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5h
revised Solve for b and d
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5h
answered Solve for b and d
22h
comment What the difference between the smallest two numbers from these numbers?
Hint: consider the Chinese remainder theorem
23h
comment Find six triples of positive integers $(a, b, c)$ such that in $ \frac{9}{a} + \frac{a}{b} + \frac{b}{9} = c$.
I checked the last two with Alpha, which does exact rational math and agrees. Given the source of the problem, I suspect one was supposed to have a non-computer approach. There are no more with both $a,b \lt 10,000$
23h
revised Find six triples of positive integers $(a, b, c)$ such that in $ \frac{9}{a} + \frac{a}{b} + \frac{b}{9} = c$.
add info
23h
answered Find six triples of positive integers $(a, b, c)$ such that in $ \frac{9}{a} + \frac{a}{b} + \frac{b}{9} = c$.
1d
revised Solve for b and d
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1d
comment Arrangement of any number of objects from $n$ objects
The -1 comes from the fact that your original sum started at $r=1$. I believe that was the correct reading of the problem. You do not count the empty set as a subset for this purpose. If you do count it, the $-1$ goes away.
1d
comment Differentiate the Function: $g(y)=ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$
@SunnyBlack: if you are using that formula, you are getting confused with $g(x)$ Given you have $g(x)=\ln h(x)$, then $g'(x)=\frac {h'(x)}{h)x)}$
1d
answered Is there a difference between these statements about natural numbers?
1d
answered Arrangement of any number of objects from $n$ objects
1d
answered Resultant Temperature
2d
answered Length of a belt?
2d
comment I'm trying to solve for a stopping time given a distance. Think I have the answer.
There shouldn't be one. The distance covered is $16t^2$ and you are given that it is $870$ feet. Your units do not match, as $870t$ is feet-seconds.
2d
answered I'm trying to solve for a stopping time given a distance. Think I have the answer.
2d
comment Two different samples from different time periods.
It is quite fair to compute the mean and standard deviation of any finite distribution. The only place the normal distribution assumption comes in is when you want to translate $n\sigma$ to a probability.
2d
comment Two different samples from different time periods.
It is quite fair to compute the mean and standard deviation of any finite distribution. The only place the normal distribution assumption comes in is when you want to translate $n\sigma$ to a probability.
2d
comment Two different samples from different time periods.
The normal distribution is convenient, but is rarely correct. Almost all real distributions have longer tails than the normal distribution. From H. Poincare "Everyone is sure of this [that errors are normally distributed], Mr. Lippman told me one day, since the experimentalists believe that it is a mathematical theorem, and the mathematicians that it is an experimentally determined fact."
2d
comment Two different samples from different time periods.
The sample size in the first case gives you more confidence that the mean and standard deviation values are close to the actual sample mean and standard deviation. That doesn't address the question of whether the samples are the same. The use of mean/standard deviation implies you are assuming a normal distribution, which is almost certainly not correct.
2d
answered Two different samples from different time periods.