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click the 2nd button then click atan

Thanks for that Rom,

I am having major problems with the system this morning.

Can someone tell me if it is my connection or if there are problems with the site please?

thanks.

If the 90 students surveyed were a completely random sample then yes, you would expect the answer to be about 80.  If the sample was not completely random then there would be no way of estimating an answer. 

while this does have a closed form it's a nasty one involving Elliptic Integral functions.

I am just going to add a little bit of trivia.   

i had never heard of PEDMAS untill I started foruming (7months back).

In Australia we call it BODMAS  

B for Brackets, O for of (meaning, power of) the rest is the same.

Parenthesis are just the curly brackets that arn't used much.

Exponents is not a word used much in Australian schools - we just call them powers.

I don't know why BEDMAS is not used - maybe it would bring on too many snickers in an adolescent classroom. (I am sure many would find it very funny - it is  bringing a smile to my face, I admit it) 

IT MIGHT BE EASY TO REMEMBER THOUGH!

Re: PEMDAS:What does it stand for?

It's an "acronym" that is supposed to remind us of  the "correct" "order of operations".......it stands for

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

However,  this "order" isn't exactly "correct." Many students think they're supposed to always perform "Multiplication" before "Division" and "Addition" before "Subtraction."

A more "correct" order of things would be

Parenthesis (or Brackets)

Exponents

(Multiplication    Division)   or  (Division  Multiplication)

(Addition    Subtraction)    or  (Subtraction  Addition)

The first two liness should be self-explanatory....work inside any parenthesis first and then apply any exponents!!

The last two lines require some explanation. Multiplication and Division are performed at what is known as the same level of "precedence." This means that we perform these operations as we encounter them from "left to right." The same thing applies to Addition and Subtraction.

Let me demonstrate what I'm talking about - first with Multiplication and Division, and then with Addition and Subtraction.

Let's supppose we have this :

36 / 9 X 4

Many students assume that they're supposed to perform the "multiplication" first and then the "division.'

So we get (wrongly).....  36 / 36   ..and then, "dividing," we have the "answer'....."1"

This is incorrect...we perform any multiplication or division as we encounter them from "left to right."

Correctly...we would have .....36 /9 = 4 and multiplying this by 4  we obtain 16. Note that that's a lot different than "1.'

Addition and Subtraction behave the same way. Suppose we have:

14 - 3 + 10

Uising the "incorrect" way, we would add the 3 + 10 to get 13, and then subtract this from 14 to get '1.'

The correct way is to evaluate (14 -3) first - (11) - and then add this to (10) getting 21. Note that that's a different answer from  "1.' 

Oddly enough, the beginning maths seem to make this more difficult by their reluctance to use any parentheses (or brackets). I don't find this to be true of the higher maths - In these, it's usually explicitly spelled out what's to be done. Take, for example, the addition and subtraction problem we just did.

The "lower maths" tend to write it just as we did, i.e.,  14 - 3 + 10. In the upper maths, it would be written as (14 -3) + 10. Note that there is no confusion about what to do here.......the operation inside the parenthesis is done first, and then the "10" is added to the result !!!!

Also.....note the confusing nature of something like the following:

3 x 6 / 10 + 8 - 4 / 2

....Ouch!!....that makes my head hurt just looking at it !!!

This is far easier to understand:

[(3 x 6) / (10) ] + 8 - (4/2).........don't you agree ???

Yep....I'm long-winded !!!  ....but, I hope this helps !!!

parenthesis, exponents, Multiplication, division, addision, subration

24/90=x/300 cross multiply and divide getting 24*300=7200/90=80 

80 students

how do i solve 2^x=7?

Let me give you an example.....then you should be able to do this one....

Suppose we had   4^x = 9

First....take the "log" of both sides......this gives us

log 4^x = log 9

By a property of logarithms, we can bring the "x" in front of the "log 4"......and we have

x log 4 = log 9......now, divide both sides by the "log 4" .....and we have

x = log 9 / log 4........ we can evaluate this on the calculator on this site, and we find that

x ≈ 1.58

Be careful !!!   Some students - mistakenly - take the log of (9/4).....this is NOT the same thing as

log 9 / log 4     

I think you have enough to get you on the right road, now.........

wow you've been on the computer for quite a while you just divide the answer you get from pi*radius and if you get something close to pi your pi*radius answer is right

Take logs (any base) and then use the rule

 $$\log (a^{n})=n\log(a)$$.

$${\mathtt{A}} = {{\mathtt{Pe}}}^{{\mathtt{rt}}} = {\mathtt{A}} = {{\left(\underset{{\tiny{\text{Error: Unknown Identifier}}}}{{\mathtt{Pe}}}\right)}}^{\underset{{\tiny{\text{Error: Unknown Identifier}}}}{{\mathtt{rt}}}}$$

You cannot.

Your answer doesn't nearly even answer my question, but I found it out, you just put an = from your keyboard into it.

$${\mathtt{42}}{\mathtt{\,\small\textbf+\,}}{\mathtt{24}} = {\mathtt{66}}$$

$${\mathtt{300}}{\mathtt{\,-\,}}{\mathtt{66}} = {\mathtt{234}}$$

From the information given, we can assume that the remaining 234 like to watch Soccer.

Unless we factor in the 3.3 increase in the amount of students, so from the first given amount

42+24=66 there's a remainder of 24, so 24 like soccer.  Factor in the 3.3 increase we get

139 Basketball

80 Baseball

And 80 Soccer

I rounded to the nearest tenth.

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{26}} = {\mathtt{78}}$$

$${\mathtt{78}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,245}} = {\mathtt{1\,323}}$$

Yes. If you mean an expression, then no.

THX!!!!!!! I got it right now!!!!!

the only question you need to ask is if they like one direction whoever says yes is the brother that tells the truth whoever says no is a liar if neither then lick their hair :)

honestly who dosent know the answer to that question STOP BEING LAZY

if you think math is stupid then whyd you join this site now whos stupid ha?

if you know how to use a computer then you should know how to multiply fractions by now smart one

a scientific notation is a notation that is scientific

Re: what is a combination

It represents the number of sets (or subsets) that we can form by selecting "r" things from "n" things. Here's an example...let's suppose we have a set of 3 objects labeled "A," "B,"  and "C."

Note that I can "choose" all three obejcts to form one set, namely, {A,B,C} The braces denote that we're talking about a set of things.

Or, I could choose any two of the objects. This would give us {A,B}, {A.C} and {B,C}.....note that the "order" of things in the set doesn't matter in combinations.

Or, I could choose  any one of the objects from the three...this gives us {A}, {B}, {C}.

There is one more way to form a "set" of these objects......this may seem strange....DON"T CHOOSE ANY OF THEM!!! This is known as the "empty" set and is just denoted as { } or as Ø.

Note the number of sets we've formed  = 2^(n things) = 2^(3) = 8.

Since it's cumbersome to try and figure how many sets can be formed by selecting some number of objects from a large number of them, we have a "formula" to do that.

It's denoted by C(n,r) or nCr.......both things are used......and it tells us how many sets can be formed by choosing 'r" things from "n" things. Note that n is always greater than or equal to r....That makes sense.....I couldn't choose 6 things from just 3 !!!!

The "formula" is.... (n!) / [(n-r)! (r!)].......where  "r" is the number of objects chosen from "n" things.

The " ! " is known as a "factorial" or just "factorial."

I'll spare you the "fancy" definition of this, but it's really just the product of "n" (or "r") and all the positive integers less than "n' or "r." For example, if "n'  (or "r') = 3, then 3! =  3 x 2 X 1 = 6.   (Note a "special case"..... 0! = 1)

So, looking at our example, let's suppose that we wanted  to know how many sets we could form by choosing 2 things from 3. By the "formula," we have  C(3,2) =

(3!) / [(3-2)! (2!)]     =  ( 3!) /  (1)! (2!)  =   (3X2x1) / [(1) * (2X1)]   =   6/2 =   3  .....which is exactly what we found!!

Combinations are used extensively in statistics and probability. Another thing that is sometimes encountered is the "permutation." It, unlike the combination, DOES take into account "order" within sets. Thus, in a permutation, {A,B} ≠ {B,A}    In general, the "permute" of something is usually greater than a "combination" of that something....but not always!!!

I hope this helps.....