Percent of Exponential Growth or Decay

Rhonda deposited $3000 in an account in the Merrick National Bank, earning 4.2% interest, compounded annually. She made no deposits or withdrawals. Write an equation that can be used to find B, her account balance after t years.

• B = 3000(1 – 4.2)^t

• B = 3000(1 + 4.2)^t

• B = 3000(1 – 0.042)^t

• B = 3000(1 + 0.042)^t

Marilyn collects old dolls. She purchases a doll for $450. Research shows this doll's value will increase by 2.5% each year. Write an equation that determines the value, V, of the doll t years after purchase.

• V = 450(1 + 0.025)^t

• V = 450(1 – 0.025)^t

• V = 450(1 + 2.5)^t

• V = 450(1 – 2.5)^t

A car was purchased for $25,000. Research shows that the car has an average yearly depreciation rate of 18.5%. Create a function that will determine the value, V(t), of the car t years after purchase.

• V(t) = 25000(1 – 0.185)^t

• V(t) = 25000(1 + 0.185)^t

• V(t) = 25000(1 – 18.5)^t

• V(t) = 25000(1 + 18.5)^t

The bear population in a given area is currently 1580. They anticipate the bear population to decrease by 2% each year. Which function represents the population of bears, B, after t years.

• B = 1580(0.02)^t

• B = 1580(1 – 0.02)^t

• B = 1580(1 + 0.02)^t

• B = 1580(1 – 0.2)^t

What is the percent growth or decay of the following function?

Which of the following could be represented by A=200(1.1)^t

• I started with $200 and lost 1% per week

• I started with $200 and lost 10% per week

• I started with $200 and gained 1% per week

• I started with $200 and gained 10% per week

The bear population in a given area is currently 1580. They anticipate the bear population to decrease by 2% each year. Which function represents the population of bears, B, after t years.

• B = 1580(0.02)

• B = 1580(1 – 0.02)

• B = 1580(1 + 0.02)

• B = 1580(1 – 0.2)

• Growth, 85%

• Decay, 85%

• Decay, 15%

What is the percentage rate of growth or decay?

• 12%

• 88%

• .12%

• 112%

• .88%

The population of a town is currently 6342 and is increasing by a rate of 1.3% each year. Which function represents the population of people, P, after t years.

• P = 6342(1 + 1.3)^t

• P = 6342(1 – 1.3)^t

• P = 6342(1 + .013)^t

• P = 6342(1 – .013)^t