|
Title: An eager graduate student at Carleton, counts geese in the marshes off the Ottawa river. In 2004, ... Post by: Katie b987 on Apr 2, 2021 An eager graduate student at Carleton, counts geese in the marshes off the Ottawa river. In 2004, they counted 2500 geese. During the next year, 500 goslings hatched, 45 immigrated, 200 died and 30 emigrated. Determine the following:
a) Population change in individuals, in % and per capita b) The expected size of the population after 2, 4 and 8 years, given unlimited resources C)The doubling time, where r = 0.1725 d)Assuming a carrying capacity of 4235 geese, when will the geese population reach carrying capacity under ideal conditions? Title: Re: An eager graduate student at Carleton, counts geese in the marshes off the Ottawa river. In ... Post by: bio_man on Apr 3, 2021 Sorry for the late reply, Katie b987 (https://biology-forums.com/index.php?action=profile;u=998797)
If you're still interested in the rest of the answer, let me know. I've gone ahead and done (a) for you. If you need the rest, reply back: (https://biology-forums.com/gallery/42/6_03_04_21_4_37_53.png) (https://biology-forums.com/index.php?action=gallery;sa=view&id=42868) Title: Re: An eager graduate student at Carleton, counts geese in the marshes off the Ottawa river. In ... Post by: lily4ever on Apr 10, 2021 Hi could you do the rest please! bio_man
Title: Re: An eager graduate student at Carleton, counts geese in the marshes off the Ottawa river. In ... Post by: bio_man on Apr 10, 2021 Quote b) The expected size of the population after 2, 4 and 8 years, given unlimited resources I think you'd need to set-up an exponential equation: y = initial population * growth factor ^ time y = 2500 (1.126) ^ t 2 years: y = 2500 (1.126) ^ 2 = 3169 4 years: y = 2500 (1.126) ^ 4 = 4018 8 years: y = 2500 (1.126) ^ 8 = 6460 Quote C)The doubling time, where r = 0.1725 I don't understand this one. I don't know know what "r" means. If I had to guess, it'd assume that r presents the 'growth rate'? In that case, y = initial population * growth factor ^ time 5000 = 2500 (1.1725) ^ t 2 = (1.1725) ^ t log [ 2 ] = log [ 1.1725 ^ t ] t = 4.4 years Again, I'm just speculating here. You might just be expected to use the formula: td = ln(2) / r td = ln(2) / 0.1725 td = 4 years Quote d)Assuming a carrying capacity of 4235 geese, when will the geese population reach carrying capacity under ideal conditions? I think you're supposed to use the formula y = 2500 (1.126) ^ t again, and set 4235 as y to solve for t 4235 = 2500 (1.126) ^ t 4235/2500 = 1.126 ^ t ln [ 4235/2500 ] = ln [ 1.126 ^ t ] ln [ 4235/2500 ] / ln [ 1.126 ] = 4.46 years or 4.5 years Title: Re: An eager graduate student at Carleton, counts geese in the marshes off the Ottawa river. In ... Post by: duddy on Apr 10, 2021 Are you sure about (a)?
Quote a) Population change in individuals, in % and per capita In 2004, it was 2500 In 2005 = 2500 + 500 - 45 - 200 - 30 = 2725 geese. Population change = 2725 - 2500= 225 % = (225/2500) x 100 = 9% Title: Re: An eager graduate student at Carleton, counts geese in the marshes off the Ottawa river. In ... Post by: bio_man on Apr 11, 2021 You did your calculation wrong:
In 2005 = 2500 + 500 + 45 - 200 - 30 That's why you got a different percentage than me. Please review this resource to help you understand this a little bit more: https://biology-forums.com/index.php?action=downloads;sa=view;down=14509 |