Asappp
a company wants to determine the level of interest among its employees for an annual summer picnic. the company wants to survey 200 employees.

what is the best way to randomly choose these 200 employees?

choose 50 people from 4 of the 7 divisions at the company.

select the first 200 employees to arrive at work monday morning.

survey the first 200 people to enter the company cafeteria.

use employee identification numbers to randomly select 200 employees.

Last one: Use employee identification numbers to randomly select 200 employees.

It's the most random.

Use employee identification numbers to randomly select 200 employees.

Step-by-step explanation:

In order to choose 200 random people for the survey, it shall be purely random and unbiased.

If we use employee identification numbers to randomly select 200 employees it would be completely random and an unbiased choice.

For all the other options, none of them are purely random.

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)