Computers and Technology, 31.10.2021 01:00, babyboogrocks5695
Based on the point and illustration, which is the best explanation or analysis of the evidence?
Answers: 1
Computers and Technology, 22.06.2019 15:10, AleciaCassidy
Consider a direct-mapped cache with 216 words in main memory. the cache has 16 blocks of 8 words each. it is a word-addressable computer (rather than a byte-addressable computer which we normally discuss). (a) how many blocks of main memory are there? (b) what is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, cache block, and block offset fields (if they apply)? (c) to which cache block will the memory reference db6316 map?
Answers: 1
Computers and Technology, 23.06.2019 00:30, lilobekker5219
Knowing that the central portion of link bd has a uniform cross sectional area of 800 mm2 , determine the magnitude of the load p for which the normal stress in link bd is 50 mpa. (hint: link bd is a two-force member.) ans: p = 62.7 kn
Answers: 2
Computers and Technology, 23.06.2019 03:10, nxusasmangaliso8780
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; }
Answers: 2
Computers and Technology, 23.06.2019 03:10, kyleereeves2007
Acomputer has a two-level cache. suppose that 60% of the memory references hit on the first level cache, 35% hit on the second level, and 5% miss. the access times are 5 nsec, 15 nsec, and 60 nsec, respectively, where the times for the level 2 cache and memory start counting at the moment it is known that they are needed (e. g., a level 2 cache access does not even start until the level 1 cache miss occurs). what is the average access time?
Answers: 1
Based on the point and illustration, which is the best explanation or analysis of the evidence?...
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