Answer:
(a)Food-12%
9b)Transportation-17%
Step-by-step explanation:
In the Year 2000
- Average Household Income (before taxes)=$44,649
- Amount Spent on Food =$5,158
- Amount Spent on Transportation=$7,417
Therefore:
(a)Percentage of the average household income in 2000 was spent on food
(b)Percentage of the average household income in 2000 was spent on transportation
Part (a)
There are 7 red out of 7+3 = 10 total
<h3>Answer: 7/10</h3>
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Part (b)
We have 3 green out of 10 total
<h3>Answer: 3/10</h3>
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Part (c)
3/10 is the probability of getting green on any selection. This is because we put the first selection back (or it is replaced with an identical copy)
So (3/10)*(3/10) = 9/100 is the probability of getting two green in a row.
<h3>Answer: 9/100</h3>
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Part (d)
Similar to part (c) we have 7/10 as the probability of getting red on each independent selection.
(7/10)*(7/10) = 49/100
<h3>Answer: 49/100</h3>
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Part (e)
7/10 is the probability of getting red and 3/10 is the probability of getting green. Each selection is independent of any others.
(7/10)*(3/10) = 21/100
<h3>Answer: 21/100</h3>
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Part (f)
We have the exact same set up as part (e). Notice how (7/10)*(3/10) is the same as (3/10)*(7/10).
<h3>Answer: 21/100</h3>
Answer:
I think the answer is F hope its right fingers crossed!
Step-by-step explanation:
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:
I believe this that it would be 1/3 chances you grab a yellow or a orange
