Suppose 80 points are placed around a circle. A line segment is drawn between each pair of points.How many line segments are drawn?

- anonymous

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- 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

you know how to compute that number?

- anonymous

I have not idea

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## More answers

- anonymous

i take it this is some combinatorics class
or maybe probability?

- anonymous

it is my MGF1107 class LIberal Art II

- anonymous

liberal arts?! lol have fun
actually it is the same idea as your previous problem

- anonymous

\[\binom{80}{2}=\frac{80\times 79}{2}\]

- anonymous

yeah I am completely lost

- anonymous

would the answer be 3160

- 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

ok

- 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

no

- 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

you certainly don't have time to count them all!

- anonymous

I am not sure because this math class is online

- anonymous

you got a book?

- anonymous

yeah online and I am lost in the book

- anonymous

I dont see the formula in the book at all

- anonymous

hmm you need it for the last problem you asked as well, the maze problem

- anonymous

maybe it is written as \(_nC_k\)

- anonymous

is there anything there about pascal's triangle?

- anonymous

I know i posted the maze i dont know if you can click on the link i posted

- anonymous

http://www.coursesmart.com/9781285221540/firstsection?portal=coursesmart&CSTenantKey=cengage&key=A9438807DBF48783826B50DD6D3AF8BF0E7FAF703C07BD9FEE00354F9C01481B6CEBCCEC35541EDEFC045B8A42CB4BF49E2E2807289730B0BEF8983EAAC74E2A5B5626976C2002694A3760266CA6BC3D23E6EF808A5754D6FD40FE6B57F0B7B9E9195060C5FDA2C9D4F8D10885DA3B79870BE547A0CCEC8CD35BC6D533CF35E6&spid=&TenantOption=cengagebrain#X2ludGVybmFsX0J2ZGVwRmxhc2hSZWFkZXI/eG1saWQ9OTc4MTI4NTIyMTU0MC8yNA==

- anonymous

let me see if i can find it, i can help you with that one as well

- anonymous

thank you I sent you the link to my book

- anonymous

yeah i got it
what page are you on?

- anonymous

problem solving

- anonymous

do you know what page it is?

- anonymous

page 13

- anonymous

page 14

- anonymous

jeez you are right, i guess you are just supposed to "figure it out"

- anonymous

let me find your last problem and see if i can help you with it

- anonymous

ok thank you

- anonymous

I am so lost thank you for helping

- anonymous

could you repost? i can't find the damned problem

- anonymous

yeah

- anonymous

you can put it here if you like

- 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

ok

- anonymous

ok now before we do this one lets redo the first one with a method that might be more suitable

- anonymous

ok

- 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

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

ok

- 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

3160 was the answer

- anonymous

yes

- anonymous

now for the next one, i cannot think of simple explanation but i can explain it

- anonymous

i have 10 questions to do. Can you help me with them

- anonymous

Thank you. Your such a big help

- anonymous

lol sure until i get tired

- anonymous

lol ok thank you. I need to pass this class for my RN degree

- 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

ok I am looking at it now

- 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

ok so I took 5 steps to the right and 5 up

- anonymous

that would give me 25 steps

- 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

ok

- 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

now I am lost

- anonymous

forget that then, it is unnecessary

- anonymous

sorry

- anonymous

the question is, how many ways can i fill ten slots with 5 \(r\) and 5 \(u\)

- anonymous

oh ok

- anonymous

now we have to use the formula i wrote above
i see no way around it

- anonymous

ok let me right that formula down

- anonymous

the question translates in to this
"how many ways can i pick 5 out of the 10 slots to put an \(r\)?"

- 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

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

ok let me solve this

- anonymous

ok

- anonymous

252 is the answer

- anonymous

yes

- anonymous

thank you!!

- anonymous

yw
next?

- anonymous

What is the 76th decimal digit in the decimal representation of

- anonymous

13/11

- anonymous

that can't be that hard, you only have two to choose from

- 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

76 is an even number right?

- anonymous

so it is 8

- anonymous

lots easier than the last two by far

- anonymous

yes that is what i got

- anonymous

lol

- anonymous

ok
i hope the rest are that easy

- 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

we are on number 5

- anonymous

you got lots of way to do this
you can write an equation if you like
or just guess and check

- anonymous

ok what way would be the easy

- 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

ok

- 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

or as we say in math,
\[2x+3.5=46.5\]

- anonymous

can you solve that one? takes two steps
subtract \(3.5\) from both sides, divide both sides by \(2\)

- anonymous

ok 21.5

- anonymous

answer would be 21.5 feet and 18 feet

- anonymous

no

- 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

and you can check that this is right, because
\[21.5+25=46.5\]

- anonymous

i took 3.5 from both

- anonymous

then I added it back to the longer side

- 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

yes you are correct

- anonymous

I see now

- anonymous

4 left?

- anonymous

yes

- anonymous

k

- 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

unless there is a trick to this, shouldn't it just be \(6\times 4\) ?

- anonymous

thats what I put too

- anonymous

ok wow these are tons easier than the first two

- anonymous

lol

- anonymous

3 to go

- anonymous

wahoo. Thanks again

- 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

would it be 4

- 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

so that is not the right answer
if there were only 4 players at the beginning the winner would only play 2 games

- anonymous

thats right because that would not make sense

- anonymous

lets try it with 16
16
8
4
2
four games

- anonymous

wanna try 32?

- anonymous

yeah

- anonymous

32
16
8
4
2

- anonymous

5 games

- anonymous

got 5

- anonymous

ok what do you think the answer might be? 32 is too small

- anonymous

I think it might be 5 games

- anonymous

\[2,4,8,16,32,64,...\]

- anonymous

we know the winner played 8 games, the question is, how many were in the tournament

- anonymous

how many people are in a chess tournament to begin with is what we dont have

- anonymous

right
is the pattern clear?

- anonymous

no

- anonymous

would it be 50

- anonymous

no they are all powers of 2

- anonymous

oh ok

- 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

so your job is to compute \(2^8\)

- anonymous

256

- anonymous

yes

- anonymous

two to go

- anonymous

i would multiply 2 to the second power which is 8 correct

- anonymous

? you would take two to the power of 8, which you did

- 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

number would not coy over

- anonymous

\[3^{3887}\]?

- anonymous

yes

- anonymous

ok lets look for a pattern, because i have no idea

- anonymous

ok
me either

- anonymous

\[3^1=3\\3^2=9\\3^3=27\\3^4=81\]

- anonymous

once we see that 1 at the end, we know it will start over

- anonymous

ok i have what you have

- 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

and so on and so on

- anonymous

\[3,7,9,1,3,7,9,1,3,7,9,1...\]

- 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

every fourth one is a 1
so figure out what the integer remainder is when you divide \(3887\) by \(4\)

- anonymous

so the answer would be 3 or 7

- 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

and so \(3^{3887}\) move back one step

- anonymous

yes

- anonymous

so the answer would be one

- anonymous

on no

- anonymous

ok

- anonymous

\[3^{3888}\] ends in a \(1\)

- anonymous

since 4 goes in to 3888 evenly

- anonymous

I see that

- anonymous

but you want the units digit of \(3^{3887}\) \[3,7,9,1\] pick the one previous to \(1\)

- anonymous

9

- anonymous

yup

- 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

5 10 17 26 ? 50 65

- anonymous

would this answer be 36

- anonymous

no

- anonymous

ugh

- anonymous

it is ok really

- anonymous

what do you add to 5 to get 10?

- anonymous

5

- anonymous

what do you add to 10 to get 17?

- anonymous

7

- anonymous

and what do you add to 17 to get 16?

- anonymous

i mean 26?

- anonymous

9

- anonymous

take a guess as to what you are going to add to 26

- anonymous

11

- anonymous

yes

- anonymous

see the patter now

- 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

but you can do it by looking at the patter of differences as well

- anonymous

oh ok now I see it

- anonymous

\(26+11=37\) and also \(6^2+1=37\)

- anonymous

your way was easy

- 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

well i guess it has to be more than 100

- anonymous

oh no it doesn't

- anonymous

no

- anonymous

i guess they could have one 98% of their games

- anonymous

*won

- anonymous

You know what I get lost on wheat formula to use to solve some of these problems

- anonymous

no formula
just thinking

- anonymous

oh ok

- anonymous

would this be a fraction question

- 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

right

- anonymous

but that is not the answer because they did not win 100% of the games, they won less

- anonymous

i know there is 162 games in mlb

- 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

lol all i know is that the phillies will be lucky if they win 75 of them

- anonymous

lol I agree

- anonymous

I live in florida and I am a phillies fan

- 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

why a phillies fan in florida? clearwater?

- anonymous

yeah
dont like the Rays

- anonymous

you done?
i mean are the problems done?

- anonymous

yes

- anonymous

Thank you

- anonymous

yw
you gonna pass this class?

- anonymous

hope so . My last class for my RN degree. Everything is online and no mid term or final

- anonymous

well you can always come here for help
i guess you put the math off until last
good luck!

- anonymous

I also have anatomy and physiology

- anonymous

bet they are harder

- anonymous

Thank you very much. You are awesome !!!

- anonymous

yeah it takes up a lot of my time

- anonymous

your welcome again
and good night

- anonymous

good night

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