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makkiz [27]
4 years ago
12

Which is not a step in most scientific investigations?

Biology
1 answer:
SIZIF [17.4K]4 years ago
3 0
Proving a theory and writing a law.
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With adequate hydration, what is the expected approximate, normal daily urine production of an adult?
kicyunya [14]

Answer:

With sufficient hydration of about 2liters daily, the normal urine volume expected of an adult is between 800-2000 milliliters

Explanation:

The normal range of fluid required by an adult is 2 liters daily. Urine is produced by the kidney and it account for the largest volume of water leaving the body.

The kidneys adjust the volume of urine ejected based on the body's water requirements; if the body is dehydrated, less urine is ejected and if there is sufficient or excessive hydration, more water is excreted as urine.

7 0
3 years ago
What is the factor affects of mutation?<br><br>​
Vlad1618 [11]

Answer:

Both the nature of the gene and its environment can influence the mutation rate.

Explanation:

Mutations are caused by environmental factors known as mutagens.

Types of mutagens include

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3 0
2 years ago
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The American Heart Association recommends eating at least _______ ounces of oily fish per week to help reduce your risk of heart
Vikki [24]
The American Heart Association recommends eating at least 7 <span>ounces of oily fish per week to help reduce your risk of heart disease.

In short, Your Answer would be Option A

Hope this helps!</span>
6 0
3 years ago
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What is a potential limiting factor for a wolf population?
lina2011 [118]

Answer:

complete your answer first

4 0
3 years ago
If 1000 f2 offspring are obtained, how many flies of each phenotype are expected?
aniked [119]
<span>In Drosophila + indicates wild-type allele for any gene, m is mahogany and e is ebony. Female parents are m+/m+ and males are +e/+e. F1 are m+/+e, all wild type. F1 females are crossed with me/me males - the test cross. Offspring will be : non recombinant m+/me, mahogany wild type or +e/me wild type ebony. OR recombinant me/me mahogany ebony or ++/++ wild type. As the two genes are 25 map units apart, the percentage of recombinants will be 25% and therefore percentage parental types will be 75%. 75% 1000 is 750. There are two parental types, so you would expect 375 of each. Therefore, you would expect 375 m+/me and 375 +e/me. 25% of 1000 is 250 split between two recombinants =125 of each. Therefore you would expect 135 me/me and 125 ++/++</span>
7 0
4 years ago
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