Larry Mitchell invested part of his advance at annual simple interest and the rest at annual simple interest. If his total yearly interest from both accounts was , find the amount invested at each rate. The amount invested at is $ nothing. The amount invested at is $ nothing.

Complete question :

Larry Mitchell invested part of his $32000 advance at 4% annual simple interest and the rest at 9% annual simple interest. If his total yearly interest from both accounts was 2430, find the amount invested at each rate. The amount invested at is $ nothing. The amount invested at is $ nothing.

Investment at 4% = $9000

Investment at 9% = $23,000

Step-by-step explanation:

Given that :

Total investment = 32000

Let;

amount invested at 4% = a

amount invested at 9% = 32000 - a

Total interest earned = 2430

Simple interest = principal * rate * time

Investment at 4%:

(a * 0.04) + [(32000 - a) * 0.09] = 2430

0.04a + 2880 - 0.09a= 2430

-0.05a + 2880 = 2430

-0.05a = 2430 - 2880

-0.05a = - 450

a = 9000

Hence,

Investment at 4% = $9000

Investment at 9% = $32,000 - $9000 = $23,000

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.