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JamenS
Group Title
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??
 one year ago
 one year ago
JamenS Group Title
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??
 one year ago
 one year ago

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Shane_B Group TitleBest ResponseYou've already chosen the best response.1
This 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:\[\Large y(36)=2e^{(\frac{ln(2)}{3}*36)}=?\]
 one year ago

JamenS Group TitleBest ResponseYou've already chosen the best response.0
i got 8192?
 one year ago

Shane_B Group TitleBest ResponseYou've already chosen the best response.1
That's correct
 one year ago
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