Let $¯ x$ be the mean of $x_{1}, x_{2}, \dots, x_{n}$ . and $¯ y$ be the mean of $y_{1}, y_{2} \dots, y_{n}$ If $¯ z$ is the mean of $x_{1}, x_{2}, \dots, x_{n}, y_{1}, y_{2}, \dots y_{n}$ , then $¯ z = ?$

(a) $(¯ x + ¯ y)$

(b) $\frac{1}{2} (¯ x + ¯ y)$

(b) $\frac{1}{2 n} (¯ x + ¯ y)$

Q. If

¯ x \cdot ¯ y = ¯ x \cdot ¯ z

and

¯ x \times ¯ y = ¯ x \times ¯ z

and

¯ x \neq ¯ 0

, then prove that

¯ y = ¯ z

¯ x, ¯ y, ¯ z

are non-zero vectors, if

(¯ x . ¯ y) = \frac{π}{2}

and

¯ z = ¯ x + ¯ y

, then

Q. If

| ¯ x | = | ¯ y | = 1

and

¯ x ⊥ ¯ y

, then

| ¯ x - ¯ y | =

_____________.

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Let ¯x be the mean of x1,x2,...,xn and ¯y the mean of y1,y2,...,yn. If ¯z is the mean of x1,x2,...,xn, y1,y2,...,yn, then ¯z is equal to

The correct option is B ¯x+¯y2¯x is the mean of x1,x2,...,xn, then ¯x = x1+x2+...+xnn ¯y is the mean of y1,y2,...,yn. then¯y = y1+y2+...+ynn¯z is the mean of x1,x2,...,xn, y1,y2,...,yn,¯z = x1+x2+...+xn+y1+y2+...+yn2n¯z = ¯x+¯y2

Let $¯ x$ be the mean of $x_{1}, x_{2}, . . ., x_{n}$ and $¯ y$ the mean of $y_{1}, y_{2}, . . ., y_{n}$ . If $¯ z$ is the mean of $x_{1}, x_{2}, . . ., x_{n}$ , $y_{1}, y_{2}, . . ., y_{n}$ , then $¯ z$ is equal to