Exponential Function Word Problems

The population of Bloom Falls, Mass. (population 937) is slowly moving to a bigger city.
Every year the population drops by 4.5%.
What is the multiplier for this problem?

• 1.045 

• .045 

• .955

• 1.955

Tags

CCSS.HSF-LE.A.1C

A sculpture is increasing in value at a rate of 9% per year, and its value in 2000 was $1500. Write an exponential growth function to model this situation.

• f(t)=1500(1.09)^t

• f(t)=1500(0.09)^t

• f(t)=2000(0.09)^t

• f(t)=1500(9)^t

Tags

CCSS.HSF.LE.A.2

The population of a school is 1000 students and is increasing at a rate of 3% each year. What will the population be in 5 years?

• 1159.27

• 2000

• 1159

• 300

Tags

CCSS.HSF-LE.A.1C

Which of the following functions shows an initial amount of $15 and an increase of 35% each year?

• y = 15(35)^x

• y = 15(0.35)^x

• y = 15(1.35)^x

• y = 35(1.15)^x

Tags

CCSS.HSF.LE.A.2

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

• y=8(15,000)^x

• y=15,000(0.92)^x

• y=15,000(1.08)^x

• y=15,000(0.08)^x

Tags

CCSS.HSF.LE.A.2

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

• y=5000(0.7)^x

• y=5000(1.3)^x

• y=30(5000)^x

• y=5000x^x

Tags

CCSS.HSF.LE.A.2

Write an equation for this problem: A population of 1200 bacteria is doubling each day for x days.

Tags

CCSS.HSF.LE.A.2

Write an equation that models the following situation:

Samantha's hair was known to grow very rapidly. It began at a length of 6 in and grew at a rate of 14% a week.

• y=6(0.14)^x

• y=6(1.14)^x

• y=6(1+14)^x

• y=6(0.86)^x

Tags

CCSS.HSF.LE.A.2

A population of 1500 deer decreases by 1.5% per year. Which function can be used to determine the number of deer, y, in this population at the end of t years?

• y = 1500(1 - 0.015)^t

• y = 1500(0.015)^t

• y = 1500(1 + 0.015)^t

• y = 1500(1.5)^t

Tags

CCSS.HSF.LE.A.2

Suppose a culture of bacteria begins with 5000 cells and dies by 30% each year. The equation that represents this situation is y = 5000(0.7)^t where t = time. How many bacteria cells will be present after 2 years?

• 24500

• 20450

• 2450

• 245