Question

1. choose the correct answer.

in 1970, a rancher stocked one of his ponds with 200 catfish. the following table shows how the population grew in his pond. write an exponential equation modeling the populations change over time. use your graphing calculator. subtract 1,970 from the year and enter the values in list 1. enter the population in list 2.

year	population
1975	322
1980	519
1985	836
1990	1,346
1995	2,167
2000	3,490

• $a(t) = 200(122)^t$
• $a(t) = 200(1.10)^t$
• $a(t) = 200(0.90)^t$
• $a(t) = 200(1.22)^t$

Explanation:

Step1: Define time variable $t$

Let $t$ = year - 1970, so $t=0,5,10,...,30$ for the given years.

Step2: Recall exponential model form

The standard exponential growth model is $A(t)=A_0(b)^t$, where $A_0=200$ (initial population at $t=0$).

Step3: Test growth factor $b$

Use $t=5$, $A(5)=322$. Substitute into model:
$$322=200(b)^5$$
Solve for $b$:
$$b=\sqrt[5]{\frac{322}{200}}=\sqrt[5]{1.61}\approx1.10$$
Verify with $t=10$: $A(10)=200(1.10)^{10}\approx200\times2.5937\approx518.74$, which matches the given 519.

Answer:

B. $A(t) = 200(1.10)^t$