anonymous
  • anonymous
Suppose 80 points are placed around a circle. A line segment is drawn between each pair of points.How many line segments are drawn?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
you have 80 points along a circle and you have to choose 2 of them to make a line segment the number of ways to do this is \(\binom{80}{2}\) some times written as \(_{80}C_2\) and read "80 choose 2"
anonymous
  • anonymous
you know how to compute that number?
anonymous
  • anonymous
I have not idea

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anonymous
  • anonymous
i take it this is some combinatorics class or maybe probability?
anonymous
  • anonymous
it is my MGF1107 class LIberal Art II
anonymous
  • anonymous
liberal arts?! lol have fun actually it is the same idea as your previous problem
anonymous
  • anonymous
\[\binom{80}{2}=\frac{80\times 79}{2}\]
anonymous
  • anonymous
yeah I am completely lost
anonymous
  • anonymous
would the answer be 3160
anonymous
  • anonymous
yeah i can see what you need to know is a formula for \(\binom{n}{k}\) the number of ways you can choose \(k\) items from a set of \(n\)
anonymous
  • anonymous
ok
anonymous
  • anonymous
the formula is \[\binom{n}{k}=\frac{n!}{k!(n-k)!}\] but you don't really use it exactly in this form have you seen this before?
anonymous
  • anonymous
no
anonymous
  • anonymous
oh then i don't know how you are expected to do the problem without it what methods are you supposed to use?
anonymous
  • anonymous
you certainly don't have time to count them all!
anonymous
  • anonymous
I am not sure because this math class is online
anonymous
  • anonymous
you got a book?
anonymous
  • anonymous
yeah online and I am lost in the book
anonymous
  • anonymous
I dont see the formula in the book at all
anonymous
  • anonymous
hmm you need it for the last problem you asked as well, the maze problem
anonymous
  • anonymous
maybe it is written as \(_nC_k\)
anonymous
  • anonymous
is there anything there about pascal's triangle?
anonymous
  • anonymous
I know i posted the maze i dont know if you can click on the link i posted
anonymous
  • anonymous
http://www.coursesmart.com/9781285221540/firstsection?portal=coursesmart&CSTenantKey=cengage&key=A9438807DBF48783826B50DD6D3AF8BF0E7FAF703C07BD9FEE00354F9C01481B6CEBCCEC35541EDEFC045B8A42CB4BF49E2E2807289730B0BEF8983EAAC74E2A5B5626976C2002694A3760266CA6BC3D23E6EF808A5754D6FD40FE6B57F0B7B9E9195060C5FDA2C9D4F8D10885DA3B79870BE547A0CCEC8CD35BC6D533CF35E6&spid=&TenantOption=cengagebrain#X2ludGVybmFsX0J2ZGVwRmxhc2hSZWFkZXI/eG1saWQ9OTc4MTI4NTIyMTU0MC8yNA==
anonymous
  • anonymous
let me see if i can find it, i can help you with that one as well
anonymous
  • anonymous
thank you I sent you the link to my book
anonymous
  • anonymous
yeah i got it what page are you on?
anonymous
  • anonymous
problem solving
anonymous
  • anonymous
do you know what page it is?
anonymous
  • anonymous
page 13
anonymous
  • anonymous
page 14
anonymous
  • anonymous
jeez you are right, i guess you are just supposed to "figure it out"
anonymous
  • anonymous
let me find your last problem and see if i can help you with it
anonymous
  • anonymous
ok thank you
anonymous
  • anonymous
I am so lost thank you for helping
anonymous
  • anonymous
could you repost? i can't find the damned problem
anonymous
  • anonymous
yeah
anonymous
  • anonymous
you can put it here if you like
anonymous
  • anonymous
Morris Mouse can easily find his way through the maze from the entrance A to the exit B. However, he only receives food if he finds the exit without going west or south. (North is towards the top of the page.) How many different paths can he take through the maze to receive food? (Note: Different paths have at least one distinct section. See the diagram for an example.) https://angel.spcollege.edu/AngelUploads/QuestionData/2bfb73e4-236a-4599-bf10-af270b97f4c8/61324442322642544448.png#{7a7875da-913f-49bb-a105-e7c68ffc0aca}
anonymous
  • anonymous
ok
anonymous
  • anonymous
ok now before we do this one lets redo the first one with a method that might be more suitable
anonymous
  • anonymous
ok
anonymous
  • anonymous
pick one of the 80 points it has 79 places to go pick the next one it has 78 place to go (because you don't want to count the first one twice) pick the next one it has 77 places to go etc
anonymous
  • anonymous
so you can solve this by adding \[79+78+77+76+...+2+1\] or \[1+2+3+...+77+78+79\] now i do notice that you have a formula for this in the book
anonymous
  • anonymous
ok
anonymous
  • anonymous
the formula is \[1+2+3+...+n=\frac{n(n+1)}{2}\] which in your case is \[\frac{79\times 80}{2}\] same exact answer as before
anonymous
  • anonymous
3160 was the answer
anonymous
  • anonymous
yes
anonymous
  • anonymous
now for the next one, i cannot think of simple explanation but i can explain it
anonymous
  • anonymous
i have 10 questions to do. Can you help me with them
anonymous
  • anonymous
Thank you. Your such a big help
anonymous
  • anonymous
lol sure until i get tired
anonymous
  • anonymous
lol ok thank you. I need to pass this class for my RN degree
anonymous
  • anonymous
in the maze problem you have, you can think of it this way: you have to go to the right 5 times and up 5 times, in some order
anonymous
  • anonymous
ok I am looking at it now
anonymous
  • anonymous
let me know when it is clear that you are going to take 5 steps right and 5 steps up to get from point A to point B
anonymous
  • anonymous
ok so I took 5 steps to the right and 5 up
anonymous
  • anonymous
that would give me 25 steps
anonymous
  • anonymous
now the question is, how many ways can you do that here is one way \[(r, r, u, r, u, u, r, u, r, u)\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
if i replace all \(r\) by \(0\) and all \(u\) by \(1\) we turn \[(r, r, u, r, u, u, r, u, r, u)\] in to \[(0,0,1,0,1,1,0,1,0,1)\]
anonymous
  • anonymous
now I am lost
anonymous
  • anonymous
forget that then, it is unnecessary
anonymous
  • anonymous
sorry
anonymous
  • anonymous
the question is, how many ways can i fill ten slots with 5 \(r\) and 5 \(u\)
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
now we have to use the formula i wrote above i see no way around it
anonymous
  • anonymous
ok let me right that formula down
anonymous
  • anonymous
the question translates in to this "how many ways can i pick 5 out of the 10 slots to put an \(r\)?"
anonymous
  • anonymous
and the answer is \[\binom{10}{5}=\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2}\]
anonymous
  • anonymous
as i said this i called "10 choose 5" and we can ask wolfram for the answer, although this is very easy to compute if you cancel first and multiply last
anonymous
  • anonymous
ok let me solve this
anonymous
  • anonymous
ok
anonymous
  • anonymous
252 is the answer
anonymous
  • anonymous
yes
anonymous
  • anonymous
thank you!!
anonymous
  • anonymous
yw next?
anonymous
  • anonymous
What is the 76th decimal digit in the decimal representation of
anonymous
  • anonymous
13/11
anonymous
  • anonymous
that can't be that hard, you only have two to choose from
anonymous
  • anonymous
What is the 76th decimal digit in the decimal representation of 13/11 https://angel.spcollege.edu/AngelUploads/QuestionData/acae6292-47fd-4c42-9c1a-f19417bc9091/62K31R313554314L4653.png#{2752460f-f064-4665-9c63-2331704bfd48}
anonymous
  • anonymous
76 is an even number right?
anonymous
  • anonymous
so it is 8
anonymous
  • anonymous
lots easier than the last two by far
anonymous
  • anonymous
yes that is what i got
anonymous
  • anonymous
lol
anonymous
  • anonymous
ok i hope the rest are that easy
anonymous
  • anonymous
If two ladders are placed end to end, their combined height is 46.5 feet.One ladder is 3.5 feet shorter than the other ladder.What are the heights of the two ladders?
anonymous
  • anonymous
we are on number 5
anonymous
  • anonymous
you got lots of way to do this you can write an equation if you like or just guess and check
anonymous
  • anonymous
ok what way would be the easy
anonymous
  • anonymous
i am always accused of doing this the not math way but if it was me i would guess if one is 20 and the other is 3.5 longer then it would be 23.5 but then the total would only be \(20+23.5=43.5\) so that is not the right answer try a bigger one
anonymous
  • anonymous
ok
anonymous
  • anonymous
if you want an equation, call one length \(x\) so the other would be \(x+3.5\) and then solve \[x+x+3.5=46.5\]
anonymous
  • anonymous
or as we say in math, \[2x+3.5=46.5\]
anonymous
  • anonymous
can you solve that one? takes two steps subtract \(3.5\) from both sides, divide both sides by \(2\)
anonymous
  • anonymous
ok 21.5
anonymous
  • anonymous
answer would be 21.5 feet and 18 feet
anonymous
  • anonymous
no
anonymous
  • anonymous
\[2x+3.5=46.5\] \[2x=43\] \[x=21.5\] is right, but \(x\) is the length of the SHORTER one, so the other one is \(21.5+3.5=25\)
anonymous
  • anonymous
and you can check that this is right, because \[21.5+25=46.5\]
anonymous
  • anonymous
i took 3.5 from both
anonymous
  • anonymous
then I added it back to the longer side
anonymous
  • anonymous
one is \(x\) and the other is \(x+3.5\) when you solve for \(x\) you get \(21.5\) so the other one is \(21.5+3.5=25\) right? the check should convince you that it is correct
anonymous
  • anonymous
yes you are correct
anonymous
  • anonymous
I see now
anonymous
  • anonymous
4 left?
anonymous
  • anonymous
yes
anonymous
  • anonymous
k
anonymous
  • anonymous
The UCSB has 6 squads of badminton players. The UCI team has 4 squads. Every squad from UCSB must play every squad from UCI at least once. What is the least number of games that must be played?
anonymous
  • anonymous
unless there is a trick to this, shouldn't it just be \(6\times 4\) ?
anonymous
  • anonymous
thats what I put too
anonymous
  • anonymous
ok wow these are tons easier than the first two
anonymous
  • anonymous
lol
anonymous
  • anonymous
3 to go
anonymous
  • anonymous
wahoo. Thanks again
anonymous
  • anonymous
In a chess tournament, each player played a game in the first round. The losers dropped out. This process continued until a winner was declared. The winner of the tournament played 8 games. How many people were in the tournament?
anonymous
  • anonymous
would it be 4
anonymous
  • anonymous
no idea lets think suppose there were 8 player then after the first round there would be 4 then 2 so the winner would have played 3 games
anonymous
  • anonymous
so that is not the right answer if there were only 4 players at the beginning the winner would only play 2 games
anonymous
  • anonymous
thats right because that would not make sense
anonymous
  • anonymous
lets try it with 16 16 8 4 2 four games
anonymous
  • anonymous
wanna try 32?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
32 16 8 4 2
anonymous
  • anonymous
5 games
anonymous
  • anonymous
got 5
anonymous
  • anonymous
ok what do you think the answer might be? 32 is too small
anonymous
  • anonymous
I think it might be 5 games
anonymous
  • anonymous
\[2,4,8,16,32,64,...\]
anonymous
  • anonymous
we know the winner played 8 games, the question is, how many were in the tournament
anonymous
  • anonymous
how many people are in a chess tournament to begin with is what we dont have
anonymous
  • anonymous
right is the pattern clear?
anonymous
  • anonymous
no
anonymous
  • anonymous
would it be 50
anonymous
  • anonymous
no they are all powers of 2
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
winner played 1 game \(2^1=2\) to start winner played 2 games \(2^2=4\) to start winner played 3 games \(2^3=8\) to start winner played 4 games \(2^4=16\) to start
anonymous
  • anonymous
so your job is to compute \(2^8\)
anonymous
  • anonymous
256
anonymous
  • anonymous
yes
anonymous
  • anonymous
two to go
anonymous
  • anonymous
i would multiply 2 to the second power which is 8 correct
anonymous
  • anonymous
? you would take two to the power of 8, which you did
anonymous
  • anonymous
Here is number 8.. Determine the units digit ofhttps://angel.spcollege.edu/AngelUploads/QuestionData/7515d5e6-1da0-40f7-a356-ba07f06634ef/12462452114315333G23.png#{0bee0263-e4d3-41d7-b1d2-7e4a22158da6}
anonymous
  • anonymous
number would not coy over
anonymous
  • anonymous
\[3^{3887}\]?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok lets look for a pattern, because i have no idea
anonymous
  • anonymous
ok me either
anonymous
  • anonymous
\[3^1=3\\3^2=9\\3^3=27\\3^4=81\]
anonymous
  • anonymous
once we see that 1 at the end, we know it will start over
anonymous
  • anonymous
ok i have what you have
anonymous
  • anonymous
\[3^1=3\to 3\\3^2=9\to 9\\3^3=27\to 7\\3^4=81\to 1\\3^5=243\to 3\\3^6=279\to 9\]
anonymous
  • anonymous
and so on and so on
anonymous
  • anonymous
\[3,7,9,1,3,7,9,1,3,7,9,1...\]
anonymous
  • anonymous
so it is a number from that list we have narrowed it down to 4 numbers, now we have to figure out which one it is
anonymous
  • anonymous
every fourth one is a 1 so figure out what the integer remainder is when you divide \(3887\) by \(4\)
anonymous
  • anonymous
so the answer would be 3 or 7
anonymous
  • anonymous
can pretty much do it in your head it is clear that 4 divides \(3888\) evenly right? so \(3^{3888}\) ends in a 1
anonymous
  • anonymous
and so \(3^{3887}\) move back one step
anonymous
  • anonymous
yes
anonymous
  • anonymous
so the answer would be one
anonymous
  • anonymous
on no
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[3^{3888}\] ends in a \(1\)
anonymous
  • anonymous
since 4 goes in to 3888 evenly
anonymous
  • anonymous
I see that
anonymous
  • anonymous
but you want the units digit of \(3^{3887}\) \[3,7,9,1\] pick the one previous to \(1\)
anonymous
  • anonymous
9
anonymous
  • anonymous
yup
anonymous
  • anonymous
ok number 9 Following the pattern shown in the number sequence below, what is the missing number? You may find it helpful to list perfect squares or cubes beneath the sequence terms to try to see how they may relate to the sequence.
anonymous
  • anonymous
5 10 17 26 ? 50 65
anonymous
  • anonymous
would this answer be 36
anonymous
  • anonymous
no
anonymous
  • anonymous
ugh
anonymous
  • anonymous
it is ok really
anonymous
  • anonymous
what do you add to 5 to get 10?
anonymous
  • anonymous
5
anonymous
  • anonymous
what do you add to 10 to get 17?
anonymous
  • anonymous
7
anonymous
  • anonymous
and what do you add to 17 to get 16?
anonymous
  • anonymous
i mean 26?
anonymous
  • anonymous
9
anonymous
  • anonymous
take a guess as to what you are going to add to 26
anonymous
  • anonymous
11
anonymous
  • anonymous
yes
anonymous
  • anonymous
see the patter now
anonymous
  • anonymous
what they want you do see is that \[5=2^2+1,10=3^2+1,17=4^2+1,26=5^2+1\]
anonymous
  • anonymous
but you can do it by looking at the patter of differences as well
anonymous
  • anonymous
oh ok now I see it
anonymous
  • anonymous
\(26+11=37\) and also \(6^2+1=37\)
anonymous
  • anonymous
your way was easy
anonymous
  • anonymous
last oneA baseball team has won more than 97% (but less than 100%) of its games in a season.What is the least possible number of games that the team could have played in the season?
anonymous
  • anonymous
well i guess it has to be more than 100
anonymous
  • anonymous
oh no it doesn't
anonymous
  • anonymous
no
anonymous
  • anonymous
i guess they could have one 98% of their games
anonymous
  • anonymous
*won
anonymous
  • anonymous
You know what I get lost on wheat formula to use to solve some of these problems
anonymous
  • anonymous
no formula just thinking
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
would this be a fraction question
anonymous
  • anonymous
they could have won any percent between 97 and 100 not including 97 or 100 if they won 100% of there games they could have only played one game and won right?
anonymous
  • anonymous
right
anonymous
  • anonymous
but that is not the answer because they did not win 100% of the games, they won less
anonymous
  • anonymous
i know there is 162 games in mlb
anonymous
  • anonymous
if they had won 98.5% of the games, they would have had to play MORE that 100 games to get a fraction of \(98\%\)
anonymous
  • anonymous
lol all i know is that the phillies will be lucky if they win 75 of them
anonymous
  • anonymous
lol I agree
anonymous
  • anonymous
I live in florida and I am a phillies fan
anonymous
  • anonymous
in any case it is getting late lets finish suppose they win 99% then they would have to play 100 games at least, because \(99\%=\frac{99}{100}\) but if they win \(98\%\) then then can have played fewer games, because \[98\%=\frac{98}{100}=\frac{49}{50}\]
anonymous
  • anonymous
why a phillies fan in florida? clearwater?
anonymous
  • anonymous
yeah dont like the Rays
anonymous
  • anonymous
you done? i mean are the problems done?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Thank you
anonymous
  • anonymous
yw you gonna pass this class?
anonymous
  • anonymous
hope so . My last class for my RN degree. Everything is online and no mid term or final
anonymous
  • anonymous
well you can always come here for help i guess you put the math off until last good luck!
anonymous
  • anonymous
I also have anatomy and physiology
anonymous
  • anonymous
bet they are harder
anonymous
  • anonymous
Thank you very much. You are awesome !!!
anonymous
  • anonymous
yeah it takes up a lot of my time
anonymous
  • anonymous
your welcome again and good night
anonymous
  • anonymous
good night

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