The independent variable is the one that causes an effect on the dependent variable. In this case, the independent variable is <u>depth.</u> Different depths lead to a variation in temperature. 2)<em> depths</em>.
<h3>What are de dependent and independent variables?</h3>
- Independent variables are those modified or changed by the researcher to study how this change affects another variable and hence the results.
This variable is not affected by any other one but influences or causes a reaction in other variables.
- The dependent variable is the one influenced by the independent variable, and is affected by any change on this last one.
Its response might be either directly proportional or inversely proportional to the change in the independent variable.
Now let us answer the question.
The aim of the study was probably to analyze how temperature changed with depth.
So, according to this, depth influenced on the temperatureof water.
- The independent influencing variable is <u>depth</u>.
- The <u>dependent affected variable</u> is water temperature.
The correct option is 2. depths.
You will learn more about variables at
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Students measured the temperature of the water at different depths in Lake Skywalker and found that the temperature varied. Which is the independent variable?
- Lake Skywalker
- depths
- temperature
- depths and temperature
Answer:
It goes to the plants or the predators that ate the organisms
Explanation:
Decomposers will break down carcasses into nutrients and provide nutrients for plants in the form of dirt
However if the organism is killed by a predator, the predator will then receive their share of energy by eating the organism.
Answer:
When the pKa is 6.0, we can determine the fraction of protonated H is by:
pH = pKa + log [A]/[HA]
Where
A = Deprotonated imidazole side
HA = Protonated side
Given, pH = 5.0
5 = 6 + log [A]/[HA]
log [A]/[HA] = -1 (take antilog of both side)
[A]/[HA] = 0.1
The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 0.1
Given, pH = 7.5
7.5 = 6 + log [A]/[HA]
log [A]/[HA] = 1.5 (take antilog of both sides)
[A]/[HA] = 31.62
The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 31.62