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 popula

Question

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
.

The solution is x =
years.

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Khang Minh 5 years 2021-09-05T12:48:32+00:00 2 Answers 119 views 0

Answers ( )

  1. RuslanHeatt
    0
    2021-09-05T12:49:51+00:00
    Reply

    Answer:

    y2 = – 92.57x + 1052 ;

    11.331 years

    Step-by-step explanation:

    Given that :

    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 boater fish is :

    Rate of change, slope = – 92.57 ; initial population, intercept = 1052

    Representing the equation in the form ;

    y = bx + c

    b = slope or rate of change ; c = intercept

    y2 = – 92.57x + 1052

    To express equality :

    y1 = y2

    – 19.76x + 227 = – 92.57x + 1052

    Collect like terms :

    -19.76x + 92.57x = 1052 – 227

    72.81x = 825

    x = 825 / 72.81

    x = 11.331 years

  2. giabao
    0
    2021-09-05T12:50:10+00:00
    Reply

    y2 =  –92.57x + 1,052

    –19.76x + 227 = –92.57x + 1052

    11.33

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