(
)we need 3 equations
1. midpoint equation which is (
) when you have 2 points
2. distance formula which is D=
3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height
so
x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=(
)
midpoint=(
)
midpoint= (15,8)
point x=(15,8)
y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=(
)
midpoint=(
)
midpoint=(13,4)
Y=(13,4)
legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D=
D=
D=
D=
D=
BC=
X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D=
D=
D=
D=
D=
XY=
the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC=
and XY=
AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D=
D=
D=
D=
D=
AD=
so we have
AD=
BC=
XY=
AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
(
) times 1/2 times
=
(
) times \sqrt{5} [/tex] =
=
=80
=252.982
X=(15,8)
Y=(13,4)
BC=
XY=
Area=80
square unit or 252.982 square units
According to Sturge's rule, number of classes or bins recommended to construct a frequency distribution is k ≈ 7
Sturge's Rule: There are no hard and fast guidelines for the size of a class interval or bin when building a frequency distribution table. However, Sturge's rule offers advice on how many intervals one can make if one is genuinely unable to choose a class width. Sturge's rule advises that the class interval number be for a set of n observations.
Given,
n = 66
We know that,
According to Sturge's rule, the optimal number of class intervals can be determined by using the equation:
Here, n is equal to 66 and by substituting the value to the equation we get:
k = 7.0444
k ≈ 7
Learn more about Sturge's rule here: brainly.com/question/28184369
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Answer:
Step-by-step explanation:
Graph is shifted 4 units to the right and reflected over the x-axis.