The fundamental step would be to determine the values of p and E which i believe is the margin of error in this case. Consider the following two equations;

$p-E=0.319\\p+E=0.488$

We simply add the two equations and solve for p;

$2p=0.807\\p=0.4035$

We then substitute this value of p in either equation and solve for E;

$0.4035-E=0.319\\E=0.0845$

Since we have both p and E, the confidence interval can thus be written as;

$0.4035+/-0.0845$

8 0

4 years ago

What is the 4 letter code?

ASHA 777 [7]

1:A
2:H
3:C
4:B
Hope this helps :)

8 0

3 years ago

Drag the dots to plot (5.5, 4), (-7,1), and (0,–9).<br> y

AURORKA [14]

5.5 always goes between 5 and 6

remember (x,y) one of the dots goes on the exactly on the axis, you have to drag it down to -9 from the y-axis

3 0

3 years ago