8 bits if unsigned, 9 bits if signed
how many bits are needed to represent decimal values ranging from 0 to 12,500?
To convert a decimal into a fraction, place a 1 under the decimal number (to represent a fraction), move the decimal point to the right of the decimal number and add as many zeroes to the bottom number as the number of places the decimal was moved on the top number. Reduce the fraction to its lowest term. Conversion of given decimal number: .7 (place a 1 underneath to make a fraction) ,7/1 (move the decimal point to the right of the decimal number and add that many zeroes to the bottom number) 7/10 (this fraction cannot be reduced so it is the answer) Other example: conversion of .375 .375 .375/1 375/1000 (now reduce the fraction t its lowest term) 3/8
None. If you move the decimal point you will change the value of the number!
The only possible answer is 0.75 since for any other decimal number, there are infinitely many decimal numbers which are closer.
5
103
8 bits if unsigned, 9 bits if signed
Count them: 643(10)=1010000011(2)
how many bits are needed to represent decimal values ranging from 0 to 12,500?
1200
8 (assuming unsigned numbers - i.e., you don't reserve a bit for the sign).
There are 16 decimal numbers that can be represented by 4-bits.
5 bits are 5 binary digits. If they represent a decimal number, then that number can be anything from zero to 31, and can have either 1 or 2 digits.
When you convert this decimal number to the binary format, we have 111001001 that has 9 digits so 9bits is required to represent it in normal case. To convert decimals to binary visit http://acc6.its.brooklyn.cuny.edu/~gurwitz/core5/nav2tool.html
10 bits would be required. 10 bits long (10 digits long) can represent up to 1024.
If this is a homework assignment, please consider trying to answer it yourself first, otherwise the value of the reinforcement of the lesson offered by the assignment will be lost on you.The largest decimal number with 25 digits is 9,999,999,999,999,999,999,999,999.The smallest decimal number in the form 2n-1 which is greater than or equal to that is 19,342,813,113,834,066,795,298,815. That corresponds to 284-1.So, the minimum number of binary bits required to represent the decimal number 25 nines in a row is 84. This is 84 ones in a row. If you want to support negative as well as positive numbers, you will need 85.Since the largest integer in most compilers is 64 bits, this will require a special library supporting 128 bits, or an arbitrary length decimal library, if you want to manipulate such large numbers in a computer and still retain the precision of an integer.