Answer:
Except C. Increase in multimedia quality
Explanation:
The increase in technological advancements has posed some ethical questions in the areas of data storage improvement, data analysis advancements and an increase in the use of mobile devices.
However, ethical issues are not raised over the increase in multimedia quality. Because the better the quality the better the service is enjoyed. The possible concern about multimedia quality is maintaining high-quality multimedia while reducing the size of the multimedia file (data compression).
When associating an event handler with a button click which event occurs the user hits the right mouse button, the outcome, if any, is That hurt.
How can the result be established?
The flow of the programme indicates that:
When the user presses the button, the application prints "That hurt" without any quotation marks.
This operation may be found in the third line of the programme.
Therefore, the outcome is (b) That hurt if the user clicks the right mouse button.
Details on the programme are available at:
brainly.com/question/16397886
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The answer is all of the above
Answer:
Question is incomplete.
Assuming the below info to complete the question
You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.
Detailed answer is written in explanation field.
Explanation:
We have to find the reachability using the directed graph G = (V, E)
In this V are boxes are considered to be non empty and it may contain key.
Edges E will have keys .
G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.
Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.
Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.
If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.
After first search from B we can start marking all other vertex of graph G.
So algorithm will be O ( V +E ) = O (n+m) time.
