A species of extremely rare, deep water fish rarely have children. if there are 821 of this type of fish and their growth rate is 2% each month, how many will there be in half of a years, in 2 years, and in 10 years?

There are 925 in half of a year

There are 1321 in 2 years

There are 8838 in 10 years

Step-by-step 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