Let assume are lettered A to E in that order. Thus, there
will be 10 potential lines: AB, AC, AD, AE, BC, BD, BE, CD, CE and DE. Each of
these potential lines has 4 possibilities. Therefore, the total number of
topologies is 4¹⁰=1,048,576. 1,048,576. At 100ms <span>it will take 104,857.6 seconds which is slightly above 29 hours to inspect
each and one of them.</span>
Answer:
im not sure
Explanation: is their supposed to be a picture or something?
This is the correct Answer <span>Attribute</span>
True.
It doesn't measure unique users, so you can just hit refresh and the counter keeps going up...
Answer:
D. The 4th one
Explanation:
The loops are nested, and the program ends when loop 1 is completed. Since loop 4 is the innermost one, that one is completed first.
