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by **Salvadoramerican** » Tue Mar 20, 2012 1:33 pm

$\int\frac{p}{q+rcos^2x}=\frac{1}{4}tan^{-1}(\frac{4}{5}tanx)+1$ . Find p+q+r

Ma Bell's telephne company assigns 10 digit phone numbers. What are the odds that the last digit of a randomly chosen phone number is 9 or 0?

by **Salvadoramerican** » Tue Mar 27, 2012 10:39 am

Can someone answer please

by **BlasianAsian** » Tue Mar 27, 2012 1:56 pm

Alright, I can do the second one.

So...10 digits.

Last one has to be a 9 or a 0.

It always helps to make a list in probability problems.

10 digits: __ __ __ __ __ __ __ __ __ __

Last on is 9 or 0

__ __ __ __ __ __ __ __ __ 9

__ __ __ __ __ __ __ __ __ 0

And the other digits can be anything:

So:

$10^{9} \cdot 2$ , in essence.

This is out of: $10^{10}$ Choices. Then you have:

$\dfrac{10^{9} \cdot 2}{10^{10} - (10^{9} \cdot 2)}$

Calculated: $\dfrac{2000000000}{8000000000} = 2.5 \cdot 10^{-1}$

Same answer? WTF?

by **88bobcat** » Tue Mar 27, 2012 2:24 pm

Salvadoramerican wrote: $\displaystyle \int\frac{p}{q+rcos^2x}=\frac{1}{4}tan^{-1}(\frac{4}{5}tanx)+1$ . Find p+q+r

Try differentiating using the Chain Rule to identify the integrand.

$\displaystyle \frac{d}{dx} \tan^{-1}(x) = \frac{1}{x^2+1}$

$\displaystyle \frac{d}{dx} \tan(x) = \sec^2(x)$

$\displaystyle \frac{d}{dx} \left[ \frac{1}{4}\tan^{-1} \left( \frac{4}{5} \tan(x) \right) + 1\right] = \frac{1}{4} \cdot \frac{1}{\displaystyle \frac{16}{25} \tan^2(x) + 1} \cdot \frac{4}{5} \sec^2(x)$

This is

$\displaystyle \frac{5}{16 \sin^2(x) + 25 \cos^2(x)} = \frac{5}{16+9\cos^2(x)}$

$p+q+r = 5+16+9 = \boxed{30}$

by **88bobcat** » Tue Mar 27, 2012 2:27 pm

BlasianAsian wrote:Alright, I can do the second one.

So...10 digits.

Last one has to be a 9 or a 0.

It always helps to make a list in probability problems.

10 digits: __ __ __ __ __ __ __ __ __ __

Last on is 9 or 0

__ __ __ __ __ __ __ __ __ 9

__ __ __ __ __ __ __ __ __ 0

And the other digits can be anything:

So:

$9! \cdot 2$ , in essence.

This is out of: 10! Choices. Then you have:

$\dfrac{9! \cdot 2}{10! - (9! \cdot 2)}$

Calculated: $\dfrac{725760}{2903040} = 2.5 \cdot 10^{-1}$

Am I right? =D

I don't think this is right. Because the numbers are random, they can be repeated. Also, the 10-digits are not being chosen from a finite set of 10 digits.

I think the probability is $\displaystyle \frac{1}{10} + \frac{1}{10} = \boxed{\frac{1}{5}}$

by **BlasianAsian** » Tue Mar 27, 2012 4:45 pm

Oh. Forgot. =D

What do you mean by finite set? ????

by **007math** » Tue Mar 27, 2012 5:08 pm

Both your answers are the same.....if you convert the 1/5 into odds (since he said that was the probability) you get 1/4.

I'm getting 1/4 odds myself whether you repeat digits or not so it doesn't seem to matter in the final answer. Although, has anyone seen a phone number that starts with "0"?

Heh, I actually remember doing this question a long time ago. I had done it both ways on the test...

by **BlasianAsian** » Tue Mar 27, 2012 7:05 pm

Wow. So I changed it to suit BobCat88...and got the same answer. No, I mean the first time, I got a giant bamboobalized fraction, which somehow equated into 1/4, and when I changed it to where you could use all ten digits for every digit space, I got a cleaner fraction, which still equaled 1/4. Hmm...lucky! =D

by **nsguy1350** » Tue Mar 27, 2012 7:13 pm

1/4 sounds right to me. What does the answer key say?
Also, I would bet that phone numbers such as 037-962-3458, with the zero first are excluded, but this changes nothing, answer-wise.

by **Salvadoramerican** » Wed Mar 28, 2012 6:43 am

thanks

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