Maths Questions

The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1, 2 and 6, then remaining will be

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Solution : As given \(\bar{x}\) = 4, n = 5 and \({\sigma}^2\) = 5.2. If the remaining observations are \(x_1\), \(x_2\) then \({\sigma}^2\) = \(\sum{(x_i โ€“ \bar{x})}^2\over n\) = 5.2 \(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2 + {(1-4)}^2 + {2-4)}^2 + {(6-4)}^2\over 5\) = 5.2 \(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2\) = 9ย  โ€ฆ..(1) Also \(\bar{x}\) = 4 \(\implies\) โ€ฆ

The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1, 2 and 6, then remaining will be Read More ยป

A student obtained 75%, 80%, 85% marks in three subjects. If the marks of another subject are added then his average marks can not be less than

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Solution : Total marks obtained from three subjects out of 300 = 75 + 80 + 85 = 240 if the marks of another subject is added then the total marks obtained out of 400 is greater than 240 if marks obtained in fourth subject is 0 then minimum average marks = \(240\over 400\)\(\times\)100 = โ€ฆ

A student obtained 75%, 80%, 85% marks in three subjects. If the marks of another subject are added then his average marks can not be less than Read More ยป

If mean of the series \(x_1\), \(x^2\), โ€ฆ.. , \(x_n\) is \(\bar{x}\), then the mean of the series \(x_i\) + 2i, i = 1, 2, โ€ฆโ€ฆ, n will be

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Solution : As given \(\bar{x}\) = \(x_1 + x_2 + โ€ฆ. + x_n\over n\) If the mean of the series \(x_i\) + 2i, i = 1, 2, โ€ฆ.., n be \(\bar{X}\), then \(\bar{X}\) = \((x_1+2) + (x_2+2.2) + (x_3+2.3) + โ€ฆ. + (x_n + 2.n)\over n\) = \(x_1 + x_2 + โ€ฆ. + x_n\over n\) โ€ฆ

If mean of the series \(x_1\), \(x^2\), โ€ฆ.. , \(x_n\) is \(\bar{x}\), then the mean of the series \(x_i\) + 2i, i = 1, 2, โ€ฆโ€ฆ, n will be Read More ยป

Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + โ€ฆโ€ฆ

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Solution : By using method of differences, The \(n^{th}\) term is (2n-1)(2n+1)(2n+3) \(T_n\) = (2n-1)(2n+1)(2n+3) \(T_n\) = \(1\over 8\)(2n-1)(2n+1)(2n+3){(2n+5) โ€“ (2n-3)} = \(1\over 8\)(\(V_n\) โ€“ \(V_{n-1}\)) \(S_n\) = \({\sum}_{r=1}^{nโ€Ž} T_n\) = \(1\over 8\)(\(V_n\) โ€“ \(V_0\)) \(\therefore\)ย  \(S_n\) = \((2n-1)(2n+1)(2n+3)(2n+5)\over 8\) + \(15\over 8\) = \(n(2n^3 + 8n^2 + 7n โ€“ 2)\) Similar Questions Find the โ€ฆ

Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + โ€ฆโ€ฆ Read More ยป

If \({\sum}_{r=1}^{nโ€Ž} T_r\) = \(n\over 8\) (n + 1)(n + 2)(n + 3), then find \({\sum}_{r=1}^{nโ€Ž} \)\(1\over T_r\)

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Solution : \(\because\) \(T_n\) = \(S_n โ€“ S_{n-1}\) = \({\sum}_{r=1}^{nโ€Ž} T_r\) โ€“ \({\sum}_{r=1}^{nโ€Ž โ€“ 1} T_r\) = \(n(n+1)(n+2)(n+3)\over 8\) โ€“ \((n-1)(n)(n+1)(n+2)\over 8\) = \(n(n+1)(n+2)\over 8\)[(n+3) โ€“ (n-1)] = \(n(n+1)(n+2)\over 8\)(4) \(T_n\) = \(n(n+1)(n+2)\over 2\) \(\implies\) \(1\over T_n\) = \(2\over n(n+1)(n+2)\) = \((n+2)-n\over n(n+1)(n+2)\) = \(1\over n(n+1)\) โ€“ \(1\over (n+1)(n+2)\) Let \(V_n\) = \(1\over n(n+1)\) \(\therefore\) โ€ฆ

If \({\sum}_{r=1}^{nโ€Ž} T_r\) = \(n\over 8\) (n + 1)(n + 2)(n + 3), then find \({\sum}_{r=1}^{nโ€Ž} \)\(1\over T_r\) Read More ยป

If A = {x,y}, then the power set of A is

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Solution : Clearly P(A) = Power set of A = set of all subsets of A = {\(\phi\), {x}, {y}, {x,y}} Similar Questions Let A and B be two sets containing 2 elements and 4 elements, respectively. The number of subsets A\(\times\)B having 3 or more elements is If aN = {ax : x \(\in\) โ€ฆ

If A = {x,y}, then the power set of A is Read More ยป

Let A = [x: x \(\in\) R, |x| < 1]; B = [x : x \(\in\) R, |x - 1| \(\ge\) 1] and A \(\cup\) B = R - D, then the set D is

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Solution : A = [x: x \(\in\) R,-1 < x < 1] B = [x : x \(\in\) R, x โ€“ 1 \(\le\) -1 or x โ€“ 1 \(\ge\) 1] [x: x \(\in\) R, x \(\le\) 0 or x \(\ge\) 2] \(\therefore\) A \(\cup\) B = R โ€“ D where D = [x : x โ€ฆ

Let A = [x: x \(\in\) R, |x| < 1]; B = [x : x \(\in\) R, |x - 1| \(\ge\) 1] and A \(\cup\) B = R - D, then the set D is Read More ยป

If N is a set of first 10 natural numbers and a relation R is defined as a + 2b = 10 where a, b \(\in\) N. find inverse of R.

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Solution : R = {(2, 4), (4, 3), (6, 2), (8, 1)} \(R^{-1}\) = {(4, 2), (3, 4), (2, 6), (1, 8)} Similar Questions If A = {x,y}, then the power set of A is If aN = {ax : x \(\in\) N}, then the set 6N \(\cap\) 8N is equal to Let A = โ€ฆ

If N is a set of first 10 natural numbers and a relation R is defined as a + 2b = 10 where a, b \(\in\) N. find inverse of R. Read More ยป
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