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# What is the decimal value of "01111111111111111101111111111102"

What is the decimal value of "01111111111111111101111111111102" (in Two's complement)

Hello there:

0(2^0) = 0

1(2^1)= 2

1(2^2)= 4

1(2^3)=8

1(2^4)= 16

1(2^5)=32

1(2^6)=64

1(2^7)=128

1(2^8)=256

1(2^9)=512

1(2^10)= 1024

1(2^11)= 2048

0(2^12)= 0

1(2^13)=8192

1(2^14)= 16384

1(2^15)=32768

1(2^16)=65536

1(2^17)=131072

1(2^18)=262144

1(2^19)=524288

1(2^20)=(NNN) NNN-NNNN/p>

1(2^21)=(NNN) NNN-NNNN/p>

1(2^22)=(NNN) NNN-NNNN/p>

1(2^23)=8388608

1(2^24)= 16777216

1(2^25)=33554432

1(2^26)=67108864

1(2^27)=134217728

1(2^28)=268435456

1(2^29)=536870912

0(2^30)= 0

________________________+

If you would like to double-check the work, you may use the calculator here: http://mistupid.com/computers/binaryconv.htm

Customer: replied 4 years ago.
After getting all the values do we need to add all the numbers
Customer: replied 4 years ago.
Is this going to be same (in Two's complement)? or we need to more calculation