All categories

3 years ago

DV

Biology

1 answer:

Georgia [21]3 years ago

4 0

That question, this incomplete. The full question is:

3. Three redwood trees are kept at different humidity levels inside a greenhouse for 12 weeks. One tree is left outside in normal conditions. The heights of the trees are measured once a week. What are the dependent variables, independent and the controlled variable of this experiment?

Answer:

Independent variable: different humidity levels
Dependent variable: the heights of the trees
controlled variable: the types of trees

Explanation:

An independent variable is one that exerts an influence within the experiment and can be measured, acting on other elements to promote the observation of a result. As the experiment seeks to observe the influence of humidity on the height of the tree, we can say that humidity is the independent variable.

Dependent variable is that which is influenced by the independent variable to generate a result that can be observed, can also be measured. As the experiment wishes to analyze how the height of the trees was affected by the level of humidity, we can say that the height of the trees is the dependent variable.

Controlled variable is that which is present in the experiment in a constant and equal way, cannot be measured. This variable is the type of trees.

You might be interested in

Which of the following molecule is NOT needed for the Calvin Cycle to take place?

Yuki888 [10]

Answer:

Explanation:

The chemical RuBP, is not even mentioned in that cycle, making it the least viable answer.

5 0

3 years ago

Read 2 more answers

You learn that a Mars lander has retrieved a bacterial sample from the polar ice caps. You obtain a sample of these bacteria and

Misha Larkins [42]

Answer:

a. conservative replication

b. semi conservative replication

c. dispersive replication

d. discontinuous replication

e. continuous replication

The CORRECT ANSWER IS C.

c. dispersive replication

Explanation:

The attached image shows a single band that becomes lighter (in volume) and increase in darker intensity (in times of color) is revealed with consecutive replications. The major pattern that would produce these outcome is dispersive replication.

Conservative replication do reveal two bands (a 14N for the wholly latest molecule, and a 15N for the totally antiquated molecule) after the initial replication. Therefore, this wholly wrong.

Semi-conservative replication reveal an individual band of transitional weight after the initial replication, followed by two bands following the second replication (one of transitional weight, and the other at the 14N position). These two bands is observed in each following replication, with the transitional weight becoming lighter in color as the 14N band is seen to appear darker in color.

The real pattern gives an individual band that pinpoints to all molecules in an individual replicative cycle possessing the same constituents.

Dispersive replication, has its molecules to be a blend of old (15N) and new (14N) strands. Each replication raises the ratio of new to old strands as it explains the reason an individual band is shown raised (lighter in weight) with every replication.

4 0

4 years ago

Monosaccharides are building blocks of larger carbohydrates.<br><br> True<br> False

ale4655 [162]

The answer to this is true.

7 0

3 years ago

Help!!!

Ymorist [56]

Answer:

its option A. bright tail feathers of peacocks during breeding season

Explanation:

4 0

3 years ago

In a maternity ward, four babies become accidentally mixed up. The ABO blood types of the four babies are known to be. Baby Jane

nydimaria [60]

Answer:

See explanation

Explanation:

We inherit our blood groups from both parents. Our blood type is controlled by a single gene. This gene displays 3 different allelic pairs which must include A, B or O. A child is expected to receive one of the three possible allelic pairs from each of the both parents. This means that there are six possible genotype and four possible phenotype for the child.

Given that Christopher's parents (Andersons) are AB and O,Christopher can have a B blood group.

Given that John's parents (Browns) are A and O, it explains how John can have blood group A.

Given that The Christiansons are A and AB, they can give birth to Robin whose blood group is AB.

Since the Dietrichs are O and O, the only possible blood group of their child is O, hence they are definitely the parents of Jane.

7 0

3 years ago