Question

Frogs have been breeding like flies at the Enormous State University (ESU) campus! Each year, the pledge class of the Epsilon Delta fraternity is instructed to tag all the frogs residing on the ESU campus. Two years ago (t = 0) they managed to tag all 46,000 of them (with little Epsilon Delta Fraternity tags). This year's pledge class discovered that all the tags had all fallen off, and they wound up tagging a total of 55,200 frogs.a) Find an exponential model for the frog population.b) Assuming exponential population growth, and that all this year's tags have fallen off, how many tags should Epsilon Delta order for next year's pledge class?

Answer 1

Answer:

Step-by-step explanation:

Part a

Exponential growth model for the frog population is given by

$P=P_{0}b^{t}\texttt{ \ where} \ P_{0}=Population \ at \ t=0 \ and \ b=Base$

The population of frog at t=0 (2 years ago) is given as 46000 which implies

$P_{0}=46000$

The population in the current year is given as 55200 which implies

 At,

t=2 years, P=55200

Therefore, substitute the above values to calculate the value of b

$55200=46000b^{2} => b^2=\frac{55200}{46000}=1.2$

$b=\sqrt{1.2}=1.095445$

Therefore, the model equation becomes

$P=46000(1.095445)^{t}$

Part b

Since all the tags for this year have fallen off which implies no of tags for the next year can be calculated by calculating the population at t=3

$P=46000(1.095445)^{3}\\\\=46000\times 1.3145\\\\=60467$