The oxygen saturation of a lake is found by dividing the amount of dissolved oxygen the lake water currently hasper liter by the dissolved oxygen capacity per liter of the water, and then converting that number into a percent. If the lake currently has 7.4 milligrams of dissolved oxygen per liter of water and the dissolved oxygen capacityis 9.4 milligrams per liter, what is the oxygen saturation level of the lake, to the nearest percent? Please help me with this one ;(

Answers

Answer 1
Answer: (7.4)/(9.4)*100\%= (7.4*100)/(9.4)\%= (740)/(9.4)\%\approx78.72\%\approx79\%
Answer 2
Answer:

Final answer:

To find the oxygen saturation level of the lake, divide the amount of dissolved oxygen in the lake's water per liter (7.4 mg/L) by the water's dissolved oxygen capacity per liter (9.4 mg/L) to get the proportion. Multiply this by 100 to convert it into a percentage (78.72%), which rounds to 79% saturation.

Explanation:

To calculate the oxygen saturation of the lake, we divide the amount of dissolved oxygen the lake water currently has per liter by the dissolved oxygen capacity per liter of the water, and then we convert that number to a percentage.

Here are the steps to solve this problem:

  1. First, divide the amount of dissolved oxygen the water currently has (7.4 milligrams per liter) by the dissolved oxygen capacity (9.4 milligrams per liter). This will give you the proportion of oxygen in the water. This step is a simple division: 7.4 ÷ 9.4 = 0.7872 (rounded to the nearest four decimal places).
  2. Next, multiply the result by 100 to convert the proportion into a percentage. This step is 0.7872 x 100 = 78.72% (rounded to the nearest hundredth).
  3. However, your question asks to round to the nearest percent, so the final answer would be 79% when rounded to the nearest percent. Thus, the oxygen saturation level of the lake is 79%.
  4. Note that this oxygen saturation is important to many aquatic organisms that need oxygen from the water to survive.

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