Question

1. an investment of $75,000 increases at a rate of 12.5% per year. find the value of the investment after 30 yr.
2. the bear population increases at a rate of 2% per year. there are 1573 bear this year. write a function that models the bear population. how many bears will there be in 10 yr?

Explanation:

Problem 2

Step1: Identify compound growth formula

The formula for compound growth is $A = P(1 + r)^t$, where $P$ is principal, $r$ is annual rate, $t$ is time in years.

Step2: Plug in given values

$P = 75000$, $r = 0.125$, $t = 30$
$A = 75000(1 + 0.125)^{30}$

Step3: Calculate the growth factor

$1 + 0.125 = 1.125$
$1.125^{30} \approx 37.481$

Step4: Compute final amount

$A = 75000 \times 37.481$

Problem 3

Step1: Define population growth function

The exponential growth function is $P(t) = P_0(1 + r)^t$, where $P_0$ is initial population, $r$ is annual rate.

Step2: Substitute initial values

$P_0 = 1573$, $r = 0.02$
$P(t) = 1573(1 + 0.02)^t = 1573(1.02)^t$

Step3: Find population at t=10

$P(10) = 1573(1.02)^{10}$

Step4: Calculate growth factor

$1.02^{10} \approx 1.2190$

Step5: Compute final population

$P(10) = 1573 \times 1.2190$

Answer:

1. The value of the investment after 30 years is approximately $\$2,811,075$
2. Population model: $P(t) = 1573(1.02)^t$

The bear population after 10 years is approximately 1917