Okay, so this is just a basic multiplication problem. first, we need to divide to figure out how much the trout population grows after 1 year. If it doubles in 4 years, then it will rise by a quarter each single year. So to put that in terms of fractions, then for every year, the equation would be 250x1/4. So then we do the equation, coming up with 62.5. This means that the population rises roughly 63 fish a year. So after 1 year, you will have roughly 313 fish. (312.5 to be exact). Then to get that by ten years, we can just multiply the number that we got for one year by ten. 62.5x10=625. So then we add the 250 original trout, and 250+625=875. So after 10 years, there will be 875 trout.
If this is for E2020 the answer is D)<span>No, his value of </span><span> should be positive because an even exponent indicates a positive value. thats why it was wrong</span>
