anonymous
  • anonymous
oil consumption is increasing at a rate of 2.2% per year. What is the doubling time? by what factor will oil consumption increase in a decade?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Have you learned about the rule of 70 when figuring out doubling time?
anonymous
  • anonymous
no b
anonymous
  • anonymous
When I learned doubling time, I was taught to take the 70 and put the growth rate (r) in the denominator. 70/r(growth rate)

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anonymous
  • anonymous
I assume that p also symbolizes growth rate, so all you have to do is write 70/2.2
anonymous
  • anonymous
I have 2.2%*2^10years/1 year for the second part
anonymous
  • anonymous
The second part of the problem?
anonymous
  • anonymous
by what factor will oil consumption increase in a decade
anonymous
  • anonymous
Well once we divide 70/2.2 we get approximately 30 years, so we can divide that in 3 to make 1 decade. Does that make sense?
anonymous
  • anonymous
no its doesn't I am sorry
anonymous
  • anonymous
I'll be honest I did not see the factor part of the equation. You may need someone else to help you with it just because i'm not quite sure i get it myself. Have a good day!
anonymous
  • anonymous
thanks for trying
jdoe0001
  • jdoe0001
ok, correct if I'm wrong if today is 2.2% in a year will be 2.2 + 2.2 the year after that will be 2.2 + 2.2 + 2.2 and then after that a year 2.2 + 2.2 + 2.2 + 2.2
jdoe0001
  • jdoe0001
so it really is $$ \sum_{n=1}^\infty 2.2^n $$
jdoe0001
  • jdoe0001
so, to know what will it be in 10 years, just make n = 10 :) so, when will it double, that is, the increment will reach 100% to make the current 100% double or 200% of the current oil price when \(2.2^n = 100\)
jdoe0001
  • jdoe0001
$$ 2.2^n = 100\\ \text{using log cancellation rule}\\ log_{2.2}(2.2^n) = log_{2.2}(100)\\ n = log_{2.2}(100)\\ \text{use log change of base rule} \implies n = \cfrac{log(100)}{log(2.2)} $$
jdoe0001
  • jdoe0001
well, not a sequence per se, just an exponential growth, with the sigma notation :)

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