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computer architecture
The sum of the two 32 bit integers may not be representable in 32 bits. In this case, we say that an overflow has occurred. W
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6)
-7.425 in simple binary => 111.01101100110011001100110011001100110011001100110011
so, -7.425 in normal binary is 111.01101100110011001100110011001100110011001100110011
=> 1.1101101100110011001100110011001100110011001100110011 * 2^2

single precision:
--------------------
sign bit is 1(-ve)
exp bits are (127+2=129) => 10000001
frac bits are 11011011001100110011001

so, -7.425 in single-precision format is 1 10000001 11011011001100110011001
in hexadecimal it is 0xC0ED9999

double precision:
--------------------
sign bit is 1(-ve)
exp bits are (1023+2=1025) => 10000000001
frac bits are 1101101100110011001100110011001100110011001100110011

so, -7.425 in double-precision format is 1 10000001 1101101100110011001100110011001100110011001100110011



as per guideltnes we are only allowed to answer one question Please make new posts for remaining questions Thanks

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computer architecture The sum of the two 32 bit integers may not be representable in 32 bits. In this case, we say that an overflow has occurred. Write MIPS instructions that adds two numbers stor...
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