Describe the distribution of wdiff in terms of its center, shape, and spread, including any plots you use
1 answer:
Answer:
Figure attached
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.
Explanation:
We can use the following R code to obtain the data for wdiff:
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"
[9] "gender"
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))
> mean(wdiff)
[1] 14.5891
> median(wdiff)
[1] 10
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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The answer is B
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Charge = number of ( electron or proton ) x charge of ( electron or proton )
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I used these equations then i putted it together.
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We can use the equation of state for an ideal gas to answer the question:
or, by rewriting it,
where p is the gas pressure, V its volume, T its temperature, n the number of moles of the gas and R the gas constant.
When the gas is sprayed from the can into the room, its volume V has increased, while n (the number of moles of the gas) stayed the same. Since R is a constant and the temperature T also stayed constant, if we look at the formula we see that the numerator didn't change, while the denominator (V) has increased, so the pressure of the gas has decreased.
or, by rewriting it,
where p is the gas pressure, V its volume, T its temperature, n the number of moles of the gas and R the gas constant.
When the gas is sprayed from the can into the room, its volume V has increased, while n (the number of moles of the gas) stayed the same. Since R is a constant and the temperature T also stayed constant, if we look at the formula we see that the numerator didn't change, while the denominator (V) has increased, so the pressure of the gas has decreased.