Respuesta :
Using the given information, we can create the following exponential function.
Let t be the number of years after 2005.
Let F be the fish population.
F = 750(1.083)^t
In this case, t = 45 (2050 - 2005 = 45)
F = 750(1.083)^(45)
= 27123.255...
Round to the nearest whole number.
F = 27123
That's the expected population in 2050.
Have an awesome day! :)
Let t be the number of years after 2005.
Let F be the fish population.
F = 750(1.083)^t
In this case, t = 45 (2050 - 2005 = 45)
F = 750(1.083)^(45)
= 27123.255...
Round to the nearest whole number.
F = 27123
That's the expected population in 2050.
Have an awesome day! :)
a = 750, the number of fish in the year 2005.
8.3% = 0.083, the growth rate
Let n = the number of years, counted from 2005.
Let a(n) = the number of fish after n years.
When n=1, a(1) = 750*1.083
When n=2, a(2) = 750*1.083²
When n=3, a(3) = 750*1.083³
Therefore, after n years, a(n) = 750*1.083ⁿ
The number of years between 2005 and 2050 is 2050-2005 = 45
Therefore in the year 2050, the number of fish will be
a(45) = 750*1.083⁴⁵ = 27123.26
We cannot have fractional fish, so a(45) =27123.
Answer: 27123
8.3% = 0.083, the growth rate
Let n = the number of years, counted from 2005.
Let a(n) = the number of fish after n years.
When n=1, a(1) = 750*1.083
When n=2, a(2) = 750*1.083²
When n=3, a(3) = 750*1.083³
Therefore, after n years, a(n) = 750*1.083ⁿ
The number of years between 2005 and 2050 is 2050-2005 = 45
Therefore in the year 2050, the number of fish will be
a(45) = 750*1.083⁴⁵ = 27123.26
We cannot have fractional fish, so a(45) =27123.
Answer: 27123