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Help me with the question

Help me with the question-example-1
User AIMABLE
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10 images per day. Since it can receive 3 mb per second for 11 hours a day, that’s up to 118,800 megabits it can receive in one day. By multiplying the amount of gigabits in a typical picture (11.2) by the amount of megabits in a gigabit (1024) you get that there’s 11,468.8 megabits in each picture. Lastly, divide the number of megs that the station receives in one day by the amount of megs in a picture, and you get 10 and some change, therefore it can receive up to ten FULL pictures in a day
User DMTintner
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