Answer:
What is the probability that a randomly selected family owns a cat? 34%
What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat? 82.4%
Step-by-step explanation: We can use a Venn (attached) diagram to describe this situation:
Imagine a community of 100 families (we can assum a number, because in the end, it does not matter)
So, 30% of the families own a dog = .30*100 = 30
20% of the families that own a dog also own a cat = 0.2*30 = 6
34% of all the families own a cat = 0.34*100 = 34
Dogs and cats: 6
Only dogs: 30 - 6 = 24
Only cats: 34 - 6 = 28
Not cat and dogs: 24+6+28 = 58; 100 - 58 = 42
What is the probability that a randomly selected family owns a cat?
34/100 = 34%
What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?
A = doesn't own a dog
B = owns a cat
P(A|B) = P(A∩B)/P(B) = 28/34 = 82.4%
Answer:
the number of hamburgers sold on Thursday were 325.
Step-by-step explanation:
The total number of hamburger and cheese burger is missing
i will replace it with any figure, you can replace it wit your given data and you will get the solution.
A local hamburger shop sold a combined total of 593 hamburgers and cheeseburgers on Thursday
There were 57 fewer cheeseburgers sold than hamburgers
How many hamburgers were sold on thursday
Let h be the number of hamburgers and c be the number of cheeseburgers.
Using this information we can set up two equation as:

Now we need to solve these two equations to get the value of number of hamburgers. For that we use substitution method as shown below:

Therefore, the number of hamburgers sold on Thursday were 325.
This to numbers are equal
Answer:
Graph One.
Step-by-step explanation:
See below. The graph is in a straight line, also known as linear. The other graphs are not. Also, if you look at the range (y) of the tables, you can see the increase is linear as it is increasing by .5 (1/2) everytime.