Asurvey of households revealed that 38% have a dog, 47% have a cat, and 15% have both a cat and a dog. given that a household owns a dog, what is the probability that it also owns a cat? p(cat│dog)= since p(cat│dog)=39.47% and p(cat)=47%, are the events independent or not independent? what is the probability of a household owning a cat or a dog? p(cat or dog)=

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

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

t= half of the arc subtended by t.

r= half of the arc subtended by r.  

t+r= half of the arc subtended by t + half of the arc subtended by r= half of 360=180

step-by-step explanation: