<span>C. proofread her work carefully, ask a coworker to look it over, and correct all mistakes
</span>
Answer:
Please check options are not given. Please check explanation for corrected version.
Explanation:
Options are not mentioned. Please post the complete question.
However, if "below" is removed, the question makes sense. I am taking it that way.
Dichotomous question means those questions which has two outcomes: true or false.
For the given condition, this is possible only if:
- coin shows head and dice shows 1
- coin shows head and dice shows 2
- head, 3
- head, 4
- head, 5
- head, 6
- tails, and all above cases
However, each time, each mentioned condition should be strictly followed.
And fewer outcome than 1 is virtually or realistically impossible, as both coin and dice will roll out one outcome in any condition certainly.
// This command takes input from the user.
Scanner input = new Scanner(System.in);
int response= input.nextInt();
// Making the variables required
int noCount =0 ;
int yesCount =0;
// Checking the response if it is 1 or 2 and reporting accordingly
if(response == 1 || response ==2){
yesCount+=1;
System.out.println("YES WAS RECORDED");
}
// Checking if the input is 3 or 4 then printing the required lines
else if (response == 3 || response ==4){
noCount+=1;
System.out.println("NO WAS RECORDED");
}
// if the input is not valid than printing INVALID
else{
System.out.println("INVALID");
}
Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
