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anonymous
 5 years ago
2) The population P in 1999 for a state is given along with r, its annual % rate of continuous growth P=10 Millions, r = 2.6% Estimate the population in 2019 = ?
anonymous
 5 years ago
2) The population P in 1999 for a state is given along with r, its annual % rate of continuous growth P=10 Millions, r = 2.6% Estimate the population in 2019 = ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0interent down, we had a bad storm roll through

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By the way, did you realize that the phrasing of yesterdays problems were not good, and I conceived something which was not correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For that problem on cumulative quadratic equation, .....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It meant that the equation itself provided the cumulative result

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So our result would be f(5)f(4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am sorry, but the phrasing was too poor, and I didn't get it at all

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry.. ok can you help me with this one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02) The population P in 1999 for a state is given along with r, its annual % rate of continuous growth P=10 Millions, r = 2.6% Estimate the population in 2019 = ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did you actually understand the problem ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean did you understand what its saying?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no because i tried to use the formula 10e^.026(20) yet I get the wrong answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Before using the formula you must understand what its being said...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is the precise meaning of the term "given along with r"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know, but I feel that it could have been better phrased

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is it from some book?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02) The population P in 1999 for a state is given along with r, its annual % rate of continuous growth P=10 Millions, r = 2.6% Estimate the population in 2019 = ? r is the 2.6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0r is the decimal notation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey! I have the feeling you are actually phrasing the problem, it can't be from some printed material. Am I right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats what it says in the book, then it says write the formula PE^rx where r is the decimal notation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and that models the population in millions of years x after 1999

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Listen the problem can't be solved if I don't understand the english. You see, I can't make out any thing from this part "its annual % rate of continuous growth P=10 Millions, r = 2.6% ". This is some kind of broken sentence. It makes no sense .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then it says the population in 2019 will be?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just a min I will write he whole thing and paste it correctly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Come here https://docs.google.com/document/d/1dYgwObyfbr4lrP_SHyZtFEUl9vvq7jruH0u6dHgXLU/edit?hl=en_GB#

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02) The population P in 1999 for a state is given along with r, its annual % rate of continuous growth P=10 Millions, r = 2.6%  use the formula f(x) = PE^rx, where r is in decimal notation, that models the population in millions x years after 1999. The population in 2019 will be approximately ? million

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Tell me what you wanted to say

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did it on the calc again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did it like this 10e^(.037(20) and it came out with 17

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I used extra parenthasis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, it is something around 16.8 ............

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get 16.8?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just calculated using the calculator, thats it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you do it cause i got 17?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Come back to that place https://docs.google.com/document/d/1dYgwObyfbr4lrP_SHyZtFEUl9vvq7jruH0u6dHgXLU/edit?hl=en_GB&pli=1#

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am sorry there was some connection problem here, I got disconnected. Please visit the link (given on the previous post) and post your email address....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I forgot your email address

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[A=10\times e^{.026\times 20}=10 e^{.52}=16.82\] rounded. so 16.82 million
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