1/3 sign 1, exponent 10, significand 12

A.  -299 

     299
     149  1  LSB
      74  1 
      37  0
      18  1
       9  0
       4  1
       2  0
       1  0
       0  1  MSB   1 0010 1011 
 
       Note we can store 12 signficant bits and we have 8 digits
       following the most significant bit. So we only need 5 more.
       4 more bits to record and one for rounding the LSB.

B. Convert decimal fraction portion.

    . .87
    1 .74  
    1 .48
    0 .96
    1 .92
    1 .84  with rounding   .1110

        
C. Adjust the binary value to scientific style notation.
  
   100101011.1110 = 1.0010 1011 1110 x 2^8

D. Calculate the bias for the exponent

  2^(10-1)-1 = 2^9-1 = 512-1 = 511

E. Add bias to exponent and covert to binary

   511+8 = 519

   519
   259 1  LSB
   129 1
    64 1
    32 0
    16 0
     8 0
     4 0
     2 0
     1 0
     0 1  MSB   10 0000 0111

F.  
Sign negative, sign bit 1

Biased exponent       10 0000 0111

Find significand

  Most significant bit 1. not stored

  And float can hold only 12 significant digits, so we store only the most 
  significant portion.

  1.0010 1011 1110
  ^ ^^^^ ^^^^ ^^^^
  h  12 digits
  i
  d
  d
  e
  n

  110 0000 0111 0010 1011 1110 
  ^^----------^ ^------------^
  s  exponent     significand
  i
  g
  n