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wariber [46]
3 years ago
6

A marine biologist is examining the effects of oil pollution on a population of birds known as seagulls (Larus canus). She is pa

rticularly concerned that oil pollution may reduce the number of eggs raised in a seagull nest. During one breeding season, she counted the number of eggs present in a sampling of six seagull nests near each of the 14 refineries throughout the state. She discovered that seagulls laid and raised an average of four eggs per season. To confirm her hypothesis, the researcher must now examine seagull nests that have not been exposed to oil pollution. The researcher believes she is correct, and so expects to find A) 0-2 eggs per nest. B) 2-4 eggs per nest. C) 1 egg per nest. D) 4-6 eggs per nest.
Biology
2 answers:
timurjin [86]3 years ago
6 0

A marine biologist is examining the effects of oil pollution on a population of birds known as seagulls (Larus canus). She is particularly concerned that oil pollution may reduce the number of eggs raised in a seagull nest. During one breeding season, she counted the number of eggs present in a sampling of six seagull nests near each of the 14 refineries throughout the state. She discovered that seagulls laid and raised an average of four eggs per season. To confirm her hypothesis, the researcher must now examine seagull nests that have not been exposed to oil pollution. The researcher believes she is correct, and so expects to find D) 4-6 eggs per nest.

Hoochie [10]3 years ago
5 0

The researcher would expect to find 4-6 eggs per nest. For the oil pollution to have a significant impact, it must substantially lower egg production.

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Guys, please. I genuinely need help! If you don't know- don't answer. And please don't look it up on g00gle, those type of answe
Mumz [18]

Answer:

I want to say <u>Substitution</u>.

Explanation:

They removed the number 5, and inserted an A, which makes me believe its substitution.

At first, I thought it was Insertion when they added the A, but i had to look back again and see that the 5 was removed.

Can't be duplication, since no number/letter was duplicated.

And it can't be Deletion, since another thing was inserted,

I hope this helps!!

<u>#teamtrees #WAP (Water And Plant) #ELM (Every Life Matters)</u>

4 0
2 years ago
How cytoplasm acts as a 'ground substance' for all cell activities?
Maslowich
The cytoplasm , because it’s like a fluid that fills in the cell .
8 0
3 years ago
The bones from an animal found at an archaeological dig have a C614 activity of 0.10 Bq per gram of carbon. The half-life of C61
erastova [34]

C14 is an isotope used in radiocarbon dating techniques to date organic matter remains. The age of these bones is approximately<u> 6890 years.</u>

<h3>What is Carbon 14?</h3>

Carbon 14, also known as radiocarbon, is a radioactive carbon isotope.

Isotopes are the atoms of the same element -carbon- that vary in neutrons and, hence, in their massic number. They are alternative forms of the same element.

The radioactive C14 nucleus contains 6 protons and 8 neutrons and has a half-life of 5730 years.

The term half-life is a reference. It means that an organism that has been dead for 5730 years has half the C14 amount or concentration than the same organism had when it was alive.

Knowing the half-life of an element is useful to determine the age of the dead matter.

C14 is used in radiocarbon dating techniques or methods to estimate the age of fossils. This is a reliable technique used for dating organic samples that are less than 50,000 years old.

<u>Available data</u>:

  • The half-life of C14 is 5730 years
  • Bones activity of 0.10 Bq per gram of carbon

To answer this question we can make use of the following equation

Ln (C14T₁/C14 T₀) = - λ T₁

Where,

  • C14 T₀ ⇒ Amount of carbon in a living body. We know, by bibliography, that living organism activity is 0.23 Bq per gram of carbon. So, C14 T₀ = 0.23 Bq/g
  • C14T₁ ⇒ Amount of carbon in the dead body. C14T₁ = 0.1 Bq/g
  • λ ⇒ radioactive decay constant = (Ln2)/T₀,₅
  • T₀,₅ ⇒ The half-life of carbon 14 = 5730 years
  • T₀ = Time when the organism was alive
  • T₁ = Age of bones

Let us first calculate the radioactive decay constant.

λ = (Ln2)/T₀,₅

λ = 0.693/5730

<u>λ = 0.0001209</u>

Now, let us calculate the first term in the equation

Ln (C14T₁/C14 T₀) = Ln (0.1/0.23) = Ln 0.4347 =<u> - 0.833</u>

Finally, let us replace the terms, clear the equation, and calculate the value of T₁.

Ln (C14T₁/C14 T₀) = - λ T₁

- 0.833 = - 0.0001209 x T₁

T₁ = - 0.833 / - 0.0001209

T₁ =  6889.99 ≅ <u>6890 years</u>

The bones are approximately<u> 6890 years.</u>

You can learn more about dating organic matter with carbon14 at

brainly.com/question/4149380

#SPJ1

6 0
2 years ago
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nordsb [41]

Answer:

imaginary audience

Explanation:

Imaginary audience is a state in which an individual, most especially growing adolescent, are preoccupied with excessive and exaggerated belief and thought that they are the main focus of other people’s attention, and as such become worried unnecessarily and overtly self-conscious. The individual often feels they are the center focus of discussion of other people.

6 0
2 years ago
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Briefly describe alternation of generations in plants. Be sure to include a description of what alternation of generations is, t
aniked [119]
Alternation of generations is the fluctuation of the diploid and haploid stages in plants. The diploid stage occurs through mitosis while the haploid stage works through meisos and asexual reproduction. In vascular plants, the diploid stage produces seeds. In non-vascular plants, the haploid stage produces the spores.
7 0
3 years ago
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