1. Splash Screen

2. Over Lesson 7–6 5-Minute Check 1

3. Over Lesson 7–6 5-Minute Check 1

4. Then/Now You have already solved problems using the percent proportion. Solve simple interest problems and apply the simple interest equation to real-world problems.

5. Example 2 Find the Interest Rate SWIMMING POOL Mr. Webster borrowed $1280 to buy a new swimming pool. He will pay $57.60 each month for the next 24 months. Find the simple interest rate for his loan. UnderstandYou need to find the simple interest rate. PlanUse the formula I = prt. SolveFirst, find the amount of interest he will pay over 24 months. 24 ● $57.60 = $1382.40Multiply to find the total amount.

6. Example 2 Find the Interest Rate $1382.40 – $1280 = $102.40Subtract to find the interest. The principal is $1280. So, p = 1280. The loan will be for 24 months or 2 years, so t = 2. I=prtWrite the simple interest formula. 102.40=1280 ● r ● 2Replace I with 102.40, p with 1280, and t with 2. 102.40= 2560rSimplify.

7. Example 2 Find the Interest Rate Divide each side by 2560. Answer: The simple interest rate is 0.04 or 4%. 0.04=r CheckUse the formula I = prt. $1280 ● 0.04 ● 2 = $102.40 

8. Example 2 pg. 372 #16 Thomas borrowed $4800 to buy a new car. He will be paying $96 each month for the next 60 months. Find the simple interest rate for his car loan.

9. Example 2 pg. 372 #23 Denise has a car loan of $8000. Over the course of the loan, she paid a total of $1680 in interest at a rate of 6%. How many months was the loan?

10. Example 2 pg. 372 #24 A certificate of deposit has an annual simple interest rate of 5.25%. If $567 in interest is earned over a 6 year period, how much was invested?