Answer:
C. Byte pair encoding is an example of a lossless transformation because an encoded string can be restored to its original version.
Explanation:
Byte pair encoding is a form of encoding in which the most common pairs of consecutive bytes of data are replaced by a single byte which does not occur within the set of data.
For example, if we has a string ZZaaAb, it can be encoded if the pairs of string ZZ are replaced by X and the second pair by Y. So, our data now becomes XYAb.
To get our original data, that is decode it, we just replace the data with the keys X = ZZ and Y = aa thus allowing our original data to be restored.
Since our original string is restored without loss of data, it implies that <u>byte pair encoding is an example of a lossless transformation because an encoded string can be restored to its original version.</u>
Answer:
When myMethod is invoked by a method in the same class as myMethod.
Explanation:
When you are in a class, it is like a different enviroment and within that class you can call a method without using the dot notation because you are still within the class.
Outside the class and despite the type of method, public, private or static, the dot notation will be required.
Answer:
The correct answer to the following question will be "Magnetic medium".
Explanation:
Some storage device that portrays details or information using magnetic patterns is called a magnetic device.
- Magnetic tape, and hard drives are the commonly used magnetic storage devices.
- This will be the method by which magnetic methods are used to hold and retrieve information on tapes.
So, it's the right answer.
Answer:
0.8488
Explanation:
Let E =error found by test 1
Let F=error found by test 2
Let G=error found by test 3
Let H=error found by test 4
Let I= error found by test 5
Given P(E)=0.1, P(F)=0.2, P(G)=0.3, P (H)= 0.4, P (I)=0.5
therefore P(notE)=0.9, P(notF)=0.8, P(notG)=0.7, P(not H)=0.6, P (notI)=0.5
Tests are independent P(not E & not F ¬ G & not H & not I=P(notE)*P(notF)*P(notG)*P (notH)*P (not I) =0.9*0.8*0.7*0.6*0.5 =0.1512
P(found by at least one test)= 1- P(not found by any test)=1-P(not E& not F & not G & not H & not I ) = 1-0.1512 = 0.8488
