Question 1036173
you are given that one kilybote of computer memory will hold about 2/3 of a page of a typical text (without formatting).


a typical book has about 400 pages.


2/3 of a page is equal to one kilobyte of memory.


if you take 400 pages and divide them by 2/3 of a page, you will get 400 / (2/3) = 400 * 3/2 = 1200/2 = 600 * 2/3 of a page being contained in a 400 page book.


since 2/3 of a page is handled by one kilobyte of memory, then 600 * 2/3 of a page is equal to 600 kilobytes of memory.


this means that one book requires 600 kilobytes of memory.


1 gigabyte is equal to 1,000 megabytes which is equivalent to 1,000 kilobytes.


that means that 1 gigabyte is equal to 1,000 * 1000 = 1 million kilobytes.


so 9 gigabytes is equal to 9 million kilobytes.


since 1 book requires 600 kilobytes, then you divide 9 million by 600 and you get 15000 books that can fit into a 9 gigabyte flash storage device.


the 9 gigabyte flash storage device can handle thousands of 400 page books.