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Liula [17]
3 years ago
15

If 25 cm' of a quantity of marbles has a mass of 45 g, calculate the density of the marble in kg/m​

Mathematics
1 answer:
Elena-2011 [213]3 years ago
7 0

The density of the marble will be "0.18 kg/m³".

According to the question,

Volume of marbles,

  • V = 25 cm³

By converting it into m, we get

  • 0.25 m³

Mass,

  • M = 45 g

By converting it into kg, we get

  • 0.045 kg

As we know, the formula:

→ Density = \frac{Mass}{Volume}

By substituting the values, we get

→               = \frac{0.045}{0.25}

→               = 0.18 \ kg/m^3  

Thus the answer above is correct.

Learn more about density here:

brainly.com/question/15553857

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aleksley [76]

9514 1404 393

Answer:

  a = 3, b = 12, c = 13

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  (a^b)/(a^c) = a^(b-c)

  (a^b)^c = a^(bc)

___

You seem to have ...

  \dfrac{2^5\times8^4}{16}=\dfrac{2^5\times(2^3)^4}{2^4}\qquad (a=3)\\\\=\dfrac{2^5\times2^{3\cdot4}}{2^4}=\dfrac{2^5\times2^{12}}{2^4}\qquad (b=12)\\\\=2^{5+12-4}=2^{13}\qquad(c=13)

_____

<em>Additional comment</em>

I find it easy to remember the rules of exponents by remembering that <em>an exponent signifies repeated multiplication</em>. It tells you how many times the base is a factor in the product.

  2\cdot2\cdot2 = 2^3\qquad\text{2 is a factor 3 times}

Multiplication increases the number of times the base is a factor.

  (2\cdot2\cdot2)\times(2\cdot2)=(2\cdot2\cdot2\cdot2\cdot2)\\\\2^3\times2^2=2^{3+2}=2^5

Similarly, division cancels factors from numerator and denominator, so decreases the number of times the base is a factor.

  \dfrac{(2\cdot2\cdot2)}{(2\cdot2)}=2\\\\\dfrac{2^3}{2^2}=2^{3-2}=2^1

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Please create an argument about a National Landmark/Monument. Do you think there is a monument/landmark out there that should no
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Explanation:

The tallest national monument in America, it is a legendary engineering triumph, designed by Finnish-born architect Eero Saarinen. As part of the Jefferson National Expansion Memorial, the Gateway Arch serves to commemorate the accomplishments of 19th-century westward pioneers and celebrate the city’s role as the ‘Gateway to the West.’

Hope This Helps :)

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If a random sample of size nequals=6 is taken from a​ population, what is required in order to say that the sampling distributio
goldenfox [79]

Answer:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

Step-by-step explanation:

For this case we have that the sample size is n =6

The sample man is defined as :

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And we want a normal distribution for the sample mean

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

So for this case we need to satisfy the following condition:

X_i \sim N(\mu , \sigma), i=1,2,...,n

Because if we find the parameters we got:

E(\bar X) =\frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{n\mu}{n}=\mu

Var(\bar X)= \frac{1}{n^2} \sum_{i=1}^n Var(X_i) = \frac{n\sigma^2}{n^2}= \frac{\sigma^2}{n}

And the deviation would be:

Sd (\bar X) = \frac{\sigma}{\sqrt{n}}

And we satisfy the condition:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

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Check the picture below.

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