Answer:
13:a. 15:c. 14:Unknown answer
Answer:
public String bananaSplit(int insertIdx, String insertText) {
return word.substring(0, insertIdx) + insertText + word.substring(insertIdx);
}
Explanation:
Do you have the other parts of the WordGames class?
Identify, evaluate, prioritize, and then control
Answer:
The receiver will not detect the error.
Explanation:
The byte sent by transmitter: 10101010
The byte received by receiver due to channel noise: 10011010
If you see the bold part of the both sent and received bytes you can see that the number of bits changed is 2.
The two communicating devices are using a single-bit even parity check. Here there are two changed bits so this error will not be detected as this single bit even parity check scheme has a limit and it detects the error when the value of changed bit is odd but here it is even.
This parity scheme basically works well with the odd number of bit errors.
