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3 years ago

The density of aluminum is 2.7 g cm3 what is its density in kilograms per cubic meter

Mathematics

1 answer:

maw [93]3 years ago

3 0

The only thing that we have to do in this problem is to convert the units.

We know that:

1000 g = 1 kg

100 cm = 1 m

Therefore:

density = (2.7 g / cm^3) * (1 kg / 1000 g) * (100 cm / m)^3

density = 2700 kg / m^3

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Find the GCF of the following set of numbers 260, 80, 50

Vinvika [58]

Answer:

Step-by-step explanation:

The factors of 50 are: 1, 2, 5, 10, 25, 50

The factors of 80 are: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

The factors of 260 are: 1, 2, 4, 5, 10, 13, 20, 26, 52, 65, 130, 260

In all of these numbers, we can see that 10 is common throughout and 10 is also the greatest factor all these numbers share.

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3 years ago

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The differences between the data points and the estimated regression line are called the

pashok25 [27]

Linear regression is a method of finding the linear equation that comes closest to fitting a collection of data points.
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The differences between the data pints (observed values) and the estimated (pedicted) regression line is called the residue.
Residue = Observed Value - Predicted Value

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3 years ago

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vovikov84 [41]

The three missing lengths are the left hypotenuse, x, the middle altitude, y, and the right hypotenuse, z.

9/y = y/16

y^2 = 9 * 16

y^2 = 144

y = 12

9^2 + 12^2 = x^2

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3 years ago

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At Burnt Mesa Pueblo, archaeological studies have used the method of tree-ring dating in an effort to determine when prehistoric

Zanzabum

Answer:

a) (1215, 1297)

b) (1174, 1338)

c) (1133, 1379)

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 1256

Standard Deviation, σ = 41

Empirical Rule:

Also known as 68-95-99.7 rule.
It states that almost all the data lies within three standard deviation for a normal data.
About 68% of data lies within 1 standard deviation of mean.
About 95% of data lies within two standard deviation of mean.
About 99.7% of data lies within 3 standard deviation of mean.

a) range of years centered about the mean in which about 68% of the data lies

$\mu \pm 1(\sigma)\\=1256 \pm 41\\=(1215, 1297)$

68% of data will be found between 1215 years and 1297 years.

b) range of years centered about the mean in which about 95% of the data lies

$\mu \pm 2(\sigma)\\=1256 \pm 2(41)\\=1256 \pm 82\\=(1174, 1338)$

95% of data will be found between 1174 years and 1338 years.

c) range of years centered about the mean in which about all of the data lies

$\mu \pm 3(\sigma)\\=1256 \pm 3(41)\\=1256 \pm 123\\=(1133, 1379)$

All of data will be found between 1133 years and 1379 years.

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3 years ago