Simplify each radical
1 answer:
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Answer:
See figure attached and explanation below.
Step-by-step explanation:
For this case we have the following dataset given:
0.74, 0.32, 1.66, 3.59, 4.55, 6.47, 9.99, 0.7, 0.37, 0.76, 1.9, 1.77, 2.42, 1.09, 2.03, 2.69,2.41, 0.54, 8.32, 5.70, 0.75, 1.96, 3.36, 4.06, 12.48
And for this case we can use the followinf R code to create the frequency histogram.
> x<-c(0.74, 0.32, 1.66, 3.59, 4.55, 6.47, 9.99,0.7, 0.37, 0.76, 1.9, 1.77, 2.42, 1.09, 2.03, 2.69,2.41, 0.54, 8.32, 5.70, 0.75, 1.96, 3.36, 4.06,12.48)
> length(x)
[1] 25
> hist(x, prob=TRUE)
And for this case we have the histogram on the figure attached. For the number of classes we use the formula of sturges:
And for this case we have approximately 6 classes. And that's what we can see on the figure attached
As we can see most of the values are on the left so then we have a right skewed to the right and the distribution is assymetrical, with most of the values between 0 and 6
Answer:
1.) scatter plot is attached below.
There is no sufficient evidence to support the claim that there is a linear correlation between overhead widths of seals from photographs and the weights of the seals.
Step-by-step explanation:
Given the data :
Overhead width : 7.2 7.5 9.7 9.3 8.7 8.2
Weight : 119 156 243 200 199 188
The linear correlation Coefficient, R = 0.946
Using the Pvalue calculator for R score ;
Pvalue = 0.0149
Since, Pvalue > Α ; WE fail to reject the Null and conclude that there is no sufficient evidence to support the claim that there is a linear correlation between overhead widths of seals from photographs and the weights of the seals.
