Suppose you invested $10,000 part at 6% annual interest and the rest at 9% annual interest. if you received $684 in interest after one year, how much did you invest at each rate?

Assuming simple interest (i.e. no compounding within first year), then
At 6%, interest = 10000*0.06=$600
At 9% interest = 10000*0.09 = $900

Two ways to find the ratio
method A. let x=proportion at 6%
then
600x+900(1-x)=684
Expand and solve
300x=900-684=216
x=216/300=0.72 or 72%
So 10000*0.72=7200 were invested at 6%
10000-7200=2800 were invested at 9%

method B: by proportions
Ratio of investments at 6% and 9%
= 900-684 : 684-600
=216 : 84
= 18 : 7
Amount invested at 6% = 18/(18+7) * 10000 = 0.72*10000 = 7200
Amount invested at 8% = 7/(18+7)*10000=0.28*10000=2800

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

Cwill be your correct answer