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Mathematics
OpenStudy (anonymous):

PLEAASE HELP!!!! After years of overhunting, environmental scientists have reintroduced mountain goats into Yellowstone National Park. The initial number of mountain goats reintroduced to the park was 1,500; after 11 years, the population is estimated to be around 8,400. Assuming an exponential growth pattern, what is the annual growth rate (rounded to the nearest tenth of a percent) of the new mountain goat population in Yellowstone National Park? Hint: A(t) = A0(1 + r)t, where A(t) is the final amount, A0 is the initial amount, r is the growth rate expressed as a decimal, and t is time.

OpenStudy (anonymous):

0.17% 1.7% 17.0% 17.9%

OpenStudy (anonymous):

8400 = 1500*(1+r)^11, so (1+r)^11 = 8400/1500 = 28/5, so 1+r = (28/5)^(1/11), so r = (28/5)^(1/11) - 1 = 0.16954... = 17.0% (to the nearest 0.1%)

OpenStudy (anonymous):

thanks! @antares I have a few more, would you be willing to help?

OpenStudy (anonymous):

The population of a particular city is given by the function P(t) = 125,000(1.015)t, where t is time in years and P(t) is the population after t years. What is the current population, the percentage growth rate, and the population size (rounded to the nearest whole person) after 25 years? Hint: Percentage Growth Rate = r ⋅ 100, in A = A0(1 + r)t 125,000, 1.5% semiannually, 263,155 125,000, 1.5% annually, 181,368 125,000, 15% annually, 468,750 125,000, 10.15% semiannually, 317,188

OpenStudy (anonymous):

anyone willing to help?

OpenStudy (anonymous):

@imqwerty can you please help me?

OpenStudy (anonymous):

P(25) = 125000*(1.015)^25 = 181368.1... = 181368 (to the nearest person) %-growth-rate = (1.015-1)*100% = 1.5% annually

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