the number of rainbow smelt in Lake Michigan had an average rate of change of −19.76 per year between 1990 and 2000. The bloater fish population had an average rate of change of −92.57 per year during the same time. If the initial population of rainbow smelt was 227 and the initial population of bloater fish was 1,052, after how many years were the two populations equal?

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symgibbs symgibbs · Answer 1 · 2020-05-04T16:51:49+02:00

Answer: For Edgenunity

The linear function that models the population of bloater fish is y2 =

-92.57 x + 1,052

The linear equation that determines when the two populations were equal is

-19.76 x + 227= -92.57x + 1052

The solution is x = 11.33 years

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ewomazinoade ewomazinoade · Answer 2 · 2021-09-21T17:53:19+02:00

The populations would be equal in 11.3 years

The population of rainbow smelt declines by 19.76 per year

In a years, the total decline in population is 19.76 x a = 19.76a

The remaining population after a years would be : 227 - 19.76a

The population of bloater fish declines by 92.57 per year

In a years, the total decline in population is 92.57 x a = 92.57a

The remaining population after a years would be : 1052 - 92.57a

Where a represents the year in which the populations would be equal

In order to determine the year in which the population would be equal, equate the linear equations

1052 - 92.57a = 227 - 19.76a

Combine similar terms

1052 - 227 = 92.57a - 19.76a

825 = 72.81a

a = 11.3 years

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