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Asses × ure https://imathas.rationalreasoning.net/assessment/showtest.plp action-skip&to; 9 Due in 3 hours, 13 minutes. Due Sat 10/21/2017 11:59 pm On May 1, 2008, the forest service stocked a lake with 120 bluegill fish. On October1, five months later, they estimated the population of bluegill in the lake to be 200. Assume the population of bluegill increased exponentially a. For this population of bluegill, find: Preview i. the 5-month percent change: 40 ii. the 5-month growth factor: 1.66 ii the 1-month growth factor: 1.107 iv. the 1-month percent change: 10.7 Preview Preview 12% Preview b. Is the following statement true or false? The 1-month percent change is of the 5-month percent change. False 0) 0) 0] 0] 50 c. Define a function f that gives the number of fish in the lake t months after May 1, 2008. Preview t( t)-200(1.1 07yt d. Define a function g that gives the number of fish in terms of n, the number of 5- month periods after May 1, 2008. ga-2001 68ynPreview e. Use a graphing calculator and the functions you defined above to determine how many months after May 1 are required for the population to surpass 750 bluegill. Your answer should be a whole number of months. months Preview Licensc Points possible: 10 Unlimited attempts Score on last attempt: 2.5. Score in gradebook: 2.5 Post this question to forum e here to search

Explanation / Answer

Po = 120

P(5) = 200

Equation is    P(t) = P(0)*a^t

Where P(t) = Population after time t

P(0) = Initial Population

a = exponential growth factor

t = time in months

a) i) the 5-month percent change = (200 - 120)*100/120 = 66.7%

ii) the 5-month growth factor = 200/120 = 1.667

iii) The 1-month growth factor = (1.667)^0.2 = 1.107

iv) The one month percent change = (132.84 - 120)*100/120 = 10.7%

b) False

c) f(t) = 120*(1.107)^t

d) g(n) = 120*(1.667)^n

e) 750 = 120*(1.107)^t

    t = 18 months

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