Wikipedia:Reference desk/Archives/Mathematics/2008 April 15

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April 15

"Math Problems to be an Easy Task"

I'm struggling with my homework and by the policy of the reference desk I've painstakingly tried (with no avail) to solve the foremost on my homework, though found necessary data willing as a factor to solve the question. Generally, it informs the letter "e" is present about 44% of all words printed in books, magazines, and newspapers. Then it asks if a history books contain 275 worded text of page 219 it asks which is the best estimate of how many words will contain the letter "e".

I was unaware of how to establish a correct answer to answer the problem.

May someone of higher mathematical intelligence assist? --Writer Cartoonist (talk) 23:01, 15 April 2008 (UTC)

For each word, there are two options. Either it has an 'e', or it doesn't. That means you're looking at a binomial distribution. --Tango (talk) 23:17, 15 April 2008 (UTC)

In general, the fewer numerical values you are given, the easier the problem will be, as there is a limit on the number of things you can do with them. I can inform you, without violating the no-homework rule, that "219" has no bearing on the answer.—81.132.235.198 (talk) 09:22, 16 April 2008 (UTC)

If I understand the question, you're told about what percentage of words tend to have a property, and then asked about what percentage of words probably have that property. The answer is the same - about 44% of the total. See Percentage. Black Carrot (talk) 07:02, 17 April 2008 (UTC)

Just multiply 275 words by 44%, or, in other words, find 275 × 0.44. The only trick here is that they tossed in an extraneous value, the page number, which you must ignore. StuRat (talk) 18:00, 17 April 2008 (UTC)

Unless it's asking how many words with e there are in the entire book, and telling you that there are 219 pages. In that case, you can assume that there are roughly 275 words on each page, hence 275*219=60225 words in total. 60225*0.44=26499, which is an estimate for the whole book. Daniel (‽) 12:47, 20 April 2008 (UTC)

Integrals

$\int _{0}^{\infty }{e}^{-x^{2}}\ \mathrm {d} x={\frac {\sqrt {\pi }}{2}}$ Therefore, $\int _{0}^{\infty }{e}^{-x^{2}}\ \mathrm {d} x=\int _{0}^{-\infty }e^{x^{2}}\ \mathrm {d} x$?

$\int _{-\infty }^{\infty }{e}^{-x^{2}}\ \mathrm {d} x={\sqrt {\pi }}$ So, $\int _{\infty }^{-\infty }e^{-x^{2}}\ \mathrm {d} x=-{\sqrt {\pi }}$ or possibly ${\sqrt {-\pi }}$? (By the way, I know an integral with an upper limit smaller than the lower limit is improprer.)

Futhermore, how does one go about evaluating integrals with infinite limits? I assume it can't be done with the method of $F(n_{u})-F(n_{l})$. Thanks in advance, Zrs 12 (talk) 23:23, 15 April 2008 (UTC)

I think you're getting confused with how minus signs and squares/square roots work. Your first "therefore" looks wrong to me - it looks like you've changed variables, x to -x, however (-x)²=x², not -x². --Tango (talk) 23:32, 15 April 2008 (UTC)

Oh, and $\mathrm {d} (-x)=-\mathrm {d} x$. --Tango (talk) 23:34, 15 April 2008 (UTC)

Oh, and to answer your last question. It can sort of by done using the F(b)-F(a) method, but you have to replace $F(\infty )$ with $\lim _{b\to \infty }F(b)$. See improper integral. --Tango (talk) 23:38, 15 April 2008 (UTC)

Yeah, I'm more confused about calculus. I'm not in the class yet and therefore just have to wonder and try to find out. Zrs 12 (talk) 23:47, 15 April 2008 (UTC)

Oh, and by the way, what does $\textstyle {:=}\,\!$ mean? Zrs 12 (talk) 23:52, 15 April 2008 (UTC)

I think it means "defined as". nneonneo ^talk 23:53, 15 April 2008 (UTC)

Yes, it does. (It's far from a universal notation, but I've never seen it used to mean anything else.) --Tango (talk) 23:56, 15 April 2008 (UTC)

I've seen some variation: a:=b can either mean 'a is defined to be b' or 'b is defined to be a'. I believe the former is more common, but both are used. Algebraist 16:20, 16 April 2008 (UTC)

The first is the only one I've seen, I expect that is the more common. --Tango (talk) 16:39, 16 April 2008 (UTC)