Question

can anyone plz help me with this question
ASAP​

Answer 1

see in the attachment .

Answer 2

Answer:

The mean of $(x_{1}+a),(x_{2}+a),...(x_{n}+a)$ is $\bar x+a$

Step-by-step explanation:

The mean of n terms is:

$\bar x=\frac{x_{1}+x_{2}+x_{3}...+x_{n}}{n}$

If the a constant, say a, is added to all the terms the mean of n terms is:

$\frac{(x_{1}+a)+(x_{2}+a)+(x_{3}+a)...+(x_{n}+a)}{n}=\frac{(x_{1}+x_{2}+x_{3}...+x_{n})+(a+a+a...+n\ times)}{n}\\=\frac{x_{1}+x_{2}+x_{3}...+x_{n}}{n}+\frac{na}{n}\\=\bar x+a$

Thus, the mean of $(x_{1}+a),(x_{2}+a),...(x_{n}+a)$ is $\bar x+a$.