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Triss [41]
1 year ago
6

I need help with this question how to solve it for Brass and Cooper

Physics
1 answer:
Ksenya-84 [330]1 year ago
3 0

Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

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Answer:

Figure attached

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Explanation:

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source("http://www.openintro.org/stat/data/cdc.R")  #obtain the info

nrow(cdc) # number of elements

names(cdc)  # obtain the name for the variable

[1] "genhlth"  "exerany"  "hlthplan" "smoke100" "height"   "weight"   "wtdesire" "age"      

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wdiff represent the difference between desired weight (wtdesire) and current weight (weight) and we can obtain the data with the following code:

wdiff <- (cdc$weight-cdc$wtdesire)

And now we can create the histogram with this code

hist(wdiff,xlim =c(-100,150))

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> median(wdiff)

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And the result is on the figure attached.

And we can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.

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