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Since $-13.5$ is negative number so $s = 1$

Exponent = $7$ bits, therefore bias = 63

$13.5$ in binary = $1101.1$ = $1.1011 * 2^3$

E = e + bias = $3 + 63$ = $66$ = $1000010$

We store number like

S (1bit) E (7 bits) M (8bits

 

1 1000010 10110000

 

$(1100001010110000)_{2}$ = $(C2B0)_{H}$

 

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Since is -13.5 is negative number so s=1

Exponent = 7 bits So, bias  = $2^{^{n-1}} – 1$ = 63 bits, therefore bias = 63 and not 64

number 1101.1 (i.e. 13.5)  after normalization is $1.1011*e^{^{3}}$

E = e + bias = 3 + 63 = 66

S (1bit) E (7 bits) M (8bits

 

1 1000010 10110000

 

in hexadecimal  = (C2B0)H

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