The mean of all the factors of $12$ is:

3 \frac{2}{3}

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4 \frac{3}{2}

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4 \frac{2}{3}

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Solution

The correct option is C $4 \frac{2}{3}$
The factor of $12$ are $1, 2, 3, 4, 6, 12$ .

We know, Mean is given by,

$Mean = \frac{Sum of the elements}{Total number of elements}$ .

Here, number of observations $= 6$ .

$\Rightarrow$ Mean $= \frac{1 + 2 + 3 + 4 + 6 + 12}{6} = \frac{28}{6} = \frac{14}{3} = 4 \frac{2}{3}$ .

Hence, option $C$ is correct.

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Similar questions

Q. Simplify:

\frac{{(3 \frac{2}{3})}^{2} - {(2 \frac{1}{2})}^{2}}{{(4 \frac{3}{4})}^{2} - {(3 \frac{1}{3})}^{2}} \div \frac{3 \frac{2}{3} - 2 \frac{1}{2}}{4 \frac{3}{4} - 3 \frac{1}{3}}

$1 + \frac{1^{2} + 2^{2}}{2!} + \frac{1^{2} + 2^{2} + 3^{2}}{3!} + \frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{4!} + . . . \infty = \frac{b}{6} e$ . Find $b$

Q. The sum of the series

1 + \frac{1^{2} + 2^{2}}{2!} + \frac{1^{2} + 2^{2} + 3^{3}}{3!} + \frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{4!} + . . . . .

Q. The value of the infinite series

\frac{1^{2} + 2^{2}}{3!} + \frac{1^{2} + 2^{2} + 3^{2}}{4!} + \frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{5!} + . . . . . .

Q. If

\frac{1 \times 2^{2} + 2 \times 3^{2} + 3 \times 4^{2} + \dots + n (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + 3^{2} \times 4 + \dots + n^{2} (n + 1)} = \frac{8}{7}

, then

n =

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The mean of all the factors of 12 is:

The correct option is C 423The factor of 12 are 1,2,3,4,6,12.We know, Mean is given by, Mean=Sum of the elementsTotal number of elements. Here, number of observations =6.⇒ Mean =1+2+3+4+6+126=286=143=423.Hence, option C is correct.

The mean of all the factors of $12$ is: