The following data points represents how many tunnels Gary the groundhog dug each year. 18,5,13,9,6,2,10 using the data to creat
e a histogram
1 answer:
Answer:
The histogram for the data is attached below.
Step-by-step explanation:
The methods commonly used for depicting a frequency distribution are:
- Histogram/Column graph
- Bar graph
- Frequency Polygon
- Pie chart
Out of these, one of the most general and extensively used devices of illustrating a frequency distribution is the histogram.
The data provided for the number of tunnels Gary the groundhog dug each year is:
S = {18, 5, 13, 9, 6, 2, 10}
The histogram for the data is attached below.
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Answer:
Hence proved that m∠AOD = m∠BCD.
Step-by-step explanation:
Given:
A Trapezoid with AC and BD as diagonals with O as intersection point to it.
Four angles formed at A and D are equal.
To Prove:
m∠AOD = m∠BCD
Proof:
Consider a Quadrilateral ABCD as trapezoid with diagonals as AC and BD
(Refer the Attachment).
And AB=CD which means the Quadrilateral ABCD isosceles in nature
i.e. angle A=angle D
Now
Consider triangle AOD and triangle BCD
Apply sum of all angles of triangle is 180.
And also
m∠2+∠AOD+m∠3=180
∠AOD=180-m∠2-m∠3..............equation(1)
And for Triangle BCD
m∠BCD=180-m∠3-m∠4,.....................Equation(2)
∠A=∠D................ as Quadrilateral is Isosceles.
i.e.∠2=∠3
That means by transitive property,
∠1=∠4=∠3=∠2
Using this values in above equation(1) we get ,
∠AOD=180-∠3-∠4.......................(3)
Now comparing equation 2 and 3 we get
RHS is equal for both equation that means LHS is also equal .
Therefore ∠AOD=∠BCD
Hence proved.
