cos x^2/ 1+sinx^2 + (1+sinx^2)/cosx^2=?


4cscx^2 -1=0 for 0<x<360
x=?

the probability of getting 6 tales out of 9 tosses?


arrangement of 6 people in a circular table?

cos x^2/ 1+sinx^2 + (1+sinx^2)/cosx^2=?


4cscx^2 -1=0 for 0<x<360
x=?

the probability of getting 6 tales out of 9 tosses?


arrangement of 6 people in a circular table?
---------------------------------------------------------------
Ok the second question is an identity: if you graph out csc +-1/2 you will see the graph and the set will come up empty.

Prob: 9*8*7*6*5*4/1*2*3*4*5*6 = 84
Then you take 1/2^6 = 1/64 1/2^3= 1/8

so.......... 84/1*1/64*1/8= 21/128

The table arrangment...... you have 6 spots ........ 6*5*4*3*2*1/6 = 120

I am pretty sure these are correct:)

---------------------------------------------------------------
Ok the second question is an identity: if you graph out csc +-1/2 you will see the graph and the set will come up empty.

Prob: 9*8*7*6*5*4/1*2*3*4*5*6 = 84
Then you take 1/2^6 = 1/64 1/2^3= 1/8

so.......... 84/1*1/64*1/8= 21/128

The table arrangment...... you have 6 spots ........ 6*5*4*3*2*1/6 = 120

I am pretty sure these are correct:)

can you explain how you calculate those numbers??

can you explain how you calculate those numbers??

Note, for a circular arrangement the formula is (N-1)! This is due to the fact that there is no end unit place around the table.

---------------------------------------------------------------
Prob: 9*8*7*6*5*4/1*2*3*4*5*6 = 84
Then you take 1/2^6 = 1/64 1/2^3= 1/8

so.......... 84/1*1/64*1/8= 21/128

The table arrangment...... you have 6 spots ........ 6*5*4*3*2*1/6 = 120

I am pretty sure these are correct:)

Can you explain these with more detail. It seems kind of arbitrary with your lack of explanation.

Can you explain these with more detail. It seems kind of arbitrary with your lack of explanation.

What dental #1 did was this:

The total possibilities of 9 coin tosses, I mean, all the possible different combinations is 2^9, right?

And, the possibilities of getting only 6 tails is, using combination,
(9!)/(9-6)!(6!). This is a combination, not permutation, because the order
doesn't matter as long as you get exactly 6 tails.

nCr = n! / (n-r)!(r!)

So, the probability is nCr / 2^9 = 21/128.

Notice, since this is a probability problem using a combination, the result would be the same as getting only 3 tails, 3 heads, 6 heads, etc because

nCr = nC(n-r)

Hope that helps. :luck:

cos x^2/ 1+sinx^2 + (1+sinx^2)/cosx^2=?




For the first trig question, I think answer is 2.

Here's how I would go about solving it:

We have two fractions, so add them up by making the common denominator, which would be (cosx)^2 (1+(sinx)^2)

Make appropriate changes to the numerators, and you'll see that the resulting fraction will have a common factors that cancel out, yielding 2 as your answer.

4cscx^2 -1=0 for 0<x<360
x=?


Since we're not allowed to use a graphing calculator,

the best way to go about solving this is by rearranging the equation:

Notice csc x = 1 / sinx

So,

4 (1 / (sinx)^2) = 1

which means (sinx)^2 has to equal 4 for this equation to make sense.
However, since the minimum value of sin is -1, and the max is +1, this equation can't be true.

thus, answer would be an empty set.

I don't think I agree with the coin problem solution.

I would simply say chances of get tails.... 1/2

(1/2)^6 = 1/64

I don't think I agree with the coin problem solution.

I would simply say chances of get tails.... 1/2

(1/2)^6 = 1/64

If you take an example w/ 3 coins, then it should be obvioius that the solution above is a correct one.

If you have 3-coin fllips, what is prob of getting exactly 2 heads?

It's not 1/2.

Let's just list all the possibilities we can have:

HHH
TTT
HTT
THT
TTH
THH
HTH
HHT

So, out of 8 you have 3 different events of exactly 2 tails.

thus, prob is 3/8 in this case.

If you do this problem, using the equation I gave you:

(3 nCr 2) / (2^3) = 3/8 Same thing. :luck:

I don't think I agree with the coin problem solution.

I would simply say chances of get tails.... 1/2

(1/2)^6 = 1/64


you're not just getting 6 tails, you're getting 6 tails AND 3 heads. very different. if any of those 3 turn up tails, you no longer have 6 tails....

(1/2)^9

If you take an example w/ 3 coins, then it should be obvioius that the solution above is a correct one.

If you have 3-coin fllips, what is prob of getting exactly 2 heads?

It's not 1/2.

Let's just list all the possibilities we can have:

HHH
TTT
HTT
THT
TTH
THH
HTH
HHT

So, out of 8 you have 3 different events of exactly 2 tails.

thus, prob is 3/8 in this case.

If you do this problem, using the equation I gave you:

(3 nCr 2) / (2^3) = 3/8 Same thing. :luck:

This analogy doesn't apply. He completely misunderstood what the prior guy was saying. the chances of getting tails on ONE coin toss is not comparable to getting it multiple times and with different combinations...

This analogy doesn't apply. He completely misunderstood what the prior guy was saying. the chances of getting tails on ONE coin toss is not comparable to getting it multiple times and with different combinations...

well yeah, i was trying to illustrate the same point with a simpler example :thumbup:

Thank you everyone for jumping in to explain what I did on the probability[# of desired/# possible]I am getting ready for the DAT on 8/16!!!!!!!!!! Yiks. I would have explained in great detail what I did, however I thought you were just asking how to solve it.

Thanks again everyone!