Answer: DoDM 5200.01, DoD Information Security Program
Explanation:
The document that provides basic guidance and regulatory requirements for derivative classification
for DoD personnel is referred to as the DoDM 5200.01, DoD Information Security Program.
The purpose of this is to help in the promotion of an effective way that can be used in the classification, protection, and application of applicable instructions.
Answer:
a. 6 bits
b. 1
Explanation:
Ans (a)
40 Characters need to be represent by binary coded Zebronian (BCZ) , So You will need 6 bits.
5 bits wold only give you 32 = 2 x 2 x 2 x 2 x 2 unique characters.
So 6 bits would allow you to represent 64 characters.
Ans(b)
BCD = Binrary Coded Decimal is very common in electronics, particularly it displays numerical data.
BCD Encodes each digit of a decimal number into 4 digit binary form.
Each decimal digit is indiviidually converted to oits binary equivalent
For Example : 146 , the decimal degits are replaced by 0001 , 0100 and 0110 respectively
Addition
1 0 = 10 is binary value of 2 2
+1 1 = 11 is binary value of 3 + 3
---------- -----------
1 0 1 5 Ans
Subtraction
1 1= binary value of 3 3
- 1 0 = binary value of 2 - 2
--------- -----------
0 1 1 Ans
Answer:
Explanation:
The following code is written in Java. It creates the Bug class with the position and direction variables. Then it creates a constructor, move method, turn method, and getPosition method. Finally, a bug object called bugsy is created in the main method, and we move it once to the right, then again to the right, and then we turn it and move it 5 times to the left, printing out the position when it is done moving. Output can be seen in the attached picture below.
class Brainly {
public static void main(String[] args) {
Bug bugsy = new Bug(10);
bugsy.move();
System.out.println("Current bug position: " + bugsy.getPosition());
bugsy.move();
System.out.println("Current bug position: " + bugsy.getPosition());
bugsy.turn();
bugsy.move();
bugsy.move();
bugsy.move();
bugsy.move();
bugsy.move();
System.out.println("Current bug position: " + bugsy.getPosition());
}
}
class Bug {
char direction = 'r';
int position = 0;
public Bug(int initialPosition) {
this.position = initialPosition;
}
public void turn() {
if (this.direction == 'r') {
this.direction = 'l';
} else {
this.direction = 'r';
}
}
public void move() {
if (this.direction == 'r') {
this.position += 1;
} else {
this.position -= 1;
}
}
public int getPosition() {
return this.position;
}
}
Answer:
A wave that has been digitized can be played back as a wave over and over, and it will be the same every time. For that reason, digital signals are a very reliable way to record information—as long as the numbers in the digital signal don’t change, the information can be reproduced exactly over and over again.
Explanation:
