Question

Algebra one help please?

Answer 1

During the first hour . . .

5% of the 1,000 bacteria die.  At the end of the hour, 95% of them are left.

                                    95%  of  1,000  =  950

Then 100 are added :          950 + 100 = 1,050

1,050 bacteria swimming around in the soup as the second hour begins.

During the second hour . . .

5% of the 1,050 bacteria die.  At the end of the hour, 95% of them are left.

                                          95%  of  1,050  =  997.5

Then 100 are added :             997.5 + 100 = 1,097.5 . . . . . 1,098 rounded

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Playing with this some more:

If the same process continues, and the result at the end of each hour
is rounded to the nearest whole number, then the number of bacteria
steadily increases, but only for 88 hours.  At the end of the 88th hour,
there are 1991 of the little critters, and after that, the population stays
constant at  1991.  That's because the 5% loss during each hour after
that is  (5% of 1,991) = 99.55 , which rounds to 100, and those are
replaced by the 100 new ones.