Suppose that a single bacterium is in a bottle at 11 am. It divides into 2 at 11:01, and the population continues to double every minute until the bottle is completely full at 12:00. Find the fraction of the bottle that is full at 11:53am.
A. 1/32
B. 1/64
C. 1/128
D. 1/512
2006-12-16 02:00:59 · 7 answers · asked by Tazzy375 3 in Science & Mathematics ➔ Mathematics
bacterial count at 11 am = 2^0 =1
bacterial count at 11:01 am = 2^1 = 2
bacterial count at 12 pm = 2^60
bacterial count at 11:53 am = 2^53
Fraction of the bottle that is full at 11:53am
=2^53/2^60
=2^7
=1/128
Answer is C.1/128
2006-12-16 02:04:29 · answer #1 · answered by Som™ 6 · 2⤊ 0⤋
Since the number doubles each minute the number of bacterium each minute is 2^minute.
So at 11:53 there are 2^53 and 12:00 2^60
So the fraction is 2^53/2^60 = 1/2^7 = 1/128
2006-12-16 10:08:45 · answer #2 · answered by mulla sadra 3 · 0⤊ 0⤋
11.53am was 7 minutes before 12:00. Therefore, during this time bacteria multiplied (2)^7 = 128 times and filled the whole bottle.
Thefore, at 11:53am it occupied 1/128'th part.
2006-12-16 10:05:25 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Full is 2^60
At 11:53 is 2^53
2^53/2^60 = 1/2^7 = 1/128
2006-12-16 12:53:37 · answer #4 · answered by Anonymous · 0⤊ 0⤋
C. 1/128
2006-12-16 10:11:02 · answer #5 · answered by math 2 · 0⤊ 0⤋
1/128
2006-12-16 10:04:04 · answer #6 · answered by raj 7 · 0⤊ 1⤋
C.
Now *you* figure out how I got that answer âº
Doug
2006-12-16 10:06:52 · answer #7 · answered by doug_donaghue 7 · 0⤊ 0⤋