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Ask QuestionPosted by Harshali Hajare 7 years, 1 month ago
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Japisher Singh 7 years, 1 month ago
Posted by Priya Yadav 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
For calculating arithmetic mean from this data, We have to first take the midpoint of each class interval, represented by 'm' in the table. Then decide on anyone midpoint as assumed mean and find out the deviations.
Calculation of Arithmetic Mean
<th scope="col">Marks</th> <th scope="col">Number of Students (f)</th> <th scope="col">Mid-Value (m)</th> <th scope="col">dm (m-A) A=45</th> <th scope="col">fdm</th> <th scope="col"> </th>| 10-20 | 2 | 15 | -30 | -60 | -300 |
| 20-30 | 7 | 25 | -20 | -140 | |
| 30-40 | 10 | 35 | -10 | -100 | |
| 40-50 | 15 | 45 | 0 | 0 | |
| 50-60 | 20 | 55 | +10 | +200 | +700 |
| 60-70 | 16 | 65 | +20 | +320 | |
| 70-80 | 6 | 75 | +30 | +180 | |
| {tex}\Sigma f = 76{/tex} | {tex}\Sigma f d m = + 400{/tex} |
After multiplying fd with m for all the values and finding the total of fdm, we apply the values to the formula of mean given below:
Now, {tex}\overline { X } = A + \frac { \Sigma f d m } { \Sigma f } \Rightarrow \overline { X } = 45 + \frac { 400 } { 76 }{/tex}
{tex}\Rightarrow \quad \overline { X } = 45 + 5.26 = 50.26{/tex}
Posted by Sunny Anurag 7 years, 1 month ago
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Yogita Ingle 7 years, 1 month ago
(x + 1) 2 = 2(x -3)
x2 + 2x + 1 = 2x - 6
x2 + 1 + 6 = 0
x2 + 7 = 0
x2 = - 7
x = - √
Posted by Noah Reji Mathew 7 years, 1 month ago
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Posted by Noah Reji Mathew 7 years, 1 month ago
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Posted by Deepanshi Mittal 7 years, 1 month ago
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Mansi Bisht 7 years, 1 month ago
Harshit Gupta 7 years, 1 month ago
Posted by Tarandeep Singh 7 years, 1 month ago
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Sia ? 6 years, 6 months ago
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Posted by Elangovan Chandran 7 years, 1 month ago
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Posted by Sonal Gandhi 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let the cost of a pen be Rs x and that of a pencil be Rs y.
As per given condition, Cost of 5 pens and 6 pencils is Rs.9
Then, 5x + 6y = 9 ............(i)
And Cost of 3 pens and 2 pencils is Rs.5.
So, 3x + 2y = 5 ...........(ii)
Multiplying equation (i) by 2 and equation (ii) by 6, we get
10x + 12y = 18 ...........(iii)
18x + 12y = 30 ...........(iv)
Subtracting equation (iii) from equation (iv), we get
(18x + 12y) -(10x + 12y) = 30 - 18
{tex}\Rightarrow{/tex} 18x - 10x + 12y - 12y = 12
{tex}\Rightarrow{/tex} 8x = 12
{tex}\Rightarrow x = \frac{{12}}{8} = \frac{3}{2} = 1.5{/tex}
Substituting x =1.5 in equation (i), we get
{tex}5 ( 1.5) + 6y = 9{/tex}
{tex}7.5 + 6y = 9{/tex}
6y = 1.5
{tex}y = \frac{{1.5}}{6} = 0.25{/tex}
Hence, the cost of one pen = Rs.1.50 and the cost of one pencil = R.s 0.25
Posted by Nitin Sharma 7 years, 1 month ago
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Posted by Shivam Kumar Jha 7 years, 1 month ago
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Posted by Nipunya Prakashan 7 years, 1 month ago
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Vishal Sagar 7 years, 1 month ago
Posted by A Ananditha 7 years, 1 month ago
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Yogita Ingle 7 years, 1 month ago
Let us assume that √3 is a rational number
That is, we can find integers a and b (≠ 0) such that √3 = (a/b)
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
√3b = a
⇒ 3b2=a2 (Squaring on both sides) → (1)
Therefore, a2 is divisible by 3
Hence a is also divisible by 3.
So, we can write a = 3c for some integer c.
Equation (1) becomes,
3b2 =(3c)2
⇒ 3b2 = 9c2
∴ b2 = 3c2
This means that b2 is divisible by 3, and so b is also divisible by 3.
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.
Posted by Kamlesh Meena 7 years, 1 month ago
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A Ananditha 7 years, 1 month ago
Posted by Jaydeep Gurjar 7 years, 1 month ago
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Sia ? 6 years, 6 months ago
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Posted by Anamika Chaturvedi 7 years, 1 month ago
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Ankush Das 7 years, 1 month ago
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Posted by Ranveer Singh 7 years, 1 month ago
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Mansi Bisht 7 years, 1 month ago
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Nikhil Rai 7 years, 1 month ago
Posted by Akshita Gupta 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let a be any positive integer.
Applying Euclid’s division lemma with divisor = 2, we get
{tex}\begin{array}{l}a=2q+r\;\;\;\;\;\;\;\;\;0\leq r<2\\So\;r=0,1\\\end{array}{/tex}
When r = 0,
a = 2q
So a2= (2q)2 = 4q2 = 4m--------(1) ( where m = q2, which is an integer)
When r = 1
Then a= 2q+1
a2= (2q + 1) 2 = 4q2 + 4q + 1 = 4(q2 + q ) +1 = 4m+1 --------(2) (where m = q2 + q, which is an integer)
From (1) and (2) We Can conclude that
The square of any positive integer is of the form 4m or 4m +1 for some integer m.
Posted by Gautham Prakash 7 years, 1 month ago
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Khadija Taju 7 years, 1 month ago
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Vijayant Talwar 7 years, 1 month ago
1Thank You