Question

Assume no syntax errors and all partial programs are correct and can be run.

Negative numbers are not accepted as final answers. You will shortly receive this final as a MS WORD document. Fill out this MS WORD document and return it to my IVC email address by May 23.

Question 2

Please answer the following computer assembly language question.

A) 0.23652₁₀ => N₁₀/M ₁₀=>N_{16 /} M₁₆ where N₁₀=____________ and M_{10 =_______________} AND

N₁₆ = _________ M₁₆ = _________ .

B) Solve 1₅ /2₅= (0.a₁ a₂ ...a_n )₅= ______________

C) The additive inverse of the dword 9EFF3A₁₆ is ______________

D) 0.2 ₁₆ => (0.a₁ a₂ ...a_n )₁₀ = __________

E) Write the following expression as a single floating point number: . (4.35678)(0.078997) = N E M ___________ where N and M are integers.

Answer 1

Solution: A) 0.2365210- 236520 5C64, 10000010 186A016 Thus, N10 23652 Mi0-100000 N6 5C64 M16 186A0 B) 15 /2s 0.222222222222223 The last digit a can be rounded to 3 all others are 2. Thus, al, a2, a3, ...an-12 a3 C) Additive inverse can be obtained by flipping all the bits and adding a one at the end 9EFF3A16flipping all the bits give-6100D5E9 Add 1to 6100DSE9 gives 6100DSEA Additive inverse is 6100D5EA D) 0.2 16 (0.125)10

E) 4.35678*0.078997 0.34417254966-34417254966 E-11 Thus, N= 34417254966 and M=-11