QUESTIONS points in working with the IEEE 754 binary representation of the number 0.5 in single precision, i.e. 32 bits, what are the sign bit and exponent bits in binary? oo 0111 1110 1 01111110 0 0 01111111 00 0011 1110 00 01111100 QUESTION Spaints From the truth table, select the right output function in term of the minterms used in Karnaugh Map. ABCD Output 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1 1 0 0 1 0 0 0101 1 0 0 0 100 0 1 0 0 0110 1001 010 1 0 0 110 OF(A, B, C, D) - 2 (0.1,3,4,5,7,8,11) Of(A, B, C, D) - (0.1.3.4.5.7.10) (A,B,C,D) = 2(0.2.4,5,6,7,8,12) OF(A, B, C, D) -2(0,3.4.5.7.8.10) of(A, B, C, D) -2(0,1,3,4,5,7,8,10) QUESTION spines What would be the change in the memory after executing all the instructions? Declaration: int a[6]; Arrays are adjacent locations in memory storing the same type of data object Indexing a[0] - 0x015; a (array name) returns the array's address a [5] = a[0]; Ga[1] is the address of a [0] plus i times No bounds a[6] - OXBAD; the element size in bytes checking: al-1) - OXBAD; O O Out 0x20x3 One 015 016 0x7 Pointers: Os8 09 OA OB OC OD Out Oxf Ox00 pa; Ox08 P- &a[0]; AD OB 00 00 *p - OxB; a [0] 0x10 SF 01 00 00 a[2] Ox18 a[4] Ox20 5F01 0000 0x28 AD OB.00.00 Ox30 Ox38 inte p; ] P Ox0 0x48 the value stored in a[0] would be the same the value stored in a(5) will change the value stored in a[0] would be 00 00 00 0B the value stored in address Ox40 will be 00 00 00 00 00 00 00 10 the value stored in address 0x40 will be 10 00 00 00 00 00 00 00 Speints QUESTION Simplify the expression using K-maps: HABC)-3(1,3,5,6,7) AB C ABC A'B.C O A BC OAB'C QUESTIONS points in working with the IEEE 754 binary representation of the number 0.5 in single precision, i.e. 32 bits, what are the sign bit and exponent bits in binary? oo 0111 1110 1 01111110 0 0 01111111 00 0011 1110 00 01111100 QUESTION Spaints From the truth table, select the right output function in term of the minterms used in Karnaugh Map. ABCD Output 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1 1 0 0 1 0 0 0101 1 0 0 0 100 0 1 0 0 0110 1001 010 1 0 0 110 OF(A, B, C, D) - 2 (0.1,3,4,5,7,8,11) Of(A, B, C, D) - (0.1.3.4.5.7.10) (A,B,C,D) = 2(0.2.4,5,6,7,8,12) OF(A, B, C, D) -2(0,3.4.5.7.8.10) of(A, B, C, D) -2(0,1,3,4,5,7,8,10) QUESTION spines What would be the change in the memory after executing all the instructions? Declaration: int a[6]; Arrays are adjacent locations in memory storing the same type of data object Indexing a[0] - 0x015; a (array name) returns the array's address a [5] = a[0]; Ga[1] is the address of a [0] plus i times No bounds a[6] - OXBAD; the element size in bytes checking: al-1) - OXBAD; O O Out 0x20x3 One 015 016 0x7 Pointers: Os8 09 OA OB OC OD Out Oxf Ox00 pa; Ox08 P- &a[0]; AD OB 00 00 *p - OxB; a [0] 0x10 SF 01 00 00 a[2] Ox18 a[4] Ox20 5F01 0000 0x28 AD OB.00.00 Ox30 Ox38 inte p; ] P Ox0 0x48 the value stored in a[0] would be the same the value stored in a(5) will change the value stored in a[0] would be 00 00 00 0B the value stored in address Ox40 will be 00 00 00 00 00 00 00 10 the value stored in address 0x40 will be 10 00 00 00 00 00 00 00 Speints QUESTION Simplify the expression using K-maps: HABC)-3(1,3,5,6,7) AB C ABC A'B.C O A BC OAB'C