Question

PLEASE HELP THIS IS THE FINAL SUBMISSION

2) Ruth borrows $440 at 0.6% simple interest per month. When Ruth pays the loan back 4 years later, how much interest does Ruth pay?=

4) Ralph borrows $360 at 0.7% simple interest per month. When Ralph pays the loan back 3 months later, how much interest does Ralph pay?=

8) Craig borrows $1000 at 9% simple interest per year. When Craig pays the loan back 11 years later, what is the total amount that Craig ends up repaying?=

Answer 1

2) Answer: I = $ 10.56

Equation:

I = Prt

Calculation:

First, converting R percent to r a decimal

r = R/100 = 0.6%/100 = 0.006 per year,

then, solving our equation

I = 440 × 0.006 × 4 = 10.56

I = $ 10.56

The simple interest accumulated

on a principal of $ 440.00

at a rate of 0.6% per year

for 4 years is $ 10.56.

4) Answer: I = $ 0.63

Equation:

I = Prt

Calculation:

First, converting R percent to r a decimal

r = R/100 = 0.7%/100 = 0.007 per year,

putting time into years for simplicity,

3 months ÷ 12 months/year = 0.25 years,

then, solving our equation

I = 360 × 0.007 × 0.25 = 0.63

I = $ 0.63

The simple interest accumulated

on a principal of $ 360.00

at a rate of 0.7% per year

for 0.25 years (3 months) is $ 0.63.

8) Answer: I = $ 990.00

Equation:

I = Prt

Calculation:

First, converting R percent to r a decimal

r = R/100 = 9%/100 = 0.09 per year,

then, solving our equation

I = 1000 × 0.09 × 11 = 990

I = $ 990.00

The simple interest accumulated

on a principal of $ 1,000.00

at a rate of 9% per year

for 11 years is $ 990.00.