so who has tried the webassign due tomorrow tonight? i've got most of it done but its the last half thats giving me trouble...if anyone can help me out, i'd appreciate it

[Edited on October 10, 2006 at 8:38 PM. Reason : /]

10/10/2006 8:33:08 PM

nobody, eh?

10/10/2006 10:11:29 PM

its only because everyone is on a higher math level. post the problem and someone might help.

10/10/2006 10:43:22 PM

Do you still need help on any of those problems. I'm taking the class just for credit. I have done all of the problems, so let me know.

10/10/2006 11:04:13 PM

seriously one of the easiest classes at ncsu

10/11/2006 12:39:28 AM

easy or not, I'm in it and at times, I think there are certain things that are a little trickier than others.

I haven't done the assignment yet, but I hope you get some help. You should try posting on the webassign discussion board on the homepage for this class, one of the other students might help you. Also, as long as it's not the night it's due, the TA, Jennifer George, will help you. She's very quick to respond to questions via email, etc.

10/11/2006 4:21:37 PM

here are some of the problems i'm having trouble with...

A and B are sets. Select all the sets that are equal to the set B U A^c

B^c U A
B^c n A
(A n B^c)^c
B U (B^c n A^c)
A^c U B
_______________________________________________________________________

Sets A, B, and C are subsets of the universal set U.

U = {13,32,34,47,55,66,88,95}
A = {13,32,66,88}
B = {47,32,34,66}
C = {88,66,55,47}

C n (A U B)^c

55
95
13
32
47
66
34
88
________________________________________________________________________

Draw a Venn diagram with sets labeled A, B, and C, and place a single number in each of the 8 regions shown in the diagram according to the information shown below:

A n B n C = {22}
A n B n C^c = {25}
A n B^c n C = {28}
A^c n B n C = {31}
A n B^c n C^c = {34}
A^c n B n C^c = {37}
A^c n B^c n C = {40}
A^c n B^c n C^c = {43}

For each of the sets below, select all the elements that belong to the set.

A U (B^c n C)

37
28
34
22
25
31
40
43

(A U B^c) n C

34
37
25
22
43
31
40
28

(A^c U B) U C^c
22
43
37
40
31
25
34
28

[Edited on October 11, 2006 at 8:00 PM. Reason : /]

10/11/2006 7:58:27 PM

I had trouble with the first one until I drew it... drawing it helps immensely... for both.

10/11/2006 8:51:03 PM

i drew it, but still cant get: (A U B^c) n C

10/11/2006 8:57:27 PM

I'm doing this now and it's very frustrating.

do you know if (B^C U C)^C is the same as BnC^C?
I can't figure out what to do if you raise the entire parenthetic expression to the power of C when one of them inside of the parenthesis is already raised to C and I can't find that on the lectures.

10/11/2006 9:01:04 PM

yeah that looks right to me, the main thing i dont get is how do you find problems that are unioned inside the parentheses with an intersection outside of the parentheses, such as: (A U B) n C

10/11/2006 9:07:56 PM

I think I finally get it. What you do for that is pull out your old algebra skills and make it like this:


(A U B) n C=

(AnC) U (BnC) and then you go and find where A intersects C and then where B intersects C. List all of those numbers under the appropriate set. Since the two sets have the union symbol between them, you will include all of those numbers.

For example, if Anc (a intersecting c) includes the numbers 3&4 and B intersecting C includes 5&6, you will include 3,4,5,6. If the intersection sign was in between the two sets, you would only include the common numbers, if there are any.

10/11/2006 9:37:05 PM

ahhh, thanks a lot! haha i cant believe i couldnt figure that out...

10/11/2006 10:30:37 PM

Has anyone finished Homework Assignment 4.5 Part (b)?

Thanks

10/17/2006 6:32:13 PM

can someone help me with 4.3 #10? thanks

10/17/2006 10:41:21 PM

sorry i havent been on here in a while...but can anyone help me out with number 7 from 4.4? i havent started 4.5 yet

10/20/2006 3:13:38 PM

I'll let you know when I get to it. I am doing this at the last-minute. I might have issues with it, too.

10/20/2006 6:18:03 PM

I'm drawing a blank on how to figure out the second part of questions 1,5 and 6 of section 4.4. If someone could point me in the correct direction, I would really appreciate it, thanks!

10/20/2006 9:07:51 PM

I just figured it out. It was much more obvious than I thought. I hate this kind of stuff.

10/20/2006 9:23:11 PM

has anyone figured out 4.5B #'s 9 and 10? thanks!

10/23/2006 7:02:54 PM

So for this class online with Page, you do all the work on the internet but go to campus to take the test's?

10/24/2006 10:49:40 PM

yeah, its pretty easy as long as you get help with stuff you dont understand. the webassigns are 50 submissions and the tests are webassign with 3 submissions. i have a 98.sumthing right now

10/24/2006 11:47:26 PM

If someone could help, I would really appreciate it.
I don't know why, but I can't figure out how to do the wineglass and the children questions on 4.5 A, question numbers 3 and 7. I tried to make a list and figure it out and it's not working.
this is how I was doing number three:
9 choices of children, 9 days. I used the combination format and then multiplied the result of each combination by the results of the others and I'm not coming up with the same answer.
(9,1) 9!/1!, 8!
8,1
7,1
6,1
5,1
4,1
3,1
2,1
1,1
Thank you!

10/25/2006 7:49:52 PM

For three, I think you just do the permutation of the number of children. So, 9! = 362,880 ways to construct the list of children and days.

Is question seven the wine glass tapping problem?

10/25/2006 8:01:05 PM

what are your number for #7, my answer was 21, i dont know how i did it tho, i just kept entering factors of 7 because there were 7 people

10/25/2006 8:40:24 PM

i cant figure out 4.5B numer 7, first part, if someone could get me started i could get it from there... ok i feel dumb...

[Edited on October 25, 2006 at 8:52 PM. Reason : nevermind i got it, i feel dumb now...]

10/25/2006 8:50:42 PM

statelygurly, did you ever figure out 9 or 10 on 4.5B?

[Edited on October 25, 2006 at 9:06 PM. Reason : /]

10/25/2006 8:58:11 PM

Thank you so much. Yes, seven was the wine tapping question. Anyway, thanks again.

10/25/2006 9:28:31 PM

hey, JH34, do you know how to do 9 or 10 on 4.5B?

10/25/2006 9:30:58 PM

GameOver4U, post the questions or the numbers related to them. I took this class last semester, so I probably don't have the same information to help out.

In case you want to know how to do question seven, draw a picture of a table and just count out how many taps happen when each of the seven people tap their glass with everyone else on the table. Then move onto the next person and count how many taps occur with them (ignoring the person who tapped their glass previously).

You'll eventually find a pattern where you'll add up the number of taps each time to get 21.

10/25/2006 9:34:12 PM

Suppose a country wins 3 medals in the Olympics. This could happen in a variety of ways, for example

3 gold
2 gold and 1 silver
1 gold and 2 bronze
3 bronze
1 gold and 1 silver and 1 bronze

is a list of 5 ways it could happen. (There are other ways it could happen, so this is not a complete list.)

If instead of winning 3 medals a country were to win 6 medals, it could happen in how many ways?

and...

Bill is ordering a meal that includes two servings of vegetables from a menu that lists 14 vegetables on the menu. Bill asks, "Can I order two servings of the same thing if I see something that I really like?".

"Yes," the waiter replies.

How many choices does Bill have with regard to the two servings of vegetables that he orders? (Do assume that he is ordering two servings, but it may or may not be two servings of the same vegetable.)

[Edited on October 25, 2006 at 9:37 PM. Reason : added number 10]

10/25/2006 9:36:50 PM

I think the medals one is 28.

When I had that one, it was if the country were to win 8 medals. I just added up 9+8+7+6+5+4+3+2+1 to get 45. On the test, it had it if country were to win 9 medals, so I just added up 10+9+8+7+6+5+4+3+2+1 to get 55.

I can only assume the problem with 6 medals uses the same "formula." Try adding 'em up like that.

10/25/2006 9:43:51 PM

I haven't done those yet

10/25/2006 9:46:39 PM

ok, that one worked for the medals, but why do you add the number to the front, like if its 6 medals, why do you do 7+6+5+4+3+2+1 and if its 7 medals why do you do 8+7+6+5+4+3+2+1?

10/25/2006 9:50:09 PM

Admittedly, when I did it on the homework, I think I got the right answer by luck.

On the test, I ended up listing out all the possibilities, and found that little trick. You have to count out the number of golds, silvers, and bronzes won in each set of 6 (in your case). So...

6 golds
5 golds, 1 silver
5 golds, 1 bronze
4 golds, 2 silvers
4 golds, 2 bronzes
4 golds, 1 silver, 1 bronze
(and so on)

So, you've got 1 possibility for 6 golds, 2 for 5 golds, 3 for 4 golds, and so on until you get to 0 golds, where you'll have 7 possibilities. The number of possible medal counts increases by 1 for every less gold medal won. After that, you just add 'em up.

I'm not too sure of how to do number ten, but in thinking about that, have you guys checked the TA's help sheet for the homework? She used to post them for us last semester. Page used the post links to them on the news page, I think.

[Edited on October 25, 2006 at 10:08 PM. Reason : Help sheet?]

10/25/2006 9:58:42 PM

yes, I just found it, thank you!

10/25/2006 10:20:51 PM

A combination lock shows 23 different numbers on the dial. To open the lock, the user must know the three-number sequence that is the "combination" for the lock. For example, to open the lock one might have to dial the three numbers 12-2-14 in the order shown here (first dialing 12, then 2, then 14). In a combination, you can't have two consecutive numbers that are the same. For example, you can have 7-10-7 but you can't have 7-7-10.

How many different combinations are there in which the combination consists of three different numbers and in which the numbers occur in order of increasing magnitude? (For example, 5-10-13 but not 10-5-13)

and...

Peg has 6 long-sleeve shirts, 6 short-sleeve shirts, and 4 pairs of pants.

How many possibilities would there be if instead of deciding the shirts should ALL be long sleeve she decides that she should take AT LEAST ONE long sleeve shirt?

10/25/2006 10:37:26 PM

For the combination lock one, I think you need to take the combination of 23 and 3.

So, C(23,3) = 1,771.

[Edited on October 25, 2006 at 10:57 PM. Reason : -]

10/25/2006 10:46:18 PM

awesome thanks

any clue on the other one about Peg??

[Edited on October 25, 2006 at 10:57 PM. Reason : /]

10/25/2006 10:56:39 PM

For the Peg one, how many shirts and pairs of pants is she taking on the trip? All 12 shirts and 4 pairs of pants?

10/25/2006 10:57:12 PM

sorry i forgot i didnt post the whole problem, here it is...

Peg has 6 long-sleeve shirts, 6 short-sleeve shirts, and 4 pairs of pants

If she is going on a trip and plans to take 3 shirts and 2 pairs of pants, how many possibilities are there for the articles of clothing she chooses?

Same as the previous question, except she decides that the shirts she takes should all be long sleeve?

How many possibilities would there be if instead of deciding the shirts should ALL be long sleeve she decides that she should take AT LEAST ONE long sleeve shirt?

i got all of it except the last question...

10/25/2006 11:00:18 PM

i got 5 mins til its due...haha

10/25/2006 11:00:41 PM

Nice.

Alright, so take the combinations of how many shirts and pants she'll take with at least one long sleeve shirt, or with two long sleeve shirts, and three long sleve shirts. Multiply each of those by the combinations of how many shirt sleeves she'll be taking along (depending on how many long sleeves she is taking) and also multiply each of them by the combination of the pants she'll be taking.

So...

C(6,2) x C(6,1) x C(4,2) + C(6,1) x C(6,2) x C(4,2) + C(6,3) x C(4,2) should do it.

10/25/2006 11:05:32 PM

ooh, thanks a lot, its due now, but i appreciate you helpin out

i got a 57/59

[Edited on October 25, 2006 at 11:06 PM. Reason : /]

10/25/2006 11:06:32 PM

Coulda' been much worse, I guess. Glad to have been able to help out.

[Edited on October 25, 2006 at 11:10 PM. Reason : -]

10/25/2006 11:09:26 PM

anyone ever figure out the answer to number 10 on 4.5 B? It's the vegetable one with two helpings. I looked at her tip sheet and still can't figure it out, I have 66 for the first combo and I can't figure out the second part.

11/1/2006 7:50:13 PM

Maybe I'm oversimplifying it, but isn't the answer to the second part just 14 (or however many options Bill has)?

[Edited on November 1, 2006 at 8:28 PM. Reason : -]

11/1/2006 8:21:35 PM

okay, I am a dork because I wrote down the wrong number of menu options from webassign and was working with that number the entire time. Hmm, no wonder I didn't get the answer!

It just ended up being the menu items combination plus the number of menu items because it was one of those "or" situations.

Thank you for the info, though!

11/1/2006 10:05:47 PM

now for help on section 5.3....this chapter is throwing me curveballs....

Solve the following questions correct to 3 decimal places. A college has 700 students. A survey has determined that

68 read French
36 read German
39 read Spanish
14 read French and German
15 read French and Spanish
6 read German and Spanish
3 read all three languages

What is the probability a randomly selected student reads exactly 2 of these 3 languages?

What is the probability a randomly selected student reads at least 1 of the 3 languages?
What is the probability a randomly selected student reads exactly 1 of the 3 languages?
What is the probability a randomly selected student does not read any of the 3 languages?

----------------------------------------------------------------------------------------------

A normal checkerboard has 8 rows and 8 columns of alternating colored squares (see picture of antique checkerboard). Suppose you have a 12 by 12 checkerboard with alternating brown and white squares. If 4 pieces are randomly placed on the checkerboard,

what is the probability that they will all be in the column #1?

what is the probability that they will all be in the same column?

what is the probability that they will all be placed on white squares?

what is the probability that they will all be placed on squares of the same color?

------------------------------------------------------------------------------------------------

Most mp3 players (like the iPod, for example) have a "shuffle" feature which allows you to randomize the order of songs played from a playlist. If your playlist consists of songs A, B, C, D, and E (in that order), then if "shuffle" is activated the songs might play in this order: B, E, A, D, C. In effect, the way that a "shuffle" works is to create a random permutation of the playlist prior to playing any of the songs.

Other players, however, have a "random" rather than "shuffle" feature. With this mode of operation, random songs are played no matter what has been played before. So for example, if the above playlist were played in a "random" mode, then the first four songs played might be B, E, B, A. (Each time a new song is chosen for play, it is equally likely to be any of the songs. So in this example with 5 songs on the playlist, song B has a 20% chance of being played each time a new song is selected, even if it has already been played.)

Suppose that a playlist has 14 songs (all different) with 3 of the songs being by Laura Cantrell. If you listen to 4 songs in shuffle mode, what is the probability that you will hear at least one song by Laura Cantrell?

Answer (to at least 4 decimal places accuracy) =

If you listen to 4 songs from this playlist on an mp3 player that has a "random" mode, what is the probability that you will hear at least one song by Laura Cantrell?

Answer (to at least 4 decimal places accuracy) =

11/6/2006 10:38:06 PM

For the first one you have listed, You list it in a Venn diagram like we have done on other assignments and just take the fractions that they ask for and calculate them. Let me know if you still have trouble and I can explain further.

--------------------------------------

For the checkerboard problem.....the first part,

C(144,4) = 17178876
C(12,4) = 495

So the ones in column #1 would be 495/17178876, simplify that.

I didn't get the 2nd answer, so if you figure it out let me know.

For the third blank, you take C(144/2,4)=1028790, and then take 1028790/17178876, simplify.

For the fourth blank, you just double the 1028790 number, and take that over the same value and divide.

---------------------------------------------

I didn't get anything yet for the MP3 problem yet either, so let me know.

---------------------------------------------

Did you get any of these problems.....
---------------
In a group of 5 people, what is the probability that they all have different birthdays? [Assume that each is equally likely to be born on any of the 365 days of the year, i.e. no one is born on Feb. 29 in a leap year.]
---------------
Suppose now that only 2 of the 7 men have hats when they enter, and that the receptionist gives the 2 hats back to 2 randomly picked men (from the 7) when they leave. What is the probability that each of the 2 hats will be returned to the correct owner?
---------------
A board game is played with wooden pieces, where each piece shows a number on one side. There are 44 pieces altogether, and each piece shows a number from 1 to 44. (Each number from 1 to 44 is on exactly one piece.) To begin the game you randomly select 4 pieces.

How many possibilities are there for the numbers you begin with, i.e. in how many different ways can you choose your 4 pieces?

The odd numbered pieces are painted blue and the even numbered pieces are painted white. In how many ways could you select your pieces so that they are all white?

What is the probability that all the pieces you select are white? (Your answer must be accurate to 4 decimal places.)
------------------

THANKS.

11/7/2006 7:59:34 AM