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Ann [662]
3 years ago
13

Unlike ionic bonds, covalent bonds involve the

Biology
1 answer:
lara [203]3 years ago
7 0
C. sharing of electrons
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What is Heruma's Genotype?
antoniya [11.8K]

Answer:

hjhhhhh

Explanation:

4 0
3 years ago
what variable is measured in a experiment? A. independent variable B. experimental variable c. Dependent variable
Damm [24]
C. Dependent variable

The independent variable is what you want to change (ex amount of light given to plants)

The dependent variable is what changes because of the independent variable (ex the height of the plants)

So Independent is what you change, Dependent is what you measure
5 0
3 years ago
PLEASE HELP! I DONT UNDERSTAND
Pachacha [2.7K]

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I would type it up, but I decided to insert a picture.

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6 0
3 years ago
A hiker crosses topography near the sea that has low elevation and low relief and then climbs up steeply to reach a
gladu [14]

Answer:

mountain and coastal plains

Explanation:

i think it would be coastal plains because the area with low elevation is near the sea. only one answer has coastal plains in it, so use process of elimination. additionally, mountain makes sense since theyre usually steep and end in a peak.

7 0
3 years ago
A population of 150 individuals has an allele frequency of 0.2 for the dominant allele (H) and a frequency of 0.8 for the recess
Amanda [17]
Hardy-Weinberg Equation (HW) states that following certain biological tenets or requirements, the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimel. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = genotype frequency of homozygous dominant individuals, 2pq = genotype frequency of heterozygous individuals, and q^2 = genotype frequency of homozygous recessive individuals.
The problem states that Ptotal = 150 individuals, H frequency (p) = 0.2, and h frequency (q) = 0.8.
So homozygous dominant individuals (HH) = p^2 = (0.2)^2 = 0.04 or 4% of 150 --> 6 people
Heterozygous individuals (Hh) = 2pq = 2(0.2)(0.8) = 0.32 or 32% of 150
--> 48 people
And homozygous recessive individuals (hh) = q^2 = (0.8)^2 = 0.64 = 64% of 150 --> 96 people
Hope that helps you to understand how to solve these types of population genetics problems!
6 0
3 years ago
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