Study guides

☆☆

Q: What is the decimal conversion of the binary number 1111 1111 1111?

Write your answer...

Submit

Still have questions?

Related questions

The binary number 1111 = 15

15 = 1111

1111 = 15

1111 in binary is 15 in decimal. 1111 in decimal is 10001010111‬ in binary.

1111 1111 base 2

It is 1111.

1111 in binary is 15 in decimal.

1111 1111

All I know is that when a number is negative, you convert the decimal into binary and if it is negative you put 1111 before the binary digits.

111111 in binary is 255 in decimal which is FF in hexadecimal (i.e. 15 units and 15 16s)

111111112 = 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 101010102 = 128 + 32 + 8 + 2 = 170

In binary: 1111 1111 1111 1111 1111 1111 1111 1111 In octal: 37777777777 In hexadecimal: FFFFFFFF in decimal: 2³² - 1 = 4,294,967,295

1111 or 00001111

The binary number 1111 is 15. The digits in a binary number are exponents of 2 rather than 10, so that for a four digit number in binary, the digit places represent 8, 4, 2, 1 instead of increasing values of 10. 1111 = 8+4+2+1 = 15

1111 can't be used for Binary Coded Decimal (BCD) because 1111=15 which is made of 2 digits 1 and 5. In BCD a 4-digit binary number is used for every decimal digit. ex. 1111 is incorrect 1 = 0001 5 = 0101 Answer: 0001 0101

1111

111110 = 10 001 010 1112

1111 1111 1111 1111 = 2^16 = 65536

1111

The 1's complement is formed by inverting every binary digit (bit) of the number - if it is a 0 it becomes a 1, otherwise it is a 1 and becomes a 0. If 10 is in base 2, then its 1's compliment is 01 or just 1. If 10 is in base 10, then in binary it is 1010 and its 1's complement is 0101 = 5 in decimal. However, if more bits are being used to store it, there would be leading 0s that get inverted to 1s and so the resultant number is different; examples: 8 bits (a byte): decimal 10 = 0000 1010 → 1111 0101 = 245 in decimal 16 bits: decimal 10 = 0000 0000 0000 1010 → 1111 1111 1111 0101 = 65525 Next, if 2s complement is being used to represent negative numbers, the binary 1111 0101 represents decimal -11; similarly 1111 1111 1111 0101 represents decimal -11.

The answer is 1 0101 0111 1110 1011 1011 0011 1111 1010 0001 0111

It is 111.

1111 converted from binary (base 2) to decimal (base 10) is 15 When you expand the steps... 1111 binary = (1 X 2^3) + (1 X 2^2) + (1 X 2^1) + (1 X 2^0) = 8 + 4 + 2 + 1 = 15

0000 0000 1111 1000F ( or 15) = 1111 in binary, and 8 = 1000 in binary, so F is 1111 1000

In binary the largest number (using IEEE binary16) representable would be: 0111 1111 1111 1111 (grouping the bits in nybbles* for easier reading). This is split as |0|111 11|11 1111 1111| which represents: 0 = sign 111 11 = exponent 11 1111 1111 = mantissa. Using IEEE style, the exponent is offset by 011 11, making the maximum exponent 100 00 This is scientific notation using binary instead of decimal. As such there must be a non-zero digit before the binary point, but in binary this can only ever be a 1, so to save storage it is not stored and the mantissa effectively has an extra bit, which for the 10 bits specified makes it 11 bits long. Thus the mantissa represents: 1.11 1111 1111 This gives the largest number as: 1.1111 1111 11 × 10^10000 (all digits are binary, not decimal.) This expands to 1 1111 1111 1100 0000 (binary) = 0x1ffc0 = 131,008 Note that this is NOT accurate in storage - there are 6 bits which are forced to be zero, making the number only accurate to ±32 (decimal): the second largest possible real would be 1 1111 1111 1000 000 = 0x1ff80 = 130,944 - the numbers are only accurate to about 4 decimal digits; the largest decimal real number would be 1.310 × 10^5, and the next 1.309 × 10^5 and so on. However, with proper IEEE, an exponent with all bits set is used to identify special numbers, which makes the largest possible 0111 1101 1111 1111 which is 1.1111 1111 11 × 10^1111 = 1111 1111 1110 0000 = 0xffe0 = 65504 accurate to ±16, ie the largest is about 6.55 × 10^4. * a nybble is half a byte which is directly representable as a single hexadecimal digit.