Question

Jacqueline works at an aquarium where she keeps a record of the number of fish. The population of fish increases at a rate of 8.3% per year. The following expression represents the number of fish in the aquarium after t years.
21(1.083)^t

The expression that reveals the approximate monthly growth rate of the population of fish can be modeled by the expression a^bt .

Rounding to the nearest thousandth, determine the values of a and b that reveal the approximate monthly growth rate of the population of fish.

a = 1.083 and b = 12

a = 12 and b = 1.083

a = 12 and b = 1.006

a = 1.006 and b = 12

Answer 1

a=12 and b=1.006 im 90% sure

Answer 2

but idk so cold be rong

Answer 3

How many months are there in a year?

Answer 4

thats incorrect sera

Answer 5

@AZ 12

Answer 6

12

Answer 7

60 in 12 months

Answer 8

or something like that

Answer 9

az you good

Answer 10

The original expression they gave us has 't' but that's in YEARS

we now want something to be in months

that means we have to divide the RATE by 12

and we multiply t by 12

this website can probably explain it better
https://mathbitsnotebook.com/Algebra2/Exponential/EXTimeRelated.html

Answer 11

So if we have an expression like

\( \Large a(1 + b)^t\)

and that's for every year

to make it into a month, it would have to become

\( \Large a\left(1 + \dfrac{b}{12}\right)^{12t}\)

Answer 12

now the 't' in the bottom one is for months

so you can calculate it when t is 2 months or 3 but plugging in t = 2 or 3

but in the first expression, if you plug in t = 2 that means 2 YEARS

Answer 13

thank u i got it now

Answer 14

What did you get as your final answer?

Remember that

1.083 = 1 + 0.083

so you have to divide the 0.083 part by 12

Answer 15

a = 1.006 and b = 12

Answer 16

Bingo!