Answer: Producers
Explanation: Producers are autotroph, which means they make their own food, like plants.
Answer:
The answer is "0.42"
Explanation:
Please find the complete question in the attached file.
Coastal community striped allele intensity
Landlocked community, strip intensity allele
Its eliminated allele frequency for movement
community And now in the sturdy vineyard optimum communities, genotype frequency of stripped characteristic (heterozygous device) in coasts
Heritable (striped) allele frecency:


(round up to the closest cent) is therefore the only frequency of coastal people.
The genome of pulling function in host population
Heterozygous feature regularly.
Therefore the inland community rate of recessed (striped), allele rate:
Following migration;
Its percentage of coastal migrants:

Coastal population Non-immigrant percentage of coastal residents:

Stripping coastal community of allele rate after immigration:

![= [q \times m]+[[q' \times(1-m)]\\\\=(0.62 \times 0.1)+(0.40 \times 0.9)\\\\=0.062+ 0.36\\\\=0.422\\\\=0.42](https://tex.z-dn.net/?f=%3D%20%5Bq%20%5Ctimes%20m%5D%2B%5B%5Bq%27%20%5Ctimes%281-m%29%5D%5C%5C%5C%5C%3D%280.62%20%5Ctimes%200.1%29%2B%280.40%20%5Ctimes%200.9%29%5C%5C%5C%5C%3D0.062%2B%200.36%5C%5C%5C%5C%3D0.422%5C%5C%5C%5C%3D0.42)
Answer:
It take ten units of water to break down oligosaccharide that contains 10 glucose units
I think the answer is no NADH is present in the cell
hope my answer is correct ☺