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If you have found the mean value, μ, and standard deviation, s, of a sample data set, and the data is distributed like a normal (Gaussian) distribution, then the confidence interval can be expressed as the interval between μ-z*s/√n and  μ+z*s/√n, where n is the number of data points in your sample and z is a number that depends on the degree of confidence you require.  For example, if you want a 95% confidence interval then z is 1.96.

Divide both sides by 16;

$$x^2=\frac{1000}{16}$$

Take the square root of both sides:

$$\\
x=\pm\sqrt{\frac{1000}{16}}\\
x=\pm\frac{\sqrt{1000}}{\sqrt16}\\
x=\pm\frac{\sqrt{1000}}{4}\\
x=\pm\frac{10\sqrt{10}}{4}\\
x=\pm\frac{5\sqrt{10}}{2}$$

7x7=49

Seven multiplied by seven (also seen as 7 squared) equals a total of 49.

Otherwise shown as 7²=49

Use the keyword factorize here:

$${factorize}{\left({\mathtt{2\,310}}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{2}}\\
{\mathtt{3}}\\
{\mathtt{5}}\\
{\mathtt{7}}\\
{\mathtt{11}}\\
\end{array} \right\}$$

This gives you the prime factors.

Mea culpa.  Hmm! Seems we can't delete posts!

I'm not sure if this is what you meant, or not

$${\frac{{\mathtt{2.925}}}{{\mathtt{1}}}}{\mathtt{\,-\,}}{\mathtt{1.003\: \!75}}{\mathtt{\,-\,}}{\mathtt{24}} = -{\mathtt{22.078\: \!75}}$$

????

So we have

(7^-3)^-7 (7^-2 7⁴)⁵

Simplifying the expression inside the second parentheses by using the exponential property,  a^m * aⁿ = a^(m+n),  we have

(7^-3)^-7 (7²)⁵  

Remember, by another property of exponents,  (a^m)ⁿ = a ^(m*n)

So we have

(7)²¹ (7)¹⁰

And by the first property we have

7³¹  ........which is a REALLY "big" number !!

The image below should help.  B is a reflection of A in the y-axis.

$${\mathtt{87}}{\mathtt{\,\times\,}}{\mathtt{1\,234\,567\,890}} = {\mathtt{107\,407\,406\,430}}$$

To keep his DOGma from being hit by a KARma

The images don't show here (I guess you need to have an account at Physics Central before you can see them), so you'll need to specify the detail in words or find a way of inserting the images using the Tiny Pic method.

Good one Chris!

That Big...............

2+2+4,000,000,000,246-1,234.2*432.1 =

$${\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4\,000\,000\,000\,246}}{\mathtt{\,-\,}}{\mathtt{1\,234.2}}{\mathtt{\,\times\,}}{\mathtt{432.1}} = {\mathtt{3\,999\,999\,466\,952.18}}$$

Nice answer, Heureka.......maybe I should have been more explicit....No calculators allowed!!!

(Besides, there's a simple way to "solve" this without a calculator!!)

But....I suppose I'll have to "grandfather" you into the "club," because you did provide a proof!!

Look at this answer.

This one still has not been answered.

And ....as a "silly" problem...see if some of the users can figure THIS one out!!

Show that:    sinx / n  = 6

Hint: It is not that hard!

 Excellent answer Heureka!

$$\frac{1*3*5*7......*99}{2*4*6*8*......*100}\\\\
\frac{1*3*5*7......*99}{1}\times \frac{1}{2*4*6*8*....*100}\\\\
\frac{100!}{2*4*6*...*100}\times \frac{1}{2*4*6*8*....*100}\\\\
\frac{100!}{2^{50}(1*2*3*4*....*50)}\times \frac{1}{2^{50}(1*2*3*4*....*50)}\\\\
\frac{100!}{2^{50}(50!)}\times \frac{1}{2^{50}(50!)}\\\\
\frac{100!}{2^{100}\times 50!\times 50!}\\\\$$

$${\frac{{\mathtt{100}}{!}}{\left({{\mathtt{2}}}^{\left({\mathtt{100}}\right)}{\mathtt{\,\times\,}}{\mathtt{50}}{!}{\mathtt{\,\times\,}}{\mathtt{50}}{!}\right)}} = {\mathtt{0.079\: \!589\: \!237\: \!600\: \!582\: \!7}}$$

$$0.08<\frac{1}{10}$$

$$0.07958923760058268712601067554654 < 1/10$$

@@ End of Day Wrap - Monday 12/5/14   Sydney, Australia Time 20:45

Hi everyone,

CPhill and I rather monopolised the answering today - sorry.  There were a few physics ones left for Alan and admin answered one, thank you.  Never mind the week should start getting busy tomorrow and then there should be enough questions for everyone. Also, every day I see many new names.  The member section is not finished yet so I can't see you all but I expect forum member and post numbers is growing very quickly.

To all those who have joined recently, welcome to web2.0calc forum.  We hope you find the site helpful, interesting and fun.

There were some really nice thank you notes and thumbs up given today by RheyB, bioschip, tyeiyei and an anonymous person.  Thank you, everyone likes to be appreciated!

 One of our members Daniavm reported an error via a post.  I forwarded the post address to admin along with some of my own observations.  Daniavm’s problem should now be fixed. 

My observations were mainly to do with the Android phone app.  I had learned to do most things.  The member section could only be accessed by the most convoluted, intriguing process imaginable.  I was really proud of myself for solving the puzzle.  Anyway, there is no puzzle any more.  Andre Massow (site owner and developer) has ‘fixed’ it, what a pity.  So, from most people’s point of view the app works much better now.  Thank you Andre.

I have deleted my answer because I would like to see another person have fun with this question.

Plus I will reference in Puzzles so it will be available at later times.

Thanks for my tick tyeiyei.   

I need to raise 24.000. Presently I have 14.867. Please tell me how much is the percentage. Also, please help me to determine the percentage I need to raise. Also please tell me how much will get me to 70%.

This is my interpetation of your question

You need $24000, and at the moment you have $14867

$${\frac{{\mathtt{14\,867}}}{{\mathtt{24\,000}}}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\mathtt{61.945\: \!833\: \!333\: \!333\: \!333\: \!3}}$$       So at the moment you have approx 61.9%

I'm not sure what the second part of your question means but  70% of 24000 = $16800

$${\frac{{\mathtt{70}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{24\,000}} = {\mathtt{16\,800}}$$

So you need another $1933 in order to have 70% of $24000

$${\mathtt{16\,800}}{\mathtt{\,-\,}}{\mathtt{14\,867}} = {\mathtt{1\,933}}$$

Monday  12/5/14

Interesting Physics question.