Two investments total $1000. If the combined annual interest is $68, how much was invested at 8% and how much at 5%?

600 was invested at 8% and $400 at 5%

Step-by-step explanation:

Let x represent the amount of $ invested at 8% and y the amount at 5%.

Then x + y = $1000 is the total investment.  Therefore, y = 10000 - x

Interest earned is then:

0.08x + 0.05(y) = 68, or (after substitution)

0.08x + 0.05(1000) - 0.05x = 68

0.08x + 50 - 0.05x = 68, or

0.03x = 18.  Dividing both sides by 0.03, we find that x = 18/0.03 = 600.

If x = $600, then y = $1000 - $600 = $400.

$600 was invested at 8% and $400 at 5%.

answer:

1. 0.04

2. 0.15

step-by-step explanation:

i just finished that assignment for properties of probability distributions.

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29