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# binary assignment

Question # 00008190
Subject: Statistics
Due on: 02/28/2014
Posted On: 02/10/2014 12:47 AM

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Question
1. Perform the following base conversions.
a. 101012 = ________ 10
b. 8110 = _________ 2
c. 76310 = ________ 5
d. 65510 = ________ 16

2. Represent the following decimal numbers in binary using 8-bit signed magnitude, one's
complement and two's complement formats.
a. 115
b. ?93

3. Show how each of the following decimal floating point values would be stored using
IEEE-754 single precision (be sure to indicate the sign bit, the exponent, and thesignificand fields):
a. -13.25
b. 0.625
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#### binary assignment

Tutorial # 00007808
Posted On: 02/10/2014 12:48 AM
Posted By:
spqr
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Tutorial Preview …bits: x 10101 xxxxx the decimal xxx moved three xxxxxx to xxx xxxxx the xxxxxxxx will be x (See next xxxxxxx ) xxx xxxxxxx 1 xxx the decimal xxxxx are dropped xx get: xxxxx xxx significand xxxx be 23 xxxx long, so xx need xx xxx 23 xxx 5 = xx bits (with xxxxx 0) xx xxx end: xxxxxxxxxxxx 10101000000000000000000 Exponent: xxx exponent is xxxxxx by xxxx xx the xxxxxxx value of xxx exponent here xxxx be x x 127 x 130 The xxxxx binary representation xx the xxxxxxx xxxxx 130 xx 1000 0010 xxxxxxx all three xxxxx (sign xxxx xxxxxxxxx and xxxxxxxxxxxx together gives: x 10000010 10101000000000000000000 xxxxxxxxx the xxxx xx groups xx four gives: xxxx 0001 0101 xxxx 0000 xxxx xxxx 0000 xxxxxxxxx Representation: 1100 xxxx 0101 0100 xxxx 0000 xxxx xxxx Sign xxxx 1 Exponent: xxxx 0010 Significand: xxx 0100 xxxx xxxx 0000 xxxx b 0 xxx Sign bit: xxx value xx xxxxxxxxx so xxx sign bit xxxx be 0 xxxxxxxxxxxx The xxxxx xxxx of xxx value is xxxx 0,…
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Binary.docx (51.23 KB)
Preview: the xxxx bit xxxx 32) will xx set Significand:The xxxxx 13 xx xxxxxxxxxxx as xxxxx so the xxxxxxxxxxx is initially xxxx , xxxx xxx decimal xxxxx representing the xxxxxxxxxx between the xxxxx number xxxxxxx xx 13 xx and the xxxxxxxxxx portion To xxxxxxx the xxxxxxxxx x 25 xxxxxxxx repeatedly multiply xx 2 and xxxx the xxxxx xxxx the xxxxxxx place as x bit value xxx the xxxxxxxxxx xxxxxxx is xx add the xxxx in order xx the xxxxx xxxxxx portion xxxx above:0 25 x 2 = x 5 xxx xx 00 x * 2 x 1 0 xxx is xxxx xxxxxxxxxx portion xx now 0 xx we’re done xxxxxx “01” xx xxx whole xxxxxx representation gives:1101 xxxxx decimal must xx moved xxxxxxx xxx first xxx second bits:1 xxxxxxxxxx the decimal xxx moved xxxxx xxxxxx to xxx left, the xxxxxxxx will be x (See xxxx xxxxxxx )The xxxxxxx 1 and xxx decimal point xxx dropped xx xxxxxxxxxxxx significand xxxx.....
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