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icang [17]
3 years ago
14

several species of warblers (bird) live in different areas of the same tree . this phenomenon is known as

Biology
1 answer:
Gekata [30.6K]3 years ago
6 0

Answer:

This phenomenon is called Niche Differentiation, it happens when several species live in the same area but different parts. Hope that helps! :)

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Which of the following is not one of the three major structures of coral reefs? a. Planulae b. Fringing c. Barrier d. Atoll Plea
nikklg [1K]

Coral reefs are part of the marine and oceanic ecosystem that are hard and made up of calcium carbonate. Planulae is not the major structure of coral reefs.

<h3>What are coral reefs and Planulae?</h3>

Coral reefs are the hard calcium structures that are of three major types barrier, fringing and atoll.  They feed on the planktons and algae that are the small structures inhabiting them.

Planulae is a freely swimming larvae of the jellyfish and other species of Ctenophores. They are solid, flat and have cilia over their body.

Therefore, option a. Planulae is not a structure or type of coral reef.

Learn more about coral reefs here:

brainly.com/question/2108897

6 0
2 years ago
As the temperature of a liquid goes up, the average speed of its particles blank
Ulleksa [173]

Answer:

the particles gain kinetic energy and vibrate faster and more strongly.

Explanation:

6 0
3 years ago
The loss of color (coral bleaching in coral reef organisms can be a result of
pav-90 [236]
Hello there ^ _ ^

The loss of color(coral bleaching in coral reef organisms can be a result of <span>loss of zooxanthellae.


Good luck!


</span> 
7 0
3 years ago
The systolic arterial blood pressure observed for 20 dogs is normally distributed with a mean of 152 mm of mercury (hg) and a st
Damm [24]

Given: The systolic arterial blood pressure observed for 20 dogs is normally distributed with a mean of 152 mm of mercury (Hg) and a standard deviation of 18 mm of Hg.  

To find: P(100 < 152)

Method: Calculation of Z-Score followed by the probability or area of the bell curve at X = 100.

Solution:

Mean u = 152, std s = 18

Z score = \frac{X-u}{s} = \frac{100-152}{18} = -2.89

The value of P(100<152) is calculated by looking at the value of Z in the Z score for the standard normal distribution given in the image.

P(Z=-2.89) = 0.0019

The P(Z = -2.89) corresponds to the area in the left tail of the bell curve.

Thus the probability of 100 mm Hg blood pressure is 0.0019.

7 0
3 years ago
Plz help!! <br> Will be marked BRAINLIEST if right
Kisachek [45]
- Increased due to being able to eat larger seeds

- False

- True

Hope this helps!
6 0
2 years ago
Read 2 more answers
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