A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is $6600 dollars. How much did he invest at each rate?

Let x represent amount invested in the higher-yielding account.

We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be .

We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.

, where,

I = Amount of interest,

P = Principal amount,

r = Annual interest rate in decimal form,

t = Time in years.

We are told that interest rates are 6% and 10%.

Amount of interest earned from lower-yielding account: .

Amount of interest earned from higher-yielding account: .

Let us solve for x.

Therefore, the man invested $30,000 at 10%.

Amount invested in the lower-yielding account would be .

Therefore, the man invested $60,000 at 6%.

1.0

step-by-step explanation:

54 km / hours

step-by-step explanation:

we need to convert seconds to hours

60 seconds = 1 minutes

60 minutes = 1hour

15 m         60 second   60 minutes

  * *         = 54000 meters/ hour

second     1 minute             1 hour

now we need to convert meters to km

1000 m = 1 km

54000 m       1 km

*   = 54 km / hours

hour               1000m