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vekshin1
3 years ago
9

Every year, 25 km 3 of sand is deposited on a beach by a nearby river, and 28 km 3 of sand is removed by wave action. Is the siz

e of the beach increasing or decreasing? Explain your answer.
Biology
1 answer:
riadik2000 [5.3K]3 years ago
4 0

If 25 km3 of sand is deposited on a beach yearly and 28 km3 of sand is removed by wave action, the size of the beach would be decreasing, not increase.

First, there is a need to understand that the size of a beach is a function of the amount of sand deposited into it. The more the sand deposited, the more the beach would increase in size.

In this case, 25 km3 of sand is deposited annually but 28 km3 of sand is washed away by waves. The net amount of sand deposited can be calculated by subtracting the amount washed away by waves from the amount deposited. Thus

Net sand deposited = 25 - 28

                                     = -3 km3

Hence, the net amount of sand deposited annually is negative, meaning that the beach would decrease in size.

More on land formation can be found here:  brainly.com/question/469070

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How many sex chromosomes are there​
a_sh-v [17]

Answer: 2

Explanation: There's X (female) and Y (male).

6 0
3 years ago
If the frequency of the recessive allele for a gene is 0.3, calculate the expected frequency of heterozygotes in the next genera
Kisachek [45]
Q = recessive allele frequency = 0.3, and thus in H-W equilibrium there are ONLY two alleles, q (recessive) and
p (dominant). Therefore all of the p and q present for this gene in a population must account for 100% of this gene's alleles. And 100% = 1.00.
So p, the dominant allele frequency, must be equal to 1 - q --> p = 1 - q
p = 1 - 0.3 = 0.7.
Since heterozygotes are a combination of the p and q, we must again look at the frequencies of each genotype: p + q = 1, then (p+q)^2 = 1^2
So multiplying out (p+q)(p+q) = 1, we get: p^2+2pq+q^2 = 1 (all genotypes), where p^2 = frequency of homozygous dominant individuals, 2pq = frequency of heterozygous individuals, and q^2 = frequency of homozygous recessive individuals.
Therefore if the population is in H-W equilibrium, then the expected frequency of heterozygous individuals = 2pq = 2(0.7)(0.3)
2pq = 2(0.21) = 0.42, or 42% of the population.
Hope that helps you to understand how to solve population genetics problems!
7 0
3 years ago
Complete the sentence. _____ is relatively lightweight and easy to transport through a series of pipes to refineries that form d
Alecsey [184]

Answer:

Crude oil i believe!!

Explanation:

3 0
2 years ago
In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant
mariarad [96]

Answer:

in the obtained options, there is no dominant gene for the crossed pairs, therefore, no answer would be correct, however, when assigning a dominant character to the P gene, the correct answer would be:

E. Rr pp *rrPp

Explanation:

According to the information obtained we deduce:

Comb shape is produced by R and P

Comb of walnuts with an R or a P

Pink comb: 1 dominant P and R at the first locus and 2 recessive at the next locus

Pea comb: 2 recessive at the start and one dominant in the next position

1 single comb: 2 recessive at both loci

Rr pp × rr Pp will produce from the interbreeding: 2 walnuts, 1 single and 1 pea offspring.

We deduce that for the pink comb the genotype is: Rrpp

and for the pea comb it is rrPp

Rr pp × rr Pp generates crosslinks containing RrPp, rrPp, rrpp as children in a 2: 1: 1 ratio.

4 0
3 years ago
During the course of successful prenatal development, a human organism begins as a(n) _____ and ends up as a(n) _____.
Olenka [21]
<span>The correct answers are zygote and fetus. This is correct because the initial joining of the sperm and the egg is called the zygote. The zygote is the first step in prenatal development that grows and transforms over months into the final form, a fetus. The fetus is the last form of the zygote prior to birth.</span>
7 0
3 years ago
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