First things first you have to do write it down which you did
step 2: then you multiply each digit of the binary number by the corresponding power of 2: 1x25 + 0x24 + 1x23 + 1x22 + 0x21 + 1x2 0
step 3: solve the powers: 1x32 + 0x16 + 1x8 + 1x4 + 0x2 + 1x1 = 32 + 0 + 8 + 4 + 0 + 1
step 4: add the numbers written above
32 + 0 + 8 + 4 + 0 + 1 = 45.
So, 45 is the decimal equivalent of the binary number 101101.
Answer:
for(i = 0 ; i < NUM_VALS; ++i)
{
cout << courseGrades[i] << " ";
}
cout << endl;
for(i = NUM_VALS-1 ; i >=0 ; --i)
{
cout << courseGrades[i] << " ";
}
cout << endl;
Explanation:
The first loop initializes i with 0, because we have to print the elements in order in which the appear in the array. We print each element, adding a space (" ") character at its end. After the loop ends, we add a new line using endl.
The second loop will print the values in a reverse order, so we initialize it from NUM_VALS-1, (since NUM_VALS = 4, and array indices are 0,1,2,3). We execute the loop till i >= 0, and we print the space character and new line in a similar way we executed in loop1.
Answer:
while(userNum>=1){
System.out.print(userNum/2+" ");
userNum--;
}
Explanation:
This is implemented in Java programming language. Below is a complete code which prompts a user for the number, receives and stores this number in the variable userNum.
<em>import java.util.Scanner;</em>
<em>public class TestClock {</em>
<em> public static void main(String[] args) {</em>
<em> Scanner in = new Scanner (System.in);</em>
<em> System.out.println("Enter the number");</em>
<em> int userNum = in.nextInt();</em>
<em> while(userNum>=1){</em>
<em> System.out.print(userNum/2+" ");</em>
<em> userNum--;</em>
<em> }</em>
<em> }</em>
<em>}</em>
The condition for the while statement is userNum>=1 and after each iteration we subtract 1 from the value of userNum until reaching 1 (Hence userNum>=1)
