Percent Exponential

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)

Tags

CCSS.HSF.LE.A.2

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

Tags

CCSS.HSF.LE.A.2

A student invests $500 for x years in a savings account that earns 4% interest per year. No further deposits or withdrawals are made during this time. Write an function f(x) to represent the amount of money earned after x years.

• f(x) = 500 (.60)^x

• f(x) = 500 (.96)^x

• f(x) = 500 (1.04)^x

• f(x) = 500 (1.4)^x

Tags

CCSS.HSF-BF.A.1A

There were 417 cell phones sold at an electronics store in January. Since then, cell phone sales at this store have increased at a rate of 3.75% per month. At this rate of growth, which function can be used to determine the monthly cell phone sales x months after January?

• f(x) = 417(3.75)^x

• f(x) = 417(0.0375)^x

• f(x) = 417(1.0375)^x

• f(x) = 417(1.375)^x

Tags

CCSS.HSF.LE.A.2

An antibiotic is introduced into a colony of 12,000 bacteria during a laboratory experiment. The colony is decreasing by 15% per minute. Which function can be used to model the number of bacteria in the colony after x minutes?

• f(x) = 12000(1.15)^x

• f(x) = 12000(0.15)^x

• f(x) = 12000(0.85)^x

• f(x) = 12000(1.85)^x

Tags

CCSS.HSF.LE.A.2

The original value of a painting is $1400, and the value increases by 9% each year. Write an exponential growth function to model this situation.

• y=1400(1.09)^x

• y=1.09(1400)^x

• y=1400(.91)^x

• y=1.09^x

Tags

CCSS.HSF.LE.A.2

The expression 12(1.02)^t models the population of elephants in a wildlife refuge after t years since 1975.

What does the value 1.02 represent?

• The number of elephants was 102 in 1975.

• The number of elephants increases by 2% each year.

• The number of elephants increases by 102% each year.

• Two extra elephants join the population each year

Tags

CCSS.HSF.LE.B.5

The value of a car is $15,000 and depreciates at a rate of 8% per year. What is the decay factor?

• ^.08

• ^1.08

• .92

• 8

Tags

CCSS.HSF-LE.A.1C

Suppose a culture of bacteria begins with 5000 cells and dies by 30% each year. Write an equation that represents this situation.

• y=5000(1 - 0.30)^x

• y=30(5000)^x

• y=5000(1 + 0.30)^x

• y=5000x^x

Tags

CCSS.HSF.LE.A.2

A population of fish starts at 8,000 and decreases by 6% per year. What is the population of fish after 10 years?

• 14327

• 4309

• 839

• 7680

Tags

CCSS.HSF-LE.A.1C