Question

0Over ten years, the population of fish in a lake increases by 3%. After the increase, there are 10,094 fish. Whichexpressions are equivalent to the number of fish ten years before? Select all that apply.9,80010,39710,094 : 1.0310,094 (100+310010,094 - 3%Brun

Answer 1

The fish population has been increasing over 10 years giving a total of 10094 as a result. We are asked to find the initial population of fish.

So we write the expression for popukation increase:

`N(t)=N_(0\, )(1+0.03)^t`

where No is the initial population, and what we want to find, while N(t) is the current population (10094) and the exponent (t) is the equivalent to ten year units (so we use "1"). The rate is 3% per ten years, and this in decimal form is 0.03.

We solve for No in the equation by dividing bby 1.03, which gives us:

No= 10094 / 1.03 = 9800

Therefore, one has to select the option 10094 divided by 1.03, (typed as 10094 : 1.03), and also the numerical answer 9800 (first option)