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2 answers:
Answer:
(i) Five number summary: Minimum value = 1, ,
,
, Maximum value = 26.
(ii) IQR = 2.5, upper fence = 24.25 and lower fence = 14.25.
(iii) 26 is an outlier of the data.
Step-by-step explanation:
The given data set is
19, 18, 23, 18, 18, 26, 19, 16, 20, 18, 17, 21
Arrange the data set in ascending order.
16, 17, 18, 18, 18, 18, 19, 19, 20, 21, 23, 26
Divide the data set in 4 equal parts.
(16, 17, 18), (18, 18, 18), (19, 19, 20), (21, 23, 26)
Minimum value = 16
Maximum value = 26
The inter quartile range (IQR) of the data is
Upper fence :
Lower fence :
If the any element of the data set lies out side the upper fence and lower fence, then it is called an outlier.
26 ∉ [14.25, 24.25]
Therefore, 26 is an outlier of the data.
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Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,
which is a cubic polynomial in with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).
Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.
f'(x) = 3x² - 1 = 0 ⇒ x = ±1/√3
So, we have three subsets over which f(x) can be considered invertible.
• (-∞, -1/√3)
• (-1/√3, 1/√3)
• (1/√3, ∞)
By the inverse function theorem,
where f(a) = b.
Solve f(x) = 2 for x :
x³ - x + 2 = 2
x³ - x = 0
x (x² - 1) = 0
x (x - 1) (x + 1) = 0
x = 0 or x = 1 or x = -1
Then can be one of
• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);
• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or
• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)
Volume of a cone=
r^2*h/3
*3^2*6/3=56.55
volume fo a sphere = 4/3r^3
4/3*pi*1.5^3=14.14
56.55-14.14=42.41
volume fo a sphere = 4/3
4/3*pi*1.5^3=14.14
56.55-14.14=42.41