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3 years ago

What number is 11 1/3% of 215?

2 answers:

6 0

You can change 215 to 2.15
because 11 1/3% can't change.
So is
2.15x34/3
=43/20x34/2
=731/30
=24 11/30
That's your answer.

gogolik [260]3 years ago

3 0

8 + 3/4 - 11 + 1/3 = - 215/12 = - 17 11/12 = - 17.9166666667; Result written as: improper fraction, mixednumber (also called mixed fraction) and decimal number

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A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

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Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

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(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

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By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

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$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

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$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

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3 years ago

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Sidana [21]

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