What is the interquartile range of the sequence 5,5,8,8,13,14,16,16,19,22,23,27,31 ?
Romashka-Z-Leto [24]
Answer:
The Interquartile range is 10.
Step-by-step explanation:
First, we will need to find the mean, the mean of this sequence is 16, you will now need to find quartile 1 and quartile 3. Quartile 1 is 13, and quartile 3 is 23. Lastly, subtract Quartile 3 and Quartile 1 will be the answer.
So, 23-13=10
The Answer will be 10, the interquartile range is 10.
Hope this helps!
Answer:
m= 1/2
Step-by-step explanation:
The question is worded poorly, but it looks like you have a lever in equilibrium, with a force x at a distance d from the fulcrum, and a force y at a distance L - d from the fulcrum. You already have the equilibrium formula for this situation:
xd = y(L - d)
If you know x, y, and d, you can solve for the length L.
Let the five terms be: a, a + d, a + 2d, a + 3d, a + 4d, then
a + a + d + a + 2d + a + 3d + a + 4d = 5a + 15d = 40
i.e. a + 3d = 8
Also, (a + 2d)(a + 3d)(a + 4d) = 224
(a + 3d - d)(a + 3d)(a + 3d + d) = 224
(8 - d)(8)(8 + d) = 224
(8 - d)(8 + d) = 224/8 = 28
64 - d^2 = 28
d^2 = 64 - 28 = 36
d = sqrt(36) = 6
But a + 3d = 8
a + 3(6) = 8
a = 8 - 18 = -10
Therefore, the term of the sequence is: -10, -10 + 6, -10 + 2(6), -10 + 3(6), -10 + 4(6)
= -10, -4, -10 + 12, -10 + 18, -10 + 24
= -10, -4, 2, 8, 14
