Determine the mean of first 10 natural numbers. -Maths 9th

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Answer :

Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 / 10 = 55 / 10 = 5.5

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Description : Determine the mean of first 10 natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 / 10 = 55 / 10 = 5.5

Description : Find the arithmetic mean of first-five natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5/5 = 15/5 = 3

Description : Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th

Last Answer : The given relation stated in words is R = {(x, y) : x is the fourth power of y; x ∈ N, y ∈ {1, 2, 3, 4}}.

Description : On a set N of all natural numbers is defined the relation R by a R b iff the GCD of a and b is 2, then R is -Maths 9th

Last Answer : (c) Symmetric only Let a ∈N. Then (a, a) ∉R as the GCD of a' and a' is a' not 2. R is not reflexive Let a, b ∈N. Then, (a, b) ∉R ⇒ GCD of a' and b' is 2 ⇒ GCD of b' and a' is 2 ⇒ (b, a) ∈R ∴ R ... , let a = 4, b = 10, c = 12 GCD of (4, 10) = 2 GCD of (10, 12) = 2 But GCD of (4, 12) = 4.

Description : If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Last Answer : (d) An equivalence relationWe can check the given properties as follows: Reflexive: Let (a, b) ∈ N x N. Then (a, b) ∈ N ⇒ a + b = b + a (Communtative law of Addition) ⇒ (a, b) R (b, a) ⇒ (a, b) R (a, ... , f) ⇒ (a, b) R (e, f) on N x N so R is transitive.Hence R is an equivalence relation on N N.

Description : The relation ‘is less than’ on a set of natural numbers is -Maths 9th

Last Answer : (c) Only transitiveLet N be the set of natural numbers. Then R = {(a, b) : a < b, a, b ∈N}A natural number is not less than itself ⇒ (a, a)∉R where a ∈N ⇒ R is not reflexive V a, b ∈N, (a, b) ∈R ⇒ a < b \( ot\ ... ∈N, (a, b) ∈R and (b, c) ∈R ⇒ a < b and b < c ⇒ a < c (a, c) ∈R ⇒ R is transitive.

Description : Out of 3n consecutive natural numbers 3 natural numbers are chosen at random without replacement. -Maths 9th

Last Answer : (c) \(rac{3n^2-3n+2}{(3n-1)(3n-2)}\)In 3n consecutive natural numbers, (i) n numbers are of the form 3p (ii) n numbers are of the form 3p + 1 (iii) n numbers are of the form 3p + 2 For the ... ) We can select one number from each set.∴ Favourable number of cases = nC3 + nC3 + nC3 + (nC1 nC1 nC1)

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Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

Description : The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

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Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

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Last Answer : Solution: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2+1+2−1 = 0 ∴By factor theorem, g(x) is a factor of p(x

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