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DiKsa [7]
3 years ago
6

What are 20 things that are found today on earth that demonstrate our connection to stars?

Biology
1 answer:
Aleksandr [31]3 years ago
7 0

Answer:

Explanation:

telliscope moon rock mars rock sky and 1 1 1 1 1 1 1 1 1 1 2n3ehd ok i cant delteat that the next thing s books and of coruse the sun and also t

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What idea is Malthus known for?
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Malthus’ most famous work, which he published in 1798, was An Essay on the Principle of Population as it affects the Future Improvement of Society. In it, Malthus raised doubts about whether a nation could ever reach a point where laws would no longer be required, and in which everyone lived prosperously and harmoniously. There was, he argued, a built-in agony to human existence, in that the growth of a population will always outrun its ability to feed itself. If every couple raised four children, the population could easily double in twenty-five years, and from then on, it would keep doubling. It would rise not arithmetically—by factors of three, four, five, and so on—but geometrically—by factors of four, eight, and sixteen.<span>
Between 1800 and 2000 the human population increased about six-fold. Has the food supply kept pace? Will there be enough food to support the projected population of 9.2 billion in 2050?</span>If a country’s population did explode this way, Malthus warned that there was no hope that the world’s food supply could keep up. Clearing new land for farming or improving the yields of crops might produce a bigger harvest, but it could only increase arithmetically, not geometrically. Unchecked population growth inevitably brought famine and misery. The only reason that humanity wasn’t already in perpetual famine was because its growth was continually checked by forces such as plagues, infanticide, and simply putting off marriage until middle age. Malthus argued that population growth doomed any efforts to improve the lot of the poor. Extra money would allow the poor to have more children, only hastening the nation’s appointment with famine.A new view of humans
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7 0
3 years ago
The arrow points to an example of
Over [174]

Answer:

Sound Barrier

Explanation:

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3 years ago
How long will r136a1 live? ​
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A star’s life expectancy depends on its mass. Generally, the more massive the star, the faster it burns up its fuel supply, and the shorter its life. The most massive stars can burn out and explode in a supernova after only a few million years of fusion. A star with a mass like the Sun, on the other hand, can continue fusing hydrogen for about 10 billion years. And if the star is very small, with a mass only a tenth that of the Sun, it can keep fusing hydrogen for up to a trillion years, longer than the current age of the universe.

4 0
3 years ago
Cystic fibrosis (CF) is one of most common recessive disorders among Caucasians it affects 1 in 1,700 newborns. What is the expe
Phantasy [73]

Answer: The expected frequency of carriers is P(Aa)=0.046.

The proportion of childs with CF is P(aa)=0.024.

25% of having a child with CF (aa).

Explanation:

Hardy-Weinberg's principle states that in a large enough population, in which mating occurs randomly and which is not subject to mutation, selection or migration, gene and genotype frequencies remain constant from one generation to the next one, once a state of equilibrium has been reached which in autosomal loci is reached after one generation. So, a population is said to be in balance when the alleles in polymorphic systems maintain their frequency in the population over generations.

Given the gene allele frequencies in the gene pool of a population, it is possible to calculate the expected frequencies of the progeny's genotypes and phenotypes. <u>If P = percentage of the allele A (dominant) and q = percentage of the allele a (recessive)</u>, the checkerboard method can be used to produce all possible random combinations of these gametes.

Note that p + q = 1, that is, the percentages of gametes A and a must equal 100% to include all gametes in the gene pool.

The genotypic frequencies added together should also equal 1 or 100%, and all the equations can be summarized as follows:

p+q=1\\(p+q)^{2}  = p^{2} +2pq+q^{2} = 1\\P(AA)=p^{2} \\P(aa)=q^{2} \\P(Aa)=2pq1

So, there are 1700 individuals and only one is affected. Since it is a recessive disorder, the genotype of that individual must be aa. So the genotypic frequency of aa is 1/1700=0.000588.

Then, P(aa)=q^{2}=0.000588. And with that we can calculate the value of q,

P(a)=q=\sqrt{0.000588}=0.024

And since we know that p+q=1, we can find out the value of p.

p+0.024=1\\1-0.024=p\\p=0.976

Next, we find out the genotypic frequency of the genotype AA:

P(A)=p=0.976\\P(AA)=p^{2} = 0.976^{2}=0.95

Now, we can find out the genotypic frequency of the genotype Aa:

P(Aa)=2pq=2 x 0.976 x 0.024 = 0.046

Notice than:

p^{2} + 2pq + q^{2} = 1\\x^{2} 0.976^{2} + 2 x 0.976 x 0.024 + 0.024^{2} = 1

Then, the expected frequency of carriers is P(Aa)=0.046

The proportion of childs with CF is P(aa)=0.024

If two parents are carriers, then their genotypes are Aa.

Gametes produced by them can only have one allele of the gene. So they can either produce A gametes, or a gametes.

In the punnett square, we can see that there genotypic ratio is 2:1:1 and the phenotypic ratio is 3:1. So, there is a probability of 25% of having an unaffected child, with both normal alleles (AA); 50% of having a carrier child (Aa) and 25% (0.25) of having a child with CF (aa).

5 0
3 years ago
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