Question

During a bean flipping experiment Fiona modeled her data using the function

f(t) = 4.127 (1.468) and Karl modeled his data with the function

k(t) = 0.897 (1.992) where t represents the number of trials and f(t) represents

the number of beans.

At what number of trials would Fiona and Karl have the same number of beans?

A.3

B4

C.5

0.6

Answer 1

Answer:

C.5

Step-by-step explanation:

Fiona:

After t trials, she would have:

$f(t) = 4.127(1.468)^t$

Karl:

After t trials, he would have:

$k(t) = 0.897(1.992)^t$

At what number of trials would Fiona and Karl have the same number of beans?

This is t for which:

$f(t) = k(t)$

So

$4.127(1.468)^t = 0.897(1.992)^t$

$0.897(1.992)^t = 4.127(1.468)^t$

$\frac{(1.992)^t}{(1.468)^t} = \frac{4.127}{0.897}$

$(\frac{1.992}{1.468})^t = 4.6$

$(1.35695)^t = 4.6$

$\log{(1.35695)^t} = \log{4.6}$

$t\log{1.35695} = \log{4.6}$

$t = \frac{\log{4.6}}{\log{1.35695}}$

$t = 5$

So the correct answer is 5 trials, option C.