CHEGG The accompanying observations are precipitation values during March over a 30-year period in Minneapolis-St. Paul. 0.32, 0.47, 0.52, 0.59, 0.77, 0.81, 0.81, 0.90, 0.96, 1.18, 1.20, 1.20, 1.31, 1.35, 1.43, 1.51, 1.62, 1.74, 1.87, 1.89, 1.95, 2.05, 2.10, 2.20, 2.48, 2.81, 3.00, 3.09, 3.37, 4.75 (a) Construct a normal probability plot for this data set. Hint: You can use the R software. Issue this command (cut and paste the 2 rows of code below in the R command line): qqnorm(c(0.32,0.47,0.52,0.59,0.77,0 .81,0.81,0.90,0.96,1.18,1.20,1.20,1 .31,1.35,1.43,1.51,1.62,1.74,1.87,1 .89,1.95, 2.05,2.10,2.20,2.48,2.81,3.00,3.09, 3.37,4.75))

Mean 1.5806666671.769333333

Standard Deviation 0.8907417991.123161779

Step-by-step explanation:

When we run the commands in the R software we get the graphs as follows.

The commands were

qqnorm(c(0.32,0.52,0.77,0.81,0.96,1.2,1.31,1.43,1.62,1.87,1.95,2.1,2.48,3,3.37))

and

qqnorm(c(0.47,0.59,0.81,0.9,1.18,1.2,1.35,1.51,1.74,1.89,2.05,2.2,2.81,3.09,4.75))

We can plot the graphs in any other plotting software .

The results would be same .

We see there's a slight difference between the two graphs.

One curve goes above ( Minneapolis) for  0.4 and the other ( for St Paul) is below 0.4

s

step-by-step explanation:

step-by-step explanation:

this can be solved first with pythagorean theorem, which tells us to square both legs of the triangle and sum it. this will be equal to the square of the hypotenuse.

the hypotenuse is the side "opposite" the 90 degree angle, which is x.

so we can write using pythagorean theorem,

simplifying:

to further simplify this radical, we need to use the property  

thus