Answer:
the volume of a cylinder = V=Ah =pirsqh
=pie × r sq × h
= 3.14×4 ×4×9 = 452.16 so thats eliminated?
Volume of a cone = V= 1÷3pirsqh
= 1/3×3.14×4×4×9 = 150.72
volume of cylinder = 2×pi×r×h+2×pi× r×r
=241.645
not sure choose the ones you think a closest?
Answer:
571.42857142857
Step-by-step explanation:
Answer: The slope is 3
Explanation:
The slope is 3 or 3/1 meaning that it rise 3 and run 1.
Okay, so this question has a lot to do with DISTINGUISHABILITY which basically means: Can you tell the difference between each of the 12 people and can you tell the difference between each of the 8 gummy worm segments?
There are 4 possibilities: The people and gummies are both distinguishable, the people and gummies are both indistinguishable, the people are distinguishable and the gummies are not, or the people are not distinguishable and the gummies are.
<u>The people and gummies are both distinguishable. (This is the easiest case)</u>
There are 8 gummies. Each gummy can go to 12 different people. So the total number of ways we can give 8 distinct gummies to 12 distinct people is 8^12.
<u>
</u><u>The people and gummies are both indistinguishable (Much harder)
</u><u />This case is not necessarily harder, but it is very tedious and requires lots of careful, boring work. You would have to list out every single possible case, 1 by 1, there are no easier shortcuts... for example
8 0 0 0 0 0 0 0 0 0 0 0
7 1 0 0 0 0 0 0 0 0 0 0
6 2 0 0 0 0 0 0 0 0 0 0
6 1 1 0 0 0 0 0 0 0 0 0
5 3 0 0 0 0 0 0 0 0 0 0
5 2 1 0 0 0 0 0 0 0 0 0
5 1 1 1 0 0 0 0 0 0 0 0
4 4 0 0 0 0 0 0 0 0 0 0
...
...
And this would keep on continuing until you're sure you have counted every single one. But remember:
4 4 0 0 0 0 0 0 0 0 0 0
is the same as
4 0 0 4 0 0 0 0 0 0 0 0
because the people are indistinguishable.
<u>Lifehack: Use Wolfram Alpha for questions like this. I tried this just now and Wolfram Alpha stubbornly only counts partitions with all members greater than 1... so for this particular problem, WA won't work
</u><u>
</u><u>The people are distinguishable but the gummies are not</u>
This requires a bit more experience with combinatorics...
My comments above replying to the question talk about this. It is essentially 19C11 = 19C8.
<u>
</u><u>The people are indistinguishable but the gummies are distinguishable</u>
This is by far the hardest case. Remember our second case (<u>The people and gummies are both indistinguishable (Much harder)).</u>
Now we have to add on an extra step on top of all that partitioning. For each and every partition we have to calculate how many ways there are to make that partition. For the
8 0 0 0 0 0 0 0 0 0 0 there is only 1 possible way, all the gummies go to one person
For the
7 1 0 0 0 0 0 0 0 0 0 there is 8 possible ways, we choose one from the 8 gummies that goes to a new person (instead of the person that already has 7 gummies).
This brings everything to an end. I covered the general techniques for each of the 4 distinguishability cases. Hopefully this helps. If it does, please consider making this the brainliest answer. It took me quite a while to write this! :)
