The architect stands 6 feet from a climbing frame , looking up at the top of the frame at an angle of 63.43*. It is 5 and a half feet from the ground to the architects eyes . The vertical distance from eye level to the top of the climbing frame is ? Feet. The height of the entire climbing frame is how many feet ?

*The dot plot is shown in the attachment below

2

Step-by-step explanation:

Interquartile range is the difference between the upper median (Q3) and the lower median (Q1).

First, let's write out each value given in the data. Each dot represents a data point.

We have:

2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 7

=>Find the median:

Our median is the middle value. The middle value is the 6th value = 4

==>Upper median Q3) = the middle value of the set of data we have from the median to our far right.

2, 3, 3, 4, 4, |4,| 4, 5, [5], 6, 7

Our upper median = 5

==>Lower median(Q1) = the middle value of the data set we have from our median to our far left.

2, 3, [3], 4, 4, |4,| 4, 5, 5, 6, 7

Lower median = 3

==>Interquartile range = Q3 - Q1 = 5-3 = 2