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Every year, a lake becomes more polluted, and 2% fewer organisms can live in it. If in 2010 there are one million organisms, determine the number of organisms in the lake 5 years later. Let t = 0 represent the year 2010. a. 922,368 organisms remaining c. 903,921 organisms remaining b. 941,192 organisms remaining d. 960,400 organisms remaining Please select the best answer from the choices provided A B C D

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myzombiegoldfish
Their are 900 thousand organisms remaining after 5 years what are A B C D PLEASE GIB BRAINLIST

Using an exponential equation, we have that there will be 903,921 organisms remaining, option c.

A decaying exponential equation has the following format:

[tex]A(t) = A(0)(1 - r)^t[/tex]

In which

  • A(0) is the initial value.
  • r is the decay rate, as a decimal.

In this problem:

  • Initially, one million organisms, thus A(0) = 1000000.
  • 2% fewer organisms each year, thus r = 0.02.

Then, the equation is:

[tex]A(t) = A(0)(1 - r)^t[/tex]

[tex]A(t) = 1000000(1 - 0.02)^t[/tex]

[tex]A(t) = 1000000(0.98)^t[/tex]

After 5 years, the number is:

[tex]A(5) = 1000000(0.98)^5 = 903,921[/tex]

903,921 organisms remaining, option c.

A similar problem is given at https://brainly.com/question/10354772

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