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Ask QuestionPosted by Rishith Bhardwaj 5 years, 5 months ago
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Posted by Kiran Devi 5 years, 5 months ago
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Posted by Kiran Devi 5 years, 5 months ago
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Posted by Priyanshi Agarwal 5 years, 5 months ago
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Posted by Sourabh Sharma 5 years, 5 months ago
- 1 answers
Yogita Ingle 5 years, 5 months ago
The mean (or average) of a number of observations is the sum of the values of all the observations divided by the total number of observations.
Median: Let’s arrange their score in ascending order in blue table. Now we will take the middle most score. Since we are talking about central tendency of data, middle most score should also reflect central tendency of data.
When the number of observations (n) is odd, the median is the value of the ((n+1)/2) Th observation.
When the number of observations (n) is even, the median is the mean of (n/2) & (n/2 + 1) Th observations.
Posted by Pankaj Tyagi 5 years, 5 months ago
- 1 answers
Gaurav Seth 5 years, 5 months ago
In point A(1, -1), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant. In point B(4, 5), both coordinates are positive, so it lies in I quadrant. On plotting these point, we get the following graph.
(i) On joining the points A and B, we get the line segment AB. Now, to find the coordinates of a point on this line segment between A and B draw a perpendicular to X-axis from x = 2 and 3.
[since, x = 2 and 3 lies between A and B] say it intersect line segment AB at P and p’. Now, draw a perpendicular to Y-axis from P and p’, they intersect Y-axis at y = 1 and 3, respectively. Thus, we get points (2,1) and (3, 3) which lie between line segment AB.
(ii) Extent the line segment AB. Now, draw a perpendicular to X-axis from x = 5, say it intersects extended line segment at Q. Again, draw a perpendicular to Y-axis from Q and it intersects Y-axis at y = 7. Thus, we get the point Q(5,7) which lies outside the line segment AB.
Posted by Chinky Bhan 5 years, 5 months ago
- 1 answers
Sia ? 5 years, 5 months ago
{tex}\frac { 4 + \sqrt { 5 } } { 4 - \sqrt { 5 } } + \frac { 4 - \sqrt { 5 } } { 4 + \sqrt { 5 } }{/tex}
{tex}= \frac { ( 4 + \sqrt { 5 } ) ( 4 + \sqrt { 5 } ) + ( 4 - \sqrt { 5 } ) ( 4 - \sqrt { 5 } ) } { ( 4 - \sqrt { 5 } ) ( 4 + \sqrt { 5 } ) }{/tex}
{tex}= \frac { ( 4 + \sqrt { 5 } ) ^ { 2 } + ( 4 - \sqrt { 5 } ) ^ { 2 } } { ( 4 - \sqrt { 5 } ) ( 4 + \sqrt { 5 } ) }{/tex}
{tex}= \frac { \left\{ ( 4 ) ^ { 2 } + 2 ( 4 ) ( \sqrt { 5 } ) + ( \sqrt { 5 } ) ^ { 2 } \right\} + \left\{ ( 4 ) ^ { 2 } - 2 ( 4 ) ( \sqrt { 5 } ) + ( \sqrt { 5 } ) ^ { 2 } \right\} } { ( 4 ) ^ { 2 } - ( \sqrt { 5 } ) ^ { 2 } }{/tex}
{tex}= \frac { ( 16 + 8 \sqrt { 5 } + 5 ) + ( 16 - 8 \sqrt { 5 } + 5 ) } { 16 - 5 } = \frac { 42 } { 11 }{/tex}
Posted by Harsh Gupta 5 years, 5 months ago
- 1 answers
Pramod Pant 5 years, 5 months ago
Posted by Shubhangi Mozarkar 5 years, 5 months ago
- 1 answers
Yogita Ingle 5 years, 5 months ago
5/7=0.7142 9/11=0.818181
first we have to change the the number into decimal or rational
decimal form of 5/7 is 0.71428571 and the decimal form of 9/11 is 81818181
hence the rational number are between 0.71428571 and 0.81818181
rational numbers between 5/7 and 9/11
1) 0.72428571
2) 0.73428571
3) 0.74428571
.
.
.
.
upto 0.80818181
you can take any number between them as irrational.
Posted by Naman Sen 5 years, 5 months ago
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Posted by Shivika Arora 5 years, 5 months ago
- 1 answers
Namrata Jindal 5 years, 5 months ago
Posted by Riya Suriyavanshi 5 years, 5 months ago
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Posted by Riya Suriyavanshi 5 years, 5 months ago
- 1 answers
Sia ? 5 years, 5 months ago
Take AB= BC = 1cm
{tex}\angle B = {90^0}{/tex}
In {tex}\triangle{/tex}ABC
AC2 = AB2 + BC2
AC2 = 1 + 1
AC2 = 2
AC = {tex}\sqrt{2}{/tex}
In {tex}\triangle{/tex}OCD
{tex}\angle C = {90^0}{/tex}
OD2 = OC2 + DC2
OD2 = ({tex}\sqrt{2}{/tex})2 + 1
OD2 = 2 + 1
OD2 = 3
OD = {tex}\sqrt{3}{/tex}
Posted by Jashanjot K Jashan 5 years, 5 months ago
- 2 answers
Sia ? 5 years, 5 months ago
First rational number between {tex}\frac{1}{2}{/tex} and {tex}\frac{1}{3}{/tex}
{tex}= \frac{1}{2}\left[ {\frac{1}{2} + \frac{1}{3}} \right] \Rightarrow \frac{1}{2}\left[ {\frac{{3 + 2}}{6}} \right] \Rightarrow \frac{5}{{12}}{/tex}
{tex} {\text{ = }}\frac{{\text{1}}}{{\text{2}}}{\text{,}}\frac{{\text{5}}}{{{\text{12}}}}{\text{ and }}\frac{{\text{1}}}{{\text{3}}}{/tex}
Second rational number between{tex}\frac{1}{2}and\frac{1}{3}{/tex}
{tex} {\text{ = }}\frac{1}{2}\left[ {\frac{1}{2} + \frac{5}{{12}}} \right] \Rightarrow \frac{1}{2}\left[ {\frac{{6 + 5}}{{12}}} \right] \Rightarrow \frac{{11}}{{24}} {/tex}
{tex} {\text{ = }}\frac{5}{{12}}and\frac{{11}}{{24}}{\text{ are two rational numbers between }}\frac{1}{2}and\frac{1}{3} {/tex}
Sachin Ahlawat 5 years, 5 months ago
Posted by Shorya Sharma 5 years, 5 months ago
- 2 answers
Rupa Goel 5 years, 5 months ago
Posted by Pradeep Yadav 5 years, 5 months ago
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Posted by Dev Bhati 5 years, 5 months ago
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Posted by Keshav Deshmukh 5 years, 5 months ago
- 1 answers
Sia ? 5 years, 5 months ago
Given: Perimeter of rectangular wall = {tex}2\left( l+b \right){/tex} = 250 m ……….(i)
Now Area of the four walls of the room
={tex}\frac{\text{Total cost to paint walls of the room}}{\text{Cost to paint 1 }{{\text{m}}^{2}}\text{ of the walls}}{/tex}
={tex}\frac{15000}{10} = 1500 {{m}^{2}}{/tex} ……….(ii)
{tex}\because{/tex} Area of the four walls = Lateral surface area = {tex}2\left( bh+hl \right){/tex}={tex}2h\left( b+l \right){/tex} = 1500
{tex}\Rightarrow {/tex} {tex}250\times h=1500{/tex} [using eq. (i) and (ii)
{tex}\Rightarrow {/tex} {tex}h=\frac{1500}{250}{/tex} = 6 m
Hence required height of the hall is 6 m.
Posted by Md Irshad 5 years, 5 months ago
- 3 answers
Sachin Ahlawat 5 years, 5 months ago
Posted by Shreya Maheshwari 5 years, 5 months ago
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Posted by Naresh Mittal 5 years, 5 months ago
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Posted by Sudha Nandakumar 5 years, 5 months ago
- 2 answers
Md Irshad 5 years, 5 months ago
Posted by Ayisha Deora 5 years, 5 months ago
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Posted by Krishna Aggarwal 5 years, 5 months ago
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Sameeksha V H 5 years, 5 months ago
Posted by Shruti Narasimhan 5 years, 5 months ago
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Posted by Nikshit Joshi 5 years, 5 months ago
- 2 answers
Sia ? 5 years, 5 months ago
A rational number lying between {tex}\frac {3} {6}{/tex} and {tex}\frac 25{/tex}
{tex}\frac{\frac{3}{6}+\frac{2}{5}}{2}{/tex}{tex}=\frac{\frac{15+12}{30}}{2}{/tex}{tex}=\frac{\frac{27}{30}}{2}=\frac{27}{60}{/tex}
Sakshi Sahu 5 years, 5 months ago
Posted by Simrandeep Kaur 5 years, 5 months ago
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Posted by Sumit Thakur 5 years, 5 months ago
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Annesha Mondal 5 years, 5 months ago
1Thank You