An environmental study monitors the populations of two fish species in a lake. At the star… — SAT

An environmental study monitors the populations of two fish species in a lake. At the start of the study ( $t = 0$ ), Species A has a population of $1,200$ and its population decreases by $40$ each year. Species B has a population of $800$ and its population increases by $5\%$ each year. To the nearest whole number, what is the positive difference in the populations of the two species at year $t = 3$ ?

Answer: 154

Answer

The positive difference in the populations of the two species at year

t = 3

is 154.

t = 3

, the population of Species A, which decreases linearly by 40 each year, is

1,200 - 40(3) = 1,080

. The population of Species B, which grows exponentially at

5\%

each year, is

800(1.05)^3 = 926.1

. The positive difference between these populations is

1,080 - 926.1 = 153.9

, which rounds to 154.

Step-by-Step Solution

Calculate the population of Species A at

t = 3

using a linear decay model.

A(3) = 1,080

Since Species A decreases by a constant number of 40 fish each year, its population follows the linear model

A(t) = 1,200 - 40t

. Substituting

t = 3

gives

1,200 - 120 = 1,080

Calculate the population of Species B at

t = 3

using an exponential growth model.

B(3) = 926.1

Since Species B increases by a constant percent of

5\%

each year, its population follows the exponential model

B(t) = 800(1.05)^t

. Substituting

t = 3

gives

800(1.05)^3 = 800(1.157625) = 926.1

Find the positive difference between the two populations and round to the nearest whole number.

154

The positive difference is

|A(3) - B(3)| = |1,080 - 926.1| = 153.9

. Rounding to the nearest whole number yields 154.

Key Concept

Distinguishing between linear models (constant rate of change per unit time) and exponential models (constant percent rate of change per unit time) to make projections.

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