Here, given-
homozygous alleles 'a' have a frequency of 0.3.
Also the alleles are in equilibrium in a Hardy-Weinberg population. The frequency of individuals that are homozygous for this allele are= 0.49.
The Hardy-Weinberg equilibrium can be defined as the principle which states that the variation in the genetic makeup of a population remains constant and unchanged till there are no external interferences, influencing the population.
Calculation-

Then to find the frequency of the individuals homozygous for this allele the following formula needs to be used-


Thus, the individuals homozygous for the allele can be calculated by 
Learn more about the Hardy-Weinberg equilibrium here-
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Answer:
Hilum.
Explanation:
Hilum may be defined as the type of the depression or the fissure and may present in kidney known as kidney hilum, present on spleen and the lungs. This is also known as hilus.
The hilum marks the entry and exit of the nerves and the blood vessels. The lung hilum is a triangular like depression through which the blood vessels, capillaries and the nerves enter and leave in the lungs and connects the other organ of the body.
Thus, the correct answer is option (d).
I believe it is a solution, sorry if wrong
N = Atomic Number
A = Atomic Mass
Atomic # = Protons/Electrons
Neutrons = M-N (25-12)
<span>
25-12 = 13 Neutrons </span>
The centrioles are the barrel shaped microtubules that organize the spindles during cell division.
:)