C; this is because the line of beat fit isn’t linear but curved in that one graph.
Try to do rise over run method
Answer:
I) P(cat│dog) =
II) These events are not independent
III) P(cat or dog)= 0.7
Step-by-step explanation:
Given : Households have dogs = 38%
So, P(dog) = 0.38
Households have cats = 47%
So, P(cats) = 0.47
Households have both dogs and cats = 15%
So, P(both dog and cat ) =
= 0.15
solution :
i) By formula P(A│B) =
P(cat│dog)= 
P(cat│dog) = 
ii) P(cat│dog)=39.47% = 0.39 and P(cat)=47% = 0.47, are the events not independent
Because condition for independent events in conditional probability is P(A|B)=P(A)
but P(cat│dog) ≠P(cat) i.e. 0.39≠0.47
So, these events are not independent
iii) P(cat or dog) = ?
"or" means union
Formula : 
P(cat or dog) = 
P(cat or dog)= 0.47 + 0.38 - 0.15
P(cat or dog)= 0.7
Answer:
4 balloons
Step-by-step explanation:
According to the problem, the data given are as follows,
Equation given = x + 8 = 12
Here X is showing the number of balloons which Dylan more needs to decorate.
So, by solving the equation, we get
X + 8 = 12
( by taking 8 to RHS )
X = 12 - 8
X = 4
Hence, Dylan need 4 more balloons to decorate for a party.