The number of fish in a lake growes steadily at a rate of 10% per year. If there are 242000 fish in the lake now, find the number of fish
(a) after 3 years
(b) 2 years ago.
maths (percentage 2)
2008-06-09 8:31 pm
回答 (2)
2008-06-09 8:43 pm
✔ 最佳答案
(a) No. after 3 years = 242000(1 +10%)(1 +10%)(1 + 10%) = 242000(1.1)(1.1)(1.1) = 322102.(b) No. 2 years ago = 242000 x 1/(1 + 10%) x 1/(1+10%) = 242000/[(1.1)(1.1)] =
242000/1.21 = 200000.
2008-06-09 11:07 pm
(a)Fish number after 3 years:
242000*(1 +10%)^3
=242000*1.1^3
=322102
(b)Fish number two years ago:
242000/(1+ 10%)^2
=242000/(1.1)^2
=242000/1.21
=200000
242000*(1 +10%)^3
=242000*1.1^3
=322102
(b)Fish number two years ago:
242000/(1+ 10%)^2
=242000/(1.1)^2
=242000/1.21
=200000
收錄日期: 2021-04-25 22:38:28
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