Answer:
There are 925 in half of a year
There are 1321 in 2 years
There are 8838 in 10 years
Explanation:
* Lets revise the exponential function
- The original exponential formula was y = ab^x, where a is the initial
amount and b is the growth factor
- The new growth and decay functions is y = a(1 ± r)^x. , the b value
(growth factor) has been replaced either by (1 + r) or by (1 - r).
- The growth rate r is determined as b = 1 + r
* Lets solve the problem
∵ The number of fish is growth every month
∴ We will use the growth equation y = a(1 + r)^x, where a is the initial
amount of the fish, r is the rate of growth every month and x is the
number of months
- There are 821 of a type of fish
∴ The initial amount is 821 fish
∴ a = 821
- Their growth rate is 2% each month
∴ The rate of growth is 2% per month
∴ r = 2/100 = 0.02
- We want to find how many of them be in half year
∵ There are 6 months in half year
∴ y = 821(1 + 0.02)^6
∴ y = 821(1.02)^6 = 924.58 ≅ 925
* There are 925 in half of a year
∵ There are 24 months in 2 years
∴ y = 821(1 + 0.02)^24
∴ y = 821(1.02)^24 = 1320.53 ≅ 1321
* There are 1321 in 2 years
∵ There are 120 months in 10 years
∴ y = 821(1 + 0.02)^120
∴ y = 821(1.02)^120 = 8838.20 ≅ 8838
* There are 8838 in 10 years