based on the following code mov al,OFFh cmp al,0 jg negative which of the following is true a. al register will be negative O b. no jump will be made to label negative c. jump will be made to label negative O if the address lines are 11 bits, then the number of memory locations these bits can address is a. 1 Mega locations b.4 kilo locations c. 2 kilo locations d. 1 Kilo locations If AL= 2 after the execution of the instruction "SHLAL,4" the value of AL becomes: Select one: O a 4 b. 8 O C. 32 O d. 16 Given AX = 0009H, DX = 0000H.CX = 0002H, ?what is the result of the instruction: DIV CX a. AX = 0004H, DX = 0001H O b. AX = 0009H, DX = 0000H O C. AX = 0000H, DX = 0009H d. AX = 0001H, DX = 0004H O