Answer:
- cylinder — 90π in³
- pyramid — 37 1/3 in³
- cone — 12.5π in³
Step-by-step explanation:
The volume of a cylinder is given by ...
V = Bh . . . . . where B is the base area and h is the height
The volume of a pyramid or cone is given by ...
V = (1/3)Bh . . . . . where B is the base area and h is the height
The area of a square of side length s is ...
A = s²
The area of a circle of radius r is ...
A = πr²
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Using these formulas, the volumes of these objects are ...
cylinder: (9π in²)(10 in) = 90π in³
square pyramid: (1/3)(4 in)²(7 in) = 37 1/3 in³
cone: (1/3)(π(2.5 in)²)(6 in) = 12.5π in³ . . . . slightly larger than the pyramid
Hii!
So I did this and I have A!
A: You cannot get the mean from the graph but you CAN get the third quartile!
B: To find the interquartile range (IQR) we subtract the third and first quartiles:
60-35 = 25
C: An outlier would be much larger than the rest of the data or much smaller than the rest of the data. An outlier would make the "whisker" portion longer and could potentially slightly shift the box.
You have to change the denominator of 6/10 into 100 then you will get 60/100 then you add 15/10+60/100
