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Posted by Swamy M 6 years, 1 month ago
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Sia ? 6 years, 1 month ago
Different types of graphs are explained below:
- Line Frequency Graph: Line diagrams consist in drawing vertical lines, length of each line being equal to frequency. The values of variate x are presented on a suitable scale along x-axis. Line diagrams facilitate comparison but they are not attractive to see.
- Histograms These graphs are used to graph grouped data. It is one of the most popular and commonly used devices for charting continuous frequency distribution. It consists of erecting a series of adjacent vertical rectangles on the section of X-axis with bases of equal width of the corresponding class intervals and the heights are so taken that the area of the rectangle are equal to the frequency of the corresponding classes.
- Frequency Polygons A frequency polygon is formed by marking the mid-point at the top of horizontal bars and then joining these dots by a series of straight lines. The frequency polygons are formed as a closed figure with the horizontal axis, therefore a series of straight lines are drawn from the mid-point of the top base of the first and the last rectangles to the mid-point falling on the horizontal axis of the next outlaying interval with zero frequency.
- Frequency Curve It is described as a smooth frequency polygon. A frequency curve is described in terms of its (i) symmetry (skewness) and (ii) degree of peakedness (kurtosis).Two frequency distributions can also be compared by superimposing two or more frequency curves provided the width of their class intervals and the total number of frequencies are equal for the given distributions. Even if the distributions to be compared differ in terms of total frequencies, they still can be compared by drawing per cent frequency curves where the vertical axis measures the per cent class frequencies and not the absolute frequencies.
- Time Series Graph: When information is arranged over a period of time, it is called time series graph. In it, time (hour, day/date, week, month, year, etc.) is plotted along x-axis and the value of the variable along y-axis. A line graph by joining these plotted points, thus, obtained is called arithmetic line graph (time series graph).
Posted by Aditya Juyal 6 years, 2 months ago
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Reyansh Agarwal 6 years, 2 months ago
Hitansh Garg 6 years, 2 months ago
Posted by Zeenat Khanam 6 years, 2 months ago
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Sia ? 6 years, 1 month ago
{tex}\Delta{/tex}ABD where AB = AC
AD bisects {tex}\angle{/tex}PAC,
& CD {tex}\|{/tex} AB
To prove : {tex}\angle{/tex}DAC = {tex}\angle{/tex}BCA
Proof:
AD bisects {tex}\angle{/tex}PAC
Hence {tex}\angle{/tex}PAD = {tex}\angle{/tex}DAC = {tex}\frac{1}{2}{/tex} {tex}\angle{/tex}PAC ...(i)
Also, given
AB = AC
{tex}\therefore{/tex} {tex}\angle{/tex}BCA = {tex}\angle{/tex}ABC (Angles opposite to equal sides are equal) ... (ii)
For {tex}\Delta{/tex}ABC,
{tex}\angle{/tex}PAC is an exterior angle
so, {tex}\angle \mathrm{PAC}=\angle \mathrm{ABC}+\angle \mathrm{BCA}{/tex} (Exterior angle is sum of interior opposite angles)
{tex}\angle \mathrm{PAC}=\angle \mathrm{BCA}+\angle \mathrm{BCA}{/tex} (From (2) : {tex}\angle \mathrm{ABC}=\angle \mathrm{BCA}{/tex})
{tex}\angle \mathrm{PAC}=2 \angle \mathrm{BCA}{/tex}
{tex}\frac{1}{2} \angle \mathrm{PAC}=\angle \mathrm{BCA}{/tex}
{tex}\angle \mathrm{BCA}=\frac{1}{2} \angle \mathrm{PAC}{/tex}
{tex}\angle B C A=\angle D A C \quad\left(\text { From }(1): \angle D A C=\frac{1}{2} \angle P A C\right){/tex}
Hence proved
Posted by Rehan Zaheer 6 years, 2 months ago
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Reyansh Agarwal 6 years, 2 months ago
Posted by Rehan Khan 6 years, 2 months ago
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Posted by Kavya Dekivadiya 6 years, 2 months ago
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Posted by Atul Yadav 6 years, 2 months ago
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Vedant Mohan 6 years, 2 months ago
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Vedant Mohan 6 years, 2 months ago
Posted by Anirudh Vats 6 years, 2 months ago
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Posted by Pranjal Salvi 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
x+y=2 ......... (i)
-x+y=5 ............ (ii)
Adding (i) and (ii)
x + y -x + y = 2 + 5
2y = 7
y = 7/2
put y= 7/2 in (i), we get
x + 7/2 = 2
x = 2 - 7/2
x = -3/2
Posted by Shubham Yadav 6 years, 2 months ago
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Posted by Vaibhav Pawar 6 years, 2 months ago
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Posted by Laxmi Kumari 5 years, 8 months ago
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Posted by Lalan Jha 6 years, 2 months ago
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Posted by Laxman Yadav 6 years, 2 months ago
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Posted by Adarsh Kumar 6 years, 2 months ago
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Posted by Rishabh Raoso Pawar 6 years, 2 months ago
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Posted by Rishabh Raoso Pawar 6 years, 2 months ago
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Posted by Ꮆ ꫝᗰᎥᮘᎶ• 『乙ΞᗫꫝI』 6 years, 2 months ago
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Sia ? 6 years, 2 months ago
Let 1st complementry angle be x
Another= x + 16
x + (x + 16) = 90
2x + 16 = 90
2x = 90 - 16
x= {tex}\frac{74}{2}{/tex}= 37
Another angle= 37 + 16 = 53
Posted by Ꮆ ꫝᗰᎥᮘᎶ• 『乙ΞᗫꫝI』 6 years, 2 months ago
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Sia ? 6 years, 2 months ago
= {tex}\frac{4}{3}\times90{/tex}
= 120
Supplementary of 120
= 120 + x = 180
x = 180 - 120
x = 60
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Posted by Karan Singh 6 years, 2 months ago
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Sia ? 6 years, 2 months ago
We have, AC = 10 cm and AD = 2{tex}\sqrt{5}{/tex} cm
{tex}\therefore{/tex} AC2 = AD2 + CD2
{tex}\Rightarrow{/tex} CD = {tex}\sqrt{AC^2 - AD^2}{/tex} = {tex}\sqrt{(10)^{2}-(2 \sqrt{5})^{2}}{/tex} = 4{tex}\sqrt{5}{/tex} cm
{tex}\therefore{/tex} ar(rect ABCE) = AD {tex}\times{/tex} CD = 2{tex}\sqrt{5}{/tex} {tex}\times{/tex} 4{tex}\sqrt{5}{/tex} cm2
= 8 {tex}\times{/tex} 5 cm2 = 40 cm2
Posted by Vinay Pandey 6 years, 2 months ago
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Posted by Kala Bhutia 6 years, 2 months ago
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Aryan Kumar 6 years, 2 months ago
Ruuzcee Zimik 6 years, 2 months ago
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Leena Minhas 6 years, 2 months ago
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Deepak Keshri 6 years, 1 month ago
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