Let us calculate the median; the 6th observation is 20, so it is 20. We need the 6th observation so that out of the 11 observations we have 5 above the median and 5 below (or equal). We also have that then Q1 is the median of the lowest 5 observations, hence 19 (14,16,19,19,20, the 3rd observation is 19). Similarly, we get that the median for the upper half of the observations, Q3 namely, is 22 (21,21,22,22,23, the 3rd observation is 22). Thus, the interquartile range is 3=Q3-Q1. According to our calculations, all observations are wrong.
Answer:
I am seeing you a LOT
Step-by-step explanation:
This deal is cheap! Well, the answer is 60 cents. To get these problems, divide 10.80 by 18, which is equal to 1.5 a dozen.
Remark
You don't have to decompose the second one, and it is better if you don't. Just find the area as you probably did: use the formula for a trapezoid. You have to assume that the 6cm line hits the 2 bases at right angles for each of them, otherwise, you don't know the height. So I'm going to assume that we are in agreement about the second one.
Problem One
The answer for this one has to be broken down and unfortunately, you answer is not right for the total area, although you might get 52 for the triangle. Let's check that out.
<em><u>Triangle</u></em>
Area = 1/2 * b * h
base = 16 cm
h = 10 - 4 = 6
Area = 1/2 * 16 * 6
Area = 48
<em><u>Area of the Rectangle</u></em>
Area = L * W
L = 16
W = 4
Area = L * W
Area = 16 * 4
Area = 64
<em><u>Total Area</u></em>
Area = 64 + 48
Area = 112 of both figures <<<< Answer
