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anonymous
 3 years ago
HELP...
Daisy purchased a rabbit couple to keep as pets. Perhaps Daisy got more than she bargained for! If the population of rabbits doubles every 3 months, how many rabbits will Daisy have at the end of 3 years? (Hints: Remember that your starting number is 2. The exponent is determined by the number of periods of doubling. Use your calculator.)
a.2048
b.512
c.256
d.8192
D??
anonymous
 3 years ago
HELP... Daisy purchased a rabbit couple to keep as pets. Perhaps Daisy got more than she bargained for! If the population of rabbits doubles every 3 months, how many rabbits will Daisy have at the end of 3 years? (Hints: Remember that your starting number is 2. The exponent is determined by the number of periods of doubling. Use your calculator.) a.2048 b.512 c.256 d.8192 D??

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Shane_B
 3 years ago
Best ResponseYou've already chosen the best response.1This is an exponent growth problem. To solve it, you first need to know the formula:\[y(t)=ae^{kt}\]where t is the time, a is the initial amount and k is the rate of growth. For this problem, the formula would start out as as follows since you know that the amount doubled from 2 to 4 in 3 months:\[4=2e^{3k}\]Now solve for k:\[2=e^{3k}\]\[ln(2)=ln(e^{3k})\]\[ln(2)=3k\]\[\frac{ln(2)}{3}=k\] Now that you know k, plug in t for 3 years (36 months) with a starting amount of 2:\[\Large y(36)=2e^{(\frac{ln(2)}{3}*36)}=?\]
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