In how many ways can a sample of 6 keyboards be selected so that exactly two have an electrical defect

15ways

Step-by-step explanation:

This is a combination question since combination has to do with selection. Hence the number of ways sample of 6 keyboards can be selected so that exactly two have an electrical defect is expressed as;

6C2 = 6!/(6-2)!2!

6C2 = 6!/4!2!

6C2 = 6×5×4!/4!×2

6C2 = 6×5/2

6C2 = 30/2

6C2 = 15

Hence this can be done in 15ways

answer: idk

step-by-step explanation:

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.