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natali 33
17 days ago
12

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? The linear function that models the population of rainbow smelt is y1 = −19.76x + 227, where x = the years since 1990 and y1 = the number of rainbow smelt. The linear function that models the population of bloater fish is y2 = . The linear equation that determines when the two populations were equal is –19.76x + 227 = –92.57x + 1052 . The solution is x = years.
Mathematics
2 answers:
Zina [9.1K]17 days ago
6 0

Answer:

x = 11.33 years

Step-by-step explanation:

The equation for rainbow smelt is Y1 = −19.76x + 227

For bloater fish, the equation is Y2 = –92.57x + 1052

To find where these populations are equal:

–19.76x + 227 = –92.57x + 1052

Combine like terms:

-19.76x + 92.57x = 1052 - 227

72.81x = 825

Now divide by 72.81:

x = 825 / 72.81

Thus,

x = 11.33 years

zzz [9K]17 days ago
5 0

Answer:

Step-by-step explanation:

The equation modeling the bloater fish population is given by y2 =

–92.57x + 1,052

.

The equation to find when both populations are equal is

–19.76x + 227 = –92.57x + 1052

.

The result is x = 11.33

years.

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Complete Question

The complete question appears in the first uploaded image

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