Answer:
fully simplified answer: x^2-2x+8
Step-by-step explanation:
Answer:
Data set 3 has the highest IQR, so the correct option is {13,17,12,21, 18,20}
Step-by-step explanation:
The question is:
Which data set has the greatest spread for the middle 50% of its data
We have given the data sets:
{18,13,22, 17, 21, 24}
{17,19,22,26,17,14}
{13,17,12,21, 18,20}
{18,21,16,22,24,15}
To find the greatest spread for the middle we need to find the IQR for all the data sets and check which one is highest.
So,
For the first set: 22-17 = 5
For the second: 22-17 = 5
For the third set = 20-13 = 7
For the fourth set = 22-16 = 6
Since data set 3 has the highest IQR, so the correct option is {13,17,12,21, 18,20} ....
The only math term I can think of is geometry and it means when a solid is cut through parallel to the base.
EDIT: After further research since it was all I could think of. This is the only math term for what you are looking for.
The wall area is the product of the room perimeter and the room height:
A₁ = (2*(12.5 ft + 10.5 ft))*(8.0 ft) = 368 ft²
The window and door area together is
A₂ = 2*((4 ft)*(3 ft)) + (7 ft)*(3 ft) = 45 ft²
The area of one roll of wallpaper is
A₃ = (2.5 ft)*(30 ft) = 75 ft²
Then the number of rolls of wallpaper required will be
1.1*(A₁ - A₂)/A₃ ≈ 4.74
5 rolls of wallpaper should be purchased.
_____
As a practical matter, not much of the window and door area can be saved. The rolls are 30 inches wide, but the openings are 36 inches wide. Some will likely have to be cut from two strips. The strips will have to be the full length of the wall, and the amount cut likely cannot be used elsewhere. If the window and door area cannot be salvaged, then likely ceiling(5.4) = 6 rolls will be needed (still allowing 10% for matching and waste).
