Question

After a devastating winter, when thousands of fish died, an environmental scientist has replenished the trout stock in a fishing pond. He started with 5,000 baby trout and has finished a count to find that, in 8 years, the population is estimated to be 33,000. Assuming an exponential growth pattern, what is the annual growth rate (rounded to the nearest tenth of a percent) of the new trout population?

Hint: A(t) = A0(1 + r)t, where A(t) is the final amount, A0 is the initial amount, r is the growth rate expressed as a decimal, and t is time. (4 points)

26.6%
2.7%
79.0%

Answer 1

The problem states that the growth can be modeled with the equation

A(t) = A0(1 + r)^t

Where A0 is the initial population (5,000), A(t) is the final population (33,000), t is time (8 yrs) and r is the growth rate. Plug in the appropriate quantities and solve for r.

Answer 2

33000 = 5000 ( 1 + r ) ^8

Answer 3

Good keep going

Answer 4

I usually just use a calculator like wolfram alpha but if you want to do this by hand, divide both sides by 5000, take the log8 of both sides to cancel out that 8 exponent, then subtract 1 from both sides

Answer 5

What would the set up be

Answer 6

You could just type the equation into a calculator and have it solve for r

Answer 7

wolfram is being slow

Answer 8

Mathway maybe?

Answer 9

1
+
8
r
+
28
r
2
+
56
r
3
+
70
r
4
+
56
r
5
+
28
r
6
+
8
r
7
+
r
8

Answer 10

thats what i got from mathway
it is showing a rate that is linear

Answer 11

the numbers after the "r" are the exponents

Answer 12

If you type 33000 = 5000(1+x)^8 it’ll solve for x to get x = 0.266 or 26.6% growth

Answer 13

hm smart

Answer 14

still the same thing

Answer 15

i tried every solution

Answer 16

26.6%
2.7%
79.0% has to be one of these

Answer 17

The solution to 33000 = 5000(1+x)^8 is x = 0.266 or 26.6% growth

Answer 18

ohhh