Question

The population of a heard of cattle numbered was 5000 to begin with and was 10,000 after 10 years. If the population was growing exponentially, what was the growth rate? show all work
A) r=2
B) r=20
C)r=0.69
D)r=6.9

Answer 1

Answer:

$r=0.0718$. The closest value from your given choices is C)r=0.69

Step-by-step explanation:

To solve this we are using the standard exponential growth equation:

$f(t)=A(1+r)^t$

where

$f(t)$ is the final population after $t$ years

$A$ is the initial population

$r$ is the growth rate in decimal form

$t$ is the time in years

We know from our problem that the initial population is 5000, the final population is 10000, and the time is 10 years, so $A=5000$, $f(t)=10000$, and $t=10$.

Let's replace the values and solve for $r$:

$f(t)=A(1+r)^t$

$10000=5000(1+r)^{10}$

Divide both sides by 5000

$\frac{10000}{5000} =(1+r)^{10}$

$2=(1+r)^{10}$

Take root of 10 to both sides

$\sqrt[10]{2} =\sqrt[10]{(1+r)^{10}}$

$\sqrt[10]{2} =1+r$

Subtract 1 from both sides

$\sqrt[10]{2}-1=r$

$r=\sqrt[10]{2}-1$

$r=1.0718-1$

$r=0.0718$

We can conclude that the growth rate of our exponential equation is $r=0.0718$. The closest value from your given choices is C)r=0.69