𝑨=𝑷(𝟏+ 𝒓 𝒏 ) 𝒏𝒕 Set up for Chapter 3 Word Problems p. 227 #53. We are investing $2500 (our P ) into an account earning 2.5% interest (.025 is our r) for 10 years (our t). How will the balance in the account differ depending on HOW many times a year our money is compounded (that is our n)? When n = 1 time A = ? This means the banker adds interest to our account once a year. 𝑨=𝑷(𝟏+ 𝒓 𝒏 ) 𝒏𝒕 𝑨=𝟐𝟓𝟎𝟎(𝟏+ .𝟎𝟐𝟓 𝟏 ) (𝟏)(𝟏𝟎) Key this into your calculator……did you get $3209.23

𝑨=𝑷(𝟏+ 𝒓 𝒏 ) 𝒏𝒕 Set up for Chapter 3 Word Problems p. 227 #53. We are investing $2500 (our P ) into an account earning 2.5% interest (.025 is our r) for 10 years (our t). How will the balance in the account differ depending on HOW many times a year our money is compounded (that is our n)? When n = 2 time A = ? This means the banker adds interest to our account twice a year. 𝑨=𝑷(𝟏+ 𝒓 𝒏 ) 𝒏𝒕 𝑨=𝟐𝟓𝟎𝟎(𝟏+ .𝟎𝟐𝟓 𝟐 ) (𝟐)(𝟏𝟎) Key this into your calculator……did you get $3205.09 or did you get $25628.91? Two options: multiply 2 and 10 first and use 20 as your exponent of key in: 2500(1 + .025/2)^(2*10)

𝑨=𝑷𝒆 𝒓𝒕 Set up for Chapter 3 Word Problems p. 227 #53. We are investing $2500 (our P ) into an account earning 2.5% interest (.025 is our r) for 10 years (our t). How will the balance in the account differ depending on HOW many times a year our money is compounded (that is our n)? When n = continuously time A = ? This means the banker never stops adding money to account. Because he/she never “leaves” our account we don’t need the “n” in the formula because we can’t count how many times he/she gives us interest! 𝑨=𝑷𝒆 𝒓𝒕 𝑨=𝟐𝟓𝟎𝟎𝒆 (.𝟎𝟐𝟓)(𝟏𝟎) Key this into your calculator……did you get $3210.06 Notice that this is the biggest profit for YOU over the same 10 year period.

Set up for Chapter 3 Word Problems #64. You don’t have to graph it in #64. a. For part b. you simply replace x with 500 in the equation and use your calculator. #67. They give you the ½ life formula to use. Notice the exponent is a fraction. To find the # of times the “stuff” reaches a ½ life we divide the total time by how long it takes it to reach a ½ life. (the 1599 is the length of 1 “1/2 life”.