-Simulate data for 1 amoeba in a glass container until you can see a definate pattern.

(I cannot show a table form, so I will use [.......] as separators)

t.............0,......1,......2,......3,......4,.. ....5,..

minutes....0,......3,......6,......9,......12,.... .15,..

amoebas...1,......2,......4,......8,......16,..... 32,..

pattern....2^0,..2^1,..2^2,...2^3,..2^4,...2^5,..

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-Creat a function of time (call it f(t).) which will show the number of amoebas at anytime.

f(t) = 2^t

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-How many amoeba does it take to fill the glass conatiner?

Since one t = 3minutes, then in one hour or 60 minutes, there are 60/3 = 20t

That means after 20t, the container is full of amoebas.

So, f(20) = 2^20 = 1,048,576 amoebas to fill the glass.

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-Simulate data for starting with 2 amoeba in a glass container until you can see a difinite patter.

t..............0,......1,......2,......3,......4,. .....5,..

minutes.....0,......3,......6,......9,......12,... ..15,..

amoebas....2,.....4,.......8,.....16,.....32,..

pattern....2^1,..2^2,..2^3,...2^4,...2^5,..

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-Create a function of time (call it g(t).) which will show the number of amoebas at anytime.

g(t) = 2^(t+1)

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-Use the function for starting with 2 amoeba and the knowledge of how many amoebas will fill the glass container to find out

how long it takes to fill the glass conatiner when starting with 2 amoebas (NOTE: it takes longer than half an hour.)

2^(t+1) = 1,048,576 = 2^20

2^(t+1) = 2^20

So,

t+1 = 20

t = 20 -1 = 19

In minutes, that is 19(3) = 57 minutes

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-show how to find the answer using data only

Compare the two tables above.

For 1 amoeba as starter, the number of amoebas at any time t is 2^t.

For 2 amoebas as starter, the number of amoebas at any time t is 2^(t+1)

That means the 2 amoebas are ahead by one t for the same number of propagated amoebas.

Since one t is 3 minutes, and since 1 amoeba can fill up the container in 1 hour or in 60 minutes, then 2 amoebas can fill the container in 3 minutes before 1 hour....and that is 57 minutes.