Assuming a cache of 4096 blocks, a four-word block size, each word contains 4 bytes, and a 64-bits memory address. Memory is byte addressable, and randomblock replacement policyand write back policy are implemented.1)[16 points] Find the total number of tag bits for caches that area)direct-mappedb)2-way set associativec)4-way set associatived)Fully associative2)[4 points]Calculate the ratio between total bits over the data storage bits for direct-mapped cache, 2-way set associative cache, 4-way set associative cache and fully associative cache.

given the options, i would say your best bet is c. the modem supports up to 56 kbps, which is too low for video streaming. dial up modems use actual phone lines, which only can go up to 56 kbps as that's more than sufficient for traditional phone calls. this is also the reason why when a dial-up modem connects, you hear all those sounds. it's literally using the phone line to establish a connection.

i do believe its (c); they are required to perform any summarization or redistribution tasks

explanation:

solution:

this situation is called an ip conflict.

data from a server is sent in small bits called packets. in an ip conflict, some packets will end up in one computer while some other packets will end up in another computer.

so…overall none of the computers get the data as the original message or picture being sent by the server cannot be retrieved with only half the packets. so it will show up as an error.

thus the final answer is d.