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A computer has a cache, main memory and a disk used for virtual memory. If reference word is in cache $15\hspace{0.1cm} ns$ are required to access it. If it is in main memory but not in cache, $50\hspace{0.1cm} nsec$ are needed to load it into cache and then reference it from cache. If word not in main memory $\text{10 msec}$ required to fetch word from disk, followed by $50\hspace{0.1cm} nsec$ to copy it into cache and then reference from cache. If the cache hit is $90\%$ and main memory hit is $50\%$, then to access $50$ words, the average access time on this system will be____________. (in μsec)

is this correct equation .$90*15 + .10*(.50*(50+15)+.50*(10000000+50))$
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