Hello!
The correct answer would be: D. Should not exceed a vehicle's rated load or life capacity.
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Bbetter I think glasses are better this year but also the dates. your eyes over tim adjust to things like glasses to help see so when yu got better glasses it got better
Answer:
Linear problems
Explanation:Grid computing is the term used in Information technology, Computer programming and Computer networks to describe the interconnection of different Computer resources in order to achieve certain specified goal. GRID COMPUTING ALLOWS COMPUTER RESOURCES TO WOK AS A SINGLE UNIT.
Grid computing can be used to solve various issues connected with the use of Computer resources such as Financial risk modelling,Gene analysis etc but least likely to be used to solve Linear problems.
Alternative 1:A small D-cache with a hit rate of 94% and a hit access time of 1 cycle (assume that no additional cycles on top of the baseline CPI are added to the execution on a cache hit in this case).Alternative 2: A larger D-cache with a hit rate of 98% and the hit access time of 2 cycles (assume that every memory instruction that hits into the cache adds one additional cycle on top of the baseline CPI). a)[10%] Estimate the CPI metric for both of these designs and determine which of these two designsprovides better performance. Explain your answers!CPI = # Cycles / # InsnLet X = # InsnCPI = # Cycles / XAlternative 1:# Cycles = 0.50*X*2 + 0.50*X(0.94*2 + 0.06*150)CPI= 0.50*X*2 + 0.50*X(0.94*2 + 0.06*150) / X1= X(0.50*2 + 0.50(0.94*2 + 0.06*150) ) / X= 0.50*2 + 0.50(0.94*2 + 0.06*150)= 6.44Alternative 2:# Cycles = 0.50*X*2 + 0.50*X(0.98*(2+1) + 0.02*150)CPI= 0.50*X*2 + 0.50*X(0.98*(2+1) + 0.02*150) / X2= X(0.50*2 + 0.50(0.98*(2+1) + 0.02*150)) / X= 0.50*2 + 0.50(0.98*(2+1) + 0.02*150)= 3.97Alternative 2 has a lower CPI, therefore Alternative 2 provides better performance.
Answer:
It is not possible.
Explanation:
In this example, we need to accommodate 473 computers for six clients that are 473 IP addresses.
For this request just we have /22 IPv4 address blocks, this mean
22 red bits 11111111111111111111110000000000 <--- 10 host bits
We must increase red bits to 25, we need these 3 bits to create 6 sub red, in this case, 2^3 = 8 sub red.
Why did we ask 3 bits? Because if we ask only 2, 2^2 = 4, and we need 6 sub red.
25 red bits 11111111111111111111111110000000 7 host bits
In this case, we need more than 260 computers, but just we have 7 bits, this means.
2^7 = 128 and just one customer needs 260, for that is impossible.
