Please help please actually answer Instructions
In this experiment, you will be using two coins as a simulation for a real-world compound event.
Suppose that a family has an equally likely chance of having a cat or a dog. If they have two pets, they could have 1 dog and 1 cat, they could have 2 dogs, or they could have 2 cats.
What is the theoretical probability that the family has two dogs or two cats?

Describe how to use two coins to simulate which two pets the family has.

Flip both coins 50 times and record your data in a table like the one below.
(picture below)

Based on your data, what is the experimental probability that the family has two dogs or two cats?

If the family has three pets, what is the theoretical probability that they have three dogs or three cats?
How could you change the simulation to generate data for three pets?

see below

Step-by-step explanation:

What is the theoretical probability that the family has two dogs or two cats? (1/2)

The choices are dd, dc, cd, cc

There are 4 choices = dd or cc/ total = 2/4 = 1/2

Let the heads of the coin be dogs and the tails of a coin be cats

Flip two coins and coin A is the first pet and coin B is the second pet

heads , heads = 10

heads, tails  14

tails heads  =13

tails tails =  13

          total 50

Experimental probability  2 dogs or 2 cats = ( hh, tt) /total = ( 10+13) /50 =                    23/50

If we had 3 pets  

what is the theoretical probability that they have three dogs or three cats?

ddd, ddc, dcd, dcc, ccc, ccd ,cdc, cdd

There are 8 options

 ddd or ccc/ total = 2/8 = 1/4

Let the heads of the coin be dogs and the tails of a coin be cats

Flip three coins and coin A is the first pet and coin B is the second pet Coin C be the third pet

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