Performing two iterations start with (po: .6,1.5) olve the following nonlinear system using fixed point iteration method by +y-3-0 exercise 2: solve the following nonlinear system using fixed point method by performing two iterations start with (po.4o) (1.2,1.2) .2-0 exercise 3 solve the following nonlinear system using fixed point method by performing two iterations start with (po,% 9) (r-1)3-y-o (1-4 exercise 4: solve the following nonlinear system using newton method by performing two iterations start with (po..6,1.5) y-30 exercise 5: solve the following nonlinear system using newton method by performing two iterations start with (po, o) (1.2,1.2) 2-y-02 0 x+-03 exercise 6 solve the following nonlinear system using newton method by performing two iterations start with (po..3,1.9) (x-1)3-y=0 (x-1)t(-24

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.

answ5

step-by-step explanation:

u have to pay atttion to me