Answer:
Written in Java
public static void printArray(int myarr[], String s){
for(int i = 0; i<myarr.length;i++){
System.out.print(myarr[i]+s);
}
}
Explanation:
This defines the static method alongside the array and the string variable
public static void printArray(int myarr[], String s){
The following iteration iterates through the elements of the array
for(int i = 0; i<myarr.length;i++){
This line prints each element of the array followed by the string literal
System.out.print(myarr[i]+s);
}
}
The method can be called from main using:
<em>printArray(myarr,s);</em>
Where myarr and s are local variables of the main
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
Answer:
Apply recommendations across the multiple layers of his advertising strategies.
Explanation:
In the given statement, Steven is an employee of an organization of auto parts and he uses own google advertisement page for the purpose of strategies to make perfect he owns advertisement campaigns for the google search. He values the scores of the optimization because Steve applies charge over the different layers of his ads strategies.
Answer:g
public static int addOddMinusEven(int start, int end){
int odd =0;
int even = 0;
for(int i =start; i<end; i++){
if(i%2==0){
even = even+i;
}
else{
odd = odd+i;
}
}
return odd-even;
}
}
Explanation:
Using Java programming language:
- The method addOddMinusEven() is created to accept two parameters of ints start and end
- Using a for loop statement we iterate from start to end but not including end
- Using a modulos operator we check for even and odds
- The method then returns odd-even
- See below a complete method with a call to the method addOddMinusEven()
public class num13 {
public static void main(String[] args) {
int start = 2;
int stop = 10;
System.out.println(addOddMinusEven(start,stop));
}
public static int addOddMinusEven(int start, int end){
int odd =0;
int even = 0;
for(int i =start; i<end; i++){
if(i%2==0){
even = even+i;
}
else{
odd = odd+i;
}
}
return odd-even;
}
}
I think it’s cropping could be wrong though
