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A tank contains 26,624 L of water. At the end of each subsequent day, half of the water is removed and not replaced. How much water is left in the tank after 11 days?

Sagot :

regalmocha17

13 L is left in the tank after 11 days

Work

To solve this problem, we can use the exponential decay formula which is:

V = P(r)^t

Where:

V = value

p = initial amount

r = rate

t = time

In this case, our values would be:

V = ?

p = 26624

r = 0.50

t = 11

Substitute the values in and solve.

V = 26624(0.50)^11

V = 13

Thus, 13 litres will remain in the tank on the 11th day.

jacob193

Answer:

[tex]13\; {\rm L}[/tex].

Step-by-step explanation:

Start by listing the amount of water in the tank after each day, relative to the original volume of water in the tank:

  • After [tex]1[/tex] day: [tex](1/2) \times 1 = (1/2)[/tex] of the original volume.
  • After [tex]2[/tex] days: [tex](1/2) \times (1/2) = (1/2^{2}) = (1/4)[/tex] of the original volume.
  • After [tex]3[/tex] days: [tex](1/2) \times ((1/2) \times (1/2)) = (1/2^{3}) = (1/8)[/tex] of the original volume.

Observe that after [tex]n[/tex] days, the volume in the tank would be [tex](1/2^{n})[/tex] the original volume. Hence, after [tex]n = 11[/tex] days, the volume in the tank would be [tex](1/2^{11}) = (1/2048)[/tex] of the original volume:

[tex]\displaystyle \frac{1}{2048} \times 26\, 624\; {\rm L} = 13\; {\rm L}[/tex].

In other words, there would be [tex]13\; {\rm L}[/tex] of water in the tank after [tex]11[/tex] days.

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