Presentation on theme: "8.6 Problem Solving: Compound Interests. 43210 In addition to level 3, students make connections to other content areas and/or contextual situations outside."— Presentation transcript:
8.6 Problem Solving: Compound Interests
43210 In addition to level 3, students make connections to other content areas and/or contextual situations outside of math. Students will construct, compare, and interpret linear and exponential function models and solve problems in context with each model. - Compare properties of 2 functions in different ways (algebraically, graphically, numerically in tables, verbal descriptions) - Describe whether a contextual situation has a linear pattern of change or an exponential pattern of change. Write an equation to model it. - Prove that linear functions change at the same rate over time. - Prove that exponential functions change by equal factors over time. - Describe growth or decay situations. - Use properties of exponents to simplify expressions. Students will construct, compare, and interpret linear function models and solve problems in context with the model. - Describe a situation where one quantity changes at a constant rate per unit interval as compared to another. Students will have partial success at a 2 or 3, with help. Even with help, the student is not successful at the learning goal. Focus 8 Learning Goal – (HS.N-RN.A.1 & 2, HS.A-SSE.B.3, HS.A-CED.A.2, HS.F-IF.B.4, HS.F- IF.C.8 & 9, and HS.F-LE.A.1) = Students will construct, compare and interpret linear and exponential function models and solve problems in context with each model.
Simple interest: I=prt I = interest p = principal: amount you start with r = rate of interest t= time in years
If you invest $3,000 at 5% for one year, how much will you make for the year? I = prt = 3000 0.05 1 = 150 You made $150 for the year.
A = p(1+r) t A = balancep = principal r = ratet = time in years Compound interest formula:
Find the total amount in your account if you start with $750 at 7.5% interest compounded annually for 2.5 years. A = p(1+r) t = 750( ) 2.5 = 750(1.075) 2.5 (use a calculator here!) = $898.63
How much should you invest at 7% compounded annually to have $200 after 5 years? A = p(1+r) t (Plug in what you know.) 200 = p(1.07) 5 ( get p alone, then use a calculator.) 200 = p (1.07) = p
If you put $100 in the bank at 4% interest compounded annually and leave it until you are 60, how much money will you have? A = p(1+r)t = 100(1.04) 46 (This assumes you are currently 14) =
What about a mutual fund that pays 10% interest compounded annually? A = p(1+r)t = 100(1.10) 46 =