What is the interquartile range of the data set? (40, 48, 62, 60, 62, 40, 25, 50, 20, 30, 55) PLEASE HURRY!!
1 answer:
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Ooh, fun
what I would do is to make it a piecewise function where the absolute value becomse 0
because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up
so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points
we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5
A.
B.
sepearte the integrals
![\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B-5%7D_%7B-4%7D%20%7Bx%5E2%2Bx-12%7D%20%5C%2C%20dx%20%3D%20%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%2B%5Cfrac%7Bx%5E2%7D%7B2%7D-12x%5D%5E%7B-5%7D_%7B-4%7D%3D%28%5Cfrac%7B-125%7D%7B3%7D%2B%5Cfrac%7B25%7D%7B2%7D%2B60%29-%28%5Cfrac%7B64%7D%7B3%7D%2B8%2B48%29%3D%5Cfrac%7B23%7D%7B6%7D)
next one
![\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B-4%7D_3%20%7B-x%5E2-x%2B12%7D%20%5C%2C%20dx%3D-1%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%2B%5Cfrac%7Bx%5E2%7D%7B2%7D-12x%5D%5E%7B-4%7D_%7B3%7D%3D-1%28%28-64%2F3%29%2B8%2B48%29-%289%2B%289%2F2%29-36%29%29%3D%5Cfrac%7B343%7D%7B6%7D)
the last one you can do yourself, it is
the sum is
so the area under the curve is
what I would do is to make it a piecewise function where the absolute value becomse 0
because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up
so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points
we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5
A.
B.
sepearte the integrals
next one
the last one you can do yourself, it is
the sum is
so the area under the curve is
NOTES:
Degree: the largest exponent in the polynomial
End Behavior:
- Coefficient is POSITIVE, then right side goes to POSITIVE infinity
- Coefficient is NEGATIVE, then right side goes to NEGATIVE infinity
- Degree is EVEN, then left side is SAME direction as right side
- Degree is ODD, then left side is OPPOSITE direction as right side
Multiplicity (M): the exponent of the zero. <em>e.g. (x - 3)² has a multiplicity of 2</em>
Relative max/min: the y-value of the vertices.
- Find the axis of symmetry <em>(the midpoint of two neighboring zeros)</em>
- Plug the x-value from 1 (above) into the given equation to find the y-value. <em>(which is the max/min)</em>
- Repeat 1 and 2 (above) for each pair of neighboring zeros.
Rate of Change: slope between the two given points.
********************************************************************************************
1. f(x) = (x-1)²(x + 6)
a) Degree = 3
b) end behavior:
- Coefficient is positive so right side goes to positive infinity
- Degree is odd so left side goes to negative infinity
c) (x - 1)²(x + 6) = 0
x - 1 = 0 x + 6 = 0
x = 1 (M=2) x = -6 (M=1)
d) The midpoint between 1 and -6 is -3.5, so axis of symmetry is at x = -3.5
y = (-3.5 - 1)²(-3.5 + 6)
= (-4.5)²(2.5)
= 50.625
50.625 is the relative max
e) see attachment #1
f) The interval at which the graph increases is: (-∞, -3.5)U(1, ∞)
g) The interval at which the graph decreases is: (-3.5, 1)
h) f(-1) = (-1 - 1)²(-1 + 6)
= (-2)²(5)
= 20
f(0) = (0 - 1)²(0 + 6)
= (-1)²(6)
= 6
Find the slope between (-1, 20) and (0, 6)
m =
=
= -14
********************************************************************************************
2. y = x³+3x²-10x
= x(x² + 3x - 10)
= x(x + 5)(x - 2)
a) Degree = 3
b) end behavior:
Coefficient is positive so right side goes to positive infinity
Degree is odd so left side goes to negative infinity
c) x(x + 5)(x - 2) = 0
x = 0 x + 5 = 0 x - 2 = 0
x = 0 (M=1) x = -5 (M=1) x = 2 (M=1)
d) The midpoint between -5 and 0 is -2.5, so axis of symmetry is at x = -2.5
y = -2.5(-2.5 + 5)(-2.5 - 2)
= -2.5(2.5)(-4.5)
= 28.125
28.125 is the relative max
The midpoint between 0 and 2 is 1, so axis of symmetry is at x = 1
y = 1(1 + 5)(1 - 2)
= 1(6)(-1)
= -6
-6 is the relative min
e) see attachment #2
f) The interval at which the graph increases is: (-∞, -2.5)U(1, ∞)
g) The interval at which the graph decreases is: (-2.5, 1)
h) f(-1) = -1(-1 + 5)(-1 - 2)
********************************************************************************************
3. y = -x(x + 2)(x - 7)(x - 3)
a) Degree = 4
b) end behavior:
Coefficient is negative so right side goes to negative infinity
Degree is even so left side goes to negative infinity
c) -x(x + 2)(x - 7)(x - 3) = 0
-x = 0 x + 2 = 0 x - 7 = 0 x - 3 = 0
x = 0 (M=1) x = -2 (M=1) x = 7 (M=1) x = 3 (M=1)
d) The midpoint between -2 and 0 is -1, so axis of symmetry is at x = -1
y = -(-1)(-1 + 2)(-1 - 7)(-1 - 3)
= 1(1)(-8)(-4)
= 32
32 is a relative max
The midpoint between 0 and 3 is 1.5, so axis of symmetry is at x = 1.5
y = -(1.5)(1.5 + 2)(1.5 - 7)(1.5 - 3)
= -1.5(3.5)(-5.5)(-1.5)
= -43.3125
-43.3125 is the relative min
The midpoint between 3 and 7 is 5, so axis of symmetry is at x = 5
y = -(5)(5 + 2)(5 - 7)(5 - 3)
= -5(7)(-2)(2)
= 140
140 is the relative max
e) see attachment #3
f) The interval at which the graph increases is: (-∞, -1)U(1.5, 5)
g) The interval at which the graph decreases is: (-1, 1.5)U(5, ∞)
h) f(-1) = -(-1)(-1 + 2)(-1 - 7)(-1 - 3)
= 1(1)(-8)(-4)
= 32
f(0) = -(0)(0 + 2)(0 - 7)(0 - 3)
= 0
Find the slope between (-1, 32) and (0, 0)
m =
=
= -32
