Answer:
a.
b. 10
Step-by-step explanation:
= 100 -> so 100*0.005= 5
if you still dont understand search up "scientific notation"
A student government organization is selling Christmas trees as a fundraiser. On Friday, they sold 5 noble fir trees and 3 dougl
1 answer:
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Answer:
Step-by-step explanation:
Base of the isosceles triangle = 4
Perpendicular of the triangle = 3
In an isosceles triangle , a perpendicular bisects a base equally. So, here the isosceles triangle consist of 2 right angled triangles.
In that right angled triangles,
Base = 4/2 = 2 (∵ A perpendicular divides a base into 2 equal parts. )
Perpendicular = 3
Hypotenuse = x
So , according to Pythagorean Theorem ,
Using all the values above into the formula gives :-
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.
