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raketka [301]
3 years ago
5

The following set of coordinates represents which figure? (0, 4), (0, 7), (1, 5), (1, 2)

Mathematics
2 answers:
BigorU [14]3 years ago
7 0

Answer with explanation:

The Coordinates of four points in two dimensional plane is  (0, 4), (0, 7), (1, 5), (1, 2).

Point (0,4) and point (0,7) lies on y axis.

And point (1,5) and point (1,2) lies in First Quadrant.

If you will join these four points , we will get a Quadrilateral, as it consist of four distinct line segments ,ends being joined with one another.

We can also check this by plotting the points in Two dimensional coordinate plane.

Marking, the points as  A(0, 4), B(0, 7), C(1, 5), D(1, 2).

We will check which kind of Quadrilateral is this

1. Parallelogram

2. Rhombus

3. Rectangle

4. Square

5. Kite

6. Trapezium

We will use distance formula to find the length of sides of Quadrilateral.

Distance formula

=\sqrt{({x_{2}-x_{1})^2+(y_{2}-y_{1})^2}}\\\\ AB=\sqrt{(0-0)^2+(7-4)^2} =3\\\\ BC=\sqrt{(1-0)^2+(5-7)^2}=\sqrt{1+4}\\\\ BC=\sqrt{5}\\\\ CD=\sqrt{(1-1)^2+(5-2)^2}=3\\\\ DA=\sqrt{(1-0)^2+(4-2)^2}\\\\ DA=\sqrt{5}

Length of Opposite sides are equal. So,it can be parallelogram or Rectangle.

Now, we will find the length of Diagonal.

AC=\sqrt{(1-0)^2+(5-4)^2}\\\\ AC=\sqrt{2} \\\\ BD=\sqrt{(1-0)^2+(7-2)^2}\\\\ BD=\sqrt{1+25}\\\\ BD=\sqrt{26}

Length of Diagonal is not equal, but opposite sides are equal.

So, the given Quadrilateral is Parallelogram.

Romashka [77]3 years ago
6 0

Answer:

The figure represent a parallelogram

Step-by-step explanation:

we have

(0, 4), (0, 7), (1, 5), (1, 2)

using a graphing tool

Plot the points

see the attached figure  

The figure has opposite sides parallel and equal in length

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Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
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{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

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MODAL CLASS

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{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

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\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

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