Round 2769.44 to the nearest tenths place
1 answer:
You might be interested in
Answer:
a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution
b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.
c) This last case the histogram is not symmetrical and the data seems to be skewed.
Step-by-step explanation:
For this case we have the following data:
(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)
We can use the following R code to get the histogram
> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)
> hist(x1,main="Histogram a)")
The result is on the first figure attached.
For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution
(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)
> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)
> hist(x2,main="Histogram b)")
The result is on the first figure attached.
For this case the data is skewed to the left and we can't assume that we have the normality assumption.
(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)
> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)
> hist(x3,main="Histogram c)")
The result is on the first figure attached.
This last case the histogram is not symmetrical and the data seems to be skewed.
Answer:
y =
Step-by-step explanation:
From the given table,
Two points are (1, 15) and (7, 47)
If the two points and
are lying on a line then slope 'm' of the line will be,
m =
=
=
=
Let the equation of a line passing through (h, k) is,
y - h = m(x - k)
If the line passes through (1, 15)
y - 1 =
y =
y =
y =