I think it's a b not sure about the rest though
Answer:
a) 0.06
b) 0.778
Step-by-step explanation:
Let's suppose a community of 100 families just to facilitate the calculation.
30% of the families own a dog
Dog = 30% of 100 = 30
20% of the families that own a dog also own a cat = 20% of 30 = 6
27% of all the families own a cat = 27% of 100 = 27
So, 6 families own a dog and a cat.
As 30 families own a dog, [30 - 6 =] 24 families own only dogs
As 27 families own a cat, [27 - 6 = ] 21 families own only cats
See picture attached.
a) What is the probability that a randomly selected family owns both a dog and a cat?
P(dog and cat) = dog ∩ cat/total = 6/100 = 0.06
b) What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?
So, only cat/total cat
P (not dog|cat) = 21/27 = 0.778
On the attachment, there is the graph for the region "R" and the calculations for the value a and b.
the theory is that you have to choose a value that is between the range and assume that this value will divide the area into two equal parts. This is done for x = a
For
"y" you have to change the integral from dx to dy and you have to divide the region into 2 parts, given that the area cannot be calculated by 1 integral equation, so you proceed to calculate the rectangular area and take this area into consideration, for the same procedure as before.
Then you calculate again the value y = b and that's it.
The midline of the sine function is 0. The +3 added to the sine function moves the midline of f(x) to f(x) = 3.
_____
You may be expected to write the equation as y = 3.