Answer:
The correct answer to the following question will be Error-detection.
Explanation:
Error-detection: The detection of errors caused during the transmission from the transmitter to the receiver by damage and other noises, known as Error-detection. This error-detection has the ability to resolute if something went wrong and if any error occurs in the program.
There are mainly three types of error-detection, these types can be followed:
- Automatic Repeat Request (ARQ)
- Forward Error Correction
- Hybrid Schemes
There are two methods for error-detection, such as:
- Single parity check
- Two-dimensional parity check
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Answer:
Option d is the correct answer for the above question.
Explanation:
- The first loop of the program has a second loop and then the statement. In this scenario, the second loop executes for the value of the first loop and the statement executes for the value of the second loop.
- The first loop executes 4 times, Then the second loop or inner loop executes n times for the n iteration of the first loop, for example, 1 time for the first iteration of the first loop, 2 times for the second iteration of the first loop and so on.
- Then the inner loop executes (1+2+3+4) iteration which gives the result 10 iterations.
- The sum initial value is 0 and the "sum++", increase the value of the sum by 1.
- So the value of the sum becomes 10 after completing 10 iterations of the inner for loop.
- Hence the 10 will be the output. So the Option d is the correct answer while the other is not.
Answer:
e(a) = 0
e(b) = 10
e(c) = 110
e(d) = 1110
Explanation:
The Worst case will happen when f(a) > 2*f(b) ; f(b) > 2*f(c) ; f(c) > 2*f(d) ; f(d) > 2*f(e) and f(e) > 2*f(f).
Where f(x) is frequency of character x.
Lets consider the scenario when
f(a) = 0.555, f(b) = 0.25, f(c) = 0.12, f(d) = 0.05, f(e) = 0.02 and f(f) = 0.005
Please see attachment for image showing the steps of construction of Huffman tree:- see attachment
From the Huffman tree created, we can see that endcoding e() of each character are as follows:-
e(a) = 0
e(b) = 10
e(c) = 110
e(d) = 1110
e(e) = 11110
e(f) = 11111
So we can see that maximum length of encoding is 5 in this case.
Answer: Tar.
Explanation:
A chemical substance made when tobacco is burned. Tar contains most of the cancer-causing and other harmful chemicals found in tobacco smoke. When tobacco smoke is inhaled, the tar can form a sticky layer on the inside of the lungs.
