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What is the largest prime number that can be stored in 4 bit memory?
Assuming the number holds an unsigned number, the range of numbers for 4 bits is 0-7. 7 is a prime number.
How many bits in a half dollar?
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How many significant numbers are in the number 0.1020?
4
How does decimal value waste memory space?
A decimal digit requires 4 bits of memory space to represent all possible values. So N decimal digits will require 4N bits to store in decimal format. On the other hand, if the same N decimal digits were stored as a binary number, they should require log2(10N) bits, that is, about 3.32N bits. So storing the decimal representation uses about 20% more memory. To put it another way, 4 bits is capable of holding one of 16 different values. By using it to store decimal digits, it is only being used to hold one of 10 different values.
What is the exact number of bits in a memory that contains?
4
How many decimal numbers can be represented by 4-bits?
There are 16 decimal numbers that can be represented by 4-bits.
How many numbers can you represent with 4 bits?
16 of them.
How many bits are stored by a 256 x 4 memory chip can this chip be specified as 128 byte memory?
256x4=1024 bits are stored.128x8=1024 ,yes this chip can be specified as 128 byte
What is the largest prime number that can be stored in 4 bit memory?
Assuming the number holds an unsigned number, the range of numbers for 4 bits is 0-7. 7 is a prime number.
What is the range of numbers that can be encoded in 4 bits using 2s complement notation?
Using 4 bits the signed range of numbers is -8 to 7. When working with signed numbers one bit is the sign bit, thus with 4 bits this leaves 3 bits for the value. With 3 bits there are 8 possible values, which when using 2s complement have ranges: for non-negative numbers these are 0 to 7; for negative numbers these are -1 to -8. Thus the range for signed 4 bit numbers is -8 to 7.
How many nibbles are 16 bits?
1 nibble = 4 bits, so 4 nibbles for 16 bits.
How many numbers can be stored in a 4-byte integer system?
If all four bytes are being used for its value (i.e. this is an unsigned integer) then you have 8 * 4 = 32 bits, so your range is from 0 to 2^32 (4,294,967,296) Remember, the size of various data types in C and C++ is architecture dependent. See limits.h (/usr/include/limits.h in Linux)
How many bits are in 256 bits?
Each hexidecimal character represents 4 bits, therefore 256 bits takes 256 / 4 = 64 characters.
How is scientific notation related to the floating point representation used by computers?
Floating point numbers are stored in scientific notation using base 2 not base 10.There are a limited number of bits so they are stored to a certain number of significant binary figures.There are various number of bytes (bits) used to store the numbers - the bits being split between the mantissa (the number) and the exponent (the power of 10 (being in the base of the storage - in binary, 10 equals 2 in decimal) by which the mantissa is multiplied to get the binary/decimal point back to where it should be), examples:Single precision (IEEE) uses 4 bytes: 8 bits for the exponent (encoding ±), 1 bit for the sign of the number and 23 bits for the number itself;Double precision (IEEE) uses 8 bytes: 11 bits for the exponent, 1 bit for the sign, 52 bits for the number;The Commodore PET used 5 bytes: 8 bits for the exponent, 1 bit for the sign and 31 bits for the number;The Sinclair QL used 6 bytes: 12 bits for the exponent (stored in 2 bytes, 16 bits, 4 bits of which were unused), 1 bit for the sign and 31 bits for the number.The numbers are stored normalised:In decimal numbers the digit before the decimal point is non-zero, ie one of {1, 2, ..., 9}.In binary numbers, the only non-zero digit is 1, so *every* floating point number in binary (except 0) has a 1 before the binary point; thus the initial 1 (before the binary point) is not stored (it is implicit).The exponent is stored by adding an offset of 2^(bits of exponent - 1), eg with 8 bit exponents it is stored by adding 2^7 = 1000 0000Zero is stored by having an exponent of zero (and mantissa of zero).Example 10 (decimal):10 (decimal) = 1010 in binary → 1.010 × 10^11 (all digits binary) which is stored in single precision as:sign = 0exponent = 1000 0000 + 0000 0011 = 1000 00011mantissa = 010 0000 0000 0000 0000 0000 (the 1 before the binary point is explicit).Example -0.75 (decimal):-0.75 decimal = -0.11 in binary (0.75 = ½ + ¼) → 1.1 × 10^-1 (all digits binary) → single precision:sign = 1exponent = 1000 0000 + (-0000 0001) = 0111 1111mantissa = 100 0000 0000 0000 0000 0000Note 0.1 in decimal is a recurring binary fraction 0.1 (decimal) = 0.0001100110011... in binary which is one reason floating point numbers have rounding issues when dealing with decimal fractions.
Can two hexadecimal digits be stored in one byte?
Yes, a byte is 8 bits, and a one hexadecimal digit takes up four bits, so two hexadecimal digits can be stored in a byte. The largest hexadecimal digit is F (which is 15 in base ten.) In base two, this converts to 1111, which takes up four bits, which is why it only takes four bits to store a hexadecimal digit. With 8 bits, two hexadecimal digits can be stored (FF would be 11111111, which is 8 bits), and 8 bits make up a byte. Generally, 4 bits are always used to store a hexadecimal digit, using leading zeros where necessary. For example, the hexadecimal digit 5 would be stored as 0101, and the hexadecimal digits 5A would be stored as 01011010.
How many bytes make up a nibble?
One nibble, or nybble, is equal to four bits.
How many bits in the FCS in an Ethernet frame?
32 bits. which is 4 bytes (8 bits in each byte).
