Question

@wio @Nnesha @Miracrown @nin,com.shootyourself

Answer 1

lol

Answer 2

@nincompoop he misspelled your name - and I find it extremely hilarious. xD

Answer 3

LOL - ITS SO FUNNYYY

Answer 4

Now back to the math problem. :-D

Answer 5

There are 20 odd cards in the whole deck ...

Answer 6

LOL^^^^

Answer 7

What do you think

Answer 8

C?

Answer 9

Yerp

Answer 10

So C is really correct? ! :)

Answer 11

Thank you everyone!!! :D

Answer 12

^^^^ O_O

Answer 13

We have ace,3,5,7, and 9 in four suits
so 20 of the original cards are odd

\[\frac{ 20 }{ 52 } \space (\frac{ 19 }{ 51 } ) \space = \frac{ 95 }{ 663 }\]

Answer 14

I'm sorry, I don't understand how to do that @Miracrown :)

Answer 15

\[(\frac{ 20 }{ 52 } )\space =( \frac{ 19 }{ 51 } )\]

Answer 16

you will be in trouble if somebody from your institution finds that you're taking help from other ppl during an exam

Answer 17

Once we pick our first card, we have only 51 cards left, of which 19 are now odd
that is why it is (20/52)(19/51)

Answer 18

@ganeshie8 is right

Answer 19

I did most of the questions by myself.

Answer 20

Yes, true ... you confirmed them with us - so that has no troubles attached with it, I think

Answer 21

I asked 9 questions on OpenStudy, some of which I was double-checking.

Answer 22

Anyhow could you go back to explaining the card problem? :) @Miracrown

Answer 23

a simple google search by your teacher will do
you should be extremely careful about the questions you post here

exam questions are not allowed.. double checking is also not ok

Answer 24

^

Answer 25

Alright, I wasn't aware of that, I will be sure to avoid that in the future.

Answer 26

Gr8! Betta watch out, policemen @ganeshie8 is always on duty, like 24/7

Answer 27

What do you mean?

Answer 28

He is a policemen on OS - did you not know

Answer 29

He is a moderator, I'm pretty sure.

Answer 30

That too

Answer 31

i remember @dumbsearch2 was part of the system so he knows how the site works already.. it seems he just forgot or something hm..

Answer 32

Alzheimer mayb - Hehe.

Answer 33

hah very funny.

Answer 34

i have lot of it too haha! i remember him as ex intern or something

Answer 35

:O

Answer 36

What do you mean? I was a summer intern. I helped program some elements of the website, and I was a designer as well.

Answer 37

What do you mean ex?

Answer 38

There is no clause in the Code of Conduct that explicitly states that help is prohibited in mathematic assignments.

http://openstudy.com/code-of-conduct

In fact, they say that anyone can help with problems as long as the person asking them are interested in getting to the solution and understand it, which I clearly attempted to do with many of the questions. Move on folks, nothing to see here.

Answer 39

I think you're confusing "assignment" with "midterm exam"

exams are meant to take a snapshot of your present knowledge. they are not for grading the collective knowledge of internet community :P

Answer 40

http://gyazo.com/cc92e6abe11912a0cf1fe3c775aaf2ff

Answer 41

It wasn't a midterm exam, I merely said that because it would sound more dramatic. That was a stretch.

Midterm exams are proctored with my school, with someone sitting next to me.

I was doing a homework assignment, it's graded, but it's not like an exam. I just wanted to check some of my answers, and I thought you served as a tutor more than a person to cheat with. I can even ask the teachers online from my online school to help me understand questions there.