Select the correct statement(s) regarding Carrier Ethernet (CE). a. the Metro Ethernet Forum (MEF) created a CE framework to ensure the interoperability of service provider CE offerings b. MEF certifies CE network providers, manufacturers and network professionals to ensure interoperability and service competencies c. MEF certified services include E-Line, E-LAN, E-Tree, E-Access, and E-Transit d. all are correct statements

3. (6 pts) internally in the computer, with few exceptions, all numerical computation is done using binary numbers. output, however, often uses ascii, which is formed by appending 011 to the left of a bcd code. thus, an algorithm that directly converts a binary integer to a bcd integer is very useful. here is one such algorithm 1) draw lines to the left of the binary number to bound the expected bcd decades. (each decade is a group of 4 bits.) move the binary number one bit to the left. add 0011 to each bcd decade containing a binary value> 0100 repeat steps 2-3 until the last bit in the binary number has been moved into the least significant decade position. (note that when the last bit has been shifted into bcd decade, step 3 is not repeated.) read the bcd result. 2) 3) 4) 5) a) execute the algorithm for the binary number 1101101 b) execute the algorithm for the binary number 01110101110 4. (4 pts) represent the decimal number 3568 in bcd; excess-3 code; ascil; and hex.

Some of your friends have gotten into the burgeoning field of time-series data mining, in which one looks for patterns in sequences of events that occur over time. purchases at stock exchanges--what’s being bought-- are one source of data with a natural ordering in time. given a long sequence s of such events, your friends want an efficient way to detect certain "patterns" in them--for example, they may want to know if the four events buy yahoo, buy ebay, buy yahoo, buy oracle occur in this sequence s, in order but not necessarily consecutively. they begin with a collection of possible events (e.g., the possible’ transactions) and a sequence s of n of these events. a given event may occur multiple times in s (e.g., yahoo stock may be bought many times in a single sequence s). we will say that a sequence s’ is a subsequence of s if there is a way to delete certain of the events from s so that the remaining events, in order, are equal to the sequence s’. so, for example, the sequence of four events above is a subsequence of the sequence buy amazon, buy yahoo, buy ebay, buy yahoo, buy yahoo, buy oracle their goal is to be able to dream up short sequences and quickly detect whether they are subsequences of s. so this is the problem they pose to you: give an algorithm that takes two sequences of even~s--s’ of length m and s of length n, each possibly containing an event more than once--and decides in time o(m n) whether s’ is a subsequence of s

Fill in the following program so that it will correctly calculate the price of the orange juice the user is buying based on the buy one get one sale.#include //main functionint main() { int cartons; float price, total; //prompt user for input information printf("what is the cost of one container of oj in dollars? \n"); scanf(" [ select ] ["%d", "%c", "%f", "%lf"] ", & price); printf("how many containers are you buying? \n"); scanf(" [ select ] ["%d", "%c", "%f", "%lf"] ", & cartons); if ( [ select ] ["cartons / 2", "cartons % 1", "cartons % 2", "cartons % price", "cartons / price", "cartons / total"] [ select ] ["=", "==", "! =", "< =", "> =", "< "] 0) total = [ select ] ["price * cartons", "cartons * price / 2 + price", "(cartons / 2) * price", "cartons / (2.0 * price)", "(cartons / 2.0) * price + price", "((cartons / 2) * price) + price"] ; else total = ((cartons / 2) * price) + price; printf("the total cost is $%.2f.\n", total); return 0; }