33.3% of 12,500 = 33.3% * 12500 = 0.333 * 12500 = 4,162.5
12500
12500
how many bits are needed to represent decimal values ranging from 0 to 12,500?
The 5 percent of 250000 is 12500.formula: {(amount/100)*given percentage}.(250000/100)*5=12500.RegardsRajpar
If you want an unsigned integer, you can calculate this as log2(12500), rounded up. You can calculate the log2 function (logarithm to the base 2), in this case, by dividing ln(12500) / ln(2).
0.008 percent:= 0.008%= 0.00008 in decimal= 8/100000 or 1/12500 in fraction
6% of 12500 = 6% * 12500 = 0.06 * 12500 = 750
25% of 12500 = 25/100*12500 = 3125
33.3% of 12,500 = 33.3% * 12500 = 0.333 * 12500 = 4,162.5
12500
36/12500*100=0.288 Hence 36 is 0.228% of 12500
12500
12500/3 = 4166.66...
It is: 12500+13500 = 26000
how many bits are needed to represent decimal values ranging from 0 to 12,500?
12500