IQR = 40
1) Put the numbers in order: 40, 45, 50, 60, 60, 75, 90, 90, 120
2) Find the median: Median is 60 (the 2nd one)
3) Place parentheses around the numbers above and below the median. For easy identification of Q1 and Q3. (40, 45, 50, 60,) 60, (75, 90, 90, 120)
4) Find the Q1 and Q3. Q1 = median of the lower half of the data; Q3 = median of the higher half of the data. Q1 and Q3 have even sets so its median cannot be defined.
5) Had both sets contain odd sets, the median of Q1 is subtracted from the median of Q3 to get the IQR.
We can then use the Alternative definition of IQR.
IQR is the difference between the largest and smallest values in the middle 50% of a set data.
40, 45, 50, 60, 60, 75, 90, 90, 120
Middle 50% is 50, 60, 60, 75, 90; IQR = Largest value - smallest value;
IQR = 90 - 50 = 40
S = 2*(6.72*12.7 + 5.78*(6.72 +12.7))
.. = 2*(85.344 +5.78*19.42)
.. = 2*197.5916)
.. = 395.1832
The surface area is about 395.2 cm^2.
Hello from MrBillDoesMath!
Answer: NO points of intersection. The two lines are parallel
Discussion:
Rewrite the equations so the variables and constants line up vertically
x - 2y + 8 = 0
x - 2y -6 = 0 (I added "x" to both sides of the second equation)
Subtract the bottom equation from the the top one:
(x - 2y + 8) - ( x - 2y -6) = 0 or
x - 2y + 8 - x + 2y + 6 = 0 or
(x-x) + (-2y +2y) + (8 + 6) = 0 or
0 + 0 + 14 = 0
14 = 0 !
No solution exists. (Another approach: What is the slope of each line? Hint: 1/2. Since the slopes are the same the lines are parallel and have no points of intersection)
Thank you,
MrB
Step-by-step explanation:
I think this would be an Isosceles triangle