Question

Over ten years, the population of fish in a lake increases by 3%. After the increase, there are 10,094 fish. Which expressions are equivalent to the number of fish ten years before? Select all that apply.

Answer 1

Answer:

9800 Fish

Step-by-step explanation:

According to The question,

• Given, the population of fish in a lake increases by 3% for over ten years, after the increase there are 10094 fish .

The number of fish ten years before is $10094 * \frac{100}{103}$ ⇔ 9800 fish