Question 1189569
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Answers:<ol><li>Discrete</li><li>Continuous</li><li>Discrete</li><li>Continuous</li><li>Discrete</li><li>Discrete</li><li>Discrete</li><li>Discrete</li></ol>In other words, problems 2 and 4 are continuous. Everything else is discrete. 


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Explanation:


Consider the real number line. This is continuous because we can pick any two different points, and find the midpoint between them. 


For instance, the midpoint of 3 and 7 is (3+7)/2 = 10/2 = 5. This process of finding the midpoint can be done as much as you like to find successive new midpoints. There's no stopping limit. 


So that's what it means when we say "continuous".


In contrast, a discrete set is where there are so called gaps in the set. Consider the set of whole numbers. There's a gap between 1 and 2, between 2 and 3, and so on. We cannot apply the midpoint formula to say 1 and 2 to get 1.5, because 1.5 is not a whole number.


It's tempting to say "a discrete variable is finite" but the set of whole numbers is infinite, so that statement in quotes isn't exactly true (it's only sometimes true for some cases).
I suppose you could argue to focus on a subset between two endpoints; at which this point the subset is finite. 
Eg: The infinite discrete set {1,2,3,4,...} has the finite discrete subset {5,6,7,8}.
In contrast the set of real numbers between any two endpoints is infinite and continuous.


Anyways, I'm digressing quite a bit. Hopefully what is discussed above makes a bit of sense. If not, then let me know. 



Let's address the eight problems given.


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Problem 1


This is discrete because we're talking about a number of something. We're using the counting numbers {1,2,3,4,...} which means we cant have gaps between adjacent whole numbers. It doesn't make sense to have 2.75 number of students for instance. 


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Problem 2


The length is something we can measure in terms of a continuous variable. 
It's possible to have a time value of say 2.5 hours or 2.2771 hours or any positive real number on your mind.


No matter what two different time values you pick, we can apply the midpoint formula to get some other time value. 
This is sufficient to show that we have a continuous variable.


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Problem 3


This is the same idea as problem 1. 
We have a number of something we want to count, so we use the set {1,2,3,...} which is a discrete counting set.


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Problem 4


Weights are similar to that of time lengths (see problem 2) or regular physical lengths in general. So we have a continuous variable.


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Problem 5


This is similar to problems 1 and 3. We have a number of something, so we use the natural numbers. 
Often the key phrase "number of" will be an indicator of a discrete set. Though this won't always be the case (see problem 6)


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Problem 6


It might be tempting to consider this a continuous set, but it's not continuous. 
Why not? Because we can't have a gap between something like $105.00 and $105.01


In other words, we can reduce the universe we're working with into the smallest atomic portion. Which in this case is the penny. 
We cannot reduce any smaller than this, so we're not working with a continuous set of numbers.


Put yet another way: Let's say the family's monthly spending is somewhere between $100 and $200 inclusive of both endpoints. 
This would mean the set {100, 100.01, 100.02, ..., 199.98, 199.99, 200} is finite. The finite nature directly leads to a discrete variable. 
Notice a value like 100.00756 isn't part of that set. So that's one narrow example showing a gap between 100 and 100.01


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Problem 7


Refer to problems 1, 3, and 5. We have a discrete set here.


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Problem 8


Similar to problems 1, 3, 5, and 7. This is also discrete.


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To summarize, problems 1, 3, 5, 6, 7 and 8 are discrete while 2 and 4 are continuous.


Furthermore, the key idea is that if sometimes finding the midpoint makes no sense (eg: having 1.5 persons or having 105.11119292 dollars) then those are indicators of a discrete set. 
Otherwise, the set is continuous if the midpoint works/makes sense for <i>any</i> two different values.

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