A person had $18,000 invested in two accounts, one paying 8% simple interest and one paying 9% simple interest. How much was invested in each account if the interest after 1 yr is $1606?

he Invested $11,000 in the 5% interest account and $ 7,000 in the 10% interest account.

Step-by-step explanation:Step-by-step explanation:

Let x and y represent the amount he Invested in the two accounts (5% and 10% respectively)

Given that the total investment is $18,000

x + y = 18000 1

If he earned a total of $1,250 in interest in 1 year.

0.05(x) + 0.10(y) = 1250 2

Solving the simultaneous equation,  

Divide equation 2 through by 0.05

x + 2y = 25000 3

Subtract equation 1 from 3

x-x + 2y-y = 25000-18000

y = 7000

From equation 1

x = 18000 - y = 18000 - 7000

x = 11000

Therefore, he Invested $11,000 in the 5% interest account and $ 7,000 in the 10% interest account.

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.

.95

step-by-step explanation: