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If dp/dt = e^1/2t −100, then integrating gives p(t) = (1/2) * e^1/2t - 100t + C. Now, with the initial condition p(0) = 1000, you can find C. Done! Don't forget to vote best answer ! Hopefully no one will spoil you the answer. That...
3 Answers · Science & Mathematics · 01/03/2021
dN/dt = -No*e^(-t). When No = 0.40*5,000,000 = 2,000,000, then at t = 1 you have -dN/dt = 2000000/e, which is much more than 270671. So I guess what is meant by "t years beyond 1990" is t years beyond the...
1 Answers · Science & Mathematics · 23/02/2021
Over population . Try going half way up the side of the Grand Canyon. Not many people there.
2 Answers · Science & Mathematics · 17/02/2021
...t is the number of years after 2000. In 2005, the population was 396,000. We can then set t = 5 and P = 396 to solve for k...
2 Answers · Science & Mathematics · 01/02/2021
...from the time line (x) And, assuming tat the y-azis is population == probably in billions ... 1 billion units up would be one billion ...
1 Answers · Science & Mathematics · 31/01/2021
(625,000/4,500,000) * 100 = 13.89%
9 Answers · Science & Mathematics · 29/01/2021
1, y= a(1/2)^(t/t_half_life) is half life formula y = 0.9a 0.9a = a (1/2)^(10/t_half_life) 0.9 = (1/2)^(10/t_half_life) take ln of both sides ln(0.9) = (10/t_half_life)ln(1/2) ln(0.9) /ln(1/2) = 10/t_half_life t_half_life = 10* ln(1/2) / ln(0.9) = 65.78813479 Years 2...
1 Answers · Science & Mathematics · 28/01/2021
P(t)=262,576e^0.25t 1,000,000 = 262,576e^0.25t Taking natural logarithms of both sides gives: ln(1,000,000) = ln(262,576e^0.25t) ln(1,000,000) = ln(262,576) + ln(e^0.25t) ln(1,000,000) = ln(262,576) + 0.25t t = [ln(1,000,000) - ln(262,576)] / 0.25 t =5.35 years
2 Answers · Science & Mathematics · 25/01/2021
x = 97,552 * 1.624 x = 158,424.448 Rounded x = 158,424 thousand (158,424,000)
3 Answers · Science & Mathematics · 23/01/2021
... get the mean of the squares. If this was a sample of a population we'd divide by (n - 1) (2 in this case, instead of 3), but since we weren't...
1 Answers · Science & Mathematics · 06/12/2020
