Can someone help me with this question? thank you very much appreciated.

Cannot see the other picture as you just sent one

And the tables will help you and so will this.

Step-by-step explanation:

Look at the value for question of getting an answer right and you have to do a table that looks like this attached  

4 games at 2/3 chance for each = 16/27

We look at 1/3 x 1/3 = 11/ 20

we look at 2/3 x 2/3 = 0.444 = 22/50 = 11/25

The other two games are

1/3 x 2/3 = 0.22 + 0.22 = 0.44

0.44  =  11/25

+ 3/20

= 16/27

It wont be as complicated as that. But im sure this will help.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

yes. we do proportions in probabilities.