Answer :
Exactly one-ninth of the original water is left after 8 pourings.
Given data,
Model the container's remaining contents as follows;
After the first pour, 1⁄2 is left, after the second, half * two-thirds, etc.
This is the final result. 1/2 * 2/3 * 3/4 * . . . . 8/9 * 9/10
Because the terms cancel out, 9/10 remains.
The numerators form an arithmetic sequence with a common difference of 1 and endpoints (1, 8), thus all that is left to do is count the terms or pour.
Therefore, 8 pourings do exactly one-ninth of the original water remain.
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