Question

In the year 2010 there were 1,500 foxes in a particular region. Since then, the population has increased by 2.5% per year. If the fox population keeps growing at the same rate, which expression represents the fox population in the year 2050?

Answer 1

Answer:

4028

Step-by-step explanation:

Fox number in 2010=1500

Rate of annual increase = 2.5 %

Fox number in 2050=?

This is compound interest problem

We use

A= P $(1+ \frac{R}{100}) ^{T}$

Where,

A= Fox number in 2050

P=fox number in 2010

R= 2.5

T= 40 years

Put values

==> A= 1500$(1+\frac{2.5}{100} )^{40}$

==> A= 1500 × $(1.025)^{40}$

==>A = 1500 ×  2.6850638383=4027.6=4028

Answer 2

Answer:

Step-by-step explanation:

We would apply the formula for exponential growth which is expressed as

A = P(1 + r)^t

Where

A represents the population after t years.

t represents the number of years.

P represents the initial population.

r represents rate of growth.

From the information given,

P = 1500

r = 2.5% = 2.5/100 = 0.025

The number of years from 2010 to 2050 is 40 years. So

t = 40

Therefore, the expression that represents the fox population in the year 2050 is

A = 1500(1 + 0.025)^40

A = 1500(1.025)^40