2) in the united states, 56% of all children get an allowance and 41% of all children get an allowance and do household chores. 15% of all children do neither. what is a) p(household chores only)

0.46

Step-by-step explanation:

Let A be the no of children who get an allowance and B the no of children who get both allowance and do household chores.

Given that P(A) = 0.56 and P(AB) = 0.41

Also P(A'B') = 0.15 where A' represents the complement of A and B' is complement of B.

We know that total children can be classified into 4 disjoint and exhaustive groups as

A-B, AB, B-A and A'B'

Consider A, A is union of (A-B)U(AB) (two disjoint sets)

Hence P(A) = P(A-B)+P(AB)

0.56 = P(A-B) + 0.41

Hence P(A-B) = 0.15

i.e. Total probability = 1

i.e. 0.56-0.15+P(B-A)+P(A'B') =1

Hence P(B-A) = 1-0.41-0.15 =0.46

P(household chores only) = 0.46

idk

step-by-step explanation:

shift 4 units up, and 1 unit left.

it will translate f(x) = 3^x to g(x) = 3^(x+1) + 4