A massless beam supports two weights as shown. 685 n w l 4 l 4 l 2 l a b find w such that the supporting force at a is zero. ans
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Answer:
{-2/3, 2/3, 3}
Step-by-step explanation:
The equation can be factored by grouping.
f(x) = (9x^3 -27x^2) +(-4x +12)
f(x) = 9x^2(x -3) -4(x -3)
f(x) = (9x^2 -4)(x -3)
We recognize the first factor as the difference of squares, so know it can be further factored to give ...
f(x) = (3x -2)(3x +2)(x -3)
The zeros are the values of x that make these factors be zero:
3x -2 = 0 ⇒ x = 2/3
3x +2 = 0 ⇒ x = -2/3
x -3 = 0 ⇒ x = 3
The zeros of the polynomial are x = -2/3, 2/3, 3.
Answer:
For school A: Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17, IQR=9.5
For school B: Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19, IQR=7.5
No, the box plots are not symmetric.
Step-by-step explanation:
Part A
The given data sets are
School A : 9,14,15,17,17,7,15,6,6
School B : 12,8,13,11,19,15,16,5,8
Arrange the data in ascending order.
School A : 6,6,7,9,14,15,15,17,17
School B : 5,8,8,11,12,13,15,16,19
Divide each data set in four equal parts.
School A : (6,6),(7,9),14,(15,15),(17,17)
School B : (5,8),(8,11),12,(13,15),(16,19)
For school A:
Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17
Interquartile range of the data is
For school B:
Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19
Interquartile range of the data is
Part B:
The box plots are not symmetric because the data values are different. Five number summary and IQR of both the data set are different.