4. Suppose that a single bacterium is in a bottle at 11:00 am. It divides into two bacteria at 11:01 am, and the population continues to double every minute until the bottle is completely full at 12:00 noon. Find how many bacteria there are at 11:26am. Show work or calculations! (4.1). (4.2) Work: 11:00-11.26 = 26 min + = 26 = 20 = 67,108,864 = 11:26 am Answer: 11:26 am = 67,108,864 Find the fraction of the bottle that is full at 11:55 am. (Hint, work backward!) Show calculations or give an explanation. Work: Answer:
4. Suppose that a single bacterium is in a bottle at 11:00 am. It divides into two bacteria at 11:01 am, and the population continues to double every minute until the bottle is completely full at 12:00 noon. Find how many bacteria there are at 11:26am. Show work or calculations! (4.1). (4.2) Work: 11:00-11.26 = 26 min + = 26 = 20 = 67,108,864 = 11:26 am Answer: 11:26 am = 67,108,864 Find the fraction of the bottle that is full at 11:55 am. (Hint, work backward!) Show calculations or give an explanation. Work: Answer:
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:4. Suppose that a single bacterium is in a bottle at 11:00 am. It divides into two bacteria at 11:01 am,
and the population continues to double every minute until the bottle is completely full at 12:00 noon.
Find how many bacteria there are at 11:26am. Show work or calculations!
(4.1).
(4.2)
Work: 11:00-11.26 = 26 min
+ = 26
= 20 = 67,108,864
= 11:26 am
Answer:
11:26 am = 67,108,864
Find the fraction of the bottle that is full at 11:55 am. (Hint, work backward!) Show
calculations or give an explanation.
Work:
Answer:
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