Find the probability of rolling a die twice, first getting a 3 and then an odd number. (This sounds simple but every answer I get isn’t the correct one)
PLEASE I DON’T UNDERSTAND

1/3 or 0.333333333

Step-by-step explanation:

Probability of getting a 3 is 1/6.

1/6

Probability of getting an odd number is 3/6.

3/6

Multiply them.

1/6*3/6 = 4/12

Simplify.

4/12 = 1/3

Correct me if I am incorrect.

10

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.