Answer:
ms word can help you to writer and print
ur document and make u more attractive
Explanation:
a
Answer:
Top down design
Explanation:
Top-down design is an approach that is used to break down the problem into the smaller subpart so that it can be manageable into more clear form.
C programming is the example of a top-down approach while C++ is the example of the bottom-up approach.
The advantages of the top-down design approach are:
1) easy to manage
2) easy to find the error
3) easy to debug
The program is an illustration of loops.
Loops are used to perform repetitive and iterative operations.
The program in C++ where comments are used to explain each line is as follows:
#include <iostream>
using namespace std;
int main(){
//This declares and initializes all variables
string star = "*", blank = " ", temp;
//The following iteration is repeated 8 times
for (int i = 1; i <= 8; i++) {
//The following iteration is repeated 8 times
for (int j = 1; j <= 8; j++) {
//This prints stars
if (j % 2 != 0) {
cout << star;
}
//This prints blanks
else if (j % 2 == 0) {
cout << blank;
}
}
//This swaps the stars and the blanks
temp = star;
star = blank;
blank = temp;
//This prints a new line
cout << endl;
}
}
Read more about similar programs at:
brainly.com/question/16240864
Answer
First part:
The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.
Second part:
The invalid bit sequence are option a. 01001000 and d. 11100111
Explanation:
Explanation for first part:
In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.
If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.
If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.
Explanation for second part:
A valid odd parity bit sequence will always have odd number of 1s.
Since in option a and d, total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.
And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.
#include using namespace std;int main(){int year = 12,value = 10,total = 0;do{year++;value *= 2;total += value;}while(value*2 < 1000);cout << "Age: " << year << endl;cout << "Last gift: " << value << endl;cout << "Total: " << total << endl;cin.get();return 0;
