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alisha [4.7K]
3 years ago
9

A group of marine scientists introduced a species of fish into an artificial habitat and wanted to determine whether it will gro

w well in the new environment. They recorded the population over several years and their data is plotted below. Analyze the data and formulate a conclusion. Does this species of fish do well in this habitat? Predict what will happen to the population over the next five years.

Physics
2 answers:
krek1111 [17]3 years ago
6 0

Enumerating the data points from the chart over ten years:

{100, 105, 95, 102, 93, 105, 98, 99, 101, 100}

Their initial, final, and the median values are all 100. The mean value is 99.8, and the standard deviation slightly less than 3.9. All of these statistics indicate stability of the population over the observed span of ten years. There is neither a significant growth, nor a decline. Assuming stability is what corresponds to a species doing well in a habitat, a conclusion can be made that the species is doing well!

Provided the conditions of the artificial habitat won't change significantly  in the next five years, the population will continue to remain close to an average of 100, with minor deviations of +/- 4 likely year by year.

valina [46]3 years ago
6 0

the population at year ends was 100, 105, 95, 102,93, 105,98,102 and 100. mean is 100. so the fish population is not growing or declining after 9 years. it means the fish is doing ok; not well enough to increase its pop'n but not bad to decrease either.


over the  next 5 yrs, pop'n likely to stay around 100.


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C

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4 0
3 years ago
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If a quasar is moving away from us at v/c = 0.8, what is the measured redshift?
Delicious77 [7]

Answer:

The measured redshift is z =2

Explanation:

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z= \sqrt{\cfrac{c+v}{c-v}}-1

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We can prepare the formula by dividing by lightspeed inside the square root to both numerator and denominator to get

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Replacing the given information

z= \sqrt{\cfrac{1+0.8}{1-0.8}}-1

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4 0
3 years ago
#1 Not sure where to start. This is for AP Physics!
yaroslaw [1]

First,

\rho=\dfrac mV

where \rho is density, m is mass, and V is volume. We can compute the volume of the roll:

2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V

\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness x. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

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V=Ax

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4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x

\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}

If we're taking significant digits into account, the volume we found would have been V=470\,\mathrm m^3, in turn making the thickness x=250\,\mathrm{mm}.

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Explanation:

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