = 1 byte\n1024 bytes = 1 KB(Kilo Bytes)\n1024 KB = 1 MB(Mega Bytes)\n1024 MB = 1 GB(Giga Bytes)\n1024 GB = 1 TB(Terra Bytes)
864328
binary 11 is 3 decimal
389147
PEEK read a 8bit value from any memory address\ninteger_variable = PEEK(address)\n\nPOKE write a 8bit value to a memory address (not all addresses may be written!)\nPOKE address, value
409800
Generaly, becouse bit numbers are on a boundy of an exponet of 2, we can calculate like this:\n\n1024 bits in 1 kilobit\n1024 kilobits in a megabit\nthus: 1024 * 1024 = 1,048,576 bits
359204
100 which is the binary equivalent of 4 if you are considering those 2 numbers to be added binary numbers
660222
1,024 Byte = 1 Kilobyte (KB) \n1,024 Kilobyte (KB) = 1 Megabyte (MB)\n1,073,741,824 Bytes = 1 Gigabyte (GB) \n1 Gigabyte (GB) = 1,024 Megabyte (MB)
644438
you can find it in a byte\nASCII - it is a 7-bit code \n\nEBCDIC - 8-bit (the usual byte)
255673
8000 x 8 bits = 64Kb/s \n24 x 64Kb/s = 1536000 - \nremenber that the signalling/timing channel will take up one of the 24 channels\n(in the U.S.A I beleive they call it a T1 circuit - 23 data 1 signalling/timing....here in the U.K. E1 32x64Kb/s thats 30 data, 1 timing, 1 signalling)\n\nas for the data storage sounds like you could look at it like this:\n3 mins ->\n3 x 60 = 180\n180 x 64000 = 11.5Mb for one 3 min message (bits)\ntherefore 11.5Mb /8 = 1.44MB for one 3 min message(Bytes)
1159041
If you are thinking of f : ? -> { 0, 1, 2, 3, ... } (i.e., the domain is mapped into the set of non-negative integers) then clearly the domain "?" contains [at least] all finite bit strings.  If you keep the same restriction on the range, you can even extend the domain to include those infinite bit strings (however you wish to define that) which contain only finitely many ones.  If you extend the range to also include infinite cardinals and/or ordinals then you can include even more infinite bit strings into the domain.