After every hour, you have 88 percent of the caffeine in your bloodstream that you had the previous hour. You're looking to get to 50 percent of your original caffeine level. So you can set the equation up as follows:
0.88^t = 0.5
Where t is the number of hours. The 2 cups and 140 mg of caffeine are irrelevant.
To solve, take the natural log of both sides:
t*ln(0.88) = ln(0.5)
t = ln(0.5)/(ln(0.88))
I don't have a calculator on me so I can't tell you what that number equals, but thats' how you solve it.
Your friend,
Bigferribunny
2007-10-09 14:07:46
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answer #1
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answered by Bigferribunny 2
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You have 140MG of caffeine, you need to metabolize until you've got under 70MG, and if you metabolize -12% a hour, just take 140 and subtract 12%, you'll then have 123.2MG remaining after one hour. Subtract another 12% for the next hour, and you've got 108.4MG after two hours. Do that until you've got under 70MG, 7 hours.
2007-10-09 14:11:51
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answer #2
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answered by Zach K 1
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I won't give you your answer, but all you have to do to solve it is to add 140mg and 140mg (for the two cups). Multiply the 280 times .12 (12%) to get how much is depleated after the first hour. Subtract that from 280mg and repeat until you get below 140mg.
2007-10-09 14:08:08
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answer #3
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answered by huskerchica 2
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the total caffeine is 280 mg
12% of 280 is 33.6mg.
if 33.6 mg takes 1 hour, lets find how long it takes for 140 mg(half of caffeine in his body)
33.6 : 140 = 1 : x
33.6 x = 140
x = 4.16 hours
2007-10-09 14:07:53
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answer #4
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answered by Anonymous
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Here a couple of hints:
How much total caffinee does he consume?
If in the first hour the body takes care of 12%, how much is left after one hour?
2007-10-09 14:06:26
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answer #5
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answered by peachie 2
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Hello,
We have 140 = 280 ( 1 - 12%)^t so .5 = .88^t now take the log of both sides and we have: log .5 = t* log(.88) divide both sides by log .88 giving us log.5/log.88 = t so t = 5.4 hours
Hope this helps!!
2007-10-09 14:14:14
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answer #6
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answered by CipherMan 5
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Where are you parents? Quit having everyone else do your homework
2007-10-09 14:03:48
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answer #7
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answered by Anonymous
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