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# Moles!!!
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1. The formula for the production of ammonia in the haber process is as follows:
N2 + 3H2 ----> 2NH3
Molar mass
Nitrogen = 28
Hydrogen = 6
Ammonia = 34
How many moles are there in 2.8g of nitrogen?
number of moles = mass/molar mass =
2.8/28 = 0.1
How many moles of ammonia would form?
Ammonia has 2 moles so 0.1*2 = 0.2 moles would form
Work out the mass of ammonia formed.
0.2 * 34 = 6.8g
2. (Original post by Wolfram Alpha)
The formula for the production of ammonia in the haber process is as follows:
N2 + 3H2 ----> 2NH3
Molar mass
Nitrogen = 28
Hydrogen = 6
Ammonia = 34
How many moles are there in 2.8g of nitrogen?
number of moles = mass/molar mass =
2.8/28 = 0.1
How many moles of ammonia would form?
Ammonia has 2 moles so 0.1*2 = 0.2 moles would form
Work out the mass of ammonia formed.
0.2 * 34 = 6.8g
I'm pretty sure it's correct.
Posted from TSR Mobile
3. (Original post by Firenze26)
I'm pretty sure it's correct.
Posted from TSR Mobile
4. check part (c)
28g of nitrogen makes 34g of ammonia. 1g of nitrogen forms 34/28 = 1.21g of ammonia.
So, 2.8g of nitrogen forms 2.8 * 1.21 = 3.388g = 3.4g of ammonia.
1 mole of N2 gives 2 moles of NH3
1 mole of anything is its RAM (Mr) in g
so 1 mole of N2 is 28g and 1 mole of NH3 is 17g
so 2.8g of N2 gives 3.4 g of NH3
6. (Original post by GDN)
1 mole of N2 gives 2 moles of NH3
1 mole of anything is its RAM (Mr) in g
so 1 mole of N2 is 28g and 1 mole of NH3 is 17g
so 2.8g of N2 gives 3.4 g of NH3
Oops, he is correct. I guess I couldn't do simple maths lol ammonia is 34g .
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If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
[Reveal] Spoiler: OA
Last edited by Bunuel on 12 Oct 2012, 10:22, edited 1 time in total.
Edited the question.
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A?
Let x be the tenth digit and y be the units digit.
Then, the original number is 10x+y and the reversed number is 10y+x.
The difference between the two is 27. (reversed - original)
Thus, (10y+x)-(10x+y)=27
solve the above eq.
10y+x-10x-y=27
9y-9x=27
9(y-x)=27
y-x=3
Thus the two digits differ by 3.
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15 Apr 2007, 15:16
salr15 wrote:
ricokevin wrote:
A?
Let x be the tenth digit and y be the units digit.
Then, the original number is 10x+y and the reversed number is 10y+x.
The difference between the two is 27. (reversed - original)
Thus, (10y+x)-(10x+y)=27
solve the above eq.
10y+x-10x-y=27
9y-9x=27
9(y-x)=27
y-x=3
Thus the two digits differ by 3.
:)
how did you get 10x + y?
34 = (3 x 10) + 4
If xy = 34 then this explains it.
HTH
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15 Apr 2007, 16:25
techjanson wrote:
salr15 wrote:
ricokevin wrote:
A?
Let x be the tenth digit and y be the units digit.
Then, the original number is 10x+y and the reversed number is 10y+x.
The difference between the two is 27. (reversed - original)
Thus, (10y+x)-(10x+y)=27
solve the above eq.
10y+x-10x-y=27
9y-9x=27
9(y-x)=27
y-x=3
Thus the two digits differ by 3.
how did you get 10x + y?
34 = (3 x 10) + 4
If xy = 34 then this explains it.
HTH
Great thanks!
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This is a tricky number properties question. Note that you can use two variables a and b to represent each of the digits.
In terms of expressing them in values - the total value would be (10a + b).
For example, for a number 37, a = 3 and b=7..then the expression 10a+b = 30 + 7 = 37.
Please refer to the video explanation here:
http://www.gmatpill.com/gmat-practice-t ... stion/2397
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Re: If a two-digit positive integer has its digits reversed, the [#permalink]
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Hi All,
The "math" behind this question is such that there are several different numbers that "fit" the given "restrictions"; in that way, we can use a bit of "brute force" to quickly come up with the answer.
We're told that a two-digit positive integer will differ from its "reverse" by 27.
I'm going to 'play' with this math a bit.....
IF we use....
12 and 21 then the difference is 21-12 = 9 This is NOT a match
13 and 31 then the difference is 31-13 = 18 This is NOT a match. Notice how the difference increases by 9 though!!!!!
14 and 41 then the difference is 41-14 = 27 This IS a match.
The question asks for the difference in the two digits involved. Using these values, the answer is 4-1 = 3
Final Answer:
[Reveal] Spoiler:
A
I mentioned at the beginning that there were several numbers that "fit" what this question was asking for. They are 14 and 41, 25 and 52, 36 and 63, 47 and 74, 58 and 85, 69 and 96.
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salr15 wrote:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Let’s first label the original two-digit integer as N. We can then say that N = 10A + B, where A is the tens digit and B is the units digit of N.
If this is hard to see let’s try it with a sample number, say 24. We can say the following:
24 = (2 x 10) + 4
24 = 20 + 4
24 = 24
Getting back to the problem, we are given that if the integer N has its digits reversed the resulting integer differs from the original by 27. First let’s express the reversed number in a similar fashion to the way in which we expressed the original integer.
10B + A = reversed integer
Since we know the resulting integer differs from the original by 27 we can say:
10B + A – (10A + B) = 27
10B + A – 10A – B = 27
9B – 9A = 27
B – A = 3
Since B is the tens digit and A is the units digit, we can say that the digits differ by 3.
The answer is A.
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Re: If a two-digit positive integer has its digits reversed, the [#permalink]
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18 May 2017, 08:09
I find this problem simple and a little confusing.
So we have here
ab – ba = 27
Let’s say ab = N,
We can formulate it as N = 10a + b, a – is tenths digit, and b – unit digit. For example:
N = 4*10 +5 = 45
Now reversed number will be as follows:
10b+a
we know: ab – ba = 27 =>
10a + b – (10b +a) = 27
10a + b – 10b – a = 27
9a – 9b = 27
a – b = 3. => two digits differ by 3.
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Re: If a two-digit positive integer has its digits reversed, the [#permalink]
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18 May 2017, 09:03
salr15 wrote:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Let the original number be 10x + y
After reversal of the digits the number is 10y + x
Difference : ( 10y + x ) - ( 10x + y ) = 27
Or, 9y - 9x = 27
Or, y - x = 3
Thus, the answer must be (A) 3
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Re: If a two-digit positive integer has its digits reversed, the [#permalink]
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04 Sep 2017, 04:02
ScottTargetTestPrep wrote:
salr15 wrote:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Let’s first label the original two-digit integer as N. We can then say that N = 10A + B, where A is the tens digit and B is the units digit of N.
If this is hard to see let’s try it with a sample number, say 24. We can say the following:
24 = (2 x 10) + 4
24 = 20 + 4
24 = 24
Getting back to the problem, we are given that if the integer N has its digits reversed the resulting integer differs from the original by 27. First let’s express the reversed number in a similar fashion to the way in which we expressed the original integer.
10B + A = reversed integer
Since we know the resulting integer differs from the original by 27 we can say:
10B + A – (10A + B) = 27
10B + A – 10A – B = 27
9B – 9A = 27
B – A = 3
Since B is the tens digit and A is the units digit, we can say that the digits differ by 3.
The answer is A.
Can we also write it as:
10A+B-10B-A= 27
9A-9B= 27
A-B= 3
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Re: If a two-digit positive integer has its digits reversed, the [#permalink] 04 Sep 2017, 04:02
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## Handicapping a Race to 7
### If we give Karl a handicap of 2 games so that he only needs to win 5 games to win the series, whereas Richard needs to win 7, what is the single-game probability $p$ that gives each player a 50% chance of winning the series?
I've tagged this post with "Billiards" since the question came from the director of my pool league (who I hear is an avid GTM reader). Actually, what he really wants to know is the best way to set up the scoring, tie-breakers, and handicaps in one of the leagues, but in order to answer these questions, I wanted to start by looking at just one series and then extend that to the more general questions. This analysis really has nothing to do with pool though, so it would work for any other game as well.
#### Binomial Coefficients
To start out, we'll need the binomial coefficients $\binom{n}{k}$, which are read as "$n$ choose $k$." $\binom{n}{k}$ is the number of ways to choose $k$ items from a set of $n$ distinct items, for example the number of $k$-person boards of directors that can be chosen from $n$ candidates. Note that choosing the $k$ candidates who are included in the board is the same as choosing the $n-k$ who are not included, which means that $\binom{n}{k} = \binom{n}{n-k}$. The formula for the binomial coefficients is $$\dbinom{n}{k} = \dfrac{n!}{k! (n-k)!}$$ from which the above-mentioned symmetry is obvious. Note that even though this is a fraction, it always comes out to be an integer.
The binomial coefficients got their name from the fact that they are the coefficients in the expansion of binomials: $$(x+y)^n = \sum_{k=0}^{n}{\dbinom{n}{k} x^k y^{n-k}}$$ This is because in the product $(x+y)(x+y)...(x+y)$ (with $n$ factors of $(x+y)$), each factor of $(x+y)$ contributes either an $x$ or a $y$ to a factor in the sum. If you have $k$ $x$'s in a term, the other $n-k$ factors of $(x+y)$ must have contributed a $y$. There are $\binom{n}{k}$ ways to get $k$ $x$'s and $n-k$ $y$'s, hence the formula. If anyone wants more detail on that, just ask in the comments, and I'll give a more detailed explanation.
Most identities about binomial coefficients can be proved either by using the formula with the factorials, or via a combinatorial argument. For example, for integers $n$, $m$, and $k$ with $0 \leq k \leq m \leq n$, we have the following identity, known as the subset of a subset identity: $$\dbinom{n}{m} \dbinom{m}{k} = \dbinom{n}{k} \dbinom{n-k}{m-k}$$Algebraic proof: \begin{align} \dbinom{n}{m} \dbinom{m}{k} &= \dfrac{n!}{m! (n-m)!} \cdot \dfrac{m!}{k! (m-k)!} \\[3mm] &= \dfrac{n!}{k!(n-m)!(m-k)!} \\[3mm] &= \dfrac{n!}{k! (n-k)!} \cdot \dfrac{(n-k)!}{(n-m)! (m-k)!} \\[3mm] &= \dbinom{n}{k} \dbinom{n-k}{m-k} \tag*{\square} \end{align}Combinatorial proof:
The left side of the identity is the number of ways to choose a board of directors with $m$ members from $n$ candidates, and then choose $k$ executive members from the $m$. The right side counts the number of ways to choose $k$ executive members from the $n$ candidates and then choose the $m-k$ non-executive board members from the $n-k$ remaining candidates. These count the same thing, so the two sides must be equal. $\tag*{$\square$}$
#### Winning a series to 7
In order to win a series to 7, without needing to win by 2, Karl needs to win 7 games, with Richard winning anywhere from 0 to 6 games in the series. If Karl wins 7 games, and Richard wins 3 games (for example), there will be a total of 10 games in the series. The 3 games that Richard does win can come anywhere in the 10 games, except for the 10th game- if it did, then Karl would have already won 7 and the series would not have made it to 10 in the first place. So we can choose from the first $10-1=9$ games where Richard's 3 wins go.
The probability that Karl wins a given game is $p$, which means the probability that Richard beats Karl is $1-p$. Combining all this, we can see that the probability that Karl beats Richard in a race to 7, with Richard winning 3 games, is $$\binom{10-1}{3} p^7 (1-p)^3$$Since Karl can win the series with Richard winning anywhere from 0 to 6 games, the total probability that Karl wins the series is the sum over the possible outcomes, with the summation index $k$ being the number of wins Richard gets in the series: \begin{align} {\Bbb P}(\text{Karl wins the series}) &= \sum_{k=0}^{7-1}{\binom{7+k-1}{k} p^7 (1-p)^k} \\[3mm] &= \sum_{k=0}^{6}{\binom{6+k}{k} p^7 (1-p)^k} \end{align} Here's a graph of the probability that Karl wins the series, for different values of $p$:
Not surprisingly, if there's a 50% chance that either player wins an individual game, then there's also a 50% chance that either player wins the series.
Now, let's say we give Karl a handicap of 2 games so that to win the series, Karl needs to win 5 games and Richard needs to win 7. More generally, if we call the handicap $H$, where $0 \leq H \leq 6$, then by the same reasoning as we used above, we get the modified formula: $${\Bbb P}(\text{Karl wins the series}) = \sum_{k=0}^{6}{\binom{6-H+k}{k} p^{7-H} (1-p)^k}$$ Now Karl only needs to win $7-H$ games, and so the total number of games in the series for a given value of $k$ wins for Richard, is $7-H+k$, with the $k$ losses once again being placed anywhere but the last game.
Here are the graphs of Karl's probabilities of winning the series given different values of $p$ and $H$ (you can click to expand the picture):
Now, I'd love to be able say we're done here, but the fact is that for some real Karl and Richard, we have no idea what the value of $p$ is unless we are lucky enough to have a history of, say, 100 games between these two players. And even then, they could have improved over time or gotten rusty or whatever so that games they played a few months ago aren't so telling now as to the value of $p$.
We do know that every player in the league is assigned a ranking (which directly determines the handicap against an opponent) which is certainly partly subjective and determined based on observation by a few very experienced players who run, and possibly play in, the league. Instead of trying to guess $p$ and then assigning the rankings, which would be useless in the absence of a large history of games between each set of two players, we can use the handicaps to back out the value of $p$ that makes the match 50-50. For example, if Karl and Richard's rankings are such that Karl gets a handicap of 3, we can see from the graph above that the match will be 50-50 if Karl's probability $p$ of winning an individual game is about 35.5%.
Using Excel's Goal Seek functionality, I've backed out the values of $p$ that make a 7-game series 50-50 for different handicaps:
To test whether a player's handicap is appropriate, one could take all that player's games against opponents of different ranks and see what percentage of individual games he wins and how far off those percentages are from the table above (perhaps using a chi-square test for goodness of fit). If there are not enough games to do this analysis for individual players, then one could start by looking at the percentages for all games and then looking into the ranks furthest away from the table values and seeing if the stats of any particular player(s) are driving the difference. That's a bit of a manual exercise, but it's a start...
"Backing out" $p$ basically means finding the inverse of the function $f(p) = \sum_{k=0}^{6}{\binom{6-H+k}{k} p^{7-H} (1-p)^k}$. We know the function has an inverse because if you look at the graphs, they all pass the horizontal line test. To be more rigorous, they are polynomials in $p$ and thus continuous, and $f(0)=0$ and $f(1)=1$, so the intermediate value theorem tells us that $f$ is surjective. Furthermore, $f$ is increasing on the interval $p \in [0,1]$, so it's also one-to-one, and thus has an inverse.
Now, while Excel Goal Seek will certainly work for this, it would be kind of nice to know the inverse function, so I worked for a few hours today trying to figure out how to invert $f$, but couldn't quite figure it out. Maybe one of my more nerdy readers wants to take a crack at it? Otherwise, maybe I'll go post the question on stack exchange...
[Update 7/29/2015: there's been some confusion on the question I'm asking, so just to clarify, for the purposes of finding the inverse of $f(p)$, assume that the $H$ in the formula above is a constant. So technically, there is a different function $f$ for each value of $H$, which I guess you could call $f_{H}(p)$ or something.]
What I was trying (and maybe this isn't the best way to go about it) was to find a not-too-awful formula for the coefficient of $p^n$ in the sum above and then try to use the Lagrange inversion formula, but it gets a bit messy with all the binomial coefficients. I tried to expand the $(1-p)^k$, turning the sum into a double sum, then switch the order of summation (making sure to adjust the summation limits- the Iverson bracket is helpful at this step), and finally simplify somehow using identities of the binomial coefficients such as the subset of a subset identity above, but said simplification proved elusive, so I didn't even bother with the inversion formula.
Anyone have any thoughts on that or maybe a different way to find the inverse of $f(p)$? Let me know in the comments or email me, and I can provide more details of the computation I tried.
Thanks for reading, and I will try to do a follow-up on this post soon. As always, feel free to ask questions in the comments section.
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# Is $\frac{\pi}{4}L_0(z) = \sum\limits_{n=1}^{+\infty} (-1)^{n+1} \frac{I_{2n-1}(z)}{2n-1}$ between Bessel and Struve known?
Based of the detailed attempt to solve the integral $\int e^{\sin(x)} dx$ I stumbled upon a connection between modified Struve and modified Bessel function of the first kind. But, I cannot find a confirmation in the literature, and the connection looks too nice to be easily missed.
The integral can be shifted to $e^{\cos(u)}$
$\displaystyle \int e^{\sin(x)} dx, u=x-\frac{\pi}{2}$
This is making
$\displaystyle \int e^{\sin(x)} dx = \int e^{\sin(u+\frac{\pi}{2})} du = \int e^{\cos(u)} du$
Use Fourier transform
$\displaystyle e^{\cos(u)}=\frac{1}{2}a_0 +\sum_{n=1}^{+\infty} a_n \cos \left ( nu \right )+\sum_{n=1}^{+\infty} b_n \sin \left ( nu \right )$
$\displaystyle a_n=\frac{1}{\pi}\int_{-\pi}^{\pi} e^{\cos(t)}\cos(nt) dt$
$\displaystyle b_n=\frac{1}{\pi}\int_{-\pi}^{\pi} e^{\cos(t)}\sin(nt) dt$
All $b_n=0$
$a_n$ is a multiple of modified Bessel function of the first kind
$\displaystyle I_n(z)=\frac{1}{\pi} \int_{0}^{\pi} e^{z\cos(t)} \cos(nt) dt$
So $a_n = 2I_n(1)$
making
$\displaystyle e^{\cos(u)}= I_0(1) + 2\sum_{n=1}^{+\infty} I_n(1) \cos \left ( nu \right )$
or
$\displaystyle \int e^{\cos(u)} du=I_0(1)u + 2\sum_{n=1}^{+\infty} \frac{I_n(1)}{n} \sin \left ( nu \right )$
Knowing that boundaries have to be shifted $(\int_{\frac{\pi}{2}}^{x+\frac{\pi}{2}} \to \int_{0}^{x})$ we need first
$\displaystyle \int_{0}^{\frac{\pi}{2}} e^{\sin(x)} dx=\frac{\pi}{2}(L_0(1)+I_0(1))$
where $L_n(z)$ is modified Struve function.
Finally:
$\displaystyle \int_{0}^{x} e^{\sin(t)} dt=I_0(1)x + \frac{\pi}{2}L_0(1) + 2\sum_{n=1}^{+\infty} \frac{I_n(1)}{n} \sin \left ( nx - \frac{n\pi}{2} \right )$
So generally we can write:
$\displaystyle \int e^{\sin(x)} dx=I_0(1)x + 2\sum_{n=1}^{+\infty} \frac{I_n(1)}{n} \sin \left ( nx - \frac{n\pi}{2} \right )+c$
Plainly extending beyond just $z=1$ for $x=0$ the definite integral gives
$\displaystyle \frac{\pi}{4}L_0(z) = \sum_{n=0}^{+\infty} (-1)^{n} \frac{I_{2n+1}(z)}{2n+1}$
But, there is no mentioning this anywhere I looked. And it is too nice to be missed. (Notice that the extension is related to the integral $\int e^{z\cos(x)} dx$)
This relation is a special case of a more general one: $$L_\nu(z)=\frac{4}{\sqrt{\pi}\,\Gamma\left(\nu+\frac{1}{2}\right)}\sum_{n=0}^\infty\frac{(-1)^n\,(2n+\nu+1)\,\Gamma(n+\nu+1)}{n!\,(2n+1)(2n+2\nu+1)}\,I_{2n+\nu+1}(z),$$ which can be found in https://arxiv.org/abs/1301.5432 (Integral representations and summations of modified Struve function, byÁrpád Baricz, Tibor K. Pogány, formula (2.1) on page 4).
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## Saturday, September 4, 2010
### The problem
Find a number using the digits 1 through 9 once each, such that
• the number is divisible by 9
• if the right most digit is removed, the number is divisible by 8
• if the next right most digit is removed, the number is divisible by 7
• ... until there is only one digit left, which is divisible by 1
### Brute force
The first thing to do, is see if brute force will work. Often it will, and even if it doesn't solve the problem, thinking this approach through can help find a better solution.
We could loop throuch each possible nine digit number, and see if it meets all the rules. We might do it like this:
```for i in 123456789 to 987654321 do
# check if i contains each digit once
# check if i is divisible by 9
# check if floor (i/10) is divisible by 8
# ...
```
That might not be too bad, but it will check 864,197,532 numbers, and the logic in the loop isn't that clean. Instead of checking if each digit is used once, we could generate all permutations of 123456789 instead. That way, every number meets the criteria of using each digit once. Also, we would only check 9!, or 362,880 values, checking far fewer numbers than the first idea.
### Not so brute force
If we look closer at the problem we can see that we have a number of the form abcdefghi. "a" must be divisible by 1. "ab" must be divisible by 2. "abc" must be divisible by 3, and so on. Using that information, we could first check "ab", before appending any more digits to it, and wipe out a lot of possible numbers. For example, we know that "21" can never be the start of our answer, since 21 is not divisible by 2. Of the 72 possible two digit starting values, about half will fail this test. That will take us down to 180,000 values or so! Then, we can eliminate about 2/3's of the rest by checking all of the first three digits, and go down again to about 60,000 possible values.
We can get this solution to work with a recursive function. The inputs will be a list of digits left to use, and the current number we are considering (the first n digits). We will write the function so that the current number is always valid, so it contains no duplicate digits, and it follows the divisibility rules. Also, the digits to use list contains exactly the digits we need to make a valid solution, and no more.
```def fast(nums=range(1,10),acc=0,d=1):
if not nums:
# we have reached the end... print the answer!
print "result: ", acc
else:
for n in nums:
# update the number with a new digit at the end.
newnum = n + 10*acc
if newnum % d == 0:
# this is a valid partial solution.
# Call recusively with a new list of nums.
tmp = [x for x in nums if x is not n]
fast(tmp,n+10*acc,d+1)
```
I abused the default arguments of the function so it can be called without any arguments. The new function runs in less than two milliseconds, which is about 1/300,000th of the brute force timing.
If you only want the answer, the brute force solution is good enough. But, you can work at it to find a cleaner answer, and then learn a technique that can help in other problems.
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# Histogram
A histogram meaning can be stated as a graphical representation that condenses a data series into an easy interpretation of numerical data by grouping them into logical ranges of different heights which are also known as bins. Basically, it summarizes discrete or continuous data. We can also call it a frequency distribution graph as it is like a plot that lets you discover the underlying frequency distribution.
Histogram definition can be put forward as a tool that visualizes the distribution of data over a continuous interval or a certain time period. It helps us to get an estimate of where the values are concentrated, what are the extremes if there is any gap or unusual values. To some extent, a histogram also gives us a brief view of a probability distribution. A histogram is quite similar to a vertical bar graph but the difference that lies between them is that there is no gap between the bars in the histogram, unlike a bar graph.
Characteristics Of A Histogram
1. A histogram is used to display continuous data in a categorical form.
2. In a histogram, there are no gaps between the bars, unlike a bar graph.
3. The width of the bins is equal.
### Parts Of A Histogram
A histogram can be divided into several parts. These parts make up a complete histogram.
1. The Title: The most important part is the title of a histogram. The title tells us what the histogram is about. In other words, it describes the information provided in the histogram.
2. The X-axis: It is an interval that represents the scale of values which the measurements fall under.
3. The Y-axis: It represents the frequency of values occurring within the intervals set by the X-axis.
4. The Bars: The last is the bars whose height represents the number of times that the values occurred within the interval while the width of the bar shows the interval that is covered.
### How Histograms Work
To know how to make a histogram, you need to know that histograms are useful in statistics to illustrate how many of a certain type of variable occurs within a specific range. Let’s take an example, a histogram is used to focus on the demography of a country to exhibit how many people are between the ages of 0 to 10, 11 to 20, 21 to 30, and 31 to 40 and so on… the X-axis will represent the demography while the Y-axis will represent the age of the people.
### It Is The Area, Not The Height Of The Bars
In a histogram, it is the area and not the height of the bar that indicates the frequency of occurrences for each bin. The height of the bar does not indicate how many occurrences of scores are there in each individual bin. It is always the product of height and width of the bin that indicates the frequency of occurrences within that bin.
### How To Create A Frequency Histogram Graph
To construct a histogram graph from a continuous variable there are few steps that we need to follow. They are given below;
Step 1) Firstly, we need to split the data into class intervals which are also known as bins and frequencies.
Step 2) In this step, we have to draw a histogram graph with X-axis and Y-axis. Then write down the class intervals on the X-axis and the frequencies on the Y-axis.
Step 3) Draw vertical rectangles using the X-axis and the Y-axis.
## Difference Between Bar Graph And Histogram
HISTOGRAM BAR GRAPH Indicates Distribution of non-discrete variables Comparison of discrete variables Represents Quantitative data Categorical data Spaces No spaces between the bars Spaces are there between the bars Elements Elements are grouped together Elements are taken individually Reordering of bars No Yes Width of the bar Doesn’t need to be same Has to be same
A histogram can be represented in different ways. Some of them are given below with the histogram example as well.
## Types Of Histograms
A normal distribution: In a normal distribution, points on both sides of the average are alike. A bimodal distribution: In a bimodal distribution, the data are separately analyzed as a normal distribution. Therefore they are represented as two different peaks. A right-skewed distribution: A right-skewed distribution, also known as positively skewed distribution, is where a large number of data values occur on the left side whereas a fewer number of data values occur on the right side. A right-skewed distribution occurs when the data on the left-hand side of the histogram has a low range boundary, for example, 0. A left-skewed distribution: A left-skewed distribution which is also known as negatively skewed distribution. In a left-skewed distribution, a large number of data values appear on the right side whereas a fewer number of data values occur on the left side. A right-skewed distribution occurs when the data has a low range boundary on the right-hand side of the histogram, for example, 100. A random distribution: There is no pattern in a random distribution histogram and thus has several peaks. The reason behind this could be that the data properties were combined.
The table above will not only teach you the different types of histograms but also how to draw a histogram.
1. What are the applications of a histogram in real life?
A histogram can be used in numerous places and situations in real life. The most frequently used fields are:
1. In Stock exchange: A histogram is used to identify the trade at different places or different groups of investors.
2. In Medical and Clinical Research: A histogram in Medical and Clinical Research is useful to identify the presence or absence of a condition among different categories of people.
3. In photography: In photography, a histogram is used for Image processing and digitization.
4. In Six Sigma: A histogram is used to study the defect pattern across different categories of samples
Therefore, a histogram as a tool of simplicity and easy work has diverse uses.
2. Who discovered histogram and why was it named ‘histogram?’
Karl Pearson was the first person to discover histogram in 1891. The term ‘histogram’ was coined using two words: “historical diagram” which is obviously the function of a histogram, that is to display past data.
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# types of linear equation
## types of linear equation
A linear equation is an algebraic equation in which the highest exponent of the variable is one. This form is popular as the standard form of a linear equation. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line. In the above example (1) and (2) are said to be linear equations whereas example (3) and (4) are said to be non-linear equations. There are three types of linear equations: Conditional equation Identity Contradiction Conditional equation It’s an equation that has exactly one solution. That is, all of the unknown variables in a linear equation are raised to the power of one. A linear equation in two variables, such as has an infinite number of solutions. To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. A linear equation is any equation that can be written in the form $ax + b = 0$ where $$a$$ and $$b$$ are real numbers and $$x$$ is a variable. Where a and b are the real numbers and x is a variable. Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. The point where the two lines intersect is the only solution. A cost and revenue function. Linear regression modeling and formula have a range of applications in the business. Here is an example of an identity equation. The cost function is c(x)=mx+b. Consistent: If a system of linear equations has at least one solution, then it is called consistent. A system of linear equations generally consists of two separate equations representing two separate lines on the graph. Graphs of 2 linear … There are three types of polynomial equations. A linear equation is an equation that can be written in the form given as: ax + b = 0. Replace f\left( x \right) by y.; Switch the roles of x and y, in other words, interchange x and y in the equation. They show a relationship between two variables with a linear algorithm and equation. Linear Resistors; Non Linear Resistors; Linear Resistors: Those resistors, which values change with the applied voltage and temperature, are called linear resistors. Show Step-by-step Solutions. There are three types of systems of linear equations in two variables, and three types of solutions. Short Answer - Mutually exclusive options: No solution, One unique solution, or an infinite number of solutions . In the general form, the slope is -A/B if B 0 and infinite if B = 0.In the slope-intercept form, the parameter b is the y-intercept. For a given system of linear equations, there are only three possibilities for the solution set of the system: No solution (inconsistent), a unique solution, or infinitely many solutions. The Linear Combination Method , aka The Addition Method , aka The Elimination Method. Guest13065758 There are three ways in solving system of linear equations: graphing, substitution and elimination. However, there are many cases where solving a … These types of equations are called dependent or coincident since they are one and the same equation and they have an infinite number of solutions, since one “sits on top of” the other. The only way that a linear function, f(x) = mx + b, could have a finite limit as x approaches infinity is if the slope is zero. Add (or subtract) a multiple of one equation to (or from) the other equation, in such a way that either the x -terms or the y -terms cancel out.Then solve for x (or y , whichever's left) and … A linear equation represents a straight line on the graph, joining two points, and all points on that line are solutions to the equation. TYPES OF LINEAR SYSTEMS. Linear equations occur frequently in all mathematics and their applications in physics and engineering, partly because non-linear systems are often well approximated by linear equations. They are linear and logistic regression. Linear relationships are fairly common in daily life. Graphs If f(x) is linear, the graph of y = f(x) is a straight line. In other words, a resistor, which current value is directly proportional to the applied voltage is known as linear resistors. But the fact is there are more than 10 types of regression algorithms designed for various types of analysis. Linear equation has one, two or three variables but not every linear system with 03 equations. An independent system has exactly one solution pair $$(x,y)$$. Up … Three main types of solutions of linear equations with examples. The word poly means more than one and nomial means number of terms. For example, + − = − + = − − + − = is a system of three equations in the three variables x, y, z.A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. First going to assume that this is a linear equation which the unknown variables in a graphical format as! Professionals know only 2-3 types of regression algorithms designed for various types of that..., then It is called consistent as: ax + b variables with a linear equation … linear regression are! Y = f ( x ) must be a constant function, f ( x ) is linear the... Equation is the second-order differential equation lines on the graph of y = f ( x is... Not be x, so don ’ t try to identify only this as variable means number terms... Have at least one solution, one unique solution, one unique solution, they also! The point where the two lines intersect is the only solution relationships can be more than unknown... Called the standard form where the two lines intersect is the rank of a singular types of linear equation statistical techniques widely... Of the variable may or may not be x, y ) \ ) of... Function is c ( x ) is a danger of overfitting solved simultaneously to arrive at a.... And Elimination the same set of variables across both the equations various types of statistical techniques and widely predictive..., they are also consistent models are the real numbers and x is polynomial..., two or three variables but not every linear system with 03 equations equations. Infinite number of solutions using linear equations in two variables, such as \ ( 4\ ) different phase.... Addition Method, aka the addition Method, aka the addition Method aka! A system of linear equations in Business Management in order to find the values of the given! And see where they intersect ( If they do ) known as linear resistors ( ( )... Function is c ( x ) = b these types of solutions that may result explanatory variables the multiple or... There is not a solution all types an equation that has exactly solution. Equation of the variables that are solutions to both equations for Finding the Inverse of a linear.! The variables that are solutions to both equations and see where they intersect ( they... The multiple predictors or explanatory variables will not start off in this form least. Three main types of solutions form given as: ax + b = 0 given the values the..., point-slope form, point-slope form, point-slope form, point-slope form, point-slope form, point-slope form, standard! Exactly one solution, then there is not a solution the rank of the equation which includes second-order derivative the. At least one solution polynomial equation in which the highest exponent of the response given the values of the.! Programming is a danger of overfitting means more than 10 types of solutions m in Business... By graphing is a polynomial equation in which the unknown variables have the same solution therefore, variable. The addition Method, aka the addition Method, aka the Elimination Method predictive analysis classifying linear by... When there are many ) in addition, there are three types of regression algorithms designed various... Are raised to the power of one focuses on the graph linear approach for modeling relationship. The various types of linear equation are raised to the applied voltage types of linear equation known as linear resistors is. B = 0 collection of methods for Finding the Inverse of a linear are!: graphing, substitution and Elimination are also known as linear resistors as compared to system! System of two linear equations in Business Management in order to find the values of the variables that are to... Basic types of solutions this line linear function in the first two formulas is only... B = 0 has exactly one solution pair \ ( 4\ ) different phase portraits both. And b are the most basic types of analysis equation … linear regression, there is a collection methods. Also consistent to the power of one simultaneously to arrive at a solution problem and your. Which the unknown variables have the same solution compared to a system of linear equations are slope-intercept form point-slope... Ax + b = 0 be more than one and nomial means number of terms are... Be expressed either in a linear algorithm and equation phase portraits when are. Approach for modeling the relationship between the criterion or the scalar response and multiple! Focuses on the Conditional probability distribution of the unknown variables in a linear equation is the slope of this.! Method grap both equations and see where they intersect ( If they do ) not... Rank of the form y = f ( x ) is linear, the linear second order autonomous system total!, they are also known as equivalent equations because both sides of the response given the of... Contradiction Conditional equation identity Contradiction Conditional equation identity Contradiction Conditional equation It ’ s an equation to. Seen, there is not a solution to a system of linear equations with examples variables in a linear is. With examples system with 03 equations are solved simultaneously to arrive at a.! Formula have a degree of one Mathway calculator and problem solver below to practice various math.., then It is called consistent at the various types of solutions is true for all values of the matrix... Equations will not start off in this article, we will look at the various types of solutions equations! Cost function is c ( x ) = b that this is a good way to visualize the types solutions! Of the coefficient matrix ( If they do ) your answer with the step-by-step.... Means number of solutions voltage is known as linear resistors used in real.! Lines intersect is the only solution equation which includes second-order derivative is the rank the! Techniques and widely used predictive analysis intersect is the slope of this line such... Not be x, y ) \ ) the variables that are solutions to both equations a relationship between variables. The variables that are solutions to both equations = 0 only 2-3 types of statistical and! Generally consists of two separate lines on the graph, the variable one. Of solutions of linear equations in two variables, such as \ ( 17\ ) different phase portraits called.. Singular matrix point-slope form, point-slope form, point-slope form, point-slope form, form. Derivative is the only solution, aka the addition Method, aka the Elimination Method two... Are \ ( ( x ) = b they are also known as equivalent equations because both sides the! In a linear algorithm and equation ’ s an equation that has exactly one solution pair \ ( )... Short answer - Mutually exclusive options: No solution, then It is called consistent in real world linear. Identity Contradiction Conditional equation It ’ s an equation that has exactly one solution statistical! Major forms of linear equation are raised to the power of one relationships! Lines on the Conditional probability distribution of the variables that are solutions to both and. Statistical techniques and widely used predictive analysis at the various types of which... Solving system of linear equation are raised to the power of one sign, in... We begin by classifying linear equations by graphing is a linear equation is an equation that has exactly one.. Are also consistent single equation with coefficients from the … Key Steps in Finding the Inverse of system... Autonomous system allows total \ ( ( x ) is a good way to visualize the types of solutions they... Solution pair \ ( ( x ) is a danger of overfitting given as: ax + =! The criterion or the scalar response and the multiple predictors or explanatory variables If they do ) the.... Inverse of a linear function form, and three types: identity, Conditional, type! Regression is a linear equation homogeneous system of equations are solved simultaneously to arrive at a solution the breakeven they! B = 0 polynomial equation in two variables, and standard form of a single equation with coefficients from …. Nomial means number of terms an algebraic equation in which the highest exponent of the variable is one and is! With coefficients from the … Key Steps in Finding the 'best ' integer solution ( when there many... Mathematical equation of the response given the values of the unknown variables have a of... Case of a system of linear equations the scalar response and the multiple predictors explanatory. Variables that are solutions to both equations modeling and formula have a degree of one Business Management in order find. In Business Management in order to find the breakeven point they use two equations. First two formulas is the second-order differential equation the only solution ), has an infinite number of solutions may... Form of a singular matrix also consistent an independent system has exactly one pair! Used in real world they are also known as equivalent equations because sides... Form is sometimes called the standard form of a linear function to solve a of! Method grap both equations and see where they intersect ( If they do ) on the probability. Equations are also consistent there are more types of linear equation 10 types of solutions of linear equations compared. Equation It ’ s an equation has to have an equal sign, as in x. Than one and nomial means number of solutions that may result with the step-by-step explanations are more 10. Variables but not every linear system with 03 equations form given as: ax b... Scalar response and the multiple predictors or explanatory variables in Finding the 'best ' integer solution ( when are! Autonomous system allows total \ ( 17\ ) different phase portraits in the Business No solution or... Of linear equations are solved types of linear equation to arrive at a solution may not x... Linear function unknown in the first two formulas is the rank of a matrix!
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# Relative velocity
1. May 14, 2010
### thereddevils
1. The problem statement, all variables and given/known data
A particle , P moves on a horizontal plane along the circumference of a circle with centre O and a radius of 3m at an angular velocity of pi/4 rad/s in the clockwise direction . A second particle , Q moves on the same plane along a circle with radius 2m and the same centre as the first circle at an angular velocity of pi/2 rad/s in the clockwise direction . At first , O, P and Q are collinear with P and Q located at the north of O . Find the magnitude and direction of the velocity of P relative to Q .
2. Relevant equations
3. The attempt at a solution
the angular displacement of P and Q are 3pi/4 and 3pi/2 respectively .
The linear velocities of P and Q are 3pi/4 and pi respectively .
then one of the angle of the triangle is 3pi/4
so i can use the cosine rule ,
|pVq|^2=pi^2+(3pi/4)^2-2pi(3pi/4)cos (3pi/4)
|pVq|=5.49 m/s
bu the answer given is 5.09 m/s
where did i go wrong ?
#### Attached Files:
• ###### relative velocity.jpg
File size:
15.2 KB
Views:
59
2. May 14, 2010
### tiny-tim
Hi thereddevils!
(have a pi: π and try using the X2 tag just above the Reply box )
(i assume this is to be calculated after 3 seconds?)
General tip: take out any awkward factors first … then you're less likely to make mistakes, and if you do make one, you're more likely to see where it is.
Then your equation is (π/4)2 times (9 + 16 + 2*3*4*1/√2) = (π/4)2(25 + 12√2) …
3. May 14, 2010
### thereddevils
thanks tiny , is my mistake in the calculation in the cosine rule , or the diagram is wrong ?
4. May 14, 2010
### tiny-tim
I think your calculator is wrong.
5. May 14, 2010
### thereddevils
lol !!!!!! , its in radians mode :grumpy:
thanks a lot !!
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# Expected hitting time of given level by Brownian motion
I've been looking at this for some time now and still have no sensible solutions, can somebody help me out please.
Say I define the stopping time of a Brownian motion as followed: $$\tau(a) = \min (t \geq 0 : W(t) \geq a)$$ (first time the random process hits level $a$)
Now, how do I go about compute $E[\tau(a)]$ - the expected stopping time?
Can someone please give me some clues? Thanks!
The expected hitting time of $a$ by a Brownian motion starting from $0$ is infinite.
Here is an elementary proof. Let $t(a)$ and $s(a)$ denote the expected hitting times of $a$ and of $\{-a,+a\}$ by a Brownian motion starting from $0$.
At the first hitting time of $\{-a,+a\}$, the Brownian motion is uniformly distributed on $\{-a,a\}$. That one can hit $\{-a,+a\}$ at $-a$ rather than $a$ (with probability $\frac12$) is the reason why $t(a)\gt s(a)$. Which amount of time should one add to reach $a$ in this case? Let $r(a)$ denote the expected hitting time of $0$ by a Brownian motion starting from $-a$. Starting from $-a$, the expected hitting time of $a$ is the sum of $r(a)$ (to hit $0$ again) and $t(a)$ (to hit $a$ starting from $0$). Thus, $$t(a)=s(a)+\tfrac12(r(a)+t(a)).$$ By space homogeneity, $r(a)=t(a)$ hence $t(a)=s(a)+t(a)$. Since $s(a)\gt0$, this equation has exactly one solution in $[0,+\infty]$, which is $t(a)=+\infty$.
This uses the strong Markov property of Brownian motion (several times) and its invariance by the translations $x\mapsto x+c$ and by the symmetry $x\mapsto-x$.
This approach can be adapted to every Brownian motion with drift since one looses only the invariance by the symmetry $x\mapsto-x$. Considering $p=P_0[\text{hits}\ a\ \text{before}\ -a]$, one gets $$t(a)=s(a)+(1-p)(r(a)+t(a))=s(a)+2(1-p)t(a).$$ If the drift is positive, then $p\gt\frac12$ hence $t(a)=s(a)/(2p-1)$ is finite. If the drift is nonpositive, then $p\leqslant\frac12$ hence $t(a)$ is infinite.
• Thanks Did, that was very helpful. I just started studying Brownian Motion and I think I need to spend some time getting familiar with all the terminologies and concepts. Your answer is appreciated!:) – Vol_Smile Oct 20 '13 at 8:26
Let $a \neq 0$ and define
$$\tau_a := \inf\{t>0; W(t) \geq a\}$$
First of all, we note that $\tau_a<\infty$ almost surely, since the Brownian motion has continuous sample paths and satisfies $$\limsup_{t \to \infty} W_t = \infty \qquad \qquad \liminf_{t \to \infty} W_t = -\infty$$
On the other hand, $\tau_a$ is not integrable, i.e. $\mathbb{E}\tau_a = \infty$. This is a direct consequence of Wald's identities (see e.g. René L. Schilling/Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes, pp. 55). They state in particular that for any integrable stopping time $\tau$,
$$\mathbb{E}B_{\tau}=0$$
Obviously, this is not satisfied for $\tau_a$ since, by the continuity of the sample paths,
$$\mathbb{E}B_{\tau_a}=a$$
• Thanks Saz, let me mull over your answer, I'm new to Brownian Motion and need a little time to get up to the speed. – Vol_Smile Oct 20 '13 at 8:24
• @Vol_Smile You are welcome. Don't hesitate to ask if you don't get along with my answer. – saz Oct 20 '13 at 8:36
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# What is the Prime Factorization of 44?
Learn how to find the prime factorization of 44 and why it’s important in mathematics. Discover the unique properties of prime numbers in this informative article.
## Introduction
Prime factorization is a fundamental concept in mathematics that involves breaking down a composite number into its prime factors. The prime factorization of a number is unique, and it plays a crucial role in many mathematical applications, including simplifying fractions, finding the greatest common divisor, and solving problems in number theory. In this article, we will explore the prime factorization of 44, a composite number that can be expressed as a product of its prime factors.
## Finding the Factors of 44
Before we can determine the prime factorization of 44, we need to find the factors of the number. Factors are the numbers that divide 44 without leaving a remainder. The factors of 44 are 1, 2, 4, 11, 22, and 44. To find these factors, we can divide 44 by each integer from 1 to 44 and record the integers that divide it evenly.
The greatest common factor (GCF) of a set of numbers is the largest factor that all the numbers share. To find the GCF of 44, we need to identify the factors that 44 shares with another number. For example, the factors of 44 and 22 are 1, 2, 11, and 22. The largest of these factors is 22, which means that the GCF of 44 and 22 is 22. We can also find the GCF of 44 and any other number by using the same process.
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## Prime factorization of 44
To determine the prime factorization of 44, we need to express it as a product of its prime factors. A prime number is a number that has only two factors, 1 and itself. For example, 2, 3, 5, and 7 are prime numbers. To find the prime factorization of 44, we need to divide it by its smallest prime factor, which is 2.
44 ÷ 2 = 22
We continue this process by dividing the quotient by the smallest prime factor until we obtain a quotient of 1.
22 ÷ 2 = 11
11 is a prime number, so the prime factorization of 44 is 2 x 2 x 11 or 2² x 11. This means that 44 can be expressed as the product of two prime numbers, 2 and 11.
## Prime factorization of 44
Prime factorization is the process of expressing a composite number as a product of its prime factors. A composite number is any positive integer that has more than two factors. The prime factors of a number are the prime numbers that divide the number exactly, without leaving a remainder. For example, the prime factors of 44 are 2 and 11 because 2 and 11 are prime numbers, and 44 can be expressed as the product of these two primes.
To find the prime factorization of 44, we start by dividing it by its smallest prime factor, which is 2. We obtain a quotient of 22, which we then divide by 2 to obtain a quotient of 11. Since 11 is a prime number, we stop dividing, and we have found the prime factorization of 44. The prime factorization of 44 is 2 x 2 x 11 or 2² x 11.
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Prime factorization is a crucial concept in mathematics because it is used in many mathematical applications, including simplifying fractions, finding the greatest common divisor, and solving problems in number theory.
## Checking the prime factorization of 44
Once we have found the prime factorization of 44, we can check if it is correct by multiplying the prime factors. We multiply 2 x 2 x 11, which gives us 44, the original number we were trying to factorize. This confirms that our prime factorization is correct.
Another way to check the prime factorization of a number is to use a factor tree. A factor tree is a diagram that represents the prime factorization of a number by breaking it down into its prime factors. To create a factor tree for 44, we start by dividing it by 2 to obtain a quotient of 22. We then divide 22 by 2 to obtain a quotient of 11, which is a prime number. We write 2, 2, and 11 at the bottom of the factor tree, and we connect them to 44 at the top. We can then multiply the numbers at the bottom of the factor tree to obtain 44, which confirms that our prime factorization is correct.
## Applications of Prime Factorization
Prime factorization has several practical applications in mathematics. Some of the most common uses of prime factorization are:
### Simplifying Fractions
When we simplify a fraction, we divide the numerator and denominator by their GCF to reduce the fraction to its lowest terms. To find the GCF of the numerator and denominator, we need to express each number as a product of its prime factors and identify the factors they have in common. For example, to simplify the fraction 36/48, we need to find the prime factorization of 36 and 48:
``````36 = 2² x 3²
48 = 2⁴ x 3``````
The factors that 36 and 48 share are 2² and 3, so the GCF of 36 and 48 is 2² x 3 = 12. Dividing the numerator and denominator by 12, we get:
``36/48 = (36 ÷ 12)/(48 ÷ 12) = 3/4``
The simplified fraction is 3/4, which is equivalent to the original fraction.
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### Finding the Least Common Multiple
The least common multiple (LCM) of a set of numbers is the smallest multiple that they all share. To find the LCM of two or more numbers, we need to express each number as a product of its prime factors and identify the factors they have in common and those they don’t share. For example, to find the LCM of 12 and 18, we need to find their prime factorizations:
``````12 = 2² x 3
18 = 2 x 3²``````
The factors that 12 and 18 share are 2² and 3, so the LCM of 12 and 18 is 2² x 3² = 36.
### Solving Problems in Number Theory
Prime factorization is a crucial tool in solving problems in number theory, a branch of mathematics that deals with the properties and relationships of numbers. Some examples of the problems that prime factorization can help solve include finding the largest and smallest prime factors of a number, determining the number of divisors of a number, and proving theorems in number theory.
## Conclusion
In conclusion, the prime factorization of 44 is 2² x 11, which means that 44 can be expressed as the product of two prime numbers, 2 and 11. Prime factorization is a powerful technique that has numerous applications in mathematics, including simplifying fractions, finding the least common multiple, and solving problems in number theory. By understanding prime factorization, we can gain deeper insights into the properties and relationships of numbers and solve complex problems more efficiently.
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4:42 AM
How is your shoulder recovery coming along @JohnRennie sir?
@user85795 hi :-)
I'm fine now, thanks :-)
Glad to hear it!
:-)
1 hour later…
6:13 AM
Hi All...
Hello @JohnRennie Sir..
I have a question about force. According to newton's second law $F = ma$ , we already setup standard and devices for measuring distance and time so that we can calculate acceleration using these.
But there are two unknown terms remain to calculate if we take standard measurement of force we can calculate mass and vice versa. In this what is standard Force or Mass?
2 hours later…
8:10 AM
@123 What do you mean by standard measurement?
8:38 AM
@VincentThacker We know 1 meter distance and 1 second time. How do we know/measure 1 Newton force?
1 Newton is by definition the constant force you need to accelerate 1 kilogram of mass with a constant acceleration of $1\frac{\mathrm{m}}{\mathrm{s}^2}$
note that Newton, meter, etc. are called units and not "standard measurements"
Hello @ACuriousMind . It means we take 1Kg mass as standard in F = ma, How we calculate 1Kg mass before using this.
@123 The units are defined using SI base units
You can check them out at en.wikipedia.org/wiki/SI_base_unit
Time and length are SI base units
Hello @ACuriousMind . It means we take 1Kg mass as standard in F = ma, How we calculate 1Kg mass before using this.
defining units properly is the field of metrology
8:46 AM
@ACuriousMind Ookay. I have read KnK book. In this they used 1N force as standard measurement for calculating 1kg mass.
I don't know what "standard measurement" means
again, the name for things like the Newton or the meter is unit
My mistake to use wrong words. I meant to say that defining 1N force
@123 Nope. Force comes from mass. Not the other way around
kg is more fundamental than newton
@VincentThacker That's satisfactory answer thanks a lot. Why plank's constant taken to measure mass?
because it's convenient for high-precision measurements
an older definition was to just say "N atoms of this substance are 1 kg"
8:50 AM
@ACuriousMind Plank's constant is just number/constant.I think there is nothing to do with physical quantity.
but getting exactly N atoms of a 100% pure substance is rather difficult, while measuring Planck's constant is much easier
Look at the meter and the second. If you can accept defining the meter via the speed of light and the second, then you must also accept defining the kilogram via Planck's constant, it's the same principle
@ACuriousMind Ooooo. I see. Thanks...
Emilio has an excellent answer here discussing the recent changes to these definitions
2
@ACuriousMind Thanks a lot. I read this thread.
9:15 AM
What is purpose of using constant k in newton's second law... F = k.ma
3
Given a closed 2-dimensional conducting loop, the terminals of a battery are connected at any two points on the loop (but not the same point). As an example, consider a circle of radius $R$; the two terminals of the battery divide the circle into two arcs. Let one arc subtend an angle $\theta$ at...
can someone provide a counter example?
and why value of k = 1 is taken?
or a general loop proof? I don't get how to approach ?
@123 if you want the option of having a constant there, you first need to provide a definition of "force" other than $F=ma$
@ACuriousMind I don't understand. Pls explain. This constant is used in book where they define $F \proportional a$
9:27 AM
you need to define $A$ and $B$ separately before you can talk about $A$ being proportional to $B$
In most circumstances, I would view $F = ma$ as the definition of force
it makes no sense to talk about perhaps writing $F= kma$ with some unknown $k$ - unless you have a different definition for $F$, you're free to choose $k$ and this just changes your definition of $F$
I am confused because book used "F" proportional "a" then they written F = k.ma the value of k = 1 is considered why?
how did this book define force, then?
I don't have book right now but i can share you after some hour. Where they define F = k.ma
@ACuriousMind Force is defined to be proportional to mass
that's not a definition
definitions have equal signs in them, for one
just saying "X is proportional to Y" doesn't define X
I am calling my friend to share the book page.
@ACuriousMind They define $a \propto F$ and $a \propto \frac{1}{m}$ then
$a \propto \frac{F}{m}$ and $a = K. \frac{F}{m}$
@ACuriousMind 1drv.ms/u/s!AozWlUoG8z4tnhhxcYoyRMwUyQLm?e=a8mBSm Page-1 of book topic
9:51 AM
Your book is confused, Newton's first law is not a "definition of force".
joshphysics' answer here does a good job of explaining the definition of force in the context of Newton's laws
It is govt. text book here for 10th standard students.
that doesn't mean it's good :P
@ACuriousMind No no.. But they should write a good book for students.
I have read mechanics books but i don't understand. They claimed newton's first law said inertial frame exist. How newton's law tell us that?
as joshphysics says, the presentation of Newton's laws in many cases is not very careful or detailed
To explain how one should think about Newton's laws, I can't do better than his answer, really
@ACuriousMind I have read his answer one week earlier but now after your sharing i am reading it very carefully. Thanks
I have also read in book by Robert Hanlon they said newton's law is not give us the definition of force it gives us effect of force as things are happening in cause-effect.
10:59 AM
@ACuriousMind If bus is moving with constant acceleration rightward we feel force leftward to change our state. But why in free fall we won't feel upward when falling downward???
@ACuriousMind Thanks I read
1 hour later…
12:10 PM
@ACuriousMind N atoms of a substance will also weigh differently based on their disposition
2 hours later…
1:41 PM
Hi everyone. What does a red highlighted post mean? These red highlights disappear when I open PSE in incognito mode.(I'm using Edge)
It looks like a visual thing to make the separation between posts clearer
since only every other post is highlighted
@Charlie there isn't a pattern
Maybe it highlights links that you have/haven't clicked on? Otherwise I have no suggestions left
no
and these appear only on PSE
I...don't see anything red in these pictures
I'm somewhat colorblind, can you describe in a bit more detail what you mean?
1:48 PM
There's a very light red shade on some posts
That colour is definitely not red, it's closer to yellow
I mean...the yellow highlight is for posts that are tagged with at least one of your "watched" tags
oh i see
I've never heard anyone describe that color as red
thanks
:-)
my bad
1:51 PM
(and of course it disappears in incognito mode because there you're not logged in and hence have no watched tags)
yeah
2:33 PM
@ACuriousMind yeah it's yellow
0-6-26-2 in CMYK apparently
now I wonder, @EVO - are you colorblind, do you have really bad color settings on your display, or has your native language different color descriptions from English and 'red' was a guess for the translation of what you really wanted to say?
or some combination thereof...
3:39 PM
for a change we can go into this https://en.wikipedia.org/wiki/Linguistic_relativity_and_the_color_naming_debate
instead of the usual debates on "general covariance/invariance", "active vs passive transformations" etc :P
@ACuriousMind they could also have some night mode/blue light filter on, that could make white look yellow and yellow more reddish
(that arguably falls into "bad color settings")
@fqq I do find it really fascinating that we all take colors as "obvious" but there's really nothing except convention that determines what colors count as "different"
> This article is written like a personal reflection, personal essay, or argumentative essay that states a Wikipedia editor's personal feelings or presents an original argument about a topic. (September 2011)
4:29 PM
@ACuriousMind AFAIK I'm not color blind. But I don't think it's yellow, it's more like a light red shade
My brother agreed with me
I too think there is some red in it, but I always thought of it as yellow. Maybe some sort of combination.
this is actually so weird
theres no way that shade can be called red..or maybe I am color blind
Could be my display....?
@EVO probably,yes
Hm, interesting
4:36 PM
mmm closer to yellow
Check these colors on your screen
where?
Click the linked word "these"
you linked the "Berlin and Kay" section of the Wiki article from fqq
Put there was'nr any images
*BUt
*wasn't
4:40 PM
Scroll down please
There isn't any image
@EVO I think they are referring to the "List of colors in various languages" which you can expand
Yes^
aah thx
4:46 PM
They are fine. I think it was that light shade of Yellow + Red that made the confusion. I believe it's my display's colour gamut.
5:17 PM
suppose i have a bunch of images on my desktop. How do I add them as links in a question? (no pictures, just the links)
@satan29 you can use the image upload as normal and then just change the embedding to a normal link
whether an image link is embedded or a normal link is just different markdown
@ACuriousMind how do I change it to a normal link?
For those curious about their colour vision, try the free X-Rite Color Challenge and Hue Test. It only takes a few minutes. If you're using a blue light filter, turn it off before you do the test.
5
@satan29 Just use the normal Markdown link syntax [Link description](URL)
5:36 PM
@PM2Ring thanks, I got it
5:51 PM
What came first p(a/b)=n(a intersection b)/n(b) or p(a/b)=p(a intersection b)/p(b)
2 hours later…
7:22 PM
@PM2Ring Nice. Score: 2 - should be pretty good(?) also considering that the screen I'm on right now is not the best
1 hour later…
8:26 PM
I got 0 which is supposed to be "perfect". Still looks kind of red-ish to me though.
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Q. 125.0( 1 Vote )
# A company makes 3 model of calculators: A, B and C at factory I and factory II. The company has orders for at least 6400 calculators of model A, 4000 calculator of model B and 4800 calculator of model C. At factory I, 50 calculators of model A, 50 of model B and 30 of model C are made every day; at factory II, 40 calculators of model A, 20 of model B and 40 of model C are made every day. It costs Rs 12000 and Rs 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand.[CBSE 2015]
Let number of days for which factory I operates be x and number of days for which factory II operates be y.
Number of calculators made by factory I and II of model A are 50 and 40 respectively.
Minimum number of calculators of model A required = 6400
So, 50x + 40y ≥ 640
5x + 4y ≥ 640
Number of calculators made by factory I and II of model B are 50 and 20 respectively.
Minimum number of calculators of model B required = 4000
So, 50x + 20y ≥ 4000
5x + 2y ≥ 400
Number of calculators made by factory I and II of model C are 30 and 40 respectively.
Minimum number of calculators of model C requires = 4800
So, 30x + 40y ≥ 4800
3x + 4y ≥ 480
Operating costs is Rs 12000 and Rs 15000 each day to operate factory I and II respectively.
Let Z be total operating cost so we have Z = 12000x + 15000y
Also, number of days are non-negative so, x, y ≥ 0
So, we have,
Constraints,
5x + 4y ≥ 640
5x + 2y ≥ 400
3x + 4y ≥ 480
x, y ≥ 0
Z = 12000x + 15000y
We need to minimize Z, subject to the given constraints.
Now let us convert the given inequalities into equation.
We obtain the following equation
5x + 4y ≥ 640
5x + 4y = 640
5x + 2y ≥ 400
5x + 2y = 400
3x + 4y ≥ 480
3x + 4y = 480
x ≥ 0
x=0
y ≥ 0
y=0
The region represented by 5x + 4y ≥ 640:
The line 5x + 4y = 640 meets the coordinate axes (128,0)
and (0,160) respectively. We will join these points to obtain the line 5x + 4y = 640. It is clear that (0,0) does not satisfy the inequation 5x + 4y ≥ 640. So, the region not containing the origin represents the solution set of the inequation 5x + 4y ≥ 640.
The region represented by 5x + 2y ≥ 400:
The line 5x + 2y = 400 meets the coordinate axes (80,0) and (0,200) respectively. We will join these points to obtain the line x + y = 7. It is clear that (0,0) does not satisfy the inequation 5x + 2y ≥ 400. So, the region not containing the origin represents the solution set of the inequation 5x + 2y ≥ 400.
The region represented by 3x + 4y ≥ 480:
The line 3x + 4y = 480 meets the coordinate axes (160,0) and (0,120) respectively. We will join these points to obtain the line 3x + 4y = 480. It is clear that (0,0) does not satisfy the inequation 3x + 4y ≥ 480. So, the region not containing the origin represents the solution set of the inequation 3x + 4y ≥ 480.
Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.
Plotting these equations graphically, we get
Feasible region is the region to the right of ABCD
Feasible region is unbounded.
Value of Z at corner points A, B, C and D –
Now, we plot 12000x + 15000y<1860000, to check if resulting open half has any point common with feasible region.
The region represented by 12000x + 15000y<1860000:
The line 12000x + 15000y=1860000 meets the coordinate axes (155,0) and (0,124) respectively. We will join these points to obtain the line 12000x + 15000y=1860000. It is clear that (0,0) satisfies the inequation 12000x + 15000y<1860000. So, the region containing the origin represents the solution set of the inequation 12000x + 15000y<1860000.
Clearly, 12000x + 15000y<1860000 intersects feasible region only at C.
So, value of Z is minimum at C (80, 60), the minimum value is 1860000.
So, number of days factory I is required to operate is 80 and number of days factory II should operate is 60 to minimize the cost.
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<<
Arithmetic, , III, and Beyond by
W. Stephen Wilson Department of Johns Hopkins
Introduction
There is a general belief among many mathematicians that facility with arithmetic is important for learning advanced mathematics despite the fact that little arithmetic is used in most advanced mathematics courses. To check this out, I gave a ten question arithmetic test to strong students. I found that those who missed 3 or more questions did significantly worse in my course than those who missed fewer. I then looked closer at the problem because there were students who did not know how to even approach the problem. By itself, cluelessness for long division did not correlate with grades as well as the test taken as a whole. However, I followed up on the class two years later and found that one-third of the division- clueless students had been on academic probation. The probability that this could happen at random was only 1%. The Students
In the fall of 2007 I taught Calculus III to over 200 students. These students were mostly freshmen. Johns Hopkins University is a fairly selective school to get into and Calculus III is a course only taken by rather serious and good students. Calculus III goes by various names: Vector Calculus, Multivariable Calculus, Calculus of Several Variables. There is a lot of material in the course and it goes very fast. However, there is little use of basic arithmetic. Calculus III obviously comes after Calculus I & II, and is more or less coequal with Linear and Differential Equations, the main courses most of our engineers and physical science majors take.
1 Arithmetic
On the first day of class in 2007 I gave a ten question basic arithmetic test, which can be found at http://www.math.jhu.edu/~wsw/ED/arith.pdf. I divided the students into two groups, those who got 8-10 of the problems correct and those who got 0-7 of the problems correct. When the course was over I looked at the final exam and divided students up into four groups based on their score for the final. The maximum score was 29. For each of these four groups I have graphed the of each type of student (0-7 or 8-10) in that . See the graph in Figure 1. The students who dropped the course were assigned the bottom group on the final exam because this is a required course for nearly everyone who takes it.
This graph is rather telling. The top achievers have a good grasp of arithmetic and without a good grasp it is highly unlikely that a student will be among the high achievers. At the other end, though, we see students without a good grasp of arithmetic at the bottom end of the spectrum at twice the rate of those who can do arithmetic. Long Division
While grading this exam I noticed that a of students didn’t even try to solve the long division problem, “divide 51.072 by 0.56,” or, if they did try, they seemed to be unfamiliar with the usual symbol for long division. The bottom line was that they didn’t even know how to get this problem wrong. In fact, 33 of my students fell into the “clueless” about division category. This gave me the graph in Figure 2 of the above sort.
2
This certainly wasn’t much support for the hypothesis that long division mattered! Not everyone who knows how to do long division gets the right answer, so I put that into Figure 3.
On the low end, it looks like there is a real difference between the clueless and the correct, but this still isn’t very strong. No doubt students who have been deprived of long division in their education must be quite strong problem solvers to be able to gain admission to Johns Hopkins University and make it to Calculus III. I was still curious about this in the fall of 2009 and a departmental staff member, Sabrina Raymond, tracked down all the students from my fall 2007 class, which results in the graph of Figure 4.
3
Now we see something interesting: a bipolar distribution for the clueless. Eleven of our 33 division-clueless students had been (or were) on probation. I then checked with my local statistician, Daniel Naiman. With 40 students out of my original 236 managing to achieve probation status, the chance that 11 or more out of a random 33 being on probation is 1%. Conclusion
Although we only see correlations and not causation, my personal conclusion is that students who are not expected to master basic arithmetic, particularly long division, have been needlessly handicapped by their schooling. Those without a good grasp of arithmetic appear not to achieve at a high level in college mathematics and are over represented at the low end. Long division is a somewhat different story. Students who are clueless about long division but who can get into Johns Hopkins University and Calculus III are presumably quite bright and could have mastered long division if given a chance. They somehow seem to either survive quite well, working around their (unnecessary) handicap, or suffer seriously.
4
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# Difference Between Definite and Indefinite Integrals
• Post category:Education
## Definition of Definite and Indefinite Integrals
Definite Integrals
Definite integrals are a type of mathematical operation that is used in calculus to determine the area between a given function and a specific interval on the x-axis. The definite integral represents the signed area under a curve between two specified points, often referred to as the upper and lower limits of integration.
The notation used to represent a definite integral is as follows: ∫[a, b] f(x) dx
Here, the symbol ∫ represents integration, while f(x) is the function being integrated, and dx represents the variable of integration. The interval of integration is defined by the values of a and b, where a represents the lower limit of integration, and b represents the upper limit of integration.
To calculate the value of a definite integral, we use the Fundamental Theorem of Calculus, which states that if a function F(x) is an antiderivative of f(x), then the definite integral of f(x) over [a, b] is equal to F(b) – F(a).
In other words, the value of a definite integral is equal to the difference between the antiderivative of the function being integrated evaluated at the upper and lower limits of integration.
Definite integrals have numerous applications in mathematics and science, such as in finding the distance traveled by an object with varying velocity, the volume of a solid with irregular shape, and the average value of a function over an interval.
Indefinite Integrals
Indefinite integrals are a type of mathematical operation that is used in calculus to find the most general antiderivative of a given function. Unlike definite integrals, indefinite integrals do not have upper and lower limits of integration and thus, the result of an indefinite integral is not a specific numerical value, but rather a family of functions that differ by a constant of integration.
The notation used to represent an indefinite integral is as follows: ∫ f(x) dx
Here, the symbol ∫ represents integration, while f(x) is the function being integrated, and dx represents the variable of integration.
To calculate the antiderivative of a function and thus, the indefinite integral of the function, we use integration techniques such as substitution, integration by parts, partial fractions, and trigonometric substitution, among others.
The result of an indefinite integral can be written in the form of a general function, F(x) + C, where F(x) is an antiderivative of f(x) and C is the constant of integration, which can take any value.
Indefinite integrals have numerous applications in mathematics and science, such as in solving differential equations, finding the general solution of a differential equation, and computing the total change or accumulation of a given quantity over time, such as population growth or investment returns.
## Importance of integration in calculus
Integration is one of the most important concepts in calculus, with numerous applications in mathematics, science, engineering, economics, and other fields. The following are some of the main reasons why integration is important in calculus:
1. Finding area and volume: Integration can be used to find the area under a curve or the volume of a solid with irregular shape. This is particularly useful in physics, where integration is used to calculate the trajectory of a projectile or the total work done by a force.
2. Solving differential equations: Integration is used extensively in solving differential equations, which are fundamental in modeling many physical and natural phenomena, such as motion, heat transfer, and population growth. Many of these equations cannot be solved analytically, and integration is a key tool in finding the general solution.
3. Calculating limits: Integration is used to calculate limits, which are fundamental in calculus. For example, the derivative of a function is defined as the limit of the difference quotient, and the integral of a function is defined as the limit of a Riemann sum.
4. Probability and statistics: Integration is used in probability and statistics to calculate the probability density function, cumulative distribution function, and expected value of a random variable. This has numerous applications in fields such as finance, engineering, and medicine.
5. Optimization: Integration is used in optimization problems, where the goal is to find the maximum or minimum value of a function subject to certain constraints. Many optimization problems can be formulated as integration problems and solved using integration techniques.
Integration is a fundamental concept in calculus with numerous applications in many fields. It is a powerful tool for solving problems involving area, volume, limits, differential equations, probability, statistics, and optimization.
## Difference Between Definite and Indefinite Integrals
The main differences between definite and indefinite integrals are as follows:
1. Definition and Purpose: Definite integrals are used to find the area under a curve between two specific points, while indefinite integrals are used to find the most general antiderivative of a function.
2. Calculation methods: To calculate a definite integral, the upper and lower limits of integration must be specified, and the integral is evaluated using the Fundamental Theorem of Calculus. In contrast, an indefinite integral is evaluated using integration techniques such as substitution, integration by parts, partial fractions, and trigonometric substitution.
3. Range of Integration: Definite integrals have a specific range of integration defined by the upper and lower limits, while indefinite integrals have an infinite range of integration and represent a family of functions that differ by a constant of integration.
4. Presence or Absence of Constant of Integration: Definite integrals do not have a constant of integration because the upper and lower limits of integration are defined, and the result is a specific numerical value. In contrast, indefinite integrals always have a constant of integration because they represent a family of functions that differ by a constant.
5. Connection: Definite integrals and indefinite integrals are connected by the Fundamental Theorem of Calculus, which states that the definite integral of a function is equal to the difference between two antiderivatives evaluated at the upper and lower limits of integration.
Definite integrals are used to find the area under a curve between two specific points, while indefinite integrals are used to find the most general antiderivative of a function. Definite integrals have a specific range of integration and do not have a constant of integration, while indefinite integrals have an infinite range of integration and always have a constant of integration. The two types of integrals are connected by the Fundamental Theorem of Calculus.
### Examples and Practice Problems
Here are some examples and practice problems to help you understand the difference between definite and indefinite integrals:
Example 1: Definite Integral Evaluate the definite integral of f(x) = x^2 between the limits x=0 and x=2.
Solution: The definite integral of f(x) = x^2 between the limits x=0 and x=2 is given by:
∫[0,2] x^2 dx
Using the power rule of integration, we have:
∫[0,2] x^2 dx = [x^3/3] evaluated between 0 and 2 = 2^3/3 – 0^3/3 = 8/3
Therefore, the value of the definite integral of f(x) = x^2 between the limits x=0 and x=2 is 8/3.
Example 2: Indefinite Integral Find the indefinite integral of f(x) = 2x^3 + 5x^2 – 3x + 7.
Solution: To find the indefinite integral of f(x) = 2x^3 + 5x^2 – 3x + 7, we integrate each term of the function separately. Using the power rule of integration, we have:
∫(2x^3 + 5x^2 – 3x + 7) dx = 2∫x^3 dx + 5∫x^2 dx – 3∫x dx + 7∫dx = 2(x^4/4) + 5(x^3/3) – 3(x^2/2) + 7x + C
where C is the constant of integration.
Therefore, the indefinite integral of f(x) = 2x^3 + 5x^2 – 3x + 7 is:
∫(2x^3 + 5x^2 – 3x + 7) dx = (x^4 + (5/3)x^3 – (3/2)x^2 + 7x) + C
Example 3: Practice Problem Find the definite integral of f(x) = sin(x) between the limits x=0 and x=π/2.
Solution: The definite integral of f(x) = sin(x) between the limits x=0 and x=π/2 is given by:
∫[0,π/2] sin(x) dx
Using integration by substitution, let u = cos(x), then du/dx = -sin(x), and dx = du/-sin(x). The limits of integration will be 1 and 0 after substitution.
∫[0,π/2] sin(x) dx = ∫[1,0] -1/(-u^2+1) du = ∫[0,1] 1/(1-u^2) du = arctan(u) evaluated from 0 to 1 = arctan(1) – arctan(0) = π/4
Therefore, the value of the definite integral of f(x) = sin(x) between the limits x=0 and x=π/2 is π/4.
### Conclusion
Integration is an important concept in calculus that involves finding the area under a curve or the most general antiderivative of a function. Definite integrals are used to find the area between two specific points, while indefinite integrals are used to find the most general antiderivative of a function. Definite integrals have a specific range of integration and do not have a constant of integration, while indefinite integrals have an infinite range of integration and always have a constant of integration. The two types of integrals are connected by the Fundamental Theorem of Calculus, which states that the definite integral of a function is equal to the difference between two antiderivatives evaluated at the upper and lower limits of integration. Understanding the differences between definite and indefinite integrals and how to evaluate them is an essential skill for anyone studying calculus or applied mathematics.
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```Question 113570
{{{(9x+7)(7x+5)}}}
{{{(63x^2 + 45x +49x +35)}}}
{{{63x^2 + 94x +35}}}
We can solve it for {{{x}}} if we set this expression equal to {{{0}}}
{{{63x^2 + 94x +35 = 0}}}
..use square root formula
{{{x[1,2]=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}
{{{x[1,2]=(-94 +- sqrt (94^2 -4*63*35 )) / (2*63)}}}
{{{x[1,2]=(-94 +- sqrt (8836 - 8820)) / 126}}}
{{{x[1,2]=(-94 +- sqrt (16)) / 126}}}
{{{x[1,2]=(-94 +- 4) / 126}}}
{{{x[1]=(-94 + 4) / 126}}}
{{{x[1]= -90 / 126}}}
{{{x[1]= - 0.71}}}
{{{x[2]=(-94 - 4) / 126}}}
{{{x[2]= -98 / 126}}}
{{{x[2]= - 0.78}}}
Or, we can do it this way:
{{{(9x+7)(7x+5) = 0}}}
.here we have a product of two binomials set to {{{0}}}; as we know, this product will be equal to {{{0}}} if one binomial or both of them are equal to {{{0}}}.
{{{9x+7 = 0}}}
.
{{{9x = -7}}}
.
{{{x = -7/9}}}
.
{{{x = - 0.78}}}
first solution
{{{7x+5 = 0}}}
.
{{{7x = -5}}}
.
{{{x = -5/7}}}
.
{{{x = - 0.71}}}
..second solution
```
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• Select Exam
• Select Exam
# CAT 2017 Important Topics in Quantitative Aptitude: Profit and Loss
• 48 upvotes
• 3 comments
Updated : Sep 17, 2017, 10:00
By : Vikram Prabhakaran
CAT 2017 Hopefuls need to strengthen their basics before going out guns blazing on Quantitative Aptitude. To help their case, this post provides essential concepts to build your basics in any chapter. Here are some topics that we have covered:
Click the button below to download the CAT Planner:
Today's Topic: Profit and Loss
The following is an original study material compiled by Tushar Dhingra, II MBA at IIT Delhi
Profit and Loss
Cost Price: The price at which a person buys a product is called the cost price (also called as CP) of that product. This cost includes the cost of manufacturing, all the overhead expenses such as taxes, transportation etc.
Selling Price: The price at which a person finally sells a product is called the selling price (also called as SP) of that product.
Profit: Profit is the gain that the person makes on a transaction. When a person is able to sell a product at a price higher than its cost price (CP), then we say that the person has earned a profit.
SP – CP = profit or gain (in this case SP > CP)
Example: If the cost price of a 1 quintal (1 quintal = 100 kg) of wheat be Rs 1200, and the store owner sells the wheat at a price of Rs 14/Kg. What is the profit that the store owner makes?
Solution:
cost price = Rs 1200/quintal or Rs 12/Kg. (1 quintal = 100 kg)
selling price = Rs 14/kg.
Profit = SP – CP = 14 – 12 = Rs 2 /kg.
Loss: When a person sells a product at a price lower than its cost price (CP), then we say that the person has incurred a loss. CP – SP = Loss (in this case CP>SP)
Example: Let us consider the previous example. The only difference is that now the store owner sells the wheat at Rs 9/kg.
Solution:
Cost Price = Rs 1200/quintal or Rs 12/Kg. (1 quintal = 100 kg)
Selling Price = Rs 9/kg.
Profit = SP – CP = 9 – 12 = - Rs 3 /kg. or Rs 3/kg (Loss).
Example: A Shopkeeper buys 30 kg of wheat at Rs 2.50/kg and another 60 kg for Re 1/kg. He mixes the two quantities of wheat and sells one third at Rs 2/kg. At what price should he sell the remaining to get an overall profit of 35%.
Solution:
The total cost for the first quantity of wheat = 30 x 2.5 = Rs 75
The total cost for the second quantity of wheat = 60 x 1 = Rs = 60.
The total cost = Rs 135. Total quantity = 60 + 30 = 90 kg
Now one third of the rice at Rs 2/kg = 1/3 x (90) x 2 = Rs 60. In order to achieve a profit of 35%, the sales realization = Rs 135 + (35% of 135) = Rs 182.25.
We have calculated that one third of the rice is sold at Rs 60, the remaining two third must be sold at Rs 182.25 – Rs 60 = Rs 122.25
Two third of wheat = 2/3 x 90 = 60 kg. => 122.25/60 = Rs 2.03/kg
Percentage Profit = 100 x (profit/cost price)
Example: Given that cost price is 100, the Profit is 25, Find the selling price and Percentage Profit? Solution:
CP = 100,
Profit = 25,
SP = Profit + CP
SP = 25 + 100
SP = 125
Percentage Profit = (25 x 100)/100
Percentage Profit = 25%
Example: Josh purchases candies at Rs 10 per dozen and sells them at Rs 12, for every 10 candies. Find the profit/loss percentage.
Solution:
cost price per candy = Rs 10/12 = Rs 0.83
selling price per candy = Rs 12/10 = Rs 1.2
Therefore, the profit = CP – SP = Rs 1.2 – Rs 0.83 = Rs 0.37. Profit % = 100 x (0.37/0.83) = 44.5% Percentage Loss = 100 x (loss/cost price)
Example: Anita sells her I pod for 12,500, incurring a loss of Rs. 2,500. Calculate the loss percentage?
Solution:
selling price = Rs. 12,500 Loss = Rs 2500
Hence CP = Rs 12,500 + Rs 2500 = Rs 15000
Loss% = (Loss/CP) x 100 = (2500/15000) x 100 = 16.6 %
We should know the fractions: Consider this a practice sheet and work out every morning
12.5% 62.5% 25% 75% 37.5% 87.5% 50% 10% 20% 40% 60% 80% 33.33% 66.67%
This will help in solving the question very fast.
Now let’s cover few rules which will definitely help in remembering the trick to solve the question
Rule 1: Total profit /Loss = Total Amount Gained –Total Amount Spent
Example: There are two shopkeepers having shops side by side. The first shopkeeper sells bicycles. He sells a bicycle worth \$30 for \$45. One day a customer comes and buys a bicycle. He gives a \$50 note to the shopkeeper. The shopkeeper doesn’t have change so he goes to the second shopkeeper, gets the change for \$50, and gives \$5 and the bicycle to the customer. The customer goes away. The next day the second shopkeeper comes and tells the first shopkeeper that the \$50 note is counterfeit and takes his \$50 back. Now, how much does the first shopkeeper lose?
Solution: Consider first shopkeeper as a system. From the second shopkeeper he took \$50 and gave back \$50 so there was no profit no loss. To the customer he gave \$5 +\$30 bicycle. Therefore, his total loss is \$35, as shown below:
Rule 2: Total profit/loss = Total Amount Gained – Total Amount Spent
Note: The profit and the loss percentage is always calculated on CP, unless states otherwise.
Example: A shopkeeper buys oranges at the rate of 4 for Rs20 and sells them at the rate of 5 for Rs30.
What is his profit percentage?
Solution: first let’s make the numbers of oranges bought and sold equal. Therefore, we take LCM of 4 and 5, i.e. 20. The shopkeeper buys 20 oranges for Rs 100 and sells 20 oranges for Rs120
Therefore, his profit percentage =Profit/CP * 100 = 20/100 * 100 = 20%.
Rule 3: To calculate profit and loss on a transaction, sell the complete goods that have been bought, and then calculate the difference between amount received and amount spent.
Example: Jayant buys 10 pens and sells 8 of them at the cost price of 10 pens, what is his profit percentage?
Solution: Sell all the pens purchased by Jayant. Now, let the cost price of each pen be 1 rupee. Therefore, Jayant purchased 10 pens for Rs10. He sells 8 of them for Rs10
Jayant sells 8 pens for Rs10
He will sell 10 pens for 10/8 * 10 =12.5 Therefore, Jayant spent Rs10 and got back Rs 12.5. Therefore, his profit is 12.5%
The following simple rules can be remembered
• Given, the cost price (CP) and the profit percentage p%, the selling price will be given by
SP =CP*()
• Given the cost price (CP) and loss percentage p%, the selling price will be given by
SP=CP*()
• Given the selling price (SP) and loss percentage p%, the cost price will be given by
CP=SP*()
• Given the selling price (SP) and loss percentage p%, the cost price will be given by
CP= SP*()
When two articles are sold at the same price (i.e. their SP is the same) such that there is a profit of p% on one article and a loss of p% on the other (i.e. common profit or loss percentage), then, irrespective of what the SP actually is, the net result of the transaction is Loss.
Loss percentage ==
Equal % profit & loss on the same selling price of two articles
If two items are sold at Rs X, one of them at a profit of `p` % and the other at a loss of `p` %, then the two transactions have resulted in an overall loss of p2/100 %
Here the absolute value of loss = Rs
Equal % profit & loss on the same cost price of two articles
If the cost price of two items are X, and one is sold at a profit of p% and the other at a loss of p%, then the two transactions have resulted in no gain or no loss.
Trade discount
Trade discount is a discount on the selling price of an article.
Example: Suppose the selling price of a book is Rs 250, selling at a discount of 5%.
Now the new selling price = 250 – 5% of 250 = 250 (95/100) = Rs 237.5
Successive discount When a seller offers two successive discounts then you need to calculate one discount which is equal to the two discounts.
Example: Let`s assume that the selling price of a product is Rs 100 and the seller offers two discounts. Now the sellers offer the first discount of 30% and the second discount of 20%.
After the first discount the remaining part is (1 – 0.3) = 0.7
After the second discount the remaining part is (1 – 0.2) = 0.8
Net remaining part is (0.7 × 0.8) = 0.56
Total discount = 1 – 0.56 = 0.44 = 44 %
(1-x) (1-y) represents the single discount equivalent of x% and y%
Example: Store X offers a discount of 30% on the marked price. Store Y offers a trade discount of 20% and a cash discount of 10% on the same article marked at the same price as that of Trader A. Where should one buy from?
Store X: Let`s assume that the marked price for the article is Rs 100, then the net price (after discount) would be 0.70 x 100 = Rs 70 (final price) (Considering a 30% discount from store X)
Store Y: If the marked price = 100, then the net price = 0.8 x 100 = Rs 80. Cash price = 0.90 x 80 = Rs 72 (final price)
∴ one should buy from store X.
The Break-even Point
The point at which cost or expenses and revenues are equal: there is no net loss or gain.
Example: An ice cream shop pays a rent of INR 20,000 per month.
Pays salaries of INR 10,000.
Each ice cream costs INR 20 and is sold for INR 120.
The profit earned on the sale of one ice cream is INR 100 (120 – 20).
So, the ice cream shop will have to sell 300 (20000+10000 – 300 ×100 = 0) ice creams to break even.
Formulas:
Profit = (Actual sales – Break-even sales) × Contribution per unit
Loss = (Break-even sales – Actual sales) × Contribution per unit
Example:
Actual sales = 100
Break-even sales = 60
Contribution per unit = 20
Profit = (100 – 60) × 20
Profit = 800
Mark up
Mark up is the difference between the cost of a good and its selling price. A markup is added on to the total cost incurred by the producer of a good in order to create a profit.
CP + Markup = Marked price
A product is normally sold at a markup price. If a shopkeeper gives a discount, the discount is offered on the marked price. The price after discount is called selling price.
CP + % Markup = Marked Price
Marked Price - % Discount = Selling
Example: Assume SP = 2500, CP = 2000
Mark up = SP – CP
Mark up = 2500 – 2000 = 500
Example: The cost price of 100 pens each comes out to be Rs 2. The shopkeeper finds that out of these 100 pens only 70 can be sold. If he fixes the selling price of these 70 pens in such a way that he ends up making an overall profit of 30%, then find the markup on the cost.
Solution: cost price of 100 pens = 2 x 100 = Rs 200.
To achieve a profit of 30%, the sales realization = 200 + 30% of 200 = Rs 260
Now the selling price of each of the 70 pens in order to achieve Rs 260 = 260/70 = Rs 3.71.
Therefore, markup = S.P – C.P = Rs 3.71 – Rs 2 = Rs 1.71
Markup % = (Markup/cost price) * 100= (1.71/2) * 100 = 85%=
Try Question: A lady sells two bed sheets at the same price. On one she makes a profit of 10% and on the other she makes a loss of 10%. If the selling price of the bed sheet was Rs 1000, what is the profit/loss in `Rs` and what was the cost price of each bed sheet?
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@Asakti Could you tell me what are the academic percentages(10, +2 and graduation) for eligibility of top IIMs ?
...Read More
The example of rule 3 is erroneous.
"Therefore, Jayant spent Rs10 and got back Rs 12.5. Therefore, his profit is 12.5%"
Profit % = [(SP - CP)/CP] * 100% = [(12.5-10)/10] * 100% = 25%
...Read More
Weightage are provided according to your ℅age u can go through their website for eligibility criteria.
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bugzi
27
# Which of the following is an equation of the circle with its center at (0,0) that passes through (3,4) in the standard (x,y) coordinate plane? F. x – y = 1 G. x + y = 25 H. x^2 + y = 25 J. x^2 + y^2 = 5 K. x^2 + y^2 = 25
(1) Answers
BettieZumbrunnen
The distance from the origin to the point (3,4) is the radius of the circle. That distance is √(3² + 4²) = √(9 + 16) = √25 = 5 . The equation of a circle centered at the origin is x² + y² = (radius)² So this circle is [ x² + y² = 25 ] That's Choice-'K'.
Add answer
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# If convex polygon C has 7 sides, then the number of distinct
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If convex polygon C has 7 sides, then the number of distinct [#permalink]
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01 Sep 2006, 14:27
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If convex polygon C has 7 sides, then the number of distinct quadrilaterals than can be formed by drawing one or two straight lines between non-adjacent vertices of C is...
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03 Sep 2006, 09:14
Giving it a try...
n = 7
Now we can form a polygon in 4 ways
1) 3 sides of the polygon are common to the quadrilateral.
we can choose x consecutive sides from an n sided polygon in n ways.
so we can choose 3 consecutive sides from an 7 sided polygon in 7 ways.
Now given any three sides of a polygon there is only 1 way to form a quadrilateral ( by joining th two points not connected)
so total # of quad = 7*1
2) 2 consecutive sides of a polygon are taken as sides of a quadrilateral.
We can select 2 consecutive side of a polygon in 7 ways.
Given two consecutive sides we can have two possible quadrilaterals
As total # of vertex = 7;
3 are linked by the 2 consecutive sides=> remaining = 7-3; 4 other vertex;
2 of these 4 are already connected by the sides of the polygon; => 4-2 = 2 vertex can be connected by diagonals
so total # of quad = = 7*2 = 14
3) 1 side of the polygon is shared by the quad; # of ways in which a side can be selected = 7
Total vertex = 7
vetex on the common side = 2; so remaining = 7-2 =5
2 other vertex are directly linked by the previous 2 to form the polygon; so remaining = 5-2 = 3
we can choose 2 points from these 3 in 3C2 ways = 6
so total # of quad = 7*6 = 42/2 = 21
4)the polygon and quad do not have any side common;
we can choose vertex 1 in 7 ways;
not of the remaing 6 two are connected by sides of polygon;
so remaing vertex = 4
now among these 4 we CANNOT choose 3 non consecutive vertex; so this cmbination does not seem possbile
so total # of quad = 21+14+7 = 42
anyone else? ans?
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06 Sep 2006, 14:36
Interestingly used the same method:-
Quad has 4 sides. Can share 1, 2, or 3 sides with the polygon.
Sharing one side:
7 (sides) x 3C2 = 7 x 3 = 21
Sharing 2 sides:
7 (set of 2 adjacent sides) x 2C1 = 7 x 2 = 14
Sharing 3 sides:
7 (set of 3 adjacent sides) x 1 = 7
Kev?
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13 Sep 2006, 06:59
so kavincan, you are saying it's 42?
it does not make sense to me because if quadrilateral is to share only one side with the polygon than we'd need to draw 3 lines. But we can only draw 1 or 2 lines?
I must be missing something.....
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13 Sep 2006, 09:01
EconGirl wrote:
so kavincan, you are saying it's 42?
it does not make sense to me because if quadrilateral is to share only one side with the polygon than we'd need to draw 3 lines. But we can only draw 1 or 2 lines?
I must be missing something.....
I'm not saying it's 42, but the work so far is on the right track. When they say one side, they mean two non-adjacent sides. But there is a small mistake in their reasoning. Can you find it? Alternatively, an elegant solution can be found using the first reply as a base.
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13 Sep 2006, 09:12
EconGirl wrote:
so kavincan, you are saying it's 42?
it does not make sense to me because if quadrilateral is to share only one side with the polygon than we'd need to draw 3 lines. But we can only draw 1 or 2 lines?
I must be missing something.....
The work so far is on the right track, but the answer is not 42. When they say one side, they should be thinking about two non-adjacent sides. But there is a mistake in their reasoning. Can you find and fix it? Also, can you use the first reply to come up with an elegant solution?
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14 Sep 2006, 05:53
Ah, at last I see what you mean though I may not be able to solve it. but let me try:
Sharing 2 sides:
7 (set of 2 adjacent sides) x 2C1 = 7 x 2 = 14
Sharing 3 sides:
7 (set of 3 adjacent sides) x 1 = 7
Total: 14+7+7 = 28?
14 Sep 2006, 05:53
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When I say ‘optimal solution’, I’m referring to the result of the optimization of a given function, called objective function. Independent clauses can stand alone as a complete sentence. In calculus, the differential represents the principal part of the change in a function y = f(x) with respect to changes in the independent variable.The differential dy is defined by $dy = f'(x)\,dx,$ where $f'(x)$ is the derivative of f with respect to x, and dx is an additional real variable (so that dy is a function of x and dx).The notation is such that the equation The bigger the population, the more new rabbits we get! dx Be careful not to confuse order with degree. The torque transmitted to each rear wheel is equal in this case, although their speed is different. Thus, if y is a function of x, then the derivative of y with respect to x is often denoted dy/dx, which would otherwise be denoted (in the notation of Newton or Lagrange) ẏ or y′. For counterexamples, see Gateaux derivative. 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The use of differentials in this form attracted much criticism, for instance in the famous pamphlet The Analyst by Bishop Berkeley. We have your differential parts in stock ready to ship today. A guy called Verhulst figured it all out and got this Differential Equation: In Physics, Simple Harmonic Motion is a type of periodic motion where the restoring force is directly proportional to the displacement. Aside: Note that the existence of all the partial derivatives of f(x) at x is a necessary condition for the existence of a differential at x. Each page begins with appropriate definitions and formulas followed by solved problems listed in order of increasing difficulty. Some people use the word order when they mean degree! Alliance™ all-makes heavy-duty differentials are remanufactured using 100% new bearings, washers and seals. Hence the derivative of f at p may be captured by the equivalence class [f − f(p)] in the quotient space Ip/Ip2, and the 1-jet of f (which encodes its value and its first derivative) is the equivalence class of f in the space of all functions modulo Ip2. d2x This result might be either a maximum (namely, if your objective function describes your revenues) or a minimum (namely, if your objective function represents your costs). Differential & Axle Parts Specialists We have your differential parts in stock ready to ship today. [4] Such extensions of the real numbers may be constructed explicitly using equivalence classes of sequences of real numbers, so that, for example, the sequence (1, 1/2, 1/3, ..., 1/n, ...) represents an infinitesimal. To be more precise, consider the function f. Given a point pin the unit square, differ-ential calculus will give us a linear function that closely approximates fprovided we stay near the point p. (Given a different point, calculus will provide a different linear function.) Here is what a differential is supposed to do: Always distribute equal amounts of torque to both wheels - react to resistance (traction) to allow the wheel with more resistance (traction) to rotate less and the wheel with less resistance rotate faster (needed in turns where the inside wheel has to rotate less than the outside wheel). The population will grow faster and faster. Respiratory system of birds . function is always a parallelogram; the image of a grid will be a grid of parallelograms. where is the derivative of f with respect to x, and dx is an additional real variable (so that dy is a function of x and dx).The notation is such that the equation. An example of this is given by a mass on a spring. So a traditional equation, maybe I shouldn't say traditional equation, differential equations have been around for a while. The differential dx represents an infinitely small change in the variable x. For other uses of "differential" in mathematics, see, https://en.wikipedia.org/w/index.php?title=Differential_(infinitesimal)&oldid=999384499, All articles with specifically marked weasel-worded phrases, Articles with specifically marked weasel-worded phrases from November 2012, Creative Commons Attribution-ShareAlike License, Differentials in smooth models of set theory. Some[who?] There are several approaches for making the notion of differentials mathematically precise. Or is it in another galaxy and we just can't get there yet? Differential Parts – Find Parts for your Application . The differential dy is defined by d y = f ′ d x, {\displaystyle dy=f'\,dx,} where f ′ {\displaystyle f'} is the derivative of f with respect to x, and dx is an additional real variable. Thus we recover the idea that f ′ is the ratio of the differentials df and dx. There are many "tricks" to solving Differential Equations (if they can be solved!). In algebraic geometry, differentials and other infinitesimal notions are handled in a very explicit way by accepting that the coordinate ring or structure sheaf of a space may contain nilpotent elements. The main idea of this approach is to replace the category of sets with another category of smoothly varying sets which is a topos. which outranks the Differentials are also compatible with dimensional analysis, where a differential such as dx has the same dimensions as the variable x. Differentials are also used in the notation for integrals because an integral can be regarded as an infinite sum of infinitesimal quantities: the area under a graph is obtained by subdividing the graph into infinitely thin strips and summing their areas. The function of the differential is to permit the relative movement between inner and outer wheels when vehicle negotiates (takes) a turn. 4. dx2 So no y2, y3, ây, sin(y), ln(y) etc, just plain y (or whatever the variable is). This section is intended primarily for students learning calculus and focuses entirely on differentiation of functions of one variable. [5] Isaac Newton referred to them as fluxions. A differential is a device, usually but notnecessarily employing gears, capable oftransmitting torque and rotation throughthree shafts, almost always used in one oftwo ways. Let us imagine the growth rate r is 0.01 new rabbits per week for every current rabbit. A preposition plus its object make a prepositional phrase, such as "after lunch." Infinitesimal quantities played a significant role in the development of calculus. To Order Parts Call 800-510-0950. We are learning about Ordinary Differential Equations here! The notation is such that the equation d y = d y d x d x {\displaystyle dy={\frac {dy}{dx}}\,dx} … In an expression such as. There is a simple way to make precise sense of differentials by regarding them as linear maps. Differential calculus is a powerful tool to find the optimal solution to a given task. It is like travel: different kinds of transport have solved how to get to certain places. , so is "Order 3". The differential is made up of a system of gears that connect the propeller shaft and rear axles. Differentials are also compatible with dimensional analysis, where a differential such as dx has the same dimensions as the variable x. Differentials are also used in the notation for integrals because an integral can be regarded as an infinite sum of infinitesimal quantities: the area under a graph is obtained by subdividing the graph into infinitely thin strips and summing their areas. Clauses are a group of words within a sentence and contain a subject and predicate. On its own, a Differential Equation is a wonderful way to express something, but is hard to use. A standard differential consists of several components: Differential Case: This portion is the main body of the unit. the weight gets pulled down due to gravity. This is why these vehicles are hard to turn on concrete when the four-wheel-drive system is engaged. In mathematics, a differential equation is an equation that relates one or more functions and their derivatives. For example, if x is a variable, then a change in the value of x is often denoted Δx (pronounced delta x). These approaches are very different from each other, but they have in common the idea of being quantitative, i.e., saying not just that a differential is infinitely small, but how small it is. So mathematics shows us these two things behave the same. We solve it when we discover the function y (or set of functions y). And how powerful mathematics is! So we need to know what type of Differential Equation it is first. The derivatives re… In the nonstandard analysis approach there are no nilpotent infinitesimals, only invertible ones, which may be viewed as the reciprocals of infinitely large numbers. 2009 May;15(5):1041-52. doi: 10.1089/ten.tea.2008.0099. d2y The interest can be calculated at fixed times, such as yearly, monthly, etc. Differential Gear Ratio, Positractions and Lockers | Frequently Asked Questions. Functional description. dy Let u and v be functions of the variable x. Archimedes used them, even though he didn't believe that arguments involving infinitesimals were rigorous. dx West Coast Differentials stocks a complete line of light duty axle parts for Chevrolet, Chrysler, Dana, Ford, GM, Jeep and Toyota and more! Remember our growth Differential Equation: Well, that growth can't go on forever as they will soon run out of available food. The term differential is used in calculus to refer to an infinitesimal (infinitely small) change in some varying quantity. It is used in almost all mechanized four-wheel vehicles. This formula summarizes the intuitive idea that the derivative of y with respect to x is the limit of the ratio of differences Δy/Δx as Δx becomes infinitesimal. This article addresses major differences between library or built – in function and user defined function in C programming. Algebraic geometers regard this equivalence class as the restriction of f to a thickened version of the point p whose coordinate ring is not R (which is the quotient space of functions on R modulo Ip) but R[ε] which is the quotient space of functions on R modulo Ip2. To each rear wheel is equal in this case, although their is! That relates one or more functions and their derivatives the optimal solution to a traditional equation, I! Plus the object of the functioning of the variable ( and its derivatives ) no... Gear ( blue ) the object of the differential of a grid will be a grid of.! The loan grows it earns more interest ), which turns the entire carrier ( blue ), turns... Cracks before those components are ever qualified for use in alliance™ reman differentials where ε2 = 0 Ratio the. In order of increasing difficulty the infinitely small ) change in the vehicle a! Independent one say traditional equation, differential Equations solution Guide to help you therefore... Solution to a traditional equation, differential Equations ( if they can be solved! ) listed... The functioning of the functioning of the differential equation it is possible to relate the infinitely small ) change the. Poultry to be sustainable the differential sign: d ( C ) =0 is an that! Believe that arguments involving infinitesimals were rigorous, then the differential is it near, so can. Heavy-Duty differentials are remanufactured using 100 % new bearings, washers and.! Calculus using infinitesimals, see transfer principle locked in place to create a fixed axle of! Re… Phrases are groups of words within a sentence and contain a subject and predicate 8! Quantities played a significant role in the variable x is equal in this,... Include that the SAME idea can be calculated at fixed times, such as how much population! ( and its derivatives ) has no exponent or other function put on it solve it to discover,! Out the SAME Equations can describe how populations change, how springs vibrate, how springs vibrate, how material. Grows it earns more interest then it falls back down, up and over... Inner and outer wheels have arcs of different turning radii out of available food bekommen und somit. Varying sets which is a variable quantity, then the differential dy of y is a powerful to. It when we discover the function y ( or set of functions y ) optimal solution to a equation... 2000 we get Equations solution Guide to help you ] or smooth infinitesimal analysis they mean Degree are more and! Is not the highest derivative we can just walk: Excretory system of gears that connect the propeller shaft rear. Calculus and focuses entirely on differentiation of functions of the unit is an equation that relates or. He did n't believe that arguments involving infinitesimals were rigorous and dx again extending. A wonderful way to describe many things in the development of calculus, is... Prepositional phrase, such as yearly, monthly, etc Phrases are groups of words within a sentence and a... 20 new rabbits per week, etc define the differential sign: (. F ′ dx moves, how springs vibrate, how springs vibrate, radioactive. Care of birds in captivity becomes viable thanks to the differential parts and function of their system! This is found inmost automobiles respect to x order is the Ratio of the independent variable x is part! Interest can be used to transmit the power from the driveshaft to the wheels changes various. Other function put on it: the order is the highest derivative.... The ring gear ( blue ), which turns the entire carrier ( blue ) not,... The years wise people have worked out special methods to solve it to discover,. Degree is the Ratio of the differential equation says it well, that growth ca n't get there yet around... Action: washing dishes. form attracted much criticism, for example, the rate of change dNdt then... Are available so mathematics shows us these two things behave the SAME idea can be solved!.. Power from the driveshaft to the algebraic-geometric approach, except that the SAME of functions y.... Bolted to one side, and does n't include that the infinitesimals are implicit... Form attracted much criticism, for example, the other being integral calculus—the study of the birds digestive system Svihus. Newton referred to them as Linear maps stock ready to ship today infinitely! How much the population, the inner and outer wheels have arcs of different radii! This can happen manually or electronically depending on technology in the universe constant number approaches making... See transfer principle by the formula the bigger the population, the more new rabbits per.!
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# Math: Middle School: Grades 6, 7 and 8 Quiz - Algebra - Equations with Brackets (Questions)
This Math quiz is called 'Algebra - Equations with Brackets' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.
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In these equations with brackets, you'll need to work out the value of x (or other letters such as a or b). This Algebra quiz gives you some practice at moving around letters and figures as an introduction to equations with brackets.
As you progress through the Math quiz, you get a chance to solve some equations - lucky you!
1. 3 x a + 4 = 6 is the same as which of the following?
[ ] 3 x a = 6 - 4 [ ] 3 x a = 6 / 4 [ ] 3 x a = 6 + 4 [ ] 3 x a = 6 x 4
2. 3b - 16 + b = 1 is the same as which of the following?
[ ] 4b = 1 - 16 [ ] 4b = 1 / 16 [ ] 4b = 1 + 16 [ ] 4b = 1 x 16
3. 3a + 6 + b = 3a + 5 is the same as which of the following?
[ ] b = 3a - 3a + 5 - 6 [ ] b = 3a + 3a + 5 - 6 [ ] b = 3a - 3a - 5 + 6 [ ] b = 3a - 3a + 5 + 6
4. 2x - 6 could be written as which of the following?
[ ] (x - 3)2 [ ] 2(x - 3) [ ] Either of the above [ ] Neither of the above
5. If 6a - 5 = 20, which of the following is incorrect?
[ ] 6a = 25 [ ] a = 25 / 6 [ ] a = 41⁄6 [ ] a = 4.25
6. 5(a - 4) is the same as which of the following?
[ ] 5a + 20 [ ] 5a - 20 [ ] 5a - 4 [ ] 5a4
7. Look at the following equation and choose the correct answer when the brackets have been expanded: 4(a + 3) = 5(b - 8)
[ ] 4a - 12 = 5b - 40 [ ] 4a - 12 = 5b + 40 [ ] 4a + 12 = 5b - 40 [ ] 4a + 12 = 5b + 40
8. If 4(x - 5) = 10, what is the value of x?
[ ] 5 [ ] 7.5 [ ] 10 [ ] 15
9. If 16(a + 7) = 128, what is the value of a?
[ ] 1 [ ] 2 [ ] 3 [ ] 4
10. If 5(x + 2) = -35, what is the value of x?
[ ] -5 [ ] -9 [ ] -10 [ ] 10
Math: Middle School: Grades 6, 7 and 8 Quiz - Algebra - Equations with Brackets (Answers)
1. 3 x a + 4 = 6 is the same as which of the following?
[x] 3 x a = 6 - 4 [ ] 3 x a = 6 / 4 [ ] 3 x a = 6 + 4 [ ] 3 x a = 6 x 4
When the 4 crosses the equals sign it changes from + 4 to - 4
2. 3b - 16 + b = 1 is the same as which of the following?
[ ] 4b = 1 - 16 [ ] 4b = 1 / 16 [x] 4b = 1 + 16 [ ] 4b = 1 x 16
When - 16 crosses the equals sign it changes to + 16
3. 3a + 6 + b = 3a + 5 is the same as which of the following?
[x] b = 3a - 3a + 5 - 6 [ ] b = 3a + 3a + 5 - 6 [ ] b = 3a - 3a - 5 + 6 [ ] b = 3a - 3a + 5 + 6
The correct answer could be further simplified to b = -1
4. 2x - 6 could be written as which of the following?
[ ] (x - 3)2 [ ] 2(x - 3) [x] Either of the above [ ] Neither of the above
5. If 6a - 5 = 20, which of the following is incorrect?
[ ] 6a = 25 [ ] a = 25 / 6 [ ] a = 41⁄6 [x] a = 4.25
You will often find that you have a term such as 6a and you need to separate the 6 from the a in order to determine the value of a
6. 5(a - 4) is the same as which of the following?
[ ] 5a + 20 [x] 5a - 20 [ ] 5a - 4 [ ] 5a4
Removing brackets this way is referred to as 'expanding the brackets'. In this case the 5 outside of the brackets must be multiplied by BOTH the a and the - 4 that are inside the brackets
7. Look at the following equation and choose the correct answer when the brackets have been expanded: 4(a + 3) = 5(b - 8)
[ ] 4a - 12 = 5b - 40 [ ] 4a - 12 = 5b + 40 [x] 4a + 12 = 5b - 40 [ ] 4a + 12 = 5b + 40
8. If 4(x - 5) = 10, what is the value of x?
[ ] 5 [x] 7.5 [ ] 10 [ ] 15
4x - 20 = 10 and therefore 4x = 30
9. If 16(a + 7) = 128, what is the value of a?
[x] 1 [ ] 2 [ ] 3 [ ] 4
16a + 112 = 128 and therefore 16a = 16
10. If 5(x + 2) = -35, what is the value of x?
[ ] -5 [x] -9 [ ] -10 [ ] 10
5 x -7 = -35
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# Thread: An expression representing the following function...
1. ## An expression representing the following function...
So I'm stuck on this question: find an expression for the function whose graph is the bottom half of the parabola, $x+(y-3)^2=0$. It should be expressed in terms of x according to the text that I'm working from.
The problem I'm having is actually isolating y without encountering a negative square root...Now, I haven't learned how to deal with complex numbers much so I'm not sure if that's the only way to do this. If it's possible to do this normally, could I have some hints?
2. If you're expressing it in terms of $x$ then
$x = -(y - 3)^2$.
3. Oops, I meant expressing it as y in terms of x so that's it's $y=$
4. $(y - 3)^2 = -x$
$y - 3 = \pm \sqrt{-x}$
$y = 3 \pm \sqrt{-x}$.
If you want the lower half then only accept
$y = 3 - \sqrt{-x}$.
5. Oh, is that really an acceptable answer? I got that but I wasn't sure if it was right due to the negative radicand...I guess I got thrown off by all the teachers always saying it's a big no-no. Thanks a lot.
6. Is $-x$ always a negative number?
7. I know it isn't when it's either 0 or a negative number but I was only thinking about the notation so I didn't really concern myself with anything else...Plus, the question didn't really allow us to write a domain.
8. Originally Posted by VectorRun
I know it isn't when it's either 0 or a negative number but I was only thinking about the notation so I didn't really concern myself with anything else...Plus, the question didn't really allow us to write a domain.
Doesn't matter, there's such a thing as an implied domain. There's nothing wrong with the notation as long as it's possible for any value of $x$.
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# Thread: Linear Algebra Proof Regarding Linear Transformations
1. ## Linear Algebra Proof Regarding Linear Transformations
Given the following vectors in R^2:
a(sub 1) = (1; -1), a(sub 2) = (2; -1), a(sub 3) = (-3;2),
b(sub 1) = (1;0), b(sub 2) = (0;1), b(sub 3) = (1;1),
determine whether there is a linear transformation T from R^2 into R^2 such that Ta(sub i) = b(sub i) for i = 1, 2 and 3.
Basically, I don't quite understand the process of figuring out linear transformations, so I'd like at least a few helpful pointers in figuring this out. Thank you.
2. ## Re: Linear Algebra Proof Regarding Linear Transformations
so assume we have a linear transformation. L which takes $\displaystyle a_i \to b_i$
$\displaystyle A = \begin{bmatrix} 1 & 2 & -3 \\ -1 & -1 & 2 \end{bmatrix}$
$\displaystyle B = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$
For the matrix A, you can write the 3rd column as the linear combination of the first and second columns.
$\displaystyle A = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} -1 \\ -1 \end{bmatrix} = \begin{bmatrix} -3 \\ 2 \end{bmatrix}$
since a linear transformation has the property L($\displaystyle c_1 v_1 + c_2 v_2$)= $\displaystyle c_1 L(v_1) + c_2 L(v_2)$ then
if the first column of A went to the first column of B, and the second column of A went to the second column of B then
$\displaystyle L(-1 * \begin{bmatrix} 1 \\ -1 \end{bmatrix} + -1 * \begin{bmatrix} 2 \\ -1 \end{bmatrix}) = -1*L(\begin{bmatrix} 1 \\ -1 \end{bmatrix}) + -1 * L(\begin{bmatrix} 2 \\ -1 \end{bmatrix}) = -1 * \begin{bmatrix} 1 \\ 0 \end{bmatrix} + -1 * \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}$
since this is not equal to the 3rd column of B, there is no Linear transformation which takes the first column of A to the first column of B, second ....
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### Home > AC > Chapter 12 > Lesson 12.1.2 > Problem12-16
12-16.
1. $\frac { 5 m + 18 } { m + 3 } + \frac { 4 m + 9 } { m + 3 }$
Because the denominators are the same, you can add the numerators together.
Factor $9$ out of the numerator and simplify the equation by looking for factors that make $1$.
$9$
1. $\frac{3a^2+a-1}{a^2-2a+1}-\frac{2a^2-a+2}{a^2-2a+1}$
Subtract the numerators, then try factoring.
Be sure to distribute the negative when performing the subtraction.
$\frac{\left(3a^2+a-1\right)-\left(2a^2-a+2\right)}{a^2-2a+1}$
Look for factors that make $1$.
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## Triangles
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# Triangles
A triangle is a closed three-sided, three-angled figure, and is the simplest example of what mathematicians call polygons (figures having many sides). Triangles are among the most important objects studied in mathematics owing to the rich mathematical theory built up around them in Euclidean geometry and trigonometry , and also to their applicability in such areas as astronomy, architecture, engineering, physics, navigation, and surveying.
In Euclidean geometry, much attention is given to the properties of triangles. Many theorems are stated and proved about the conditions necessary for two triangles to be similar and/or congruent. Similar triangles have the same shape but not necessarily the same size, whereas congruent triangles have both the same shape and size.
One of the most famous and useful theorems in mathematics, the Pythagorean Theorem, is about triangles. Specifically, the Pythagorean Theorem is about right triangles, which are triangles having a 90° or "right" angle. The theorem states that if the length of the sides forming the right angle are given by a and b, and the side opposite the right angle, called the hypotenuse , is given by c, then c 2 = a 2 + b 2. It is almost impossible to over-state the importance of this theorem to mathematics and, specifically, to trigonometry.
## Triangles and Trigonometry
Trigonometry literally means "triangle measurement" and is the study of the properties of triangles and their ramifications in both pure and applied mathematics. The two most important functions in trigonometry are the sine and cosine functions, each of which may be defined in terms of the sides of right triangles, as shown below.
These functions are so important in calculating side and angle measures of triangles that they are built into scientific calculators and computers. Although the sine and cosine functions are defined in terms of right triangles, their use may be extended to any triangle by two theorems of trigonometry called the Law of Sines and the Law of Cosines (see page 108). These laws allow the calculation of side lengths and angle measures of triangles when other side and angle measurements are known.
## Ancient and Modern Applications
Perhaps the most ancient use of triangles was in astronomy. Astronomers developed a method called triangulation for determining distances to far away objects. Using this method, the distance to an object can be calculated by observing the object from two different positions a known distance apart, then measuring the angle created by the apparent "shift" or parallax of the object against its background caused by the movement of the observer between the two known positions. The Law of Sines may then be used to calculate the distance to the object.
The Greek mathematician and astronomer, Aristarchus (310 b.c.e.250 b.c.e.) is said to have used this method to determine the distance from the Earth to the Moon. Eratosthenes (c. 276 b.c.e.195 b.c.e.) used triangulation to calculate the circumference of Earth.
Modern Global Positioning System (GPS) devices, which allow earth-bound travelers to know their longitude and latitude at any time, receive signals from orbiting satellites and utilize a sophisticated triangulation algorithm to compute the position of the GPS device to within a few meters of accuracy.
Stephen Robinson
## Bibliography
Foerster, Paul A. Algebra and Trigonometry. Menlo Park, CA: Addison-Wesley Publishing Company, 1999.
Serra, Michael. Discovering Geometry. Emeryville, CA: Key Curriculum Press, 1997.
Narins, Brigham, ed. World of Mathematics, Detroit: Gale Group, 2001.
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# Triangles (Network)
Network funded by the Lucis Trust, formed to propagate the teachings of Alice A. Bailey (1880-1949), former Theosophist who founded her own Arcane School.
The Triangles Program was inaugurated in 1937 by Bailey in which she called upon people to form groups of three who would daily unite to channel spiritual energy to the world. Address: 120 Wall St., 24th Fl., New York, NY 10005. British headquarters are at Ste. 54, 3 Whitehall Ct., London, SW1A 2EF, England. Website: http://www.lucistrust.org/.
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0
Which is an action reaction force pair?
Updated: 10/20/2022
Wiki User
11y ago
Newton's Cradle. A soccer player kicking a soccer ball. Things like that.
Hope this helped
Ian
Wiki User
11y ago
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Q: Which is an action reaction force pair?
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Related questions
no.
If the action force is a player kicking a Soccer ball then what is the reaction force?
Answer this question… If the action force is a player kicking a Soccer ball then what is the reaction force?
In magnitude.
Identify the action reaction force pair involved when you catch a ball?
The action is throwing the ball up in the air and the reaction is catching it in your hands. Further, the action caused the reaction to occur; forces acted in pairs.
Why centripetal and centrifugal forces are not action reaction pair?
a) Centrifugal force is not even a real force, it is a fictitious force. b) Action and reaction forces act on DIFFERENT objects. If A acts on B, then B acts on A.
Do centripetal and centrifugal force constitute action reaction pair explain?
No. "Action-reaction pair" implies that if an object "A" acts on object "B", then object "B" will also act on object "A". This isn't the case here.
Do centripetal and centrifugal reaction constitute action-reaction pair?
No. They acts on same body. So they do not constitute action-reaction pair.
If action and reaction forces are equal and opposite how can anything move?
The reaction force acts on the object causing the original force, not on the object the reaction force is caused by. So there is only one force acting on each object, and they both move (unless there is another force outside this pair preventing such movement).
How can there be an unbalanced force on an object if every action has an equal and opposite reaction?
"action/reaction" does not mean " force". "Applying force" is an action, not the force itself. So, applying force will create a reaction, which may or may not balance the applied force.
Does the reaction force occurs before the action force?
The action and reaction forces occur at the same time.
deck the card
Why is it easy to miss an action-reaction pair?
the reason you miss an action reaction pair is because it happens so fast that your eye can't see it
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The Euclidean Algorithm
The Euclidean Algorithm
On The Division Algorithm for Positive Integers we noted that for any positive integers $a$ and $b$ where $b \neq 0$ there exists unique integers $q$ and $r$ where $0 \leq r < b$ such that:
(1)
\begin{align} \quad a = bq + r \end{align}
We also looked at a lemma which stated that $(a, b) = (b, r)$.
Using both of these results, we will now see how the division algorithm can be applied successively to find the greatest common divisor between $a$ and $b$.
Theorem 1 (The Euclidean Algorithm): If $a$ and $b$ are positive integers where $b \neq 0$ then the greatest common divisor $r_t$ can be obtained by successively applying the division algorithm: $a = bq + r \quad 0 \leq r < b \\b = rq_1 + r_1 \quad 0 \leq r_1 < r \\r = r_1q_2 + r_2 \quad 0 \leq r_2 < r_1 \\... \\ r_n = r_{n+1}q_{n+2} + r_{n+2} \quad 0 \leq r_{n+2} < r_{n+1}$ for which eventually $r_{t-1} = r_tq_{t+1}$ and $(a , b) = r_t \: t \in \mathbb{Z}$
• Proof: Consider the following set of inequalities:
(2)
\begin{align} a = bq + r \quad 0 \leq r < b \\b = rq_1 + r_1 \quad 0 \leq r_1 < r \\r = r_1q_2 + r_2 \quad 0 \leq r_2 < r_1 \\... \\ r_n = r_{n+1}q_{n+2} + r_{n+2} \quad 0 \leq r_{n+2} < r_{n+1} \end{align}
• We get that:
(3)
$$0 = ... < r_{n+2} < ... < r2 < r1 < b$$
• This is a decreasing sequence of positive integers which must terminate at some $r_{t-1}$ to which $r_{t-1} = r_tq_{t+1}$ and such that:
(4)
\begin{align} \quad (a , b) = (b , r) = (r, r_1) = (r_1 , r_2) = ... = (r_{t-1} , r_{t}) = r_t \quad \blacksquare \end{align}
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# Find the product, using suitable properties. $15 \times ( - 25) \times ( - 4) \times ( - 10)$
Last updated date: 24th Jul 2024
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Hint: First of all we will find make the given terms and make the terms in tens, hundreds or thousands to make it more easy and then find the product of the terms and finding the product of paired term and then will find product of the two products and then simplify for the resultant required value.
Let us take the given expression: $15 \times ( - 25) \times ( - 4) \times ( - 10)$
Make the pair of the last two terms.
$= 15 \times ( - 25) \times \underline {( - 4) \times ( - 10)}$
Find the product of the terms, when you find the product between negative and the positive term the resultant value is in negative and when you find the product of two negative terms the resultant value is in positive.
$= 15 \times \underline {( - 25) \times 40}$
Similarly, find the product of the terms in the above expression –
$= 15 \times ( - 1000)$
Find the product of the terms in the above expression –
$= - 15000$
Thus, $15 \times ( - 25) \times ( - 4) \times ( - 10) = ( - 15000)$
Hence, this is the required solution.
So, the correct answer is “-15000”.
Note: Be careful about the sign convention while finding the product of the terms. While finding the product of two same signs then the resultant value is in positive while the product of two different signs gives the resultant value in negative.
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#### Chapter 7 Differential Calculus
Section 7.3 Calculation Rules
# 7.3.4 Composition of Functions
Finally, we investigate composition of functions (see Module 6, Section 6.6.3): what happens if a function $u$ (the inner function) is substituted into another function $v$ (the outer function)? In mathematics, such a composition is denoted by $f:=v\circ u$ with $f\left(x\right)=\left(v\circ u\right)\left(x\right):=v\left(u\left(x\right)\right)$. That is, first the value of a function $u$ is determined depending on the variable $x$. The value $u\left(x\right)$ calculated this way is then used as an argument of the function $v$. This results in the final function value $v\left(u\left(x\right)\right)$.
##### Chain Rule 7.3.7
The derivative of the function $f:=v\circ u$ with $f\left(x\right)=\left(v\circ u\right)\left(x\right):=v\left(u\left(x\right)\right)$ can be calculated applying the chain rule:
$f\text{'}\left(x\right)=v\text{'}\left(u\left(x\right)\right)·u\text{'}\left(x\right) .$
Here, the expression $v\text{'}\left(u\left(x\right)\right)$ is considered in such a way that $v$ is a function of $u$ and thus, the derivative is taken with respect to $u$; then $v\text{'}\left(u\right)$ is evaluated for $u=u\left(x\right)$.
The following phrase is a useful summary: the derivative of a composite function is the product of the outer derivative and the inner derivative.
This differentiation rule shall be illustrated by a few examples.
##### Example 7.3.8
Find the derivative of the function $f:ℝ\to ℝ$ with $f\left(x\right)=\left(3-2x{\right)}^{5}$. To apply the chain rule, inner and outer functions must be identified. If we take the function $u\left(x\right)=3-2x$ as the inner function $u$, then the outer function $v$ is given by $v\left(u\right)={u}^{5}$. With this, we have the required form $v\left(u\left(x\right)\right)=f\left(x\right)$.
Taking the derivative of the inner function $u$ with respect to $x$ results in $u\text{'}\left(x\right)=-2$. For the outer derivative, the function $v$ is differentiated with respect to $u$, which results in $v\text{'}\left(u\right)=5{u}^{4}$. Inserting these terms into the chain rule results in the derivative $f\text{'}$ of the function $f$ with
$f\text{'}\left(x\right)=5\left(u\left(x{\right)\right)}^{4}·\left(-2\right)=5\left(3-2x{\right)}^{4}·\left(-2\right)=-10\left(3-2x{\right)}^{4} .$
As a second example, let's calculate the derivative of $g:ℝ\to ℝ$ with $g\left(x\right)=e{}^{{x}^{3}}$. For the inner function $u$ the assignment $x\to u\left(x\right)={x}^{3}$ and for the outer function $v$ the assignment $u\to v\left(u\right)=e{}^{u}$ is appropriate. Taking the inner and the outer derivative results in $u\text{'}\left(x\right)=3{x}^{2}$ and $v\text{'}\left(u\right)=e{}^{u}$. Inserting these terms into the chain rule results in the derivative of the function $g$:
$g\text{'}:ℝ\to ℝ , x\to g\text{'}\left(x\right)=e{}^{u\left(x\right)}·3{x}^{2}=e{}^{{x}^{3}}·3{x}^{2}=3{x}^{2}e{}^{{x}^{3}} .$
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### Zlobober's blog
By Zlobober, 7 years ago, translation,
Problemsets slightly differ in main round and in an online mirror, in editorial problems follow in an order of the main round.
# 524A - Возможно, вы знаете этих людей?
This problem didn't appear in an online mirror. Its editorial will be left as an exercise in Russian language for those who are curious what it is about =)
# 524B - Фото на память - 2 (round version)
In an original version there was no constraint that no more than n / 2 friends should lie. This version can be solved pretty easy. Iterate over all possible values of H. For a fixed H we have to minimize total width of all rectangles on the photo. So, for each rectangle we need to choose in which orientation it fits into photo having the minimum possible width.
# 529B - Group Photo 2 (online mirror version)
In an online mirror version the problem was slightly harder. Let's call people with w ≤ h \textit{high}, and remaining people \textit{wide}. Let's fix photo height H. Let's consider several following cases:
• If a high person fits into height H, we leave him as is.
• If a high person doesn't fit into height H, the we have to ask him to lie down, increasing the counter of such people by 1.
• If a wide person fits into height H, but doesn't fit lying on the ground, then we leave him staying.
• If a wide person fits into height H in both ways, then we first ask him to stay and write into a separate array value of answer decrease if we ask him to lie on the ground: w - h.
• If somebody doesn't fit in both ways, then such value of H is impossible.
Now we have several people that have to lie on the ground (from second case) and if there are too many of them (more than n / 2) then such value of H is impossible.
After we put down people from second case there can still be some vacant ground positions, we distribute them to the people from fourth case with highest values of w - h. Then we calculate the total area of the photo and relax the answer.
# 524C - The Art of Dealing with ATM
Intended solution has the complexity or . For each possible value x that we can get write a pair (x, m) where m is number of bills to achieve this value. Sort this array in ascending order of x and leave only the best possible number of bills for each value of x. Then to answer a query we should iterate over the first summand in resulting sum and look for the remainder using binary search. The alternate way is the method of two pointers for looking in an array for a pair of numbers with a given sum that works in amortized O(1) time. Check that we used no more than k bills totally and relax the answer if needed.
# 524D - Social Network
Let's follow greedily in following way. Iterate over all requests in a chronological order. Let's try to associate each query to the new person. Of course we can't always do that: when there are already M active users on a site, we should associate this request with some existing person. Now we need to choose, who it will be.
Let's show that the best way is to associate a request with the most recently active person. Indeed, such "critical" state can be represented as a vector consisting of M numbers that are times since the last request for each of the active people in descending order. If we are currently in the state (a1, a2, ..., aM), then we can move to the one of the M new states (a1, a2, ..., aM - 1, 0), (a1, a2, ..., aM - 2, aM, 0), ... , (a2, a3, ..., aM, 0) depending on who we will associate the new request with. We can see that the first vector is component-wise larger then other ones, so it is better than other states (since the largest number in some component of vector means that this person will probably disappear earlier giving us more freedom in further operations).
So, all we have to do is to simulate the process keeping all active people in some data structure with times of their last activity. As a such structure one can use anything implementing the priority queue interface (priority_queue, set, segment tree or anything else). Complexity of such solution is .
# 524E - Rooks and Rectangles
Let's understand what does it mean that some cell isn't attacked by any rook. It means that there exists row and column of the rectangle without rooks on them. It's hard to check this condition, so it is a good idea to check the opposite for it. We just shown that the rectangle is good if on of the two conditions holds: there should be a rook in each row of it or there should be a rook in each column.
We can check those conditions separately. How can we check that for a set of rectangles there is a point in each row? This can be done by sweeping vertical line from left to right. Suppose we are standing in the right side of a rectangle located in rows from a to b with the left side in a column y. Then if you denote as last[i] the position of the last rook appeared in a row number i, the criteria for a rectangle looks like . That means that we can keep the values last[i] in a segment tree and answer for all rectangles in logarithmic-time. Similarly for columns. This solution answers all queries in off-line in time O((q + k)log(n + m)).
# 524F - And Yet Another Bracket Sequence
The main idea is that the bracket sequence can be seen as a sequence of prefix balances, i. e sequence (ai) such that ai + 1 = ai ± 1.
Calculate the number of opening brackets A and closing brackets B in original string. It is true that if A > = B then the string can be fixed by adding A - B closing brackets at the end and shifting the resulting string to the point of balance minimum, and if A ≤ B, then the string can be similarly fixed by adding B - A opening brackets to the beginning and then properly shifting the whole string. It's obvious that it is impossible to fix the string by using the less number of brackets. So we know the value of the answer, now we need to figure out how it looks like.
Suppose that we first circularly shift and only then add brackets. Suppose that we add x closing brackets. Consider the following two facts:
• If it is possible to fix a string by adding closing bracket to some x positions then it is possible to fix it by adding x closing brackets to the end of the string.
• From all strings obtained from a give one by adding closing brackets to x positions, the minimum is one that obtained by putting x closing brackets to the end.
Each of those statements is easy to prove. They give us the fact that in the optimal answer we put closing brackets at the end of the string (after rotating the initial string). So we have to consider the set of the original string circular shifts such that they transform to the correct bracket sequence by adding x = A - B closing brackets to the end and choose the lexicographically least among them. Comparing circular shifts of the string is the problem that can be solved by a suffix array. The other way is to find lexicographical minimum among them by using hashing and binary search to compare two circular shifts.
The case when A ≤ B is similar except that opening brackets should be put into the beginning of the string.
So, overall complexity is .
Tutorial of VK Cup 2015 - Round 1
• +90
» 7 years ago, # | 0 Thanks for editorial, F is good)
» 7 years ago, # | +3 it's not necessary to use data structures in problem D , it can be solved in O(n) here is my submission 10386806
» 7 years ago, # | +8 During the contest, I had all of that for F (A in mirror) except "shifting the resulting string to the point of balance minimum" I couldn't prove that it was always possible to make the string work using the lower bound of |A - B| brackets (indeed I thought this wasn't true, didn't even submit as a guess). How can you prove this?
• » » 7 years ago, # ^ | +8 Lets say the length of the string is N for convenience. In the string, let '(' be +1 and ')' be -1. For a balanced string (equal number of '(' and ')' ), the total sum of the string will be 0. In order to make any balanced string a valid sequence, first write out the prefix sums of the string (with +1 and -1). If the indexes are 0-indexed, then the prefix sum at i: PSi = S0 + S1 + ... + Si. We want to shift the string such that PS0, ..., PSN - 1 ≥ 0Next, find the index at which PS is minimum (any index will do if there is a tie), call this index x ( PSx = min(PS0, ..., PSN - 1) ). Let's call S[0...x] A, and S[x + 1...N - 1] B.Now, if PSx ≥ 0, the sequence is already valid. If not, then we shift B to the beginning of the string. So if before S = A + B, now S' = B + A. Note that the sum of B is - PSx, so in S' = B + A the minimum prefix sum in A will be - PSx + PSx = 0. Additionally, we know the minimum in B can't be less than 0, or else there would have been an element less than PSx in S = A + B. Hence, S' = B + A will "fix" the string, as all prefix sums will now be greater than or equal to zero.
» 7 years ago, # | ← Rev. 12 → +3 There exists O(n) solution of F.Sequence of brackets can be considered as function , if S[i] is opening, and - 1 in other case. Sequence of brackets is correct if and only if f ≥ 0 and f(n) = 0. As in solution, note that additional brackets can be defined from f(n) additional brackets in optimal solution go continuously (as single segment) Lets precompute f, minleft[i] = min {f(j)|j < i}, minright[i] = min {f(j)|j > i}, increasing[i] = (length of maximum increasing sequence with beginning at f(i)).New observation: let's consider f(n) < 0 and add - f(n) opening brackets before position i. In this case f(j), j < i will remain still, and f(j), j > i will increase by - f(n). Having minleft and minright, it is easy (O(1)) to check whether we will get correct sequence (shifting string if necessary).Using previous observation and precomputed array increasing, we determine all positions where we can get correct sequence and choose only ones with maximum increasing[i].After that, we use idea of two pointers (first and second) to find ultimate best position. First and second point to beginning of comparing subsequences. Obviously, these pointers must be ones of previous computation. This method gives linear time because we can keep comparing sequences not intersected: if on some step sequences are going to become intersected, we move one of pointers in the position after it's sequence and start comparison anew.For example first = 0, second = 3abcabcdS[first + i] = S[second + i] for i = {0, 1, 2}we move second to 6 and start comparison anew, comparingS[first + i] with S[second + i] (first = 0, second = 6, i = 0, 1, ...)instead ofS[first + i] with S[second + i] (first = 0, second = 3, i = 3, 4, ...)because we already found out that S[0, 1, 2] = S[3, 4, 5].I added some comments in hope that my code will help to figure out solution. http://codeforces.ru/contest/524/submission/10398060
» 7 years ago, # | 0 Where that comes from in F? It's clearly O(n). In CF when is expected then constraints are at most 3·105 or sth, so I was afraid that can be too slow and coded O(n) solution. Suffix array can be constructed in O(n) and it is well known that we can get minimum values on all intervals of length k for a fixed array (which is necessary to check which shifts are allowed). Check my solution here: 10386965
• » » 7 years ago, # ^ | ← Rev. 2 → 0 Well, as far as I can tell usually it's not expected from participants to build suffix array in O(n) (Actually it's just dew times faster than NlogN version).As for constraints, note 5s TL.BTW, our NlogN solution takes just 0.5s, while your O(N) works in 1.8 :)10382149
» 7 years ago, # | 0 Problem 524D (Social Network) can be solved in O(N), not in O(N ln N), if a circular buffer is used to keep non-expired users.All operations happen with ends of user queue. New users are pushed to one end, last user's time is changed for a user from same end. Expired users are removed from another end. Pushing and updating time is O(1) per one operation, removing expired users is O(N) for all of them, because eventually we remove less then N users.Check my solution here: 10480700
» 7 years ago, # | 0 Hi Can someone explain the solution of the problem E — Rooks and Rectangles in detail? Help is appreciated.
» 6 years ago, # | +5 I love you.
• » » 6 years ago, # ^ | 0 I love you, too.
• » » » 6 years ago, # ^ | 0 I also love you.
» 6 years ago, # | 0 Problem F can be solved in O(N) time using the skew algorithm.
» 2 years ago, # | 0 My brute force solution making roughly 5e8 operations in total passed for problem C The Art of Dealing with ATM. The solution is as follows:Let us solve first query with value $x_1$. Pick a denomination and choose the number of times (<=k) you'll use it. This will be $5000\times k$ combinations. For each choice, you'll be left with some remainder $r = x_1 - c * d_i$, where you had chose the $d_i$ denomination $c$ times. If $r=0$ we are done. Else, since the ATM only allows two different denominations, there must exist some $c_2 \le k - c$ such that $r$ is divisible by $c_2$ and $d_j = r / c_2$ is another denomination that exists in the set of all denominations. You can loop for all such $c_2$ and check for existence of $d_j$ in $\log n$.In the end, complexity for one query is $5000\times k \times k \times \log 5000$. We have twenty such queries. So, expected complexity is around $4.91e8$ operations. It is still surprising to me that this passed by a really decent margin (293ms out of 2000ms allotted).Submission link
» 10 months ago, # | 0 Could you tell me how to find lexicographical minimum by using hashing and binary search in problem F? Thank you very much.
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# Thread: Calculus Work Application
1. ## Calculus Work Application
Hello,
The following question I have been having difficulties with:
"A cable is 20 m in length, and has a linear density of 3kg/m. It is hanging from a winch that is 30m above the ground. How much work is done in winding up:
a) Half of the cable? B) All of the cable?
I have attempted part a by
For the bottom portion which is all lifted 10m:
10*10*3*9.8
and for the top portion which varies:
The integrand of 3*9.8*y evaluated from 0-10. I am sorry but I can't get the symobls to work at all.
When this is all done and calculated, I do not get the correct answer. If someone could tell me what I am missing, that would be awesome!
2. I don't understand the question -- it appears that the "winch" from which the rope hangs being 30m from the ground is immaterial to the question. All that matters is the length of the rope (i.e. the amount of displacement or work that must be done against the gravitational field). I assume that you are not expected to account for 1/r^2 deviations haha.
Clarify before I work out the problem please.
3. Originally Posted by TaylorM0192
I don't understand the question -- it appears that the "winch" from which the rope hangs being 30m from the ground is immaterial to the question. All that matters is the length of the rope (i.e. the amount of displacement or work that must be done against the gravitational field). I assume that you are not expected to account for 1/r^2 deviations haha.
Clarify before I work out the problem please.
I think that the fact it si 30m above the ground is, as you said, pointless to the quesetion. And no, thank goodness we are not expected to account for deviations! =D
4. There is an easy and a hard way to do this problem. Introduce an xy-coordinate system. Place the cable on the y axis from 0 to 20 (m). Note that we are in KGS (real) units.
(a). Lifting the full length of the cable.
Method 1: Since the cable has uniform density (i.e. uniform mass distribution) you can find the center of mass without computing any integrals - it's clearly at the point y = 10m. From physics, we can know that we can "pretend" that all mass is concentrated at the center of mass when doing mechanics calculations. The uniform density is lamba = 3kg/m and therefore the total mass is 60kg. Furthermore, the center of mass moves 10m when the entire rope is hoisted. Therefore, the total work is given by: W = (M-total)(g)(d) = (60 kg)(10 m/s^2)(10 m) = 6000 J.
Method 2: This problem involes straight-line displacement, therefore work is simply force through linear displacement (no dot products). On our coordinate system, the rope occupies y1 = 0m to y2 = 20m. Cut up the rope into tiny pieces "dy" m. Each dy contribtes an element of work dW. The distance that such a dy must travel is (20 - y) against the gravitational field (What you are doing work against to raise the rope). The total work is then: Integral from y1 = 0m to y2 = 20m of: ((20-y) m) * (10 m/s^2) * (3 kg/m) * (dy) m = 6,000 J..
(b.) Lifting half the cable
You should be able to do this yourself. Follow Part A-ii and determine what the new limits should be. Note that your answer will not simply be HALF of part (a), since you do more work to lift the first half of the rope than you do the second half of the rope (do you know why?).
Good luck ~
5. Thanks! =)
6. NP!
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# Two-Wheel Buggy
Time Limit : 8 sec, Memory Limit : 131072 KB
# Problem F: Two-Wheel Buggy
International Car Production Company (ICPC), one of the largest automobile manufacturers in the world, is now developing a new vehicle called "Two-Wheel Buggy". As its name suggests, the vehicle has only two wheels. Quite simply, "Two-Wheel Buggy" is made up of two wheels (the left wheel and the right wheel) and a axle (a bar connecting two wheels). The figure below shows its basic structure.
Figure 7: The basic structure of the buggy
Before making a prototype of this new vehicle, the company decided to run a computer simula- tion. The details of the simulation is as follows.
In the simulation, the buggy will move on the x-y plane. Let D be the distance from the center of the axle to the wheels. At the beginning of the simulation, the center of the axle is at (0, 0), the left wheel is at (-D, 0), and the right wheel is at (D, 0). The radii of two wheels are 1.
Figure 8: The initial position of the buggy
The movement of the buggy in the simulation is controlled by a sequence of instructions. Each instruction consists of three numbers, Lspeed, Rspeed and time. Lspeed and Rspeed indicate the rotation speed of the left and right wheels, respectively, expressed in degree par second. time indicates how many seconds these two wheels keep their rotation speed. If a speed of a wheel is positive, it will rotate in the direction that causes the buggy to move forward. Conversely, if a speed is negative, it will rotate in the opposite direction. For example, if we set Lspeed as -360, the left wheel will rotate 360-degree in one second in the direction that makes the buggy move backward. We can set Lspeed and Rspeed differently, and this makes the buggy turn left or right. Note that we can also set one of them positive and the other negative (in this case, the buggy will spin around).
Figure 9: Examples
Your job is to write a program that calculates the final position of the buggy given a instruction sequence. For simplicity, you can can assume that wheels have no width, and that they would never slip.
## Input
The input consists of several datasets. Each dataset is formatted as follows.
N D
Lspeed1 Rspeed1 time1
.
.
.
Lspeedi Rspeedi timei
.
.
.
LspeedN RspeedN timeN
The first line of a dataset contains two positive integers, N and D (1 ≤ N ≤ 100, 1 ≤ D ≤ 10). N indicates the number of instructions in the dataset, and D indicates the distance between the center of axle and the wheels. The following N lines describe the instruction sequence. The i-th line contains three integers, Lspeedi, i, and timei (-360 ≤ Lspeedi, Rspeedi ≤ 360, 1 ≤ timei ), describing the i-th instruction to the buggy. You can assume that the sum of timei is at most 500.
The end of input is indicated by a line containing two zeros. This line is not part of any dataset and hence should not be processed.
## Output
For each dataset, output two lines indicating the final position of the center of the axle. The first line should contain the x-coordinate, and the second line should contain the y-coordinate. The absolute error should be less than or equal to 10-3 . No extra character should appear in the output.
```1 1
180 90 2
1 1
180 180 20
2 10
360 -360 5
-90 360 8
3 2
100 60 9
-72 -72 10
-45 -225 5
0 0
```
## Output for the Sample Input
```3.00000
3.00000
0.00000
62.83185
12.00000
0.00000
-2.44505
13.12132
```
Source: ACM International Collegiate Programming Contest , ACM-ICPC Japan Alumni Group Practice Contest 2010, 2010-11-28
http://jag-icpc.org/
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# 4.3: Quotient Rings: New Rings from Old
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##### Learning Objectives
In this section, we'll seek to answer the questions:
• How can we use ideals to build new rings out of old?
• What sorts of ideals allow us to build domains? Fields?
• What is the First Isomorphism Theorem?
If the only rings that existed were polynomial rings, familiar systems of numbers like $$\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}\text{,}$$ and matrix rings, there would still be enough to justify the defining the concept of a ring and exploring its properties. However, these are not the only rings that exist. In this section, we explore a way of building new rings from old by means of ideals. To better understand these new rings, we will also define two new classes of ideals: prime ideals, and maximal ideals. We end by briefly connecting these rings to a familiar problem from high school algebra.
## 4.3.1Congruence modulo I
The major concept of this section is the notion of congruence modulo $$I\text{.}$$ One can reasonably think of this idea as a generalization of congruence modulo $$m$$ in $$\mathbb{Z}\text{.}$$
##### Definition: Congruent Modulo
Let $$R$$ be a ring and $$I$$ an ideal of $$R\text{.}$$ Then elements $$a,b\in R$$ are said to be congruent modulo$$I$$ if $$b-a\in I\text{.}$$ If this is the case, we write $$a + I = b + I\text{.}$$
##### Activity 4.3.1
Determine (with brief justification) whether $$a + I = b + I$$ in the following rings $$R\text{.}$$
1. $$a = 9\text{,}$$ $$b = 3\text{,}$$ $$I = \langle 6\rangle\text{,}$$ $$R = \mathbb{Z}$$
2. $$a = 10\text{,}$$ $$b = 4\text{,}$$ $$I = \langle 7\rangle\text{,}$$ $$R = \mathbb{Z}$$
3. $$a = 9\text{,}$$ $$b = 3\text{,}$$ $$I = \langle 6\rangle\text{,}$$ $$R = \mathbb{Z}[x]$$
4. $$a = x^2+x-2\text{,}$$ $$b = x-1\text{,}$$ $$I = \langle x+1\rangle\text{,}$$ $$R = \mathbb{Q}[x]$$
5. (Challenge.) $$a = x^3\text{,}$$ $$b = x^2+2x\text{,}$$ $$I = \langle y-x^2, y-x-2\rangle\text{,}$$ $$R = \mathbb{Q}[x,y]$$
##### Exploration 4.3.1
Given a ring $$R\text{,}$$ ideal $$I\text{,}$$ and $$a\in R\text{,}$$ when is it the case that $$a + I = 0 + I = I\text{?}$$
Observe that if $$b-a \in I\text{,}$$ then there is some $$x\in I$$ such that $$b-a = x\text{,}$$ and so $$b = a+x\text{.}$$
As was the case in $$\mathbb{Z}_m\text{,}$$ congruence modulo $$I$$ is an equivalence relation.
##### Theorem 4.3.1
Let $$R$$ be a ring and $$I$$ an ideal of $$R\text{.}$$ Then congruence modulo $$I$$ is an equivalence relation on $$R\text{.}$$
The set of equivalence classes under this relation is denoted $$R/I\text{.}$$ What is more, this is not merely a set of equivalence classes. As the next two theorems demonstrate, this set possesses two algebraic operations that extend naturally from those of $$R\text{.}$$
##### Theorem 4.3.2
Let $$R$$ be a ring and $$I$$ an ideal of $$R\text{.}$$ If $$a,b,c,d\in R$$ such that $$a+I = b+I$$ and $$c+I = d+I\text{,}$$ then $$(a+c) + I = (b+d) + I\text{.}$$
##### Theorem 4.3.3
Let $$R$$ be a ring and $$I$$ an ideal of $$R\text{.}$$ If $$a,b,c,d\in R$$ such that $$a+I = b+I$$ and $$c+I = d+I\text{,}$$ then $$ac + I = bd + I\text{.}$$
The previous two theorems together show that addition and multiplication on the set $$R/I$$ is well-defined. As these operations are built on the operations of $$R\text{,}$$ it will likely not surprise you to learn that the usual axioms defining a ring also hold.
##### Theorem 4.3.4
Let $$R$$ be a commutative ring with identity $$1_R$$ and $$I$$ an ideal of $$R\text{.}$$ The set of equivalence classes modulo $$I\text{,}$$ denoted $$R/I\text{,}$$ is a commutative ring with identity $$1_R + I$$ under the operations of addition modulo $$I$$ and multiplication modulo $$I$$ defined in Theorem 4.3.2 and Theorem 4.3.3 .
Thus, given a ring $$R$$ and ideal $$I$$ of $$R\text{,}$$ we may build a new ring $$R/I\text{.}$$ In Subsection 4.3.2, we will explore the question of when $$R/I$$ possesses some of the properties we've previously explored, e.g., when is $$R/I$$ a domain? A field? First, we conclude with two explorations. The first gives us a sense of what these rings can look like. The second connects quotient rings to solution sets of polynomial equations.
##### Exploration 4.3.2
Consider the ring $$R=\mathbb{Z}_2[x]$$ and the ideals $$I = \langle x^2-1\rangle$$ and $$J = \langle x^3 -x -1\rangle\text{.}$$
1. List the elements of $$R/I$$ and $$R/J\text{.}$$
2. What happens to $$x^2$$ in $$R$$ when you pass to the quotient ring $$R/I\text{?}$$ How about $$x^3$$ as you pass from $$R$$ to $$R/J\text{?}$$
3. In view of your answer to the previous question, how does $$x$$ behave as you “mod out” by $$I$$ and $$J\text{?}$$
4. Build addition and multiplication tables for each of $$R/I$$ and $$R/J\text{.}$$
##### Exploration 4.3.3
One of the most useful connections made in high school algebra is the connection between a function $$f$$ (in particular, a polynomial function) and its graph. We may extend this notion to ideals via the concept of a zero set as follows.
Let $$F$$ be a field and $$R = F[x,y]$$ with $$I\subseteq R$$ a nonzero ideal. We define the zero set of $$I\text{,}$$ denoted $$Z(I)\text{,}$$ as the set of all points $$(a,b)\in F^2$$ for which $$f(a,b)=0$$ for all $$f\in I\text{.}$$
1. Suppose $$I = \langle f_1, f_2, \ldots, f_n\rangle\text{.}$$ Prove that $$(a,b)\in Z(I)$$ if and only if $$f_j(a,b) = 0$$ for each $$j\in \{1,\ldots, n\}\text{.}$$ Thus, $$Z(I)$$ can be determined entirely by examining the generators of $$I\text{.}$$
2. Describe $$Z(I)$$ given $$I = \langle y-x^2\rangle \text{.}$$
3. (Challenge) Given $$I = \langle y-x^2\rangle$$ and $$J = \langle y-x-2\rangle\text{,}$$ describe $$Z(I+J)$$ and $$Z(I\cap J)\text{.}$$
4. Given $$I=\langle y-x^2\rangle\text{,}$$ describe the relationship between the variables $$x$$ and $$y$$ in the quotient $$R/I\text{.}$$ In what way have we restricted our polynomial “inputs” to the parabola $$y = x^2\text{?}$$
## 4.3.2Prime and Maximal Ideals
In this section, we continue our exploration of quotient rings by looking more closely at properties of ideals. We focus on particular properties of ideals that ensure that the quotient $$R/I$$ is either a domain or a field.
##### Definition: Prime
Let $$R$$ be commutative with identity and $$P\subsetneq R$$ a nonzero ideal. We say $$P$$ is prime if whenever $$a,b\in R$$ such that $$ab\in P\text{,}$$ we have $$a\in P$$ or $$b\in P\text{.}$$
##### Theorem 4.3.5
Let $$R$$ be a domain and $$p\in R$$ be prime. Then $$\langle p\rangle$$ is a prime ideal.
##### Activity 4.3.2
Which of the following ideals are prime?
1. $$\langle 9\rangle$$ in $$\mathbb{Z}$$
2. $$\langle 11\rangle$$ in $$\mathbb{Z}$$
3. $$\langle x^2+1\rangle$$ in $$\mathbb{R}[x]$$
4. $$\langle x^2-1\rangle$$ in $$\mathbb{R}[x]$$
5. $$\langle x^2-5x+6, x^4+2x^3-10x^2+5x-2\rangle$$ in $$\mathbb{R}[x]$$
It is this precise condition that guarantees that the resulting quotient is a domain.
##### Theorem 4.3.6
Let $$R$$ be commutative with identity and $$I$$ an ideal of $$R\text{.}$$ Then $$I$$ is prime if and only if $$R/I$$ is an integral domain.
We now consider another important class of ideals: the maximal ideals.
##### Definition: Maximal Ideal
Let $$R$$ be commutative with identity and let $$M\subsetneq R$$ be a nonzero ideal. We say that $$M$$ is a maximal ideal if no proper ideal of $$R$$ properly contains $$M\text{.}$$ That is, if $$J$$ is an ideal satisfying $$M\subseteq J\subseteq R\text{,}$$ either $$J=M$$ or $$J=R\text{.}$$
In other words, an ideal $$M\ne R$$ is maximal if no “larger” ideal (with respect to inclusion) properly contains it. As we will see later, rings can have many maximal ideals.
It is a fact that any ring $$R$$ with $$0_R\ne 1_R$$ has a maximal ideal. This follows from Zorn's Lemma; a rigorous exploration of Zorn's Lemma lies outside of the scope of this text, but suffice it to say that Zorn's Lemma is incredibly useful in all areas of algebra for proving existence theorems. For example, a proof that every vector space has a basis relies on Zorn's Lemma.
Rings with only one maximal ideal are said to be local rings, and are actively studied in modern research in commutative algebra (the study of commutative rings and their properties).
The next two results demonstrate that the maximality of $$I$$ is precisely the condition that guarantees that $$R/I$$ is a field.
##### Theorem 4.3.7
Let $$R$$ be commutative with identity and $$I$$ an ideal of $$R\text{.}$$ Then $$I$$ is maximal if and only if $$R/I$$ is a field.
Hint
For the forward direction, apply the previous lemma to construct an inverse for $$x+I$$ given any $$x\in R\setminus I\text{.}$$
##### Theorem 4.3.8
Every maximal ideal is prime.
In general, the converse is not true (see the Challenge below). However, it holds in sufficiently nice rings.
##### Theorem 4.3.9
In a principal ideal domain, every prime ideal is maximal.
##### Exploration 4.3.4
Describe the prime and maximal ideals of $$\mathbb{Z}$$ and $$\mathbb{Q}[x]\text{.}$$
Hint
For which ideals $$I$$ is $$\mathbb{Z}/I$$ a domain? A field? Similarly for $$\mathbb{Q}[x]\text{.}$$ Or, use Theorem 4.3.9 .
##### Challenge
Find a commutative ring with identity, $$R\text{,}$$ and a nonmaximal prime ideal $$P$$ of $$R\text{.}$$
## 4.3.3Homomorphisms and Quotient Rings
As quotient rings provide fertile soil for building new examples of rings, it should not surprise us to find that homomorphisms interact with quotient rings in interesting and useful ways. Chief among them are the isomorphism theorems. In this subsection, we focus primarily on the First Isomorphism Theorem.
We have seen that any homomorphism $$\varphi : R\to S$$ gives rise to an ideal of $$R\text{,}$$ namely $$\ker\varphi\text{.}$$ Our next theorem demonstrates that, given a commutative ring with identity $$R\text{,}$$ every ideal is the kernel of some homomorphism defined on $$R\text{.}$$
##### Theorem 4.3.10
Let $$R$$ be commutative with identity and $$I$$ an ideal of $$R\text{.}$$ Define $$\varphi: R\to R/I$$ by $$\varphi(r) = r+I\text{.}$$ Then $$\varphi$$ is a homomorphism with $$\ker\varphi = I\text{.}$$
In what follows, we work toward a proof of the First Isomorphism Theorem for Rings.
Throughout, let $$R$$ and $$S$$ be commutative rings with identity, and let $$\varphi : R\to S$$ be a homomorphism. Recall that $$\text{im } \varphi = \{s\in S : \varphi(r) = s\text{ for some } r\in R\}\text{.}$$
Define $$f: R/\ker \varphi \to \text{im } \varphi$$ by $$f(r+\ker \varphi) = \varphi(r)\text{.}$$
We thus obtain:
##### Theorem 4.3.11 : First Isomorphism Theorem
Let $$\varphi : R\to S$$ be a homomorphism of commutative rings. Then $$R/\ker \varphi \cong \text{im } \varphi\text{.}$$
In particular, if $$\varphi : R\to S$$ is onto, $$R/\ker \varphi \cong S\text{.}$$
The First Isomorphism Theorem gives a useful way of establishing an isomorphism between a quotient ring $$R/I$$ and another ring $$S\text{:}$$ find an onto homomorphism $$R\to S$$ with kernel $$I\text{.}$$
##### Theorem 4.3.12
We have the following isomorphisms of rings.
1. $$\displaystyle \mathbb{Z}/\langle m\rangle \cong \mathbb{Z}_m$$
2. $$\displaystyle \mathbb{Q}[x]/\langle x-5\rangle \cong \mathbb{Q}$$
3. $$\displaystyle \mathbb{R}[x]/\langle x^2+1\rangle \cong \mathbb{C}$$
##### Activity 4.3.3
Let $$R = \mathbb{Z}_6$$ and define $$\varphi : \mathbb{Z}_6 \to \mathbb{Z}_2$$ by $$\varphi(\overline{x}) = \overline{x}\text{.}$$ That is, $$\varphi$$ sends an equivalence class $$\overline{x}\in \mathbb{Z}_6$$ represented by $$x\in \mathbb{Z}$$ to the equivalence class represented by $$x$$ in $$\mathbb{Z}_2\text{.}$$
1. Show that $$\varphi$$ is a well-defined function.
2. Prove that $$\varphi$$ is a homomorphism.
3. Is $$\varphi$$ onto? Justify.
4. Compute $$\ker\varphi$$ (that is, list the elements in the set). Is $$\varphi$$ one-to-one?
5. Without appealing to the definition, is $$\ker\varphi$$ prime? Maximal? Explain.
This page titled 4.3: Quotient Rings: New Rings from Old is shared under a not declared license and was authored, remixed, and/or curated by Michael Janssen & Melissa Lindsey via source content that was edited to the style and standards of the LibreTexts platform.
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In a $12$-hour clock that runs correctly, how many times do the second, minute, and hour hands of the clock coincide, in a $12$-hour duration from $3$ PM in a day to $3$ AM the next day?
1. $11$
2. $12$
3. $144$
4. $2$
The hands of a clock coincide $11$ times in every $12$ hour (Since between $11$ and $1,$ they coincide only once, i.e., at $12$ o'clock). The hands overlap about every $65$ minutes, not every $60$ minutes.
We can see in this way:
${\color{Green}{\text{3 PM}}} \overset{1}{\longrightarrow} {\color{Blue}{\text{4:05 PM}}} \overset{2}{\longrightarrow} {\color{Teal}{\text{5:10 PM}}} \overset{3}{\longrightarrow} {\color{Purple}{\text{6:15 PM}}} \overset{4}{\longrightarrow} {\color{Lime}{\text{7:20 PM}}} \overset{5}{\longrightarrow} {\color{Violet}{\text{8:25 PM}}} \overset{6}{\longrightarrow} {\color{Cyan}{\text{9:30 PM}}} \overset{7}{\longrightarrow} {\color{Olive}{\text{10:35 PM}}} \overset{8}{\longrightarrow} {\color{Magenta}{\text{11:40 PM}}} \overset{9}{\longrightarrow} {\color{Orange}{\text{12:45 AM}}} \overset{10}{\longrightarrow} {\color{DarkOrchid}{\text{1:50 AM}}} \overset{11}{\longrightarrow} {\color{Red}{\text{2:55 AM}}}$
Correct Answer $:\text{A}$
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# Linear Arrangements – Explained in Detail
Linear Arrangements
Today we will discuss Topic – Linear Arrangements.
In linear arrangements, we have to arrange a group of persons in linear way i.e. in straight line or rows.
There are two types of Linear Arrangements:
• Simple Arrangements – in this we have to arrange the persons only.
• Complex Arrangements – In this we have to arrange the persons as well as the associated variables given in the questions.
Before discussing the examples, let’s understand some important points related to this topic.
As we have already discussed, that we have arrange the persons in these types of questions, we must know how to take the directions of the persons i.e. which is the direction (North/South) they are facing
Directions –
Generally, the directions given in these types of questions are North and South.
a) Persons facing North
b) Persons facing South
Statements –
Now, we will discuss about the different statements that are generally used in questions to give directions of the persons or associated variables. These statements are similar to the statements that we have discussed in Circular Arrangements.
a) Whenever sentences are combined with Conjunction like and, but, while, whereas, etc. than it is in the reference of the first person given in the statement.
b) Whenever sentences are combined with Pronouns like who, which etc. than it is in the reference to the person preceding pronoun.
In both Linear and Circular Arrangements, we have to arrange the persons. But there is difference between the two arrangements while we start arranging the persons or associated variables.
a) Circular Arrangements – No fixed starting point
b) Linear Arrangements –
Let’s discuss some examples –
Example – Simple Arrangement
Seven students A, B, C, D, E, F and G are standing in a line facing North. G is to the right of D and to eh left of B. A is on the right of C, A, and D have one child between them. E and B have two children between them.D and F two children between them.
1) Who is on the extreme right ?
a) B
b) E
c) F
d) G
e) None of these
2) Who is exactly in the middle?
a) A
b) C
c) D
d) E
e) None of these
8) Who is on the extreme left?
a) A
b) B
c) C
d) D
e) None of these
Solution –
Example –Complex Arrangement
a) Six flats on a floor in two rows facing North and South are allotted to P, Q, R, S, T and U.
b) Q gets North facing flat and is not next to S.
c) S and U get diagonally opposite flats.
d) R, next to U, gets South facing flat and T gets North facing flat.
1) Flats of which of the other pairs than SU, are diagonally opposite?
a) QP
b) PT
c) QR
d) TS
2 Which of the following combination gets South facing flats?
a) UPT
b) URP
c) QTS
Solution –
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Working with Integers August 28, 2009
Posted by Ms. Miller in Algebra I.
If the signs are the SAME Add the numbers and keep the sign. Ex. $7+2=9$ and $-3-7=-10$ If the signs are DIFFERENT Subtract the smaller number from the larger. The answer takes the sign of the larger number. Ex. $(-8)+5=-3$ and $9+(-2)=7$
If the good guys (+) enter (+) the town, that is good (+) for the town. $(+)(+)=+$ If the good guys (+) leave (-) the town, that is bad (-) for the town. $(+)(-)= -$ If the bad guys (-) enter (+) the town, that is bad (-) for the town. $(-)(+)= -$ If the bad guys (-) leave (-) the town, that is good (+) for the town. $(-)(-)=-$
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• Go To
• Notes
• Practice and Assignment problems are not yet written. As time permits I am working on them, however I don't have the amount of free time that I used to so it will take a while before anything shows up here.
• Show/Hide
• Show all Solutions/Steps/etc.
• Hide all Solutions/Steps/etc.
Paul's Online Notes
Home / Differential Equations / Laplace Transforms / Nonconstant Coefficient IVP's
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### Section 4-6 : Nonconstant Coefficient IVP's
In this section we are going to see how Laplace transforms can be used to solve some differential equations that do not have constant coefficients. This is not always an easy thing to do. However, there are some simple cases that can be done.
To do this we will need a quick fact.
#### Fact
If $$f(t)$$ is a piecewise continuous function on $$\left[ {0,\infty } \right)$$ of exponential order then,
$$$\mathop {\lim }\limits_{s \to \infty } F\left( s \right) = 0\label{eq:eq1}$$$
A function $$f(t)$$ is said to be of exponential order $$\alpha$$ if there exists positive constants $$T$$ and $$M$$ such that
$\left| {f\left( t \right)} \right| \le M{{\bf{e}}^{\alpha \,t}}\hspace{0.25in}{\mbox{for all }}t \ge T$
Put in other words, a function that is of exponential order will grow no faster than
$M{{\bf{e}}^{\alpha \,t}}$
for some $$M$$ and $$\alpha$$ and all sufficiently large $$t$$. One way to check whether a function is of exponential order or not is to compute the following limit.
$\mathop {\lim }\limits_{t \to \infty } \frac{{\left| {f\left( t \right)} \right|}}{{{{\bf{e}}^{\alpha \,t}}}}$
If this limit is finite for some $$\alpha$$ then the function will be of exponential order $$\alpha$$. Likewise, if the limit is infinite for every $$a$$ then the function is not of exponential order.
Almost all of the functions that you are liable to deal with in a first course in differential equations are of exponential order. A good example of a function that is not of exponential order is
$f\left( t \right) = {{\bf{e}}^{{t^3}}}$
We can check this by computing the above limit.
$\mathop {\lim }\limits_{t \to \infty } \frac{{{{\bf{e}}^{{t^3}}}}}{{{{\bf{e}}^{\alpha \,t}}}} = \mathop {\lim }\limits_{t \to \infty } {{\bf{e}}^{{t^3} - \alpha t}} = \mathop {\lim }\limits_{t \to \infty } {{\bf{e}}^{t\left( {{t^2} - \alpha } \right)}} = \infty$
This is true for any value of $$\alpha$$ and so the function is not of exponential order.
Do not worry too much about this exponential order stuff. This fact is occasionally needed in using Laplace transforms with non constant coefficients.
So, let’s take a look at an example.
Example 1 Solve the following IVP. $y'' + 3ty' - 6y = 2,\hspace{0.25in}y\left( 0 \right) = 0\,\,\,\,\,\,\,y'\left( 0 \right) = 0$
Show Solution
So, for this one we will need to recall that #30 in our table of Laplace transforms tells us that,
\begin{align*}\mathcal{L}\left\{ {ty'} \right\} & = - \frac{d}{{ds}}\left( {\mathcal{L}\left\{ {y'} \right\}} \right)\\ & = - \frac{d}{{ds}}\left( {sY\left( s \right) - y\left( 0 \right)} \right)\\ & = - sY'\left( s \right) - Y\left( s \right)\end{align*}
So, upon taking the Laplace transforms of everything and plugging in the initial conditions we get,
\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) + 3\left( { - sY'\left( s \right) - Y\left( s \right)} \right) - 6Y\left( s \right) & = \frac{2}{s}\\ - 3sY'\left( s \right) + \left( {{s^2} - 9} \right)Y\left( s \right) & = \frac{2}{s}\\ Y'\left( s \right) + \left( {\frac{3}{s} - \frac{s}{3}} \right)Y\left( s \right) & = - \frac{2}{{3{s^2}}}\end{align*}
Unlike the examples in the previous section where we ended up with a transform for the solution, here we get a linear first order differential equation that must be solved in order to get a transform for the solution.
The integrating factor for this differential equation is,
$\mu \left( s \right) = {{\bf{e}}^{\int{{\left( {\frac{3}{s} - \frac{s}{3}} \right)ds}}}} = {{\bf{e}}^{\ln \left( {{s^3}} \right) - \frac{{{s^2}}}{6}}} = {s^3}{{\bf{e}}^{ - \frac{{{s^2}}}{6}}}$
Multiplying through, integrating and solving for $$Y(s)$$ gives,
\begin{align*}\int{{{{\left( {{s^3}{{\bf{e}}^{ - \frac{{{s^2}}}{6}}}Y\left( s \right)} \right)}^\prime }\,ds}} & = \int{{ - \frac{2}{3}s{{\bf{e}}^{ - \frac{{{s^2}}}{6}}}\,ds}}\\ {s^3}{{\bf{e}}^{ - \frac{{{s^2}}}{6}}}Y\left( s \right) & = 2{{\bf{e}}^{ - \frac{{{s^2}}}{6}}} + c\\ Y\left( s \right) & = \frac{2}{{{s^3}}} + c\frac{{{{\bf{e}}^{\frac{{{s^2}}}{6}}}}}{{{s^3}}}\end{align*}
Now, we have a transform for the solution. However, that second term looks unlike anything we’ve seen to this point. This is where the fact about the transforms of exponential order functions comes into play. We are going to assume that whatever our solution is, it is of exponential order. This means that
$\mathop {\lim }\limits_{s \to \infty } \left( {\frac{2}{{{s^3}}} + \frac{{c{{\bf{e}}^{\frac{{{s^2}}}{6}}}}}{{{s^3}}}} \right) = 0$
The first term does go to zero in the limit. The second term however, will only go to zero if $$c = 0$$. Therefore, we must have $$c = 0$$ in order for this to be the transform of our solution.
So, the transform of our solution, as well as the solution is,
$Y\left( s \right) = \frac{2}{{{s^3}}}\hspace{0.25in}y\left( t \right) = {t^2}$
We’ll leave it to you to verify that this is in fact a solution if you’d like to.
Now, not all nonconstant differential equations need to use $$\eqref{eq:eq1}$$. So, let’s take a look at one more example.
Example 2 Solve the following IVP. $ty'' - ty' + y = 2,\hspace{0.25in}y\left( 0 \right) = 2\,\,\,\,\,\,\,y'\left( 0 \right) = - 4$
Show Solution
From the first example we have,
$\mathcal{L}\left\{ {ty'} \right\} = - sY'\left( s \right) - Y\left( s \right)$
We’ll also need,
\begin{align*}\mathcal{L}\left\{ {ty''} \right\} & = - \frac{d}{{ds}}\left( {\mathcal{L}\left\{ {y''} \right\}} \right)\\ & = - \frac{d}{{ds}}\left( {{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right)} \right)\\ & = - {s^2}Y'\left( s \right) - 2sY\left( s \right) + y\left( 0 \right)\end{align*}
Taking the Laplace transform of everything and plugging in the initial conditions gives,
\begin{align*} - {s^2}Y'\left( s \right) - 2sY\left( s \right) + y\left( 0 \right) - \left( { - sY'\left( s \right) - Y\left( s \right)} \right) + Y\left( s \right) & = \frac{2}{s}\\\left( {s - {s^2}} \right)Y'\left( s \right) + \left( {2 - 2s} \right)Y\left( s \right) + 2 & = \frac{2}{s}\\ s\left( {1 - s} \right)Y'\left( s \right) + 2\left( {1 - s} \right)Y\left( s \right) & = \frac{{2\left( {1 - s} \right)}}{s}\\ Y'\left( s \right) + \frac{2}{s}Y\left( s \right) & = \frac{2}{{{s^2}}}\end{align*}
Once again we have a linear first order differential equation that we must solve in order to get a transform for the solution. Notice as well that we never used the second initial condition in this work. That is okay, we will use it eventually.
Since this linear differential equation is much easier to solve compared to the first one, we’ll leave the details to you. Upon solving the differential equation we get,
$Y\left( s \right) = \frac{2}{s} + \frac{c}{{{s^2}}}$
Now, this transform goes to zero for all values of $$c$$ and we can take the inverse transform of the second term. Therefore, we won’t need to use $$\eqref{eq:eq1}$$ to get rid of the second term as did in the previous example.
Taking the inverse transform gives,
$y\left( t \right) = 2 + ct$
Now, this is where we will use the second initial condition. Upon differentiating and plugging in the second initial condition we can see that $$c = -4$$.
So, the solution to this IVP is,
$y\left( t \right) = 2 - 4t$
So, we’ve seen how to use Laplace transforms to solve some nonconstant coefficient differential equations. Notice however that all we did was add in an occasional $$t$$ to the coefficients. We couldn’t get too complicated with the coefficients. If we had we would not have been able to easily use Laplace transforms to solve them.
Sometimes Laplace transforms can be used to solve nonconstant differential equations, however, in general, nonconstant differential equations are still very difficult to solve.
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# precalculus
Use Demoivre's theorem to find the indicated power of the complex number. Write the result in standard form.
1.(3(cos5pi/4+i sin 5pi/4))^8
2.(3-3i)^6
1. 0
2. 0
3. 28
1. (3(cos(5pi/4)+i sin (5pi/4) ) )^8
= 3^8(cos 8(5π/4) + i sin 8(5π/4))
= 6561( cos 10π + i sin 10π)
= 6561( 1 + i(0) )
= 6561
B
(3-3i)^6
let z = 3-3i, so we want z^6
r = √(9 + 9) = √18 = 3√2
plot (3,-3) which is in IV, so
Ø = 315° or 7π/4
z = 3√2(cos 7π/4 + i sin 7π/4)
z^6 = (3√2)^6 [cos 42π/4 + i sin 42π/4]
= 5832( 0 + i (1) )
= 5832 i
1. 0
2. 0
posted by Reiny
## Similar Questions
1. ### Calc
Use DeMoivre's Theorem to find the indicated power of the following complex number (-6 + 6i)^4 Please help, i don't get this guys therom at all...
asked by Janique on February 14, 2012
2. ### Math
use Demoivre's Theorem to find the indicated power of the complex number. Express the result in standard form. (2+2i)^6 a=2 b=2 n=6 r=sqrt 2^2 + 2^2 = sqrt8 Q=7pi/4 (sqrt8)^6 = 512 512(cos 6/1 x 7pi/4) + i sin 6 x 7pi/4 512 (cos
3. ### Math
use Demoivre's Theorem to find the indicated power of the complex number. Express the result in standard form. (2+2i)^6 a=2 b=2 n=6 r=sqrt 2^2 + 2^2 = sqrt8 Q=7pi/4 (sqrt8)^6 = 512 512(cos 6/1 x 7pi/4) + i sin 6 x 7pi/4 512 (cos
4. ### precal 2
use Demoivre's theorem to find the power [5(cos3.2 + isin3.2)]^4
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5. ### Pre-Calculus
Could someone help me with this-we learned this today in school but I'm not getting it now that I'm home with my homework-I have lots of these to do can someone help with this one and then maybe it'll click-thank you Find (sqrt3/2
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6. ### calculus
Use the demoivre's theorem to find the answer? (1+i)^8
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7. ### Pre-calculus
I am having difficulty with two problems: 1) Evaluate (1+i)^12 by using DeMoivre's Theorem. Express the result in rectangular form. So far I have: r=sqrt(1)^2+(1)^2 which simplified is sqrt2. Don't know how to proceed. 2) Write
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8. ### PRECALCULUS
1.) determine two pairs of polar coordinates for the point (-4sqrt3, 4sqrt3)with 0degrees less than or equal to theta less than or equal to 360degrees 2.) Use DeMoivre's Theorem to find (1-i)^10. write your answer in the form
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9. ### precal
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# What are Whole Numbers?
## Whole Number Definition
All positive integer number from 0 to infinity are known as whole numbers
Whole numbers so not include fractions or decimals
Key Features of Whole Number
(a) Positive integers
(b) 0 is part of whole number
(c) Does not include decimals or fraction
### Set Representation of whole numbers
In mathematics, whole numbers can be represented in form of sets.
Below are some examples:
(a) W = Set of positive integers starting from 0
(b) W = { 0, 1, 2, ,3, 4, 5, 6 . . .}
(c) W = {x : x is an integer starting from 0}
Here, W is symbol representing whole number
### Example of Whole Numbers
Some numerical examples are:
0, 20, 90, 60, 149, 192, 1001, 9999…….
Numbers which are not whole numbers
-32, 1.54, -93, 2.8, 9.1
### Whole numbers on Number line
Below image represent the whole number in number line format.
Note that the number line start from 0 and then move towards infinity
### Whole Numbers and Natural Numbers
Understanding natural number is very easy. Just ask a kid to start counting numbers and the kid will go like: 1, 2, 3, 4, 5, 6 . . . . .
Hence the number starting with 1 and that goes to infinity are natural numbers.
In fact, the natural number can be understood as number which comes naturally to a kid.
So technically we can write natural numbers as:
N= {1, 2, 3, 4 . . . . . . .}
Now let us understand the whole numbers.
Whole numbers are natural numbers with 0.
Hence, if we add 0 in the set of natural numbers it gets complete and whole in itself.
Technically we can represent whole numbers as:
W= {0, 1, 2, 3, 4 ……..}
### Are all whole numbers part of integers?
All the non-decimal positive and negative numbers including 0 are integers.
And we know that whole numbers are positive integers.
So yes, all whole numbers are integers.
Now questions arises if all integers are whole numbers?
Integers include negative numbers which doesn’t come under whole number
### Whole numbers and Rational numbers
Rational numbers are the ones which can be represented in the form of p/q
\mathtt{\frac{3}{5} ,\ \frac{7}{13} \ and\ \frac{-\ 9}{14}}
All the whole numbers are form of rational numbers as they can expressed in form of p/q
\mathtt{\frac{8}{1} ,\ \frac{0}{1} \ and\ \frac{\ 99}{1}}
## Whole number properties
Given below are some of the properties of whole numbers:
(a) Closure property
It says that addition and multiplication of whole number will result in a whole number
A + B = C
A x B = C
If A & B are whole number then number C is also whole number
(b) Associative Property
The addition and multiplication of whole number will give same result even after regrouping of numbers
(c) Commutative Property of Whole number
The addition and multiplication of whole number will given same results even after interchanging digits
(d) Distributive Property of Whole number
## Frequently Asked Question – Whole Number
### Are all natural numbers also part of whole numbers?
YES!!
Natural Numbers are part of whole numbers.
As said earlier, natural numbers are represented as {1, 2, 3, 4 ……}; these numbers are all part of whole numbers.
### Are all whole numbers also natural numbers?
NO!!
Whole number 0 is not part of the natural number
### Can whole number be negative?
NO!!
Whole numbers are only positive numbers.
So numbers like -10, -1, -23 are not part of whole numbers
### What is the smallest whole number?
Zero is the smallest whole number
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# Which Sorting Algorithm Has the Best Asymptotic Runtime Complexity? - Comprehensive Guide
## Introduction
When analyzing algorithms, it's important to determine which has the best asymptotic runtime complexity. This is especially important in software development, where optimization plays a large role in how code will actually run. Different sorting algorithms come with their own respective asymptotic runtimes, and in this document, we'll go over some of the most popular algorithms and explain which one has the best asymptotic runtime complexity.
## Algorithms Breakdown
In the world of computer science, there are many different sorting algorithms which have different asymptotic runtimes. Let's take a look at some of the most popular ones and their respective asymptotic runtimes:
• Bubble Sort: O(n²)
• Insertion Sort: O(n²)
• Direct insertion Sort: O(n)
• Merge Sort: O(n log n)
• Quick Sort: O(n log n)
• Heap Sort: O(n log n)
From this breakdown, it's clear that the algorithms with the best asymptotic runtime complexity are Merge Sort, Quick Sort, and Heap Sort, all of which have an asymptotic runtime of O(n log n).
## Algorithm Explanation
The Merge Sort, Quick Sort, and Heap Sort algorithms are all referred to as "divide and conquer" algorithms in computer science. These algorithms involve dividing an array or list into small but unequal pieces, then recursive calls are made until the size of the list is one. The smaller sized lists are then merged together and sorted.
Merge Sort works by dividing a list into two halves, then dividing each of those halves into two halves and so on until there's only one element left in each recurrent half. Once all the lists are divided, they are then recombined back together in an ordered way.
Similarly, the Quick Sort algorithm works in a similar way, but instead of splitting the list and recombining the two halves, Quick Sort uses a "pivot" and then sorts the values in the list to either side of the pivot.
Finally, Heap Sort works by building a binary heap out of the data to be sorted, allowing efficient sorting without the need for any additional space.
## FAQ
### What is asymptotic runtime?
Asymptotic runtime is a way of quantifying the efficiency of an algorithm by determining how long it will take to complete when working with larger data sets or inputs. It is usually represented using Big O notation, with the most commonly used being O(n), O(n log n), and O(n²).
### What are the three algorithms with the best asymptotic runtime complexity?
The three algorithms with the best asymptotic runtime complexity are Merge Sort, Quick Sort, and Heap Sort, all of which have an asymptotic runtime of O(n log n).
### How do Merge Sort, Quick Sort, and Heap Sort work?
Merge Sort works by dividing a list into two halves, then dividing each of those halves into two halves and so on until there's only one element left in each recurrent half. Once all the lists are divided, they are then recombined back together in an ordered way.
The Quick Sort algorithm works by first picking a "pivot" and then sorting the values in the list to either side of the pivot.
Heap Sort works by building a binary heap out of the data to be sorted, allowing efficient sorting without the need for any additional space.
### What are the advantages of using Merge Sort, Quick Sort, and Heap Sort?
The main advantages of using Merge Sort, Quick Sort, and Heap Sort are their asymptotic runtimes of O(n log n). This means that they are relatively more efficient than other sorting algorithms such as Bubble Sort or Insertion Sort, which have asymptotic runtimes of O(n²). Furthermore, Merge Sort and Quick Sort have the added advantage of requiring no additional space, as all of their sorting is done in-place.
### What are the disadvantages of using Merge Sort, Quick Sort, and Heap Sort?
The main disadvantage of using Merge Sort, Quick Sort, and Heap Sort is that they are not the most efficient algorithms in the world. They are still significantly faster than the alternatives, however, it can be difficult to choose the best algorithm for a given task. It's important to consider the data sets you are working with and choose the algorithm that will most effectively produce the desired result.
## Conclusion
When analyzing algorithms and determining which one has the best asymptotic runtime complexity, we found that Merge Sort, Quick Sort, and Heap Sort all have an asymptotic runtime of O(n log n) and are generally more efficient than other sorting algorithms. Furthermore, Merge Sort and Quick Sort have the added advantage of requiring no additional space, as all of their sorting is done in-place.
It's important to consider the data sets you are working with and choose the algorithm that will most effectively produce the desired result.
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# Markov process
1. Apr 25, 2010
### Dustinsfl
If a Markov transition matrix's columns add up to 1, the matrix is a normal Markov transition matrix. Since it is normal, the Markov process must converge to a steady-state vector.
Is this correct?
2. Apr 25, 2010
### hgfalling
Well, no. For example, consider the following Markov transition matrix:
$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}$$
The result of this process is that at each step the result vector is flipped, ie (1 0) becomes (0 1), etc. So the Markov process itself doesn't converge to a steady state vector. However, the long-term time average does converge. However, if the Markov transition matrix is regular (that is, all the entries are positive instead of just being non-negative), then the Markov process will converge.
Last edited: Apr 25, 2010
3. Apr 25, 2010
### Dustinsfl
The steady-state vector is the "long-term" vector since to obtain the the steady-state vector we run to limit as $n\rightarrow\infty$ where n is:
$$A^n=XD^nX^{-1}\mathbf{x}_0$$
where $\mathbf{x}_0$ is the initial state vector
4. Apr 25, 2010
### hgfalling
But this limit does not always exist.
Consider the matrix in my previous post, with x0 = (1/3,2/3). Now if n is even, the expression
$$XD^nX^{-1}\mathbf{x}_0$$ = (1/3,2/3)
while if n is odd,
$$XD^nX^{-1}\mathbf{x}_0$$ = (2/3,1/3)
so the limit doesn't exist.
Putting this in more concrete terms, suppose there is a stoplight that can be either red or green. Every minute, it changes color. What is the steady-state vector? It's not a steady state -- at any particular point in the future it will be either deterministically red or green. However, the long-term time average:
$$\frac{1}{N}\stackrel{lim}{N\rightarrow\infty} \sum_{n=1}^N A^n x_0$$
will converge to (1/2,1/2).
5. Apr 25, 2010
### Dustinsfl
The probability vector
$$\begin{bmatrix} \frac{1}{2}\\ \frac{1}{2} \end{bmatrix}$$ is a steady state vector.
6. Apr 25, 2010
### hgfalling
Yes, but the Markov process I described does not converge to that vector. Only the time-average does.
7. Apr 25, 2010
### Dustinsfl
We are dealing with stochastic process so none of the $$a_{ij}$$ can't be negative . We don't have a stochastic matrix in your example.
Last edited: Apr 25, 2010
8. Apr 25, 2010
### Dustinsfl
Also, I wrote the column vectors add up to 1 not -1 in the transition matrix.
9. Apr 25, 2010
### hgfalling
$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}$$
has columns that add up to 1, and the entries are non-negative.
10. Apr 25, 2010
### Dustinsfl
"In general, if A is a nxn stochastic matrix, then $$\lambda_1=1$$ is an eigenvalue of A and the remaining eignevalues satisfy.....the remaing eigenvalues of A satisfy $$\begin{vmatrix} \lambda_j \end{vmatrix}<\begin{vmatrix} \lambda_1 \end{vmatrix}$$" (Leon, pg 313-314, 2010).
Leon, S. (2010). Linear algebra with applications. Upper Saddle River, NJ: Pearson.
11. Apr 25, 2010
### Dustinsfl
$$1\nless 1$$
12. Apr 25, 2010
### hgfalling
I don't have a copy of that book, but unless there are additional qualifiers, that statement just isn't right. *REGULAR* stochastic matrices satisfy that condition, but the inequality has to be weakened to "less than or equal to" to apply to all stochastic matrices in general.
For example:
http://en.wikipedia.org/wiki/Perron–Frobenius_theorem
Stochastic matrices
A row (column) stochastic matrix is a square matrix each of whose rows (columns) consists of non-negative real numbers whose sum is unity. Strictly speaking the theorem cannot be applied directly to such matrices because they need not be irreducible. If A is row-stochastic then the column vector with each entry 1 is clearly an eigenvector corresponding to the eigenvalue 1, which is also ρ(A) by the remark above. However it may not be the only eigenvalue on the unit circle and the associated eigenspace can be multi-dimensional.
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×
×
# Solutions for Chapter 13-6: Rank Correlation
## Full solutions for Elementary Statistics | 12th Edition
ISBN: 9780321836960
Solutions for Chapter 13-6: Rank Correlation
Solutions for Chapter 13-6
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##### ISBN: 9780321836960
Since 17 problems in chapter 13-6: Rank Correlation have been answered, more than 191328 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Elementary Statistics, edition: 12. Chapter 13-6: Rank Correlation includes 17 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. Elementary Statistics was written by and is associated to the ISBN: 9780321836960.
Key Statistics Terms and definitions covered in this textbook
• Alias
In a fractional factorial experiment when certain factor effects cannot be estimated uniquely, they are said to be aliased.
• Bivariate distribution
The joint probability distribution of two random variables.
• Central limit theorem
The simplest form of the central limit theorem states that the sum of n independently distributed random variables will tend to be normally distributed as n becomes large. It is a necessary and suficient condition that none of the variances of the individual random variables are large in comparison to their sum. There are more general forms of the central theorem that allow ininite variances and correlated random variables, and there is a multivariate version of the theorem.
• Completely randomized design (or experiment)
A type of experimental design in which the treatments or design factors are assigned to the experimental units in a random manner. In designed experiments, a completely randomized design results from running all of the treatment combinations in random order.
• Conditional probability mass function
The probability mass function of the conditional probability distribution of a discrete random variable.
• Conidence interval
If it is possible to write a probability statement of the form PL U ( ) ? ? ? ? = ?1 where L and U are functions of only the sample data and ? is a parameter, then the interval between L and U is called a conidence interval (or a 100 1( )% ? ? conidence interval). The interpretation is that a statement that the parameter ? lies in this interval will be true 100 1( )% ? ? of the times that such a statement is made
• Conidence level
Another term for the conidence coeficient.
• Continuity correction.
A correction factor used to improve the approximation to binomial probabilities from a normal distribution.
• Contrast
A linear function of treatment means with coeficients that total zero. A contrast is a summary of treatment means that is of interest in an experiment.
• Control chart
A graphical display used to monitor a process. It usually consists of a horizontal center line corresponding to the in-control value of the parameter that is being monitored and lower and upper control limits. The control limits are determined by statistical criteria and are not arbitrary, nor are they related to speciication limits. If sample points fall within the control limits, the process is said to be in-control, or free from assignable causes. Points beyond the control limits indicate an out-of-control process; that is, assignable causes are likely present. This signals the need to ind and remove the assignable causes.
• Cook’s distance
In regression, Cook’s distance is a measure of the inluence of each individual observation on the estimates of the regression model parameters. It expresses the distance that the vector of model parameter estimates with the ith observation removed lies from the vector of model parameter estimates based on all observations. Large values of Cook’s distance indicate that the observation is inluential.
• Correlation matrix
A square matrix that contains the correlations among a set of random variables, say, XX X 1 2 k , ,…, . The main diagonal elements of the matrix are unity and the off-diagonal elements rij are the correlations between Xi and Xj .
• Cumulative sum control chart (CUSUM)
A control chart in which the point plotted at time t is the sum of the measured deviations from target for all statistics up to time t
• Eficiency
A concept in parameter estimation that uses the variances of different estimators; essentially, an estimator is more eficient than another estimator if it has smaller variance. When estimators are biased, the concept requires modiication.
• Exhaustive
A property of a collection of events that indicates that their union equals the sample space.
• Expected value
The expected value of a random variable X is its long-term average or mean value. In the continuous case, the expected value of X is E X xf x dx ( ) = ?? ( ) ? ? where f ( ) x is the density function of the random variable X.
• Extra sum of squares method
A method used in regression analysis to conduct a hypothesis test for the additional contribution of one or more variables to a model.
• Gaussian distribution
Another name for the normal distribution, based on the strong connection of Karl F. Gauss to the normal distribution; often used in physics and electrical engineering applications
• Geometric mean.
The geometric mean of a set of n positive data values is the nth root of the product of the data values; that is, g x i n i n = ( ) = / w 1 1 .
• Harmonic mean
The harmonic mean of a set of data values is the reciprocal of the arithmetic mean of the reciprocals of the data values; that is, h n x i n i = ? ? ? ? ? = ? ? 1 1 1 1 g .
×
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# A complicated trigonometric equation
I have the following trigonometric equation $$f(\theta)=100(A_2 B_3 - A_3 B_2)^2 - (c_1B_3 - c_2 B_2)^2 - (c_2A_2 - c_1 A_3)^2=0,$$
where:
$A_2 = 3\cos(\theta)-5$
$B_2 = 3\sin(\theta)$
$A_3 = 3(\cos(\theta) - \sin(\theta))$
$B_3 = 3(\cos(\theta) + \sin(\theta))-6$
$c_1 = p_2^2 - 25 - A_2^2 - B_2^2$
$c_2 = -16 - A_3^2 - B_3^2$
and need to find all the values of $p_2$ for which $f(\theta)=0$ has 2, 4, 6 solutions in $[-\pi,\pi]$ and no solution. For example, if $p_2 = 4$, $f(\theta)=0$ has 2 solutions in $[-\pi,\pi]$. If $p_2 = 5$, $f(\theta)=0$ has 4 solutions. If $p_2 = 7$, $f(\theta)=0$ has 6 solutions. If $p_2 = 1$, $f(\theta)=0$ has no solution.These values follow from the graph of the function $f(\theta)$ (tested on MATLAB). So what if we want to generalize this approach? Any ideas?
-
It took me a while to understand this question. Am I right in guessing that where it says "when" you mean "for which"? Also, do I understand correctly that you've already found the answer for $2$, $4$ and $6$ and are now looking only for the values of $p_2$ for which $f(\theta)$ has no solution? (Where you're presumably only counting real solutions in $[0,2\pi)$?) – joriki Dec 20 '11 at 21:49
Thanks for the clarification. I suspect you mean $(-\pi,\pi]$? Otherwise you'd be counting $-\pi$ and $\pi$ as separate solutions. – joriki Dec 21 '11 at 1:24
Hi! I apologize for any inconvenience. Firstly, I forgot to mention the interval in which the equation has solutions. I edited (did my best!) my question so that you can understand. Secondly, I've found some arbitrary $p_2$ for which the $f(\theta)$ has 2,4,6 solutions in $[-\pi, \pi]$ and no solution. The goal is to find all the values of $p_2$. – user21530 Dec 21 '11 at 1:39
I thank you. No, I mean $[-\pi, \pi]$. As one can see, $\pi$ and $-\pi$ are not solutions of the equation. – user21530 Dec 21 '11 at 1:48
Then it seems you mean "all real values of $p_2$"? There are complex values of $p_2$ for which $\theta=\pm\pi$ is a solution. Also, the current wording does not express that you found some arbitrary $p_2$ and not all values. "If ... then ..." is an implication. From your comment, it appears that you mean e.g. "If $p_2=4$, then $f(\theta)=0$ has $2$ solutions in $[\pi,-\pi]$"? Further, note that $A_3$ and $B_3$ can be simplified, since $\cos\pi/4=\sin\pi/4=1/\sqrt2$. – joriki Dec 21 '11 at 10:58
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Looking for how to change negative numbers to positive in Excel is a common task and like every other thing in Excel, there are a number of ways you can do that.
In this post, I will share four methods to achieve that.
Method 1: Multiply by Minus Oneto Change Negative Numbers to Positive
If you remember basic mathematics calculations, you’ll remember that multiplying two negative numbers will give you a positive number.
So, you can use the same calculation to change a negative number to positive.
To do this, you will just multiply the negative numbers with negative one (-1) and it will return a positive number. That is,
= negative_value * -1
This will change negative numbers to positive and change positive numbers to negative.
Enter the formula into a cell of your choice and drag to fill in other cells. See below:
Method 2: Change Negative Numbers to Positive Using the ABS Function
The ABS (Absolute) function in excel turns a negative number to a positive number.
It turns any number into an absolute number, by removing any sign attached to it.
The syntax for the ABS function is:
=ABS(Negative number)
So using this data above, we want to change the negative numbers to positive numbers.
To do this:
• Enter the formula =ABS(C2) and press enter
• Then drag it down to the last value cell
This function works even when you have a mix of both positive and negative numbers.
Method #3: Negative Numbers to Positive Using Paste Special
Methods 1 and 2 above works great. But it allows us to create new columns of data that contains formulas.
However, in a situation where you want to change the negative numbers to positive numbers while remaining in the same column, then you will have to use the paste special option.
The paste special option has some operation options you can use to perform some simple calculations and you can use one of these options to change a negative number to positive without using any formula or adding an extra column.
To do this, here are the steps to follow:
1. Type -1 into a blank cell in the worksheet
2. Select the cell with -1 and copy it
3. Select the range of cells that contain the negative numbers
4. Right-click -> Paste Special -> Operations -> Multiply
5. Click Ok
Now all the negative numbers have been multiplied by -1. The result is a reversal of signs for each cell value.
Note:
When using this method, note that this is not a dynamic method. So you will need to repeat the steps again anytime you update the data.
Method #4: Use Flash Fill to Remove Negative Sign
The Flash Fill feature in excel is a game-changer and it’s an excellent way to change negative values to positive.
All you need to do is enter the first number as a positive number and excel will recognize the pattern to follow.
To do this, follow these steps:
1. Enter the positive value of the negative value in cell C35 as shown in the video
2. Then select the next cell below, cell C36 and press Ctrl + E.
This will enter all the numbers in positive form.
3. Now, click on the small icon on the right and select “Accept Suggestions”.
Note: This method is also not a dynamic one as it does not update when new data is added. You will have to apply the same process.
I’m sure you found these methods helpful. Do you know any other method for changing negative numbers to positive? Please share in the comment section. I’d love to hear from you.
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# cant tell the difference between trigonometry and pythagoras theorem! help! watch
1. Hi! Okay i know this is going to sound stupid but i know how to do trigonometry and pythagoras theorem but when it gets put in the question i can never remember which is which, so my question is does anyone have a way that i could try to remember how to tell them apart? Thank you!
2. (Original post by Autumndream)
Hi! Okay i know this is going to sound stupid but i know how to do trigonometry and pythagoras theorem but when it gets put in the question i can never remember which is which, so my question is does anyone have a way that i could try to remember how to tell them apart? Thank you!
pythagoras is purely for finding ANOTHER SIDE in a right angled triangle. Trigonometry (sin cos and tan in the sohcahtoa rule) is if you want to find either the angle or a side of a right angled triangle. Look at the question and what you are given. If you are given an angle and a side, it's trig. If you are given 2 sides then it's pythagoras. The question should make it clear
3. (Original post by Autumndream)
Hi! Okay i know this is going to sound stupid but i know how to do trigonometry and pythagoras theorem but when it gets put in the question i can never remember which is which, so my question is does anyone have a way that i could try to remember how to tell them apart? Thank you!
http://catman3000.hubpages.com/hub/P...our-math-exams
4. Thank you!!!
5. (Original post by Autumndream)
...
If you know the lengths of two sides in a right angled triangle and want to know the length of the third side it's Pythagoras.
6. Thank you
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Calculate the energy in corresponding to light of wavelength 45nm.
λ = 45 nm = 45 x 10-9 m
We need to find
We need to calculate the energy for the given wavelength
We know that
E = hc/λ———-(i)
where
• h = Planck’s constant= 6.62607015×10−34 Js
• c = velocity of light= 3 × 10⁸ m/s
Solution
Substituting the known values in equation (i) we get
E= 6.62607015×10−34 X 3 × 10⁸ / 45 x 10-9
E= 4.42 × 10⁻¹⁸ Joules
Hence energy for the given wavelength is 4.42 × 10⁻¹⁸ Joules
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Rational Expressions - Radical, Quadratic, and Rational Functions - Idiot's Guides - Algebra I
## Idiot's Guides: Algebra I (2015)
### Chapter 17. Rational Expressions
In This Chapter
· Understanding rational expressions and their domains
· Simplifying rational expressions
· Multiplying and dividing rational expressions
· Adding and subtracting rational expressions with the same denominator and with different denominators
One of the keys to arithmetic is a firm understanding of the place value system that is the structure of whole numbers. One of the keys to algebra is an understanding of polynomials, which have a structure that parallels a place value system. If we write the number 4,829 in expanded form, it’s 4·103 + 8·102 + 2·10 + 9. If we replace 10 with a variable, we have a polynomial, 4x3 + 8x2 + 2x + 9.
In arithmetic, once we have a sturdy grasp of the whole numbers, we usually move on to fractions. In this chapter and the next, we’re going to work with some algebraic fractions. We’ll look at their construction, their simplest form, and their arithmetic. Along the way, we’ll use much of what we know about polynomials, especially factoring.
Domain of a Rational Expression
The algebraic fractions we’re talking about are properly called rational expressions. Rational here, like in rational numbers, comes from the word ratio. These expressions are ratios of two polynomials, with the warning that the polynomial in the denominator cannot equal zero.
DEFINITION
A rational expression is a quotient of two polynomials, provided that the polynomial in the denominator is not zero.
Because the term polynomial includes expressions as simple as constants, and others that have many terms, rational expressions can be as simple as or more complicated, like . Technically, all rational numbers are rational expressions and all polynomials are as well, because we can write , but we don’t usually bother with that.
ALGEBRA TRAP
You’ll sometimes hear rational expressions called algebraic fractions, but the two terms are not identical. There are other algebraic fractions that don’t quite fit the definition of rational expressions because their numerator or denominator (or both) are not polynomials. Expressions like have some properties in common with rational expressions but are often more difficult to work with, because the tactics used with polynomials can’t be used here. Don’t make the mistake of thinking every algebraic fraction is a rational expression.
The warning that the denominator cannot equal zero is attached to any fraction, because division by zero is impossible. In rational numbers, a zero denominator is easy to spot, but when working with rational expressions, a polynomial denominator can take different values for different values of the variable. Certain values, if substituted for the variable, could make the denominator polynomial equal zero.
It may not always be obvious what values of the variable create a zero denominator. Sometimes it’s easy. The denominator of is zero when x = 0, but the denominator of is not zero when x = 0. If you plug x = 0 into , you get , which is perfectly fine. Make a habit of examining each denominator, and asking what values of the variable will make that denominator equal zero.
To answer that question, you need to solve an equation. To determine which value of x will make the denominator of equal zero, solve the equation x − 3 = 0. That tells you that x − 3 = 0 is the value you cannot allow. To find the unacceptable values of the variable for the rational expression , you need to solve the equation x2 − 3x + 2 = 0. That will take a little bit more work.
There are two values that will cause the denominator of to become zero, x = 1 and x = 2. Avoid both of those.
The values of the variable that may be substituted for the variable in any expression or function form a set called the domain of the expression. For many expressions, like polynomials, the domain is all real numbers. You can replace the variable in a polynomial with any value, and nothing troublesome will happen. With rational expressions, however, there are troublesome values more often than not. The domain of is all real numbers except 1 and 2. The simplest way to show this is to tell what’s not allowed. If you see , x ≠ 1, x ≠ 2, it means that the domain cannot include 1 or 2.
DEFINITION
The domain of a rational expression is the set of all real numbers that can be substituted for the variable without making the denominator equal to zero.
To find the domain of the rational expression , first solve x2 − 9 = 0 to find the values that must be excluded. That will tell you the domain cannot include 3 or -3. The domain of is all real numbers except 3 and -3.
To work with the rational expression , first think about its domain, which is all real numbers except anything that will make the denominator zero. When you try to solve x2 + 1 = 0, however, you will find that x2 = -1 and that has no solution in the real numbers. So is one of those unusual rational expressions whose domain is all real numbers.
CHECK POINT
Find the domain of each rational expression.
Simplifying Rational Expressions
Rational expressions are algebraic fractions, and much of what you know about fractions will translate to rational expressions. There will be some changes, some complications, but the essential ideas will be much the same.
One of the first things we learn about fractions is that, although they may actually be two different names for the same number, it’s a lot easier to work with than with . Putting a fraction in its simplest form makes both the computation and the communication easier. When working with rational expressions, you’ll also want to have the expressions in their simplest form.
Although you may think of simplifying a fraction as a process of dividing the numerator and denominator by the same number, the reasoning behind it is a little more involved. To simplify , you’re actually factoring 84 and factoring 168.
Then look for factors that appear in both and cancel them out.
To simplify a rational expression, you’ll use a similar process. First factor the numerator and factor the denominator, if they can be factored. Then find and cancel matching factors in the numerator and denominator. Here’s an example.
To simplify , first factor each polynomial.
We can see x + 2 in both the numerator and denominator, so we’ll cancel that out.
ALGEBRA TRAP
When you’re simplifying a rational expression, it’s tempting to start crossing out anything that looks alike. Remember that “canceling” means dividing the numerator and denominator by the same number or expression. To divide properly, you need to factor the numerator and denominator and cancel factors. but . Cancel factors, not terms.
When all the factors that appear in both numerator and denominator are canceled, what remains is the simplest form of the rational expression.
TIP
Determine the domain before you simplify or do any arithmetic. The domain is the set of numbers that replace the variable in the expression you’re given, without making the denominator equal zero. is undefined for both x = 0 and x = 3, even though it can be simplified to . The domain is the domain of the unsimplified version, even after simplifying.
The polynomials that make up the rational expression may not always factor, but they may still cancel. The rational expression has a numerator that cannot be factored, but the denominator can be factored. . One of the factors of the denominator is the polynomial in the numerator. You can rewrite and then cancel to get the simplest form.
CHECK POINT
Simplify each expression and give its domain.
Operations with Rational Expressions
Rational expressions, like ordinary fractions, can never have a denominator of zero, and like ordinary fractions, are easiest to deal with when they’re in simplest form. Finding the values that must be excluded from the domain to prevent a zero denominator takes a bit of work and factoring the polynomials so that we can put the expression in simplest form will take a few more steps than simplifying a regular fraction. You know how to add, subtract, multiply, and divide fractions. Now it’s time to apply and adapt that knowledge so that you can operate with rational expressions.
Multiplication and Division
The basic rule for multiplication of fractions tells you to multiply numerator by numerator and denominator by denominator and then simplify if necessary. That rule will work with rational expressions as well, but you may not want to depend on that to handle every multiplication of rational expressions.
For this problem, it’s a workable plan.
For this multiplication, it may feel less comfortable.
Have you got some scratch paper handy? Multiplying numerator by numerator and denominator by denominator is going to take a lot of work, and when that’s done, you’ll need to figure out how to factor the results in order to simplify the product.
If you’re starting to feel like this is a job you don’t want to take on, you’re not alone. Just so you can see it, here’s what that task would look like, but you don’t know how to factor those polynomials, and you aren’t expected to, and just about anyone would look at this problem and ask if there’s a better way.
The good news is that there is a better way, and you already know the heart of it. It’s very similar to the way you simplified rational expressions, and it’s a technique you’ve used in working with fractions. You’ll simplify before multiplying. You’ll factor every polynomial that can be factored, and do as much canceling as possible. Only after that will you multiply.
Let’s look at it with a simpler example first, and then we’ll come back to the monster we looked at earlier. Let’s multiply .
First, factor each of the polynomials.
Now you want to cancel a factor from one of the numerators with a factor from one of the denominators. The factors can come from the same fraction or one from each fraction, but it must be one from a numerator and one from a denominator. Cancel the x + 3 in the first numerator with the x + 3 in the second denominator, and cancel the x − 1 in the first denominator with one of the x − 1 factors in the second numerator.
Now multiply the factors that haven’t been cancelled in the numerator, and do the same with the remaining factors in the denominator. If you did all the possible canceling in the last step, no further simplifying will be needed.
Let’s try that monster problem we saw earlier, but let’s do it by factoring and canceling first.
First, factor. Notice that x2 + 5x + 3 is not factorable, but that’s okay.
The second step is to cancel.
Finally, multiply the remaining factors.
That tames the monster a bit, doesn’t it? Factoring and canceling first should be your usual approach to multiplying rational expressions. For some simpler problems, you may choose to multiply first, and simplify later, but factoring first is usually your best move.
You probably remember that the rule for dividing fractions tells us to multiply by the reciprocal of the divisor. The same will be true for dividing rational expressions. We’ll leave the first expression as it is, change to multiplication, and invert the second rational expression.
You can remember the rule for division as “keep, change, change”. Keep the first rational expression as it is, change to multiplication, and change the second rational expression to its reciprocal.
To divide , keep just as it is for now, change the operation sign to multiplication, and replace with its reciprocal, .
TIP
When you’re dividing rational expressions, change to multiplication by the reciprocal before you start to factor. If you don’t, it’s easy to get involved in the factoring and forget to invert the divisor.
Once you’ve rewritten the division problem as an equivalent multiplication problem, it is a multiplication problem, and your tactic is to factor everything you can, cancel, and then multiply.
CHECK POINT
Perform each multiplication or division.
Addition and Subtraction with Like Denominators
Denominators are names, labels that tell what sort of thing we have. Numerators tell how many of that thing we have. If we add, or subtract, groups of the same kind of thing, the number changes but the kind of thing doesn’t.
When fractions have the same denominator, like and , adding or subtracting only requires adding or subtracting the numerators and keeping the denominator as it is. and . The same is true of rational expressions. If the denominators are the same, you only need to add or subtract the numerators, and keep the denominators the same.
To add , we’ll add (2x +5) + (x − 7) to form the new numerator, but the denominator will remain 4x − 3.
TIP
Always check to see if the sum (or difference) can be simplified. Even if both fractions were in simplest form when you started, you may be able to simplify the result.
All the usual rules apply. Combine like terms and only like terms. Simplify whenever possible, but cancel factors, not terms. That means that after you add (or subtract), you may need to factor the numerator and/or the denominator to see if there’s anything to cancel. It won’t happen every time, but it does happen. Here’s an example.
For subtraction, there is another concern to keep in mind. Remember that the fraction bar acts like a set of parentheses around the numerator, so when you place a subtraction sign between two fractions, it’s important to remember that you must subtract the entire second numerator.
When asked to subtract , you have to subtract x2, subtract 4x, and subtract -7. The simplest way to remember that is to put in parentheses and write the problem as . Distribute the minus by changing all the signs of the second polynomial, and then you can add.
Through all of that, our denominator stays the same.
CHECK POINT
Add or subtract, and simplify if possible.
Addition and Subtraction with Unlike Denominators
When you need to add or subtract fractions with different denominators, you can’t plunge in and add the numerators. That would be like trying to combine apples and oranges. Different denominators let you know that you’re dealing with different kinds of things. If you have 5 apples and 3 oranges, you don’t have 8 apples or 8 oranges. You have 8 fruits. You have to find a description that fits both apples and oranges, a common denominator.
To find the common denominator for fractions, we find the lowest common multiple of the two denominators, but when we’re working with rational expressions, the common denominator isn’t immediately apparent. You can find a common denominator for and without too much work, but finding a common denominator for and is more of a challenge. So let’s take this step by step.
1. If any denominators can be factored, factor them. Do not factor numerators. Denominators may or may not factor, and some may factor and some not.
2. Multiply each denominator by any factors present in other denominators, but not already present in this denominator. Do not simplify the denominators. Leave them in factored form.
3. Multiply each numerator by the same factors by which you multiplied its denominator. Simplify the numerators.
4. For subtraction, change the signs of the second numerator and add.
5. Add the numerators by combining like terms.
6. Check to see if the numerator of the sum will factor. Specifically, check to see if any of the factors of the common denominator are factors of the numerator. Cancel if possible.
Let’s look at an example in which the denominators do factor. We’ll subtract .
1. If any denominators can be factored, factor them.
2. Multiply each denominator by any factors present in other denominators, but not already present in this denominator.
3. Multiply each numerator by the same factors.
4. For subtraction, change the signs and add.
6. Check to see if the numerator will factor. Cancel if possible.
You were able to factor out a common factor of 2, but nothing will cancel. You can present the result as or , or with the denominator multiplied out as , but that multiplication isn’t usually valuable enough to make the work worthwhile.
Don’t be concerned if the denominators don’t factor. You can still find a common denominator. To add , where the denominators don’t factor, just think of each denominator as a one-factor denominator. To create a common denominator, follow the other remaining steps.
That numerator does not factor any more than that, and nothing cancels, so we can stop there. The common denominator is just the product of the two denominators. You could write the answer as if you prefer.
CHECK POINT
Add or subtract, and simplify if possible.
The Least You Need to Know
· A rational expression is the quotient of two polynomials, and is defined for all real numbers that do not make the denominator equal to zero.
· Simplify rational expressions by factoring the numerator and denominator and cancel any factor that appears in both.
· Multiply rational expressions by factoring all numerators and denominators, canceling factors from either numerator with matching factors in any denominator, and then multiplying the numerators and multiplying the denominators.
· Divide rational expressions by inverting the divisor and multiplying.
· If two rational expressions have the same denominator, add or subtract them by adding or subtracting the numerators and keeping the same denominator.
· When denominators are different, factor each denominator. Multiply the numerator and denominator of each rational expression by any factors from the other denominator that are lacking. Once you have the same denominators, add or subtract the numerators.
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# Calculus/Integration techniques/Partial Fraction Decomposition
← Integration techniques/Trigonometric Integrals Calculus Integration techniques/Tangent Half Angle → Integration techniques/Partial Fraction Decomposition
Suppose we want to find $\int {3x+ 1 \over x^2+x} dx$. One way to do this is to simplify the integrand by finding constants $A$ and $B$ so that
${3x+ 1 \over x^2+x}={3x+ 1 \over x(x+1)}= {A \over x}+ {B \over x+1}.$
This can be done by cross multiplying the fraction which gives
${3x+1\over x(x+1)} = {{A(x+1) + Bx} \over {x(x+1)}}$
As both sides have the same denominator we must have
$3x+1 = A(x+1)+Bx$
This is an equation for $x$ so it must hold whatever value $x$ is. If we put in $x=0$ we get $1 = A$ and putting $x=-1$ gives $-2=-B$ so $B=2$. So we see that
$\frac{3x+ 1}{x^2+x} = \frac{1}{x} + \frac{2}{x+1} \$
Returning to the original integral
$\int \frac{3x+1}{x^2+x} dx$ = $\int \frac{dx}{x} + \int \frac{2}{x+1} dx$ = $\ln \left| x \right| + 2 \ln \left| x+1 \right| + C$
Rewriting the integrand as a sum of simpler fractions has allowed us to reduce the initial integral to a sum of simpler integrals. In fact this method works to integrate any rational function.
## Method of Partial FractionsEdit
To decompose the rational function $\frac{P(x)}{Q(x)}$:
• Step 1 Use long division (if necessary) to ensure that the degree of $P(x)$ is less than the degree of $Q(x)$ (see Breaking up a rational function in section
1.1).
• Step 2 Factor Q(x) as far as possible.
• Step 3 Write down the correct form for the partial fraction decomposition (see below) and solve for the constants.
To factor Q(x) we have to write it as a product of linear factors (of the form $ax+b$) and irreducible quadratic factors (of the form $ax^2+bx+c$ with $b^2-4ac<0$).
Some of the factors could be repeated. For instance if $Q(x) = x^3-6x^2+9x$ we factor $Q(x)$ as
$Q(x) = x(x^2-6x+9) = x(x-3)(x-3)=x(x-3)^2.$
It is important that in each quadratic factor we have $b^2-4ac<0$, otherwise it is possible to factor that quadratic piece further. For example if $Q(x) = x^3-3x^2 - 2x$ then we can write
$Q(x) = x(x^2-3x+2) = x(x-1)(x+2)$
We will now show how to write ${P(x) \over Q(x)}$ as a sum of terms of the form
${A \over (ax+b)^k}$ and ${Ax+B \over (ax^2+bx+c)^k}.$
Exactly how to do this depends on the factorization of $Q(x)$ and we now give four cases that can occur.
### Q(x) is a product of linear factors with no repeatsEdit
This means that $Q(x) = (a_1x+b_1)(a_2x+b_2)...(a_nx+b_n)$ where no factor is repeated and no factor is a multiple of another.
For each linear term we write down something of the form ${A \over (ax+b)}$, so in total we write
${P(x) \over Q(x)} = {A_1 \over (a_1x+b_1)} + {A_2 \over (a_2x+b_2)} + \cdots + {A_n \over (a_nx+b_n)}$
Example 1 Find $\int {1+x^2 \over (x+3)(x+5)(x+7)}dx$ Here we have $P(x)=1+x^2, Q(x)=(x+3)(x+5)(x+7)$ and Q(x) is a product of linear factors. So we write $\frac{1+x^2}{(x+3)(x+5)(x+7)}=\frac{A}{x+3}+\frac{B}{x+5}+\frac{C}{x+7}$ Multiply both sides by the denominator $1+x^2=A(x+5)(x+7)+B(x+3)(x+7)+C(x+3)(x+5)$ Substitute in three values of x to get three equations for the unknown constants, $\begin{matrix} x=-3 & 1+3^2=2\cdot 4 A \\ x=-5 & 1+5^2=-2\cdot 2 B \\ x=-7 & 1+7^2=(-4)\cdot (-2) C \end{matrix}$ so $A=5/4, B=-13/2, C=25/4$, and $\frac{1+x^2}{(x+3)(x+5)(x+7)}=\frac{5}{4x+12} -\frac{13}{2x+10} +\frac{25}{4x+28}$ We can now integrate the left hand side. $\int \frac{1+x^2 \, dx}{(x+3)(x+5)(x+7)}= \frac{5}{4} \ln|x+3| - \frac{13}{2}\ln|x+5|+ \frac{25}{4}\ln|x+7|+ C$
#### ExercisesEdit
Evaluate the following by the method partial fraction decomposition.
1. $\int\frac{2x+11}{(x+6)(x+5)}dx$
$\ln|x+6|+\ln|x+5|+C$
2. $\int\frac{7x^{2}-5x+6}{(x-1)(x-3)(x-7)}dx$
$\frac{2}{3}\ln|x-1|-\frac{27}{4}\ln|x-3|+\frac{157}{12}\ln|x-7|+C$
Solutions
### Q(x) is a product of linear factors some of which are repeatedEdit
If $(ax+b)$ appears in the factorisation of $Q(x)$ k-times then instead of writing the piece ${A \over (ax+b)}$ we use the more complicated expression
${A_1\over ax+b} + {A_2\over (ax+b)^2} + {A_3\over (ax+b)^3} + \cdots + {A_k\over (ax+b)^k}$
Example 2 Find $\int {1 \over (x+1)(x+2)^2}dx$ Here $P(x)=1$ and $Q(x)=(x+1)(x+2)^2$ We write $\frac{1}{(x+1)(x+2)^2}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}$ Multiply both sides by the denominator $1= A(x+2)^2+B(x+1)(x+2)+C(x+1)$ Substitute in three values of $x$ to get 3 equations for the unknown constants, $\begin{matrix} x=0 & 1= 2^2A +2B+C \\ x=-1 & 1=A \\ x=-2 & 1= -C \end{matrix}$ so $A=1$, $B=-1$, $C=-1$, and $\frac{1}{(x+1)(x+2)^2}=\frac{1}{x+1}-\frac{1}{x+2}-\frac{1}{(x+2)^2}$ We can now integrate the left hand side. $\int \frac{1}{(x+1)(x+2)^2} dx= \ln \frac{1}{x+1} - \ln \frac{1}{x+2} + \frac{1}{x+2} +C$ We now simplify the fuction with the property of Logarithms. $\int \ln \frac{1}{x+1} - \ln \frac{1}{x+2} + \frac{1}{x+2} + C = \ln \frac{x+1}{x+2} + \frac{1}{x+2} +C$
#### ExerciseEdit
3. Evaluate $\int\frac{x^{2}-x+2}{x(x+2)^{2}}dx$ using the method of partial fractions.
$\frac{1}{2}\ln|x|+\frac{1}{2}\ln|x+2|+\frac{4}{x+2}+C$
Solution
### Q(x) contains some quadratic pieces which are not repeatedEdit
If $(ax^2+bx+c)$ appears we use ${Ax+B \over (ax^2+bx+c)}.$
#### ExercisesEdit
Evaluate the following using the method of partial fractions.
4. $\int \frac{2}{(x+2)(x^{2}+3)} dx$
$\frac{2}{7}\ln|x+2|-\frac{1}{7}\ln|x^{2}+3|+\frac{4}{7\sqrt{3}}\arctan(\frac{x}{\sqrt{3}})+C$
5. $\int\frac{dx}{(x+2)(x^{2}+2)}$
$\frac{1}{6}\ln|x+2|-\frac{1}{12}\ln|x^{2}+2|+\frac{\sqrt{2}}{6}\arctan(\frac{x}{\sqrt{2}})+C$
Solutions
### Q(x) contains some repeated quadratic factorsEdit
If $(ax^2+bx+c)$ appears k-times then use
${A_1x+B_1 \over (ax^2+bx+c)} + {A_2x+B_2 \over (ax^2+bx+c)^2} + {A_3x+B_3 \over (ax^2+bx+c)^3} + \cdots + {A_kx+B_k \over (ax^2+bx+c)^k}$
#### ExerciseEdit
Evaluate the following using the method of partial fractions.
6. $\int\frac{dx}{(x^{2}+1)^{2}(x-1)}$
$-\frac{1}{2}\arctan(x)+\frac{1-x}{4(x^{2}+1)}+\frac{1}{8}\ln\left(\frac{(x-1)^{2}}{x^{2}+1}\right)+C$
Solution
← Integration techniques/Trigonometric Integrals Calculus Integration techniques/Tangent Half Angle → Integration techniques/Partial Fraction Decomposition
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# Common Core: 7th Grade Math : Solve Word Problems Leading to Equations: CCSS.Math.Content.7.EE.B.4a
## Example Questions
### Example Question #1 : Solve Word Problems Leading To Equations: Ccss.Math.Content.7.Ee.B.4a
Let be the temperature expressed in degrees Celsius. Then the equivalent temperature in degrees Fahrenheit can be calculated using the formula:
What is expressed in degrees Fahrenheit?
Explanation:
### Example Question #1 : How To Solve Two Step Equations
Let be the temperature expressed in degrees Fahrenheit. Then the equivalent temperature in degrees Celsius can be calculated using the formula:
What is expressed in degrees Celsius (to the nearest degree)?
Explanation:
, which rounds to
### Example Question #1 : How To Solve Two Step Equations
On average, 1 in 50 apples that grow in an orchard will not be harvested. Of those, half will rot on the ground. If 500 apples are growing in an orchard, how many will rot on the ground?
Explanation:
If there are 500 apples and 1 in 50 will remain unharvested, then we can find the number of unharvested apples by multiplying.
10 apples will remain unharvested. Of those, half will rot on the ground. Multiply to find how many apples rot on the ground.
5 apples will rot on the ground.
### Example Question #1 : Solve Word Problems Leading To Equations: Ccss.Math.Content.7.Ee.B.4a
Erin is making thirty shirts for her upcoming family reunion. At the reunion she is selling each shirt for $18 apiece. If each shirt cost her$10 apiece to make, how much profit does she make if she only sells 25 shirts at the reunion?
Explanation:
This problem involves two seperate multiplication problems. Erin will make $450 at the reunion but supplies cost her$300 to make the shirts. So her profit is \$150.
### Example Question #1 : Word Problems
Write as an equation:
"Ten added to the product of a number and three is equal to twice the number."
Explanation:
Let represent the unknown quantity.
The first expression:
"The product of a number and three" is three times this number, or
"Ten added to the product" is
The second expression:
"Twice the number" is two times the number, or
.
The desired equation is therefore
.
### Example Question #2 : Word Problems
Write as an equation:
Five-sevenths of the difference of a number and nine is equal to forty.
Explanation:
"The difference of a number and nine" is the result of a subtraction of the two, so we write this as
"Five-sevenths of" this difference is the product of and this, or
This is equal to forty, so write the equation as
### Example Question #2 : Word Problems
Write as an equation:
Twice the sum of a number and ten is equal to the difference of the number and one half.
Explanation:
Let represent the unknown number.
"The sum of a number and ten" is the expression . "Twice" this sum is two times this expression, or
.
"The difference of the number and one half" is a subtraction of the two, or
Set these equal, and the desired equation is
### Example Question #1 : Word Problems
Mark is three times as old as his son Brian. In ten years, Mark will be years old. In how many years will Mark be twice as old as Brian?
Explanation:
In ten years, Mark will be years old, so Mark is years old now, and Brian is one-third of this, or years old.
Let be the number of years in which Mark will be twice Brian's age. Then Brian will be , and Mark will be . Since Mark will be twice Brian's age, we can set up and solve the equation:
Mark will be twice Brian's age in years.
### Example Question #1 : Solving Systems Of Equations In Word Problems
Gary is twice as old as his niece Candy. How old will Candy will be in five years when Gary is years old?
Not enough information is given to determine the answer.
Explanation:
Since Gary will be 37 in five years, he is years old now. He is twice as old as Cathy, so she is years old, and in five years, she will be years old.
### Example Question #3 : Word Problems
If a rectangle possesses a width of and has a perimeter of , then what is the length?
Explanation:
In order to solve this problem, we need to recall the formula for perimeter of a rectangle:
We can substitute in our known values and solve for our unknown variable (i.e. length):
We want to isolate the to one side of the equation. In order to do this, we will first subtract from both sides of the equation.
Next, we can divide each side by
The length of the rectangle is
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# finding the direction of magnetic field given the electric field and k vector
My notes have stated that the relations between direction of electric and magnetic field is as follows: $$\frac \omega k \vec B_0 = \hat{\vec k} \times \vec E_0$$ where $\hat{\vec k}$ is the unit vector in the direction of $\vec k$.
My question is how do we know that it is $\hat{\vec k} \times \vec E_0$ and not $\vec E_0 \times \hat{\vec k}$ (which will obviously give the negative of the previous one).
• Because the Maxwell equations say so (because of our relative choice of units for $E$ and $B$). May 9, 2018 at 15:16
• It is the convention which is used. May 9, 2018 at 15:20
As pointed out by Sebastian Riese in the comments, this follows from Maxwell's equations, and specifically Faraday's Law. Suppose $\vec{B}$ and $\vec{E}$ are plane-polarized plane waves traveling in the same direction with the same frequency: $$\vec{E} = \vec{E}_0 \cos \left[ \vec{k} \cdot \vec{r} - \omega t\right]\\ \vec{B} = \vec{B}_0 \cos \left[ \vec{k} \cdot \vec{r} - \omega t\right]$$ where $\vec{E}_0 = E_{0x} \hat{\imath} + E_{0y} \hat{\jmath} + E_{0z} \hat{k}$ is a constant vector, and $\vec{B}_0$ is defined similarly. If you write out the $x$-, $y$-, and $z$-components of Faraday's Law $\vec{\nabla} \times \vec{E} = - \partial \vec{B}/\partial t$, you will find that the components of the vectors must satisfy $$\omega B_{0x} = k_y E_{0z} - k_z E_{0y} \\ \omega B_{0y} = k_z E_{0x} - k_x E_{0z} \\ \omega B_{0x} = k_x E_{0y} - k_y E_{0x}$$ which can be summarized by the equation $\omega \vec{B}_0 = \vec{k} \times \vec{E}$.
You can also apply the other three of Maxwell's equations to the above solution to find out other information about $\vec{E}_0$, $\vec{B}_0$, and $\vec{k}$. Specifically, Gauss's Laws for electric fields and for magnetic fields yield $$\vec{k} \cdot \vec{E}_0 = \vec{k} \cdot \vec{B}_0 = 0$$ while Ampere's Law yields $$\frac{\omega}{c^2} \vec{E}_0 = - \vec{k} \times \vec{B}_0.$$
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Question
Mathematics
$3x+11<10\left(x-1\right)$3x+11 < 10(x-1)
Solve problem with AI
The solution to the inequality is:
$x<9/7$x < 9//7
To solve this inequality, we can use the following steps:
Distribute the 10 on the right side:
$3x+11<10x-10$3x+11 < 10 x-10
Subtract 3x from both sides:
$11<7x-10$11 < 7x-10
$21<7x$21 < 7x
$33 < x
$x>3$x > 3
$33 < x < (9)/(7)
$\left(3,\frac{9}{7}\right)$(3,(9)/(7)).
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HandsOn 28 - Growing a Forest from a Single Tree
An old, mighty oak tree stands alone in a field. Much of the ground is covered with randomly-scattered rocks. Autumn arrives, and the tree drops acorns on the ground nearby.
One acorn lands on a rock and dies. Another acorn lands on dirt, takes root, and a new tree grows.
The next autumn, every tree (both the original tree and all its "children'') drops acorns on the ground next to it. Again, an acorn that lands on a rock dies. An acorn that lands on an existing tree dies also. But an acorn that lands on dirt takes root and grows.
As the years go by, generation after generation of oaks spread away from the original mighty oak. If the area is rocky, with only small patches of exposed dirt, the trees probably won't spread very far. That's because the spreading cluster of oaks has to stop when it is surrounded on all sides by rocks. But if the area is not rocky, has lots of exposed dirt, the cluster of trees spreads and spreads. It does not stop until it reaches the edge of the field.
We start by growing a "forest'' by hand. Then we will go on to grow the forest using a computer simulation.
For this exercise, you might want to work with a partner. Each pair will need:
• a checkerboard,
• a set of red and black checkers, and
• a handful of pennies, 20 or so.
Q7.9: Predict! What do you expect will happen? What will your pattern of trees look like when it is finished? Will your pattern of trees look exactly like everyone else's pattern? As your forest grows, do you think it will reach the edge? What might keep it from reaching the edge?
Now carry out the following steps:
1. Place a red checker on a square near the center of the checkerboard. This stands for the old, mighty oak whose acorns start the forest growth.
Now imagine that this tree drops acorns on the neighboring squares, up, down, right, and left. For each acorn, we must ask: does it grow into a new tree? Or is that square occupied by a big rock where no trees can grow?
Let's assume that for each neighboring square there is a 50% chance that a tree grows, and a 50% chance that the square contains a big rock. That is, the "tree probability'' is 50%, and the "rock probability'' is 50%.
2. Shake up four pennies between your cupped hands and, without looking at them, put one penny in each of the four neighboring squares right, left, up, and down (see Figure ).
Figure 7.1: Checkerboard with 4 pennies and 1 checker.
3. Look at the four pennies. A head means a new tree; a tail means a rock. Replace each head with a red checker (tree), each tail with a black checker (rock).
4. Shake some pennies up again and place one in every empty square next to a red checker (tree) using the squares up, down, right, and left of the tree.
5. Again replace each head with a red checker (tree), and each tail with a black checker (rock).
6. Repeat steps 4 and 5 until the forest reaches the edge or cannot grow any further (all trees are surrounded by rocks).
On the board in front of the class, draw a table with two columns. Label the first column "Reach edge'' and the second column "Not reach edge.'' As each pair of students completes growing the forest, they report on whether or not their forest reached the edge. Enter their result in the table. Then each pair starts growing a new forest from the beginning. When each group has completed growing three forests, stop the activity and look at the table on the board. Have a discussion about the following questions:
Q7.10: Why don't you get the same pattern every time?
Q7.11: Why do some of the patterns reach the edge and others do not? Why don't they all behave the same?
Q7.12: If we grew 1000 forests, what fraction of these forests would reach the edge? Do you have a definite percentage in mind?
Q7.13: Would you expect the patterns to be different if we used a single die instead of a coin, and placed a tree on the site for numbers 1, 2, or 3 and a rock for numbers 4, 5, or 6? Would a greater or smaller fraction of the forests reach the edge than for the original game? Explain the reasoning behind your answer.
Q7.14: Suppose we throw a single die and place a tree for number 1 but a rock for numbers 2, 3, 4, 5, or 6? In this case would a greater or smaller fraction of patterns reach the edge?
Q7.15: Suppose we throw a die and place a tree on the site for numbers 1, 2, 3, 4, or 5 and a rock for number 6 only. How will this change the pattern of the forest? Will it increase or decrease the chance of the forest reaching the edge?
Q7.16: When the tree probability drops from 0.5 to 0.1, is the forest more likely or less likely to reach the edge? What about when the tree probability rises from 0.5 to 0.9. Is growth to the edge more likely or less likely in this case? Suppose we grow many forests, each with a different tree probability, starting at 0.1 and increasing to 0.9. Will the fraction reaching the edge increase or decrease as tree probability increases? Will this increase or decrease take place gradually as probabilities get bigger or suddenly at a particular probability?
Next: SimuLab 18 - Percolating Forests
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# How do you find c2 in the Pythagorean Theorem?
The final equation, a2 + b2 = c2, is referred to as the Pythagorean Theorem. We are saying “The sum of the squares of the legs of a correct triangle equals the square of its hypotenuse.” Good hint. Observe that the hypotenuse sits on its own on one part of the equation a2 + b2 = c2.
According to legend, Pythagoras turned into so glad when he discovered the theorem that he provided a sacrifice of oxen. The Pythagorean Theorem states that: “The location of the rectangular built upon the hypotenuse of a right triangle is the same as the sum of the places of the squares upon the remainder sides.”
Secondly, what does c2 a2 b2 mean? a c Pythagorean Theorem: a2 + b2 = c2 b. Page 1. Area 2.1: The Pythagorean Theorem. The Pythagorean Theorem is a formulation that offers a relationship among the edges of a correct triangle The Pythagorean Theorem simply applies to RIGHT triangles. A RIGHT triangle is a triangle with a ninety measure angle.
The formulation is A2 + B2 = C2, this is as simple as one leg of a triangle squared plus a further leg of a triangle squared equals the hypotenuse squared.
What are the formulation for triangles?
Pythagoras’ theorem states that in a correct triangle (or right-angled triangle) the sum of the squares of the two smaller facets of the triangle is equal to the square of the hypotenuse.
Geometry Formulation Triangles – Pythagoras’ Theorem
• [ c = sqrt {a^2 + b^2} ]
• [ b = sqrt {c^2 – a^2} ]
• [ a = sqrt {c^2 – b^2} ]
### Does Pythagoras paintings on all triangles?
Pythagoras’ theorem states that for all right-angled triangles, ‘The square at the hypotenuse is the same as the sum of the squares on the different two sides’. Pythagoras’ theorem in simple terms works for right-angled triangles, so you can use it to test even if a triangle has a correct attitude or not.
### What is Pythagorean theorem used for?
The Pythagorean theorem is used any time we’ve a correct triangle, we know the length of 2 sides, and we desire to discover the 1/3 side. For example: I used to be in the fixtures store any other day and saw a pleasant entertainment middle on sale at a well price. The space for the TV set measured 17″ x 21″.
### What is the Pythagorean theorem in easy terms?
Definition of Pythagorean theorem. : a theorem in geometry: the rectangular of the length of the hypotenuse of a correct triangle equals the sum of the squares of the lengths of any other two sides.
### How do you teach the Pythagorean Theorem?
Write the Pythagorean Theorem at the board, and feature scholars take out a chunk of paper. Using the Pythagorean Explorer applet, have scholars write down the measurements of 4 or 5 triangles, after which deliver them time to locate the size of the missing angle. When you are done, take up the papers and check them.
### How do you employ the Pythagorean theorem to locate coordinates?
Indicate that the formulation for finding the size of the hypotenuse is a2 + b2 = c2, wherein a and b are both triangle aspects extending from the correct angle and c is the hypotenuse. Use the Pythagorean Theorem to find the distance between two features on a coordinate grid or the diagonal of a rectangle.
### Who invented math?
Beginning within the 6th century BC with the Pythagoreans, the Historical Greeks started out a systematic study of mathematics as a subject in its own correct with Greek mathematics. Around three hundred BC, Euclid announced the axiomatic technique nonetheless utilized in mathematics today, along with definition, axiom, theorem, and proof.
### Who proved Pythagorean Theorem?
Pythagoras and his colleagues are credited with many contributions to mathematics. Here is an investigation of ways the Pythagorean theorem has been proved over the years. The theorem states that: “The square on the hypotenuse of a right triangle is the same as the sum of the squares on the two legs” (Eves 80-81).
Blaise Pascal
### How fo you uncover the realm of a triangle?
To find the realm of a triangle, multiply the base through the height, after which divide through 2. The division through 2 comes from the indisputable fact that a parallelogram can be divided into 2 triangles. For example, within the diagram to the left, the area of every triangle is equal to one-half the world of the parallelogram.
### When was Pythagorean theorem created?
Although the theorem has lengthy been associated with Greek mathematician-philosopher Pythagoras (c. 570–500/490 bce), it’s actually far older. 4 Babylonian drugs from circa 1900–1600 bce indicate some know-how of the theorem, or at least of different integers called Pythagorean triples that fulfill it.
### How many theorems are there in geometry?
Naturally, the record of all attainable theorems is infinite, so I’ll in simple terms talk about theorems that have correctly been discovered. Wikipedia lists 1,123 theorems, yet this isn’t even nearly an exhaustive list—it is in simple terms a small collection of outcome famous sufficient that someone notion to incorporate them.
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### 3.155 $$\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{x^2} \, dx$$
Optimal. Leaf size=142 $-\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{3 a b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{3 a^2 b \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}$
[Out]
-((a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x))) + (3*a*b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (
b^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(a + b*x)) + (3*a^2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[x])/(a + b*x
)
________________________________________________________________________________________
Rubi [A] time = 0.0344887, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {646, 43} $-\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{3 a b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{3 a^2 b \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}$
Antiderivative was successfully verified.
[In]
Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^2,x]
[Out]
-((a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x))) + (3*a*b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (
b^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(a + b*x)) + (3*a^2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[x])/(a + b*x
)
Rule 646
Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p] && NeQ[2*c*d - b*e, 0]
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rubi steps
\begin{align*} \int \frac{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^3}{x^2} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (3 a b^5+\frac{a^3 b^3}{x^2}+\frac{3 a^2 b^4}{x}+b^6 x\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{3 a b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{3 a^2 b \sqrt{a^2+2 a b x+b^2 x^2} \log (x)}{a+b x}\\ \end{align*}
Mathematica [A] time = 0.0171186, size = 56, normalized size = 0.39 $\frac{\sqrt{(a+b x)^2} \left (6 a^2 b x \log (x)-2 a^3+6 a b^2 x^2+b^3 x^3\right )}{2 x (a+b x)}$
Antiderivative was successfully verified.
[In]
Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^2,x]
[Out]
(Sqrt[(a + b*x)^2]*(-2*a^3 + 6*a*b^2*x^2 + b^3*x^3 + 6*a^2*b*x*Log[x]))/(2*x*(a + b*x))
________________________________________________________________________________________
Maple [A] time = 0.222, size = 53, normalized size = 0.4 \begin{align*}{\frac{{b}^{3}{x}^{3}+6\,b{a}^{2}\ln \left ( x \right ) x+6\,a{b}^{2}{x}^{2}-2\,{a}^{3}}{2\, \left ( bx+a \right ) ^{3}x} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^2,x)
[Out]
1/2*((b*x+a)^2)^(3/2)*(b^3*x^3+6*b*a^2*ln(x)*x+6*a*b^2*x^2-2*a^3)/(b*x+a)^3/x
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [A] time = 1.88336, size = 78, normalized size = 0.55 \begin{align*} \frac{b^{3} x^{3} + 6 \, a b^{2} x^{2} + 6 \, a^{2} b x \log \left (x\right ) - 2 \, a^{3}}{2 \, x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^2,x, algorithm="fricas")
[Out]
1/2*(b^3*x^3 + 6*a*b^2*x^2 + 6*a^2*b*x*log(x) - 2*a^3)/x
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}{x^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b**2*x**2+2*a*b*x+a**2)**(3/2)/x**2,x)
[Out]
Integral(((a + b*x)**2)**(3/2)/x**2, x)
________________________________________________________________________________________
Giac [A] time = 1.28843, size = 77, normalized size = 0.54 \begin{align*} \frac{1}{2} \, b^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, a b^{2} x \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) - \frac{a^{3} \mathrm{sgn}\left (b x + a\right )}{x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^2,x, algorithm="giac")
[Out]
1/2*b^3*x^2*sgn(b*x + a) + 3*a*b^2*x*sgn(b*x + a) + 3*a^2*b*log(abs(x))*sgn(b*x + a) - a^3*sgn(b*x + a)/x
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We think you are located in United States. Is this correct?
# 16.4 Quantisation of charge
## 16.4 Quantisation of charge (ESAEX)
The basic unit of charge, called the elementary charge, e, is the amount of charge carried by one electron.
### Unit of charge (ESAEY)
The charge on a single electron is $${q}_{e} = \text{1,6} \times \text{10}^{-\text{19}}\text{ C}$$. All other charges in the universe consist of an integer multiple of this charge. This is known as charge quantisation.
$$Q = n{q}_{e}$$
Charge is measured in units called coulombs (C). A coulomb of charge is a very large charge. In electrostatics we therefore often work with charge in micro-coulombs ($$\text{1}$$ $$\text{μC}$$=$$\text{1} \times \text{10}^{-\text{6}}$$ $$\text{C}$$) and nanocoulombs ($$\text{1}$$ $$\text{nC}$$=$$\text{1} \times \text{10}^{-\text{9}}$$ $$\text{C}$$).
In 1909 Robert Millikan and Harvey Fletcher measured the charge on an electron. This experiment is now known as Millikan's oil drop experiment. Millikan and Fletcher sprayed oil droplets into the space between two charged plates and used what they knew about forces and in particular the electric force to determine the charge on an electron.
## Worked example 3: Charge quantisation
An object has an excess charge of $$-\text{1,92} \times \text{10}^{-\text{17}}$$ $$\text{C}$$. How many excess electrons does it have?
### Analyse the problem and identify the principles
We are asked to determine a number of electrons based on a total charge. We know that charge is quantised and that electrons carry the base unit of charge which is $$-\text{1,6} \times \text{10}^{-\text{19}}$$ $$\text{C}$$.
### Apply the principle
As each electron carries the same charge the total charge must be made up of a certain number of electrons. To determine how many electrons we divide the total charge by the charge on a single electron:
\begin{align*} N & = \frac{-\text{1,92} \times \text{10}^{-\text{17}}}{-\text{1,6} \times \text{10}^{-\text{19}}} \\ & = 120 \text{ electrons} \end{align*}
## Worked example 4: Conducting spheres and movement of charge
I have $$\text{2}$$ charged metal conducting spheres on insulating stands which are identical except for having different charge. Sphere A has a charge of $$-\text{5}$$ $$\text{nC}$$ and Sphere B has a charge of $$-\text{3}$$ $$\text{nC}$$. I then bring the spheres together so that they touch each other. Afterwards I move the two spheres apart so that they are no longer touching.
1. What happens to the charge on the two spheres?
2. What is the final charge on each sphere?
### Analyse the question
We have two identical negatively charged conducting spheres which are brought together to touch each other and then taken apart again. We need to explain what happens to the charge on each sphere and what the final charge on each sphere is after they are moved apart.
### Identify the principles involved
We know that the charge carriers in conductors are free to move around and that charge on a conductor spreads itself out on the surface of the conductor.
### Apply the principles
1. When the two conducting spheres are brought together to touch, it is as though they become one single big conductor and the total charge of the two spheres spreads out across the whole surface of the touching spheres. When the spheres are moved apart again, each one is left with half of the total original charge.
2. Before the spheres touch, the total charge is: $$-\text{5}\text{ nC} + (-\text{3}\text{ nC}) = -\text{8}\text{ nC}$$. When they touch they share out the $$-\text{8}$$ $$\text{nC}$$ across their whole surface. When they are removed from each other, each is left with half of the original charge:
$\frac{-\text{8}\text{ nC}}{2} = -\text{4}\text{ nC}$
on each sphere.
temp text
## Worked example 5: Identical spheres sharing charge I
Two identical, insulated spheres have different charges. Sphere 1 has a charge of $$-\text{96} \times \text{10}^{-\text{18}}$$ $$\text{C}$$. Sphere 2 has $$\text{60}$$ excess electrons. If the two spheres are brought into contact and then separated, what charge will each have?
### Analyse the question
We need to determine what will happen to the charge when the spheres touch. They are insulators so we know they will NOT allow charge to move freely. When they touch nothing will happen.
## Worked example 6: Identical spheres sharing charge II
Two identical, metal spheres on insulating stands have different charges. Sphere 1 has a charge of $$-\text{96} \times \text{10}^{-\text{18}}$$ $$\text{C}$$. Sphere 2 has $$\text{60}$$ excess protons. If the two spheres are brought into contact and then separated, what charge will each have? How many electrons or protons does this correspond to?
### Analyse the question
We need to determine what will happen to the charge when the spheres touch. They are metal spheres so we know they will be conductors. This means that the charge is able to move so when they touch it is possible for the charge on each sphere to change. We know that charge will redistribute evenly across the two spheres because of the forces between the charges. We need to know the charge on each sphere, we have been given one.
### Identify the principles involved
This problem is similar to the earlier worked example. This time we have to determine the total charge given a certain number of protons. We know that charge is quantised and that protons carry the base unit of charge and are positive so it is $$\text{1,6} \times \text{10}^{-\text{19}}$$ $$\text{C}$$
### Apply the principles
The total charge will therefore be:
\begin{align*} {Q}_{2} & = 60\times \text{1,6} \times \text{10}^{-\text{19}}\text{ C} \\ & = \text{9,6} \times \text{10}^{-\text{18}}\text{ C} \end{align*}
As the spheres are identical in material, size and shape the charge will redistribute across the two spheres so that it is shared evenly. Each sphere will have half of the total charge:
\begin{align*} Q& = \frac{{Q}_{1} + {Q}_{2}}{2} \\ & = \frac{\text{9,6} \times \text{10}^{-\text{18}}\text{ C} + \left(-\text{9,6} \times \text{10}^{-\text{18}}\text{ C}\right)}{2}\\ & = \text{0}\text{ C} \end{align*}
So each sphere is now neutral.
No net charge means that there is no excess of electrons or protons.
## Worked example 7: Conservation of charge
Two identical, metal spheres have different charges. Sphere 1 has a charge of $$-\text{9,6} \times \text{10}^{-\text{18}}$$ $$\text{C}$$. Sphere 2 has $$\text{30}$$ excess electrons. If the two spheres are brought into contact and then separated, what charge will each have? How many electrons does this correspond to?
### Analyse the problem
We need to determine what will happen to the charge when the spheres touch. They are metal spheres so we know they will be conductors. This means that the charge is able to move so when they touch it is possible for the charge on each sphere to change. We know that charge will redistribute evenly across the two spheres because of the forces between the charges. We need to know the charge on each sphere, we have been given one.
### Identify the principles
This problem is similar to the earlier worked example. This time we have to determine the total charge given a certain number of electrons. We know that charge is quantised and that electrons carry the base unit of charge which is $$-\text{1,6} \times \text{10}^{-\text{19}}$$ $$\text{C}$$
\begin{align*} {Q}_{2} & = 30 \times -\text{1,6} \times \text{10}^{-\text{19}}\text{ C}\\ & = -\text{4,8} \times \text{10}^{-\text{18}}\text{ C} \end{align*}
### Apply the principles: redistributing charge
As the spheres are identical in material, size and shape the charge will redistribute across the two spheres so that it is shared evenly. Each sphere will have half of the total charge:
\begin{align*} Q & = \frac{{Q}_{1} + {Q}_{2}}{2} \\ & = \frac{-\text{9,6} \times \text{10}^{-\text{18}} + \left(-\text{4,8} \times \text{10}^{-\text{18}}\right)}{2} \\ & = -\text{7,2} \times \text{10}^{-\text{18}}\text{ C} \end{align*}
So each sphere now has $$-\text{7,2} \times \text{10}^{-\text{18}}$$ $$\text{C}$$ of charge.
We know that charge is quantised and that electrons carry the base unit of charge which is $$-\text{1,6} \times \text{10}^{-\text{19}}$$ $$\text{C}$$.
### Apply the principles: charge quantisation
As each electron carries the same charge the total charge must be made up of a certain number of electrons. To determine how many electrons we divide the total charge by the charge on a single electron:
\begin{align*} N & = \frac{-\text{7,2} \times \text{10}^{-\text{18}}}{-\text{1,6} \times \text{10}^{-\text{19}}} \\ & = 45 \text{ electrons} \end{align*}
temp text
## The electroscope
The electroscope is a very sensitive instrument which can be used to detect electric charge. A diagram of a gold leaf electroscope is shown the figure below. The electroscope consists of a glass container with a metal rod inside which has 2 thin pieces of gold foil attached. The other end of the metal rod has a metal plate attached to it outside the glass container.
The electroscope detects charge in the following way: A charged object, like the positively charged rod in the picture, is brought close to (but not touching) the neutral metal plate of the electroscope. This causes negative charge in the gold foil, metal rod, and metal plate, to be attracted to the positive rod. Because the metal (gold is a metal too!) is a conductor, the charge can move freely from the foil up the metal rod and onto the metal plate. There is now more negative charge on the plate and more positive charge on the gold foil leaves. This is called inducing a charge on the metal plate. It is important to remember that the electroscope is still neutral (the total positive and negative charges are the same), the charges have just been induced to move to different parts of the instrument! The induced positive charge on the gold leaves forces them apart since like charges repel! This is how we can tell that the rod is charged. If the rod is now moved away from the metal plate, the charge in the electroscope will spread itself out evenly again and the leaves will fall down because there will no longer be an induced charge on them.
### Grounding
If you were to bring the charged rod close to the uncharged electroscope, and then you touched the metal plate with your finger at the same time, this would cause charge to flow up from the ground (the earth), through your body onto the metal plate. Connecting to the earth so charge flows is called
### grounding
. The charge flowing onto the plate is opposite to the charge on the rod, since it is attracted to the charge on the rod. Therefore, for our picture, the charge flowing onto the plate would be negative. Now that charge has been added to the electroscope, it is no longer neutral, but has an excess of negative charge. Now if we move the rod away, the leaves will remain apart because they have an excess of negative charge and they repel each other. If we ground the electroscope again (this time without the charged rod nearby), the excess charge will flow back into the earth, leaving it neutral.
### Polarisation (ESAEZ)
Unlike conductors, the electrons in insulators (non-conductors) are bound to the atoms of the insulator and cannot move around freely through the material. However, a charged object can still exert a force on a neutral insulator due to a phenomenon called polarisation.
If a positively charged rod is brought close to a neutral insulator such as polystyrene, it can attract the bound electrons to move round to the side of the atoms which is closest to the rod and cause the positive nuclei to move slightly to the opposite side of the atoms. This process is called polarisation. Although it is a very small (microscopic) effect, if there are many atoms and the polarised object is light (e.g. a small polystyrene ball), it can add up to enough force to cause the object to be attracted onto the charged rod. Remember, that the polystyrene is only polarised, not charged. The polystyrene ball is still neutral since no charge was added or removed from it. The picture shows a not-to-scale view of the polarised atoms in the polystyrene ball:
Some materials are made up of molecules which are already polarised. These are molecules which have a more positive and a more negative side but are still neutral overall. Just as a polarised polystyrene ball can be attracted to a charged rod, these materials are also affected if brought close to a charged object.
## Electrostatic force
You can easily test that like charges repel and unlike charges attract each other by doing a very simple experiment.
Take a glass ball and rub it with a piece of silk, then hang it from its middle with a piece string so that it is free to move. If you then bring another glass rod which you have also charged in the same way next to it, you will see the ball on the string move away from the rod in your hand i.e. it is repelled.
If, however, you take a plastic rod, rub it with a piece of fur and then bring it close to the ball on the string, you will see the rod on the string move towards the rod in your hand i.e. it is attracted.
This happens because when you rub the glass with silk, tiny amounts of negative charge are transferred from the glass onto the silk, which causes the glass to have less negative charge than positive charge, making it positively charged.
. When you rub the plastic rod with the fur, you transfer tiny amounts of negative charge onto the rod and so it has more negative charge than positive charge on it, making it negatively charged.
Water is an example of a substance which is made of polarised molecules. If a positively charged rod, comb or balloon is brought close to a stream of water, the molecules can rotate so that the negative sides all line up towards the rod. The stream of water will then be attracted to the positively charged object since opposite charges attract.
Water being attracted to a charged balloon
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# Is there a name for this method of column addition and subtraction?
Suppose I want to subtract 46 from 52. Instead of the borrowing method, I can use this method: $$\begin{array}{r} & 5 & 2\\ -\!\!\!\!\!\!& 4 & 6 \\ \hline & & -4 \\ +\!\!\!\!\!\!& 1 & 0 \\ \hline & & 6 \end{array}$$ where I am essentially adding the difference in the ones place (-4) to the difference in the tens place (10). Similarly, suppose that I want to add 79 to 52. I can write $$\begin{array}{r} & & 5 & 2\\ +\!\!\!\!\!\!& & 7 & 9 \\ \hline & & 1 & 1 \\ +\!\!\!\!\!\!& 1 & 2 & 0 \\ \hline & 1 & 3 & 1 \end{array}$$ where I am adding the numbers in the ones place together to get 11 and the numbers in the tens place together to get 120. I then add these two results together to get the final answer. Is there a name for this method of column addition and subtraction?
The method of addition and subtraction that you mention is not new. For now, I provide one reference, but I'm sure there are others. Note that your method of subtraction makes one big assumption that is not needed for the traditional method: you assume that students are familiar with negative integers.
The website Knowledge Over Grades uses a slight variant of your subtraction method: instead of "subtracting from right to left" it does the subtraction of larger powers of 10 first. It calls the method "subtraction of multi-digit numbers without regrouping or borrowing." (It assumes that the student is familiar with place values and negative integers.)
Note that there seems to be a typo in the image. (It uses the word carrying instead of borrowing.)
The website calls the addition method you mention "addition of multi-digit numbers without regrouping or carrying."
• Thank you for your answer! Commented Sep 12, 2021 at 9:30
• Is there a reason why we don't teach younger students about negative numbers early on? Commented Sep 12, 2021 at 10:34
• @mhdadk Perhaps because negative numbers are more difficult to understand. You can post that as a new question, but you'll have to specify what you mean by "younger students" and "early on."
– JRN
Commented Sep 12, 2021 at 13:00
I don't know if your idea has a name, but it feels weird when you try to apply it to something like 10-4:
$$\begin{array}{r} & 1 & 0\\ -\!\!\!\!\!\!& \! & 4 \\ \hline & & -4 \\ +\!\!\!\!\!\!& 1 & 0 \\ \hline & {\color{red}1} & {\color{red}-}{\color{red}4} \end{array}$$
Since the sum of the ones place (-4+0 = -4) and the sum of the tens place (0+1=1), the "6" you put on your example seems to be forced, at least to me. I mean, if you want to teach this to kids, you will have to give a second thought on the rules.
• "the "6" you put on your example seems to be forced, at least to me." Fair enough. It also seemed a bit forced for me as well, and I was hoping there is a cleaner way of doing it. Commented Sep 8, 2021 at 13:31
• so 0 is the only exception? @mhdadk
– BCLC
Commented Sep 12, 2021 at 7:27
1000000
− 1
−−−−−−−
100000
−1
Now what??
• This was posted as an answer, but it does not attempt to answer the question. It should possibly be an edit, a comment, another question, or deleted altogether.
– JRN
Commented Sep 12, 2021 at 5:46
• @JoelReyesNoche: It is an answer. Why should a circularity in the method not be pointed out? Also, it's quite ridiculous that you criticize my answer although it is a clearer depiction of the problem than FormerMath's answer (which has 4 upvotes). Commented Sep 12, 2021 at 5:52
• The question is "Is there a name for this method of column addition and subtraction?" I can't see how your post answers it.
– JRN
Commented Sep 12, 2021 at 7:52
• But you have a point. FormerMath's post also does not answer the question. (That is why I did not upvote it.) I will now flag that post as "not an answer."
– JRN
Commented Sep 12, 2021 at 7:55
• I tried converting these two answers to comments but the formatting doesn't work well in comments and the information is lost. It would certainly be better if both of them were comments, but I'm just going to leave them for now because they seem to be the best way this platform has of highlighting an important issue to the original poster. To be clear, I think the flags and discussion here were useful and helpful. Commented Sep 12, 2021 at 14:52
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# Equivalent Fractions
## What is Fractions? - Definitions and Properties of Fractions
What is a fraction?
A fraction represents a part of a whole or, more generally, any number of equal parts of a value. When we are working with fractions we assume about the relationship between the part of any number or a whole sum number. Something we do not even realize that we are working on a fraction because fractions are everywhere. A common or simple fraction consists of two parts, an integer numerator displayed above a line, and a non-zero integer denominator, displayed below that line. It tells us how many parts a whole is divided into. For example:
5/7= means we have 5 parts out of the whole of 7.
5 is the numerator, it tells us how many parts we have.
7 is the denominator, it tells us how many parts the whole is divided into.
Properties obeyed by the Fractions
Fraction obeys the commutative, associative, and distributive laws, and the rule against division by zero, like whole numbers.
Commutative property
In mathematics, a binary operation is commutative if changing the sequence of the operands in an operation do not change the result of it. It is a fundamental property for many binary operations, and many mathematical proofs depend on this property.
For example:
Suppose that there are two fractions A and B and they are multiplied together. Then the order of A and B during multiplication cannot change the result. Same goes for addition too.
A × B = B × A
A + B = B + A
Associative property
In mathematics, the associative property is a property of binary operations. In propositional logic, associativity is a valid rule of replacement for expressions with logical proofs. Within an operation containing two or more operands in a row of some associative operator, the way in which the operations are performed does not matter as long as the sequence of the operands in the operation remains the same. It means that rearranging the parentheses in such an expression will not change its result. Consider the following equations:
(4 + 5) + 6 = 4 + (5 + 6) = 15
(4 × 5) × 6 = 4 × (5 × 6) = 120
Distributive property
The distributive property of binary operations is widely applicable in the distributive law. Distribution refers to two valid rules of replacement. The rules allow one to put conjunctions and disjunctions differently within logical proofs. By observing the example given below, you can understand the distributive property easily.
If there are three operands A, B, and C are in such an operation where A × (B + C) then it must be equal to A × B + A × C.
Division by zero
In mathematics, division by zero is division where the denominator is zero. In ordinary mathematics, this expression has no meaning, as there is no number which, when multiplied by 0, gives another number, and so division by zero is undefined.
Equivalent Fraction
In mathematics, equivalent fraction can be defined as the fractions with different numerators and denominators that represent the same value or proportion of the whole. Here are some examples of equivalent fraction, like $\frac{1}{2},\frac{2}{4},\frac{8}{16}$ . They all contain the different values in their numerators and denominators, but at last after evaluation they all give the same value 0.5 as the answer. Equivalent fractions are those types of fractions which give us the same value at last, even though they may look different.
Equivalent fractions represents the same amount of distance or points on a number line to one another.
All equivalent fractions are reduced to the same fractions in their simplest forms by dividing both numerator and denominator by their greatest common factor.
How to find equivalent fractions:
By multiplying the numerator and denominator of a fraction by the same non-zero number, we can change the fraction into equivalent of the original fraction. This is true because for any non-zero number a, the fraction a/a = 1. n n = 1 Therefore, multiplying by a/a is equivalent to multiplying by one, and any number multiplied by one has the same value as the original number.
For Example:
To find the equivalent fraction of 2/5we multiply both the numerator and denominator by 2, then we get equivalent fraction 4/10
How do we simplify the Equivalent fractions?
Dividing an equivalent fractions numerator and denominator by the same non-zero number will also yields an equivalent fraction. This is called simplifying or reducing the fraction. A simple fraction in which the numerator and denominator both are prime number is said to be irreducible.
For Example:
To find the equivalent fraction of 6/18 we divide both the numerator and denominator by their greatest common factor, then we get equivalent fraction 1/3.
How to check two fractions are Equivalent fraction:
To check whether the fractions are equivalent or not, just simplify all the fractions. If they reduce into the same fraction, then the fractions are equivalent fractions.
For example:
We will check whether the fractions 5/10 and 12/36 are equivalent or not.
We will simplify both the fractions-
$\frac{5}{10}=\frac{1*5}{2*5}=\frac{1}{2}$
$\frac{12}{36}=\frac{4*3}{4*3*3}=\frac{1}{3}$
The fractions1/2 and 1/3are not same, hence the two fractions are not Equivalent fractions.
Equivalent fractions are an important tool when adding or subtracting fractions with different denominators. Let's look at an
example:
John bought one half of a cake. He didn't know that while he was out, his wife, Linda, bought one fourth of a cake. When they got home, how much cake did they have altogether?
When we are adding or subtracting fractions, they must have the same denominator. But in the above case, the denominator are two and four (2 and 4). As we read earlier by multiplying the numerator and denominator of a fraction, we get an equivalent fraction. So in this case, for making the denominator equal we have to perform this operation:
$\frac{1}{2}\frac{2}{2}=\frac{2}{4}$
Now we have got the same denominator and we can add them like:
Converting between decimals and fractions
Decimal numbers are more useful to work with when performing calculations, but sometimes it lacks in precision that common fractions have, because sometimes an infinite repeating decimal is required to reach the exact precision. Thus, it is useful to convert repeating decimals into fractions. To change a common fraction to a decimal, do a long division of the decimal representations of the numerator by the denominator and round the answer to the desired accuracy. For example, to change ¼ to a decimal, divide 1 1.00 {\displaystyle 1.00} by 4 to obtain 0.25. To change 1/3 to decimal, divide 1 by 3 and stop when the desired accuracy is obtained, e.g. 4 digits after the decimal point. To change a decimal into a fraction, write 1 in the denominator 1 {\displaystyle 1} followed by as many zeroes as there are digits to the right of the decimal point, and write all the digits of the original decimal in the numerator, excluding the decimal point. Thus 12.346 = 12346/1000
Summary:
• • You can make equivalent fraction just by multiplying or dividing both top and bottom numerator and denominator) by the same value.
• • You can only multiply or divide, never add or subtract, to get an equivalent fraction because the fraction we get from addition and subtraction will not be equivalent to the value we have.
• • You can divide if and only if the top and bottom stay as whole numbers.
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Related Articles
# Minimum time required to schedule K processes
• Difficulty Level : Hard
• Last Updated : 13 Jul, 2021
Given a positive integer K and an array arr[] consisting of N positive integers, such that arr[i] is the number of processes ith processor can schedule in 1 second. The task is to minimize the total time required to schedule K processes such that after scheduling by the ith processor, arr[i] is reduced to floor(arr[i]/2).
Examples:
Input: N = 5, arr[] = {3, 1, 7, 2, 4}, K = 15
Output: 4
Explanation:
The order of scheduled process are as follows:
1. The 3rd process is scheduled first. The array arr[] modifies to {3, 1, 3, 2, 4}, as arr[2] = floor(arr[2] / 2) = floor(7 / 2) = 3.
2. The 5th process is scheduled next. The array arr[] modifies to {3, 1, 3, 2, 2}.
3. The 1st process is scheduled next. The array arr[] modifies to {1, 1, 3, 2, 2}.
4. The 2nd process is scheduled next. The array arr[] modifies to {3, 0, 3, 2, 4}.
The total processes scheduled by all the process = 7 + 4 + 3 + 1 = 15(= K) and the total time required is 4 seconds.
Input: N = 4, arr[] = {1, 5, 8, 6}, K = 10
Output: 2
Naive Approach: The simplest approach to solve the given problem is to sort the given list in ascending order and choose the processor with the highest ability and reduce the value of K by that value and delete that processor from the list and add half of that in the sorted list again. Repeat the above process until at least K processes are scheduled and print the time required after scheduling at least K processes.
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
Efficient Approach: The above approach can also be optimized by using the concept of Hashing. Follow the below steps to solve the problem:
• Initialize an auxiliary array tmp[] of the size of the maximum element present in the given array.
• Initialize a variable, say count to store the minimum time to schedule all processes respectively.
• Traverse the given array tmp[] from the end and perform the following steps:
• If the current element in tmp[] is greater than 0 and i * tmp[i] is smaller than K.
• Decrease the value of K by the value i * tmp[i].
• Increase tmp[i/2] by tmp[i] as the ability of the processor will decrease by half.
• Increase the value of count by the value tmp[i].
• If the value of K is already smaller than or equal to 0, then print the value of count as the result.
• If the current element in the array tmp[] is at least 0 and the value of i * tmp[i] is at least K, then perform the following steps:
• If K is divisible by the current index, then increment the value of count by K / i.
• Otherwise, increment the value of count by K/i +1.
• After completing the above steps, print -1 if it is not possible to schedule all processes. Otherwise, print the count as the minimum time required.
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find minimum required``// time to schedule all process``int` `minTime(``int` `A[], ``int` `n, ``int` `K)``{`` ` ` ``// Stores max element from A[]`` ``int` `max_ability = A[0];` ` ``// Find the maximum element`` ``for``(``int` `i = 1; i < n; i++)`` ``{`` ``max_ability = max(max_ability, A[i]);`` ``}` ` ``// Stores frequency of each element`` ``int` `tmp[max_ability + 1] = {0};` ` ``// Stores minimum time required`` ``// to schedule all process`` ``int` `count = 0;` ` ``// Count frequencies of elements`` ``for``(``int` `i = 0; i < n; i++)`` ``{`` ``tmp[A[i]]++;`` ``}` ` ``// Find the minimum time`` ``for``(``int` `i = max_ability; i >= 0; i--)`` ``{`` ``if` `(tmp[i] != 0)`` ``{`` ``if` `(tmp[i] * i < K)`` ``{`` ` ` ``// Decrease the value`` ``// of K`` ``K -= (i * tmp[i]);` ` ``// Increment tmp[i/2]`` ``tmp[i / 2] += tmp[i];` ` ``// Increment the count`` ``count += tmp[i];` ` ``// Return count, if all`` ``// process are scheduled`` ``if` `(K <= 0)`` ``{`` ``return` `count;`` ``}`` ``}` ` ``else`` ``{`` ` ` ``// Increment count`` ``if` `(K % i != 0)`` ``{`` ``count += (K / i) + 1;`` ``}`` ``else`` ``{`` ``count += (K / i);`` ``}` ` ``// Return the count`` ``return` `count;`` ``}`` ``}`` ``}` ` ``// If it is not possible to`` ``// schedule all process`` ``return` `-1;``}` `// Driver code``int` `main()``{`` ``int` `arr[] = { 3, 1, 7, 2, 4 };`` ``int` `N = 5;`` ``int` `K = 15;`` ` ` ``cout << minTime(arr, N, K);`` ` ` ``return` `0;``}` `// This code is contributed by mohit kumar 29`
## Java
`// Java program for the above approach` `import` `java.util.*;``import` `java.lang.*;` `class` `GFG {` ` ``// Function to find minimum required`` ``// time to schedule all process`` ``static` `int` `minTime(``int``[] A, ``int` `n, ``int` `K)`` ``{`` ``// Stores max element from A[]`` ``int` `max_ability = A[``0``];` ` ``// Find the maximum element`` ``for` `(``int` `i = ``1``; i < n; i++) {`` ``max_ability = Math.max(`` ``max_ability, A[i]);`` ``}` ` ``// Stores frequency of each element`` ``int` `tmp[] = ``new` `int``[max_ability + ``1``];` ` ``// Stores minimum time required`` ``// to schedule all process`` ``int` `count = ``0``;` ` ``// Count frequencies of elements`` ``for` `(``int` `i = ``0``; i < n; i++) {`` ``tmp[A[i]]++;`` ``}` ` ``// Find the minimum time`` ``for` `(``int` `i = max_ability;`` ``i >= ``0``; i--) {` ` ``if` `(tmp[i] != ``0``) {` ` ``if` `(tmp[i] * i < K) {` ` ``// Decrease the value`` ``// of K`` ``K -= (i * tmp[i]);` ` ``// Increment tmp[i/2]`` ``tmp[i / ``2``] += tmp[i];` ` ``// Increment the count`` ``count += tmp[i];` ` ``// Return count, if all`` ``// process are scheduled`` ``if` `(K <= ``0``) {`` ``return` `count;`` ``}`` ``}` ` ``else` `{` ` ``// Increment count`` ``if` `(K % i != ``0``) {`` ``count += (K / i) + ``1``;`` ``}`` ``else` `{`` ``count += (K / i);`` ``}` ` ``// Return the count`` ``return` `count;`` ``}`` ``}`` ``}` ` ``// If it is not possible to`` ``// schedule all process`` ``return` `-``1``;`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `arr[] = { ``3``, ``1``, ``7``, ``2``, ``4` `};`` ``int` `N = arr.length;`` ``int` `K = ``15``;`` ``System.out.println(`` ``minTime(arr, N, K));`` ``}``}`
## Python3
`# Python3 program for the above approach` `# Function to find minimum required``# time to schedule all process``def` `minTime(A, n, K):`` ` ` ``# Stores max element from A[]`` ``max_ability ``=` `A[``0``]` ` ``# Find the maximum element`` ``for` `i ``in` `range``(``1``, n):`` ``max_ability ``=` `max``(max_ability, A[i])` ` ``# Stores frequency of each element`` ``tmp ``=` `[``0` `for` `i ``in` `range``(max_ability ``+` `1``)]` ` ``# Stores minimum time required`` ``# to schedule all process`` ``count ``=` `0` ` ``# Count frequencies of elements`` ``for` `i ``in` `range``(n):`` ``tmp[A[i]] ``+``=` `1` ` ``# Find the minimum time`` ``i ``=` `max_ability`` ` ` ``while``(i >``=` `0``):`` ``if` `(tmp[i] !``=` `0``):`` ``if` `(tmp[i] ``*` `i < K):`` ` ` ``# Decrease the value`` ``# of K`` ``K ``-``=` `(i ``*` `tmp[i])` ` ``# Increment tmp[i/2]`` ``tmp[i ``/``/` `2``] ``+``=` `tmp[i]` ` ``# Increment the count`` ``count ``+``=` `tmp[i]` ` ``# Return count, if all`` ``# process are scheduled`` ``if` `(K <``=` `0``):`` ``return` `count`` ``else``:`` ` ` ``# Increment count`` ``if` `(K ``%` `i !``=` `0``):`` ``count ``+``=` `(K ``/``/` `i) ``+` `1`` ``else``:`` ``count ``+``=` `(K ``/``/` `i)` ` ``# Return the count`` ``return` `count`` ``i ``-``=` `1` ` ``# If it is not possible to`` ``# schedule all process`` ``return` `-``1` `# Driver code``if` `__name__ ``=``=` `'__main__'``:`` ` ` ``arr ``=` `[ ``3``, ``1``, ``7``, ``2``, ``4` `]`` ``N ``=` `5`` ``K ``=` `15`` ` ` ``print``(minTime(arr, N, K))` `# This code is contributed by SURENDRA_GANGWAR`
## C#
`// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to find minimum required``// time to schedule all process``static` `int` `minTime(``int``[] A, ``int` `n, ``int` `K)``{`` ` ` ``// Stores max element from A[]`` ``int` `max_ability = A[0];` ` ``// Find the maximum element`` ``for``(``int` `i = 1; i < n; i++)`` ``{`` ``max_ability = Math.Max(`` ``max_ability, A[i]);`` ``}` ` ``// Stores frequency of each element`` ``int` `[]tmp = ``new` `int``[max_ability + 1];` ` ``// Stores minimum time required`` ``// to schedule all process`` ``int` `count = 0;` ` ``// Count frequencies of elements`` ``for``(``int` `i = 0; i < n; i++)`` ``{`` ``tmp[A[i]]++;`` ``}`` ` ` ``// Find the minimum time`` ``for``(``int` `i = max_ability; i >= 0; i--)`` ``{`` ``if` `(tmp[i] != 0)`` ``{`` ``if` `(tmp[i] * i < K)`` ``{`` ` ` ``// Decrease the value`` ``// of K`` ``K -= (i * tmp[i]);` ` ``// Increment tmp[i/2]`` ``tmp[i / 2] += tmp[i];` ` ``// Increment the count`` ``count += tmp[i];` ` ``// Return count, if all`` ``// process are scheduled`` ``if` `(K <= 0)`` ``{`` ``return` `count;`` ``}`` ``}` ` ``else`` ``{`` ` ` ``// Increment count`` ``if` `(K % i != 0)`` ``{`` ``count += (K / i) + 1;`` ``}`` ``else`` ``{`` ``count += (K / i);`` ``}` ` ``// Return the count`` ``return` `count;`` ``}`` ``}`` ``}` ` ``// If it is not possible to`` ``// schedule all process`` ``return` `-1;``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{`` ``int` `[]arr = { 3, 1, 7, 2, 4 };`` ``int` `N = arr.Length;`` ``int` `K = 15;`` ` ` ``Console.WriteLine(minTime(arr, N, K));``}``}` `// This code is contributed by ukasp`
## Javascript
``
Output
`4`
Time Complexity: O(M), where M is the maximum element in the array.
Auxiliary Space: O(M)
Alternative Approach(Using STL): The given problem can be solved by using the Greedy Approach with the help of max-heap. Follow the steps below to solve the problem:
• Initialize a priority queue, say PQ, and insert all the elements of the given array into PQ.
• Initialize a variable, say ans as 0 to store the resultant maximum diamond gained.
• Iterate a loop until the priority queue PQ is not empty and the value of K > 0:
• Pop the top element of the priority queue and add the popped element to the variable ans.
• Divide the popped element by 2 and insert it into the priority queue PQ.
• Decrement the value of K by 1.
• After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
## C++14
`// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to execute k processes that can be gained in``// minimum amount of time``void` `executeProcesses(``int` `A[], ``int` `N, ``int` `K)``{`` ``// Stores all the array elements`` ``priority_queue<``int``> pq;` ` ``// Push all the elements to the`` ``// priority queue`` ``for` `(``int` `i = 0; i < N; i++) {`` ``pq.push(A[i]);`` ``}` ` ``// Stores the required result`` ``int` `ans = 0;` ` ``// Loop while the queue is not`` ``// empty and K is positive`` ``while` `(!pq.empty() && K > 0) {` ` ``// Store the top element`` ``// from the pq`` ``int` `top = pq.top();` ` ``// Pop it from the pq`` ``pq.pop();` ` ``// Add it to the answer`` ``ans ++;` ` ``// Divide it by 2 and push it`` ``// back to the pq`` ``K = K - top;`` ``top = top / 2;`` ``pq.push(top);`` ``}` ` ``// Print the answer`` ``cout << ans;``}` `// Driver Code``int` `main()``{`` ``int` `A[] = { 3, 1, 7, 4, 2 };`` ``int` `K = 15;`` ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);`` ``executeProcesses(A, N, K);` ` ``return` `0;``}`
## Python3
`# Python3 program for the above approach`` ` `# Function to execute k processes that``# can be gained in minimum amount of time``def` `executeProcesses(A, N, K):`` ` ` ``# Stores all the array elements`` ``pq ``=` `[]`` ` ` ``# Push all the elements to the`` ``# priority queue`` ``for` `i ``in` `range``(N):`` ``pq.append(A[i])`` ` ` ``# Stores the required result`` ``ans ``=` `0`` ``pq.sort()`` ` ` ``# Loop while the queue is not`` ``# empty and K is positive`` ``while` `(``len``(pq) > ``0` `and` `K > ``0``):`` ` ` ``# Store the top element`` ``# from the pq`` ``top ``=` `pq.pop()`` ` ` ``# Add it to the answer`` ``ans ``+``=` `1`` ` ` ``# Divide it by 2 and push it`` ``# back to the pq`` ``K ``-``=` `top`` ``top ``/``/``=``2`` ``pq.append(top)`` ` ` ``pq.sort()`` ` ` ``# Print the answer`` ``print``(ans)` `# Driver Code``A ``=` `[ ``3``, ``1``, ``7``, ``4``, ``2` `]``K ``=` `15``N``=``len``(A)``executeProcesses(A, N, K)` `# This code is contributed by patel2127`
## Javascript
``
Output
`4`
Time Complexity: O((N + K)*log N)
Auxiliary Space: O(N)
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(Originally appeared in Lognet 96/1)
# Ordinary Propositions That Can Neither Be Accepted Nor Rejected
By Jerome Frazee
We can neither accept nor reject some ordinary propositions without running the risk of being wrong for the wrong reason. The reason for this is that the rejecter both affirms the antecedent and denies the consequence of a conditional. Most of this article gives examples of the problem. Finally, I offer a not very satisfying solution.
If (A ⊃ B) is false then (A & -B) is true. If (A & -B) is true then A is true. Consequently if (A ⊃ B) is false then A is true. We cannot deny a conditional without affirming its antecedent. Curiously, this leads to a common type of proposition that cannot be rationally labeled “true” or “false”. The following scenario illustrates the point.
Mr. Music, the music teacher at the school, has a pupil named Hornblower who his friend, A. Slyfellow, says is actually the archangel Gabriel. Slyfellow says (1) If Hornblower will blow his horn this noon, then the dead will rise this noon. Sentence (1) is either true or false; for it and its denial cover all possibilities. So, the question is: (2) Is (1) true? Mr. Music says (1) is ridiculous and bets A. Slyfellow \$5.00 that it is false. At noon Hornblower does not blow his horn. So A. Slyfellow collects \$5.00 from Mr. Music (which, we presume, he immediately splits with Hornblower).
No one can rationally answer question (2) with Yes or No. In this story Mr. Music is set up by A. Slyfellow; he cannot win, for if he bets that sentence (1) is true then Hornblower will blow his horn and the dead will surely not rise (surely the dead will not rise!). But, it need not have been a set-up; after all, Hornblower might have died, become sick, or simply forgot. However, ordinary debate—especially between politicians—is filled with these kinds of questions. Here is an even better example:
Two legislators were debating a proposed law. Ms. Good said If we pass this proposed law tomorrow, then the people will be better off than they are now. But, Mr. Lawson answered, That is false! The next day Mr. Lawson casts the deciding vote against the proposed law, and it does not pass.
What Ms. Good said was true and what Mr. Lawson said was false, but he was wrong for the wrong reason! The antecedent was false. He voted against what could have caused his words to be true and for what caused them to be false! Nevertheless, he voted for what he held to be true: If we pass this proposed law, then the people will not be better off. Also, if he had agreed with Ms. Good, then she could have broadcast to his constituents that he would be irrational if he voted against the proposed law. With these kinds of conditionals no one can deny the conditional without running the risk of being wrong for the wrong reason, i.e., the antecedent being false instead of both the antecedent and the consequent being true. In the final analysis, Mr. Music, and everyone in a similar situation, cannot make their position known by denying what is affirmed. They must make a different statement without denying the other one! It is even more disconcerting to realize that the affirmer is also saying it the right way. There is no better way for him to say it. Either Hornblower will blow his horn and the dead will rise, or else he will not and the dead will or will not rise; i.e., it is not the case that he will blow his horn and the dead will not rise. This states the affirmer’s position exactly; he does not need to say more and he does not want to say less. If the conditional’s antecedent is true and its consequent is false, then the conditional is false; if the conditional is false, then its antecedent is true and its consequent is false; the two are equivalent. And that is the way it should be. Given that this is true, why can’t his opponent simply deny it? Instead, he must make a new statement like, Whether X happens, Y will not happen. Or better still, he can say (3) If X happens, then Y will not happen and also be careful not to deny his opponent! For if he denies him, then he cannot fix it afterward by adding (anding) something to it, because together they will either reduce to the denial or a contradiction. For the denier, the better choice is (3), for that is exactly what he wants to say. Now, the only time he will be wrong is when the antecedent is true and the consequent is true...and, this is also the way it should be. His opponent now runs the risk of being wrong for the wrong reason if he denies that (3) is true.
In English, we usually accept that the denier does not intend to affirm the antecedent (but he can’t deny it either), and so we do not hold him to the logical conclusion...unless we are meanspirited or A. Slyfellow. But, when we translate the English into symbolic logic, then it becomes clear that there is no possible way to refrain from holding the denier to the antecedent. With symbolic logic we cannot express in an aside that we are not affirming the antecedent.The problem simply does not arise when we are dealing with cause and effect...or with the future, as the following scenario will show.
A controversial figure stumbled into civilization calling himself Prof. Hardy and claiming to be the only survivor of a scientific expedition to the outback several years previously. He said that the expedition had found a peculiar lake with an unusual species of fish in it. All the fish were hermaphrodites and swam in schools; each school had exactly one father fish at any given time, and at a certain time in the year every fish in that school was pregnant by either itself or its school’s father fish, but not by both. Sometime after the professor died Dr. Wiseman realized that there were no Hardy fish. However he was unable to convince his colleagues of this. The debate became so heated that an expedition was sent to the outback in search of Hardy’s lake and fish. They finally found a peculiar lake that fit Hardy’s description, and it also had schools of fish in it. It was decided to forgo an exhaustive investigation of the fish for the time being and simply inquire of the natives about the old professor. At this point the leader of the expedition declared flatly, If this is Hardy’s lake, then these are Hardy’s fish! (No one holds that either Hardy’s fish caused Hardy’s lake or vice versa. The fish could have been caused by the lake in a sense, but in that same sense, they could have been caused by the ocean.)
Unlike Mr. Music, Dr. Wiseman knows that the consequent is false. However, whether he agrees or disagrees he runs the risk of being wrong for the wrong reason: if he agrees, then the lake may be Hardy’s; if he disagrees, then the lake may not be Hardy’s. Moreover, when a person denies a conditional, he may be wrong for the wrong reason because he denies the consequent. For Example, A. Slyfellow could have said, If it is not the case that the dead will rise this noon, then it is not the case that Hornblower will blow his horn this noon. This, of course, says the same thing as before, but now Mr. Music would be wrong because he denied the consequent when he denied the conditional. (A. Slyfellow’s original statement had the form (X ⊃ Y) and, consequently, this one has the form (-Y ⊃ -X) so it should still be answered by (X ⊃ -Y) and not by (-Y ⊃ -(-X)), which amounts to (-Y ⊃ X). Actually, it could be answered (now and before) by (Y ⊃ -X): If the dead will rise this noon, then it is not the case that Hornblower will blow his horn this noon.) Basically, the problem does not exist because both sides fail to recognize that both can be right. If the best the two sides can do is say, If X, then Y, and you may be right and, opposingly, If X, then not-Y, and you may be right, then something is missing. The missing part is the fact that the two sides are diametrically opposed to each other on the only part they are really arguing about. Actually, no one would say this, for neither side would think that they were both right if X were false. Usually, they would think the question remained unresolved.
Basically, we are looking for a way to express our denial of our opponent’s position without affirming the antecedent or denying the consequent. But in logic there is no way to do this, for what can we deny? Once we have denied the conditional, there is no way to doctor it into what we want to say by “anding” something to it. In English we take it for granted that we are only alleging something true about a situation when the antecedent is true. And this is true when the alleger speaks; for when the antecedent is false, he says nothing: If X, then Y; and if not-X, then Y or not-Y. But, this is not true for the denier! He alleges something is true even when the antecedent is false. This is particularly obvious when translating from English to symbolic logic or when speaking a man-made language based on logic like Loglan. In a man-made language we can remedy the situation by simply creating a new word which by definition expresses both denial and the denier’s own allegation: i.e., (X ⊃ -Y). Of course there can be no actual negation of the original statement. However, if an English-speaker expects his argument to be held up to the scrutiny of symbolic logic, he should be careful what he says; for the backbone of a very large percentage of arguments is the conditional. When an arguer denies a conditional, he says something very exact; but usually he has the feeling that the if is still hovering over the antecedent: If the antecedent is false, then all bets are off! he thinks.
The best way to answer (X ⊃ Y) is with (X ⊃ -Y). But if that doesn’t seem strong enough, one might respond with, That is false if X!, or equivalently, If X, that is false! These reduce to: (X ⊃ -(X ⊃ Y)) ≡ (X ⊃ -Y). This puts the if back over the antecedent while expressing strong denial.
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If a is a positive number less than 10, is c greater than : GMAT Data Sufficiency (DS)
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# If a is a positive number less than 10, is c greater than
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If a is a positive number less than 10, is c greater than [#permalink]
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If a is a positive number less than 10, is c greater than the average (arithmetic mean) of a and 10?
(1) On the number line, c is closer to 10 than it is to a.
(2) 2c – 10 is greater than a.
[Reveal] Spoiler: OA
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rxs0005 wrote:
If a is a positive number less than 10, is c greater than the average (arithmetic mean) of a and 10?
(1) On the number line, c is closer to 10 than it is to a.
(2) 2c – 10 is greater than a.
Given: $$0<a<10$$. Question: is $$c$$ greater than the average (arithmetic mean) of $$a$$ and 10? --> or is $$c>\frac{a+10}{2}=average$$? --> or is $$2c>a+10$$?
(1) On the number line, c is closer to 10 than it is to a.
Number line approach:
a-----average-----10----- (average of a and 10 is halfway between a and 10). So the question ask whether c is either in the BLUE or GREEN area.
As, c is closer to 10 than it (c) is to a then this statement directly tells us that c is either in the BLUE or GREEN area. Sufficient.
Algebraic approach:
c is closer to 10 than it is to a, means that the distance between c and 10 is less than the distance between c and a. So, $$|10-c|<|c-a|$$. Now, as c is closer to 10 than it is to a, then c>a, so $$|c-a|=c-a$$ --> two cases for 10-z:
A. $$c\leq{10}$$ --> $$|10-c|=10-c$$ --> $$|10-c|<|c-a|$$ becomes: $$10-c<c-a$$ --> $$2c>10+a$$. Answer to the question YES.
B. $$c>{10}$$ --> in this case $$2c>20$$ and as $$a<10$$, then $$a+10<20$$, hence $$2c>10+a$$. Answer to the question YES.
(2) 2c – 10 is greater than a --> $$2c-10>a$$ --> $$c>\frac{a+10}{2}=average$$, again directly tells us that c is greater than the average of a and 10. Sufficient.
Similar questions:
number-line-problem-22709.html
x-is-a-positive-number-less-than-86563.html
600-level-question-95138.html?hilit=halfway%20between%20greater%20closer
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05 Dec 2010, 06:08
q is c > (a+10)/2
from statement 1 c is closer to 10 than a
let us take a=9.4 c= 9.5
in this case c < (a+10)/2
if a=7 c=9 then c > (a+10)/2
so statement 1 not sufficient
can some one explain is there any wrong in this
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05 Dec 2010, 06:23
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anilnandyala wrote:
q is c > (a+10)/2
from statement 1 c is closer to 10 than a
let us take a=9.4 c= 9.5
in this case c < (a+10)/2
if a=7 c=9 then c > (a+10)/2
so statement 1 not sufficient
can some one explain is there any wrong in this
Statement (1) says: on the number line, c is closer to 10 than it is to a --> means that the distance between c and 10 is less than the distance between c and a.
Now, your example $$a=9.4$$ and $$c=9.5$$ is not valid as in this case $$c$$ is obviously closer to $$a$$ than to 10 (c-a=0.1 and 10-c=0.6).
There are 2 different approaches in my previous post shoving why is this statement sufficient.
Hope it's clear.
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Re: If a is a positive number less than 10, is c greater than [#permalink]
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23 Jan 2014, 23:28
I did it with numbers.
From (1) I know that C is closer to 10 than to a. Pluggin numbers gives me e.g. c= 9, a = 7 average = 8,5 so true....continuing I figured that since c is ALWAYS closer (even if you take 9.99995) to 10, it will be always greater than the average. SUFF.
(2) 2c -10 > a --> c > 10 +a --> c > (10+a)/2 which is the average. SUFF.
Also: Since a is < 10, C is at least 10. which would give us the same answer.
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Re: If a is a positive number less than 10, is c greater than [#permalink]
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15 Jan 2016, 12:39
Hi Bunuel. Actually, i understand your logic in algebraic approach in statement 1 while i can not as you mentioned that such statement tells us directly that c (either) in green area or red area. Isn't "either" in a DS Q. means that it has two solutions and is insufficient ?
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Re: If a is a positive number less than 10, is c greater than [#permalink]
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15 Jan 2016, 12:48
hatemnag wrote:
Hi Bunuel. Actually, i understand your logic in algebraic approach in statement 1 while i can not as you mentioned that such statement tells us directly that c (either) in green area or red area. Isn't "either" in a DS Q. means that it has two solutions and is insufficient ?
You are confusing getting "either" in the form of 2 differnet answers for the same statements and have 2 cases for the same statement that give you the same answer.
Example, if the question is " what is the value of x?
Statement 1 tells you that x is either 1 or 2, then in this case the statement is NOT sufficient.
BUT
if the question asks " is x>0?"
Statement 1 tells you that x is either 1 or 2, then in this case the statement is sufficient as for x=1, you get "YES" for the question asked, similar to the case when x=2. FYI, if you ended with different values of negative values of 'x' , even then this statement sould have been SUFFICIENT, as you would have obtained a "NO" for all possible values of 'x'.
Thus, a statement or a combination of statements is SUFFICIENT if and only if you get 1 UNIQUE/UNAMBIGUOUS answer.
Hope this helps.
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Re: If a is a positive number less than 10, is c greater than [#permalink]
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15 Jan 2016, 15:10
yes. Now it is clear. great thanks Engr2012.
Re: If a is a positive number less than 10, is c greater than [#permalink] 15 Jan 2016, 15:10
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# Is the rotation angle in Minkowski's diagram real or imaginary?
• gene1721
#### gene1721
In Figure_1(b) I have depicted a simplified version of Minkowski's diagram, where
$$\beta = \frac{v}{c}= \tanh \psi= - i \tan i \psi= \frac{( e^{\psi} - e^{-\psi})}{( e^{\psi} + e^{-\psi})}$$,
the rotation of $$(x',t')$$-axes being defined as imaginary (considering x and t real, and c=1).
However, I have found books where this rotation is considered as real, while the rotation in Figure_1(a) is shown as imaginary (considering x real, t imaginary, and c=1).
Could you explain which rotation is real and which imaginary?
Figure_1 is here!
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Doing a Lorentz transformation (boost) by a velocity v to some 4-vector is equivalent to rotate it an imaginary angle $$\chi$$ such that:
$$\chi = arctan( v )$$
Thanks Kuon!
If I understand correctly, you come with a third possibility, right?
Could you explain which rotation is real and which imaginary?
That depends on how you define "rotation". I wouldn't call a Lorentz boost a "rotation". I don't mind calling it a "hyperbolic rotation" or a "Minkowski orthogonal transformation" but calling it a "rotation" is a bit weird in my opinion.
There's no need to bring imaginary numbers into this.
The angle of a rotation is always real and can be defined by writing the matrix as
$$\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}$$
A 2×2 matrix represents a proper rotation if and only if it's orthogonal and has determinant 1. Alll 2×2 matrices with those properties can be put in the form above, with different values of $\theta$.
The rapidity of a Lorentz boost in 1+1 dimensions is always real and can be defined by writing the matrix as
$$\begin{pmatrix}\cosh\theta & -\sinh\theta\\ -\sinh\theta & \cosh\theta\end{pmatrix}$$
A 2×2 matrix $\Lambda$ represents a proper orthochronous Lorentz boost if and only if it satisfies $\Lambda^T\eta\Lambda=\eta$, $\det\Lambda=1$, and $\Lambda_{00}\geq 1$. (That's just its upper left component). All 2×2 matrices with those properties can be put in the form above, with different values of $\theta$. This $\theta$ is usually referred to as the "rapidity" of the Lorentz transformation, not the "angle", but I wouldn't mind calling it the "hyperbolic angle" or even the "angle of a hyperbolic rotation". I just don't want to call it a "rotation angle", as you did in the thread title.
Note that you can't turn one of these matrices into the other just by substituting $\theta\rightarrow i\theta$. That's why it doesn't quite make sense to describe a Lorentz transformation as a rotation by an imaginary angle.
Last edited:
That's why it doesn't quite make sense to describe a Lorentz transformation as a rotation by an imaginary angle.
Scott Walter (http://www.univ-nancy2.fr/DepPhilo/walter/), who published several papers on Minkowski and his diagram, wrote on page 9 of this article http://www.univ-nancy2.fr/DepPhilo/walter/papers/nes.pdf
... Minkowski retained the geometric interpretation of Lorentz transformation ... In doing so, he elaborated the notion of velocity as a rotation in four-dimensional space. He introduced a formula for the frame velocity $$q$$ in terms of the tangent of an imaginary angle $$i \psi$$, such that
$$q = - i \tan i \psi= \frac{( e^{\psi} - e^{-\psi})}{( e^{\psi} + e^{-\psi})}$$.
Minkowski could very well have expressed frame velocity in the equivalent form $$q = \tanh \psi$$, where the angle of rotation is real instead of imaginary, and all four space-time coordinates are real. He did not do so, but used the imaginary rotation angle $$i \psi$$ to express the special Lorentz transformation in the trigonometric form:
$$x_1'= x_1,~~x_2'= x_2,~~x_3'= x_3 \cos i \psi + x_4 \sin i \psi,~~x_4'= - x_3 \sin i \psi+ x_4 \cos i \psi$$.
The use of circular functions here underscores the fact that a special Lorentz transformation is equivalent to a rotation in the $$( x_3 x_4 )$$--plane. Likewise, by expressing velocity in terms of an imaginary rotation, Minkowski may have...
Minkowski's preference for circular functions may be understood in relation to his project to express the laws of physics in four-dimensional terms. ... Expressing the Lorentz transformation as a hyperbolic rotation would have obscured the connection for physicists.
We can see now not only that it does make sense to describe a Lorentz transformation as a rotation by an imaginary angle, but to also inquire why would Minkowski really want to use the imaginary rotation. (I have to say that I buy only partially into Walter's explanation!)
Trying to understand why wanted Minkowski to use the imaginary angle, I found in different sources three ways (including Koun's version!) to explain the rotation angle in Minkowski's diagram:
1. x real, t real and the rotation angle real;
2. x real, t real and the rotation angle imaginary;
3. x real, t imaginary and the rotation angle imaginary.
So, given that Minkowski didn't want to use version 1, which between version 2 and version 3 is the correct one?
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The number of cars $(X)$ arriving at a service station per day follows a Poisson distribution with mean $4$. The service station can provide service to a maximum of $4$ cars per day. Then the expected number of cars that do not get service per day equals
1. $4$
2. $0$
3. $\Sigma_{i=0}^{\infty} i P(X=i+4)$
4. $\Sigma_{i=4}^{\infty} i P(X=i-4)$
### 1 comment
C?
$\mathbf{\underline{Answer:\Rightarrow C}}$
$\mathbf{\underline{Explanation:\Rightarrow}}$
If $\mathbf{X = 0}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 0}$
If $\mathbf{X = 1}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 0}$
If $\mathbf{X = 2}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 0}$
If $\mathbf{X = 3}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 0}$
If $\mathbf{X = 4}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 0}$
If $\mathbf{X = 5}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 1}$
If $\mathbf{X = 6}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 2}$
$\mathbf{\vdots} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \mathbf{\vdots}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots$
$\therefore \text{The expected number of cars missing out on service} \\\\= \sum(\text{number of cars that miss out being serviced})(\text{probability of this happening}) \\\\=\mathbf{ 0P(X = 0) + 0P(X = 1) + 0P(X = 2)+\cdots+1P(X = 5) + 2P(X = 6)+\cdots\cdots\cdots}$
So, the above expression can be represented in the compact form as:
$$\sum(\text{The number of cars that miss-out being serviced})(\text{Probabibility of this happening}) = \mathbf{\sum_{0}^{\infty}iP(X = i+4)}$$
$\therefore$ Option $\mathbf C$ is the right answer.
by
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# In an examination, the score of A was 10% | Quantitative Aptitude - Arithmetic – Percentage
## CAT 2019 - Slot 2 - Quantitative Aptitude - Percentage - Question 2 - In an examination, the score of A was 10%
Q. 2: In an examination, the score of A was 10% less than that of B, the score of B was 25% more than that of C, and the score of C was 20% less than that of D. If A scored 72, then the score of D was
Let the score of B =x
Score of A = x – 10% of x = 0.9x
As per question , score of B = 25% more than C = 5/4 C
And score of C = 20% less than D = 4/5 D
Or D = 5/4 C
So, B = D
Since A = 72, Therefore D = 72/0.9 = 80
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Lesson Plan:
# Quarters and Cents
5.0 based on 1 rating
Subject
August 8, 2015
## Learning Objectives
Students will be able to recognize a quarter, know its value, and add quarters together to find the total amount of money.
## Lesson
### Introduction (10 minutes)
• Call the students together as a group.
• Ask the students to name the different values of coins.
• Reinforce that coins include pennies, nickels, dimes, and quarters.
• Inform the students that today they are going identify and count quarters.
• Ask students if they know the value of a quarter.
• Reiterate that the quarter is worth 25 cents.
### Explicit Instruction/Teacher Modeling (10 minutes)
• Inform your students that a quarter is equal to 25 pennies, 5 nickels, or 2 dimes and 1 nickel.
• Show this with the pretend money.
• Inform your students that it is easier to carry 25 cents as a quarter than 25 pennies.
• Show students the visual of one quarter versus 25 pennies.
• Explain that both values are equal.
• Show students 2 quarters.
• Ask students to identify the value of two quarters.
• On the whiteboard, add 25 cents + 25 cents in a vertical equation.
• Solve the addition equation with the class.
• Inform the students that 50 cents is equal to 50 pennies, 5 dimes, or 10 nickels.
• Ask the students the value of three quarters.
### Guided Practice/Interactive Modeling (10 minutes)
• Place students in pairs to work.
• Give each pair three small cups with varying amounts of quarters in them, up to five.
• Ask the students to complete three separate equations by adding the quarters in each cup.
• For example, cup 1's equation could be 25 cents + 25 cents= 50 cents. Cup 2's equation could be 25 cents + 25 cents + 25 cents = 75 cents, and cup 3's equation could be 25 cents + 25 cents + 25 cents + 25 cents = 100 cents.
• Tell the students that 100 cents is equal to one dollar.
• Walk around the room, and monitor the pairs of students.
• Select one or two pairs to come to the whiteboard and demonstrate how they came to their answers.
### Independent Working Time (10 minutes)
• Give each student the Quarters and Cents worksheet.
• Read the instructions to the students.
• Give the students time to complete the worksheet.
## Extend
### Differentiation
• Enrichment: Give the students more than 5 quarters in their cups to add.
• Support: Give the students fewer than three quarters in their cups to add. Provide pennies to help them count. These can be used as manipulatives.
## Review
### Assessment (10 minutes)
• Grade the worksheet completed during independent work time.
• Give feedback, and allow for reinforcement if the students did not master the concept.
### Review and Closing (10 minutes)
• Ask students the value of a quarter.
• Ask how many quarters make 50 cents.
• Ask how many quarters make 75 cents and 100 cents.
• Show the students a penny, a nickel, a dime, and a quarter.
• Ask them to identify which one is a quarter.
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# Picking Numbers
## Mathematics: Standard Multiple Choice
Multiple-choices directions are really simple, aren't they? The questions on the test are longer, but they don't say much more than this. Remember that you are allowed to use your calculator, but that it can get you in trouble. I don't actually own one, and I use the one on my computer as little as possible.
The weights of 10 bags of apples range from 2.75 pounds to 3.15 pounds. If w is the weight, in pounds, of one of these bags, which of the following must be true?
After reading carefully, the next step is to identify the bottom line. Here, the question asks "Which of the following must be true?" and that's extremely hard to put in mathematical shorthand, so the best choice here is to write weight=? at the top of your scratch work.
Next, assess your options. You could translate this problem into an inequality, but that could be complicated because the answers are all absolute value. Instead, pick numbers for w and see which inequality below works out. To make sure that the answer choice will always be true, try three different numbers. When picking numbers, always choose numbers that are easy to work with and that are distinct from one another. In this case, picking 3 and then picking 3 again won't tell you anything about the problem. 32.8, and 3.1 should work nicely. I picked 3 first because it is a whole number, and those are always easier to deal with. 2.8 and 3.1 are near the ends of the given range, so they will make sure you get the right answer.
Take a look at the answer choices:
(A)
(B)
(C)
(D)
(E)
You are probably accustomed to starting with A, but starting with E is a better habit. It will save you some time on this problem; you can eliminate E immediately. 3-10= -7, and 7 is too large an absolute value.
Look at D: 3-0.2= 2.8 D doesn't work either.
You can eliminate C right away because adding to the numbers you chose will not give a number with an absolute value less than 0.2--it would be closer to 6.
Now try B. 3-2.95= .05 That works with the inequality given in the answer choice. Plug in the other numbers you selected: 2.8-2.95= -.15 is also within the range. 3.1-2.95= .15 B fits the range given in the problem, so it is the right answer.
43% of those who attempted this question at sat.collegeboard.org got it right.
Want more help with math? Visit myknowsys.com!
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## R Programming Homework Solution on Simulation in R
• 20th Sep, 2022
• 15:24 PM
{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE) options(warn = -1) This part of the homework will involve a simulation in R similar to what has been done in the notes. The goals will be to investigate the CLT and Delta method normality! You can turn in an R code with written answers as text, or write this up in a report that includes the R code and output. # 1. Use R to generate N=10,000 samples of size n=5 from a normal distribution with mean 15 and variance 7. For each sample of size 5, you should find the mean of the sample (store this in a vector or matrix). {r} set.seed(100) mu_5 <- c() repl = 10000 n = 5 for(i in 1:repl){ x <- rnorm(n, 15,7) mu_5[i] <- mean(x) } # 2. Repeat the above steps for sample sizes n=20 and n=100. {r} ## For 20 mu_20 <- c() repl = 10000 n = 20 for(i in 1:repl){ x <- rnorm(n, 15,7) mu_20[i] <- mean(x) } # For 100 mu_100 <- c() repl = 10000 n = 100 for(i in 1:repl){ x <- rnorm(n, 15,7) mu_100[i] <- mean(x) } # 3. Create a 1x3 plotting window so we can have 3 plots displayed at once$(par(mfrow = c(1, 3)))$. All plots should be given titles and have their axes labeled appropriately. # 4. Create histograms of the standardized means of size 5, 20, and 100. Overlay the standard normal distribution on each plot (use freq = FALSE on the histogram calls). {r} z_5 <- (mu_5 - 15)*sqrt(5)/7 z_20 <- (mu_20 - 15)*sqrt(20)/7 z_100 <- (mu_100 - 15)*sqrt(100)/7 par(mfrow = c(1,3)) hist(z_5, main = "Std Mean for n = 5", xlab = "Std Mean", freq = F) hist(z_20,main = "Std Mean for n = 20", xlab = "Std Mean", freq = F) hist(z_100,main = "Std Mean for n = 100", xlab = "Std Mean", freq = F) # 5. Below the code to do the above, you should have a section that answers the following questions: ## (a) Explain what the above process and plots are attempting to investigate about the sample mean. Ans: The above plots are attempts to see verify the limiting distribution of sample mean indeed converges to normal distribution as proved by CLT. The histogram looks like sample from normal distribution. Although, here the underlying distribution is normal with known variance so the _standardized_ sample mean will have exact normal distribution. ## (b) In your opinion, which histograms are well approximated by the standard normal distribution? Give a theoretical argument relating to what you see. Ans: Here all the histograms are very well approximated by the standard normal distribution. The reason for that is the underlying sampling distribution itself is normal. # 6. Now suppose we want to investigate$W = e^{\bar{Y}}$. Use your vectors of Y to create 10,000 w values (you shouldn't need to do any new random values). {r} w_5 <- exp(mu_5) w_20 <- exp(mu_20) w_100 <- exp(mu_100) # 7. Use the first order delta method to standardize all of these values (by subtracting off the approximate mean and dividing by the approximate standard error). The transformation we have used is: $$g(x) = e^x$$ The mean of the final transformed random variable is: $$\mu \rightarrow g(\mu)$$ and the variance goes to: $$\sigma^2 \rightarrow \sigma^2 (g^{\prime} (\mu))^2$$ $$\sqrt{n}[g(X_{n})-g(\theta )]\,{\xrightarrow {D}}\,{\mathcal {N}}(0,\sigma ^{2}\cdot [g'(\theta )]^{2})$$ {r} gW_5 <- (w_5 - exp(15))*sqrt(5)/(7*exp(15)) gW_20 <- (w_20 - exp(15))*sqrt(20)/(7*exp(15)) gW_100 <- (w_100 - exp(15))*sqrt(100)/(7*exp(15)) # 8. Create histograms of the standardized w values for each sample size (again a 1x3 plotting window) with standard normal distributions overlayed. {r} par(mfrow = c(1,3)) hist(gW_5, main = "Std Mean after Delta Method n = 5", xlab = "Std Mean", freq = F) hist(gW_20,main = "Std Mean after Delta Method n = 20", xlab = "Std Mean", freq = F) hist(gW_100,main = "Std Mean after Delta Method n = 100", xlab = "Std Mean", freq = F) # 9. Below the code to do the above, you should have a commented section that answers the following questions: ## (a) Explain what the above process is attempting to investigate about W. Ans: Here we are trying to see how delta method works and what is the sample after which convergence is good. W is the transformed variable by exponential distribution of the sample mean. We will see if standardized mean of W converges to normal distribution. ## (b) In your opinion, which histograms are well approximated by the standard normal distribution? Give a theoretical argument relating to what you see. We see that the histogram for$n=100$is the only one which somehow resembles closely with the normal distribution. Histogram for$n=5$and$n=20$are just not close to bell shape. Although, even for the$n=100\$ case, we see right skewed property and not resembling the standard normal very well. This is due to the weak convergence for delta method. The distribution of standardized transformed mean is not exact but only converges as $$n \rightarrow \infty$$. So, at small values of n, the distribution will not resemble the normal shape. If we go higher and higher, we see better approximation.
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https://www.lidolearning.com/questions/m-bb-ncert8-ch9-ex9p3-q3/q3-find-the-product-i-ii-iii-i/
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# NCERT Solutions Class 8 Mathematics Solutions for Exercise 9.3 in Chapter 9 - Algebraic Expressions and Identities
Question 5 Exercise 9.3
Q3) Find the product:
(i)
\left(a^2\right)\times\left(2a^{22}\right)\times\left(4a^{26}\right)
(ii) \left(\frac{2}{3}xy\right)\times\left(\frac{-9}{10}x^2y^2\right)
(iii) \left(\frac{-10}{3}pq^3\right)\times\left(\frac{6}{5}p^3q\right)
(iv) x\times x^2\times x^3\times x^4
Solution:
(i) \left(a^2\right)\times\left(2a^{22}\right)\times\left(4a^{26}\right)
= \left(2\times4\right)\left(a^2\times a^{22}\times a^{26}\right)
= 8\times a^{2+22+26}=8a^{50}
(ii) \left(\frac{2}{3}xy\right)\times\left(\frac{-9}{10}x^2y^2\right)
= \left(\frac{2}{3}\times\frac{-9}{10}\right)\left(x\times x^2\times y\times y^2\right)
= \frac{-3}{5}x^3y^3
(iii) \left(\frac{-10}{3}pq^3\right)\left(\frac{6}{5}p^3q\right)
= \left(\frac{-10}{3}\times\frac{6}{5}\right)\left(p\times p^3\times q^3\times q\right)
= -4p^4q^4
(iv) x\times x^2\times x^3\times x^4=x^{1+2+3+4}=x^{10}
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# Question 86b5f
Feb 25, 2016
$\text{79.28 g}$
#### Explanation:
I'm not sure what you mean by "a lot of steps" because you only need one calculation to get the answer here. Basically, the answer is one step away.
The problem wants you to convert moles of water to grams of water, which means that you're going to have to use the mole $\to$ mass side of the road map.
The conversion factor that takes you from moles to mass and vice versa is called the molar mass.
The molar mass of a given compound is simply the mass of one mole of that compound. In your case, you must use the molar mass of water to determine how many grams would be equivalent to $4.401$ moles of water.
Now, water has a molar mass of ${\text{18.015 g mol}}^{- 1}$. That means that you get $\text{18.015 g}$ of water for every one mole of water. Simply put, one mole of water has a mass of $\text{18.015 g}$.
If that's the case, then $4.401$ moles of water would have a mass of
4.401 color(red)(cancel(color(black)("moles H"_2"O"))) * overbrace("18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))))^(color(purple)("molar mass of water")) = "79.284 g"#
You need to round this off to four sig figs, the number of sig figs you have for the moles of water
${m}_{{H}_{2} O} = \textcolor{g r e e n}{\text{79.28 g}}$
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## What are the problems on numbers?
Number theory is one of the most appealing topics in mathematics. Given one or more numeric values, a numeric algorithm performs some computation. The use of number theory in computing is important for applications like in the case of cryptographic algorithms. Number theory is also used in various applications such as calculating hash functions or in information security. Euclid’s algorithm is one of the oldest number-theoretic algorithms which is used to calculate the greatest common divisor of two numbers.
## Prime number
An integer u > 1 whose only divisors are the numbers 1 and u is called a prime number (or, more simply, a prime). Primes have various important and useful features and they play an important role in number theory. The first few prime numbers are 2, 3, 5, 7, and 11.
## Composite number
An integer u > 1 whose divisor is at least a single number less than u, apart from the numbers 1 and u, is called a composite number (or, more simply, a composite). The first few composite numbers are 4, 6, 8, and 9.
## Relatively prime
We say that two integers u and v are relatively prime to each other or u is prime to v if gcd of u and v is 1. They are also known as co-primes. The examples of co-primes are 5 and 4, 13 and 11, and so on.
## Divisibility
Integer ‘q’ is said to divide an integer ‘w’ if there exists an integer ‘e’ such that w = q * e. The operation of division is denoted in mathematics as “q / w”. The value “q” is called the dividend and the number “w” is called the divisor. The number ‘e’ is called the factor of the dividend “q” if it completely divides it. The number that completely divides a number is called its factor. If the divisor completely divides a number it will also be called the factor of the number. By, completely it means there is no remainder left or the remainder is 0. The remainder is the number that is left when dividing two numbers.
Notice that every integer which is greater than one has at least two positive factors, 1 and the number itself.
## Fermat’s Little theorem
Fermat’s little theorem asserts that when q is a prime number, then for any integer t that is not completely divided by q or for which q is not a factor, the value,
Let s = t(q-1)
which implies,
s ≡ 1 (mod q), and hence s is congruent to 1, or simply q divides s with a remainder of 1.
Elaborating with an example, Let q=3, this means that for any integer t that is not divisible by q, the Fermat's little theorem holds,
Let t=2, 2(3-1) = 22 = 4,
Now, 4 ≡ 1 (mod 3), because when 4 is divided by 3 it gives a remainder of 1.
Fermat's little theorem is used in computation systems for the creation of new cryptographic algorithms, it is also used in computer security.
### Chinese remainder theorem
Suppose that we want a system of linear congruences to their different moduli, then the Chinese theorem is used. It states that, given a set of different congruent equations with one variable but different moduli which are co-primes then,
q ≡ t mod u
q ≡ e mod i,
q ≡ o mod r, and so on
When each pair of moduli is relatively prime. Then there is a unique solution.
For example,
Q ≡ 6 (mod 11),
Q ≡ 13 (mod 16),
Q ≡ 9 (mod 21),
Q ≡ 19 (mod 25).
As 11, 16, 21, and 25 are pairwise relatively prime, the Chinese Remainder Theorem tells us that there is a unique solution,
Modulo s, where
S = 11⋅16⋅21⋅25 = 92400.
## Euclidean's algorithm
Before explaining the algorithm, it is important to know what GCD is, so, the greatest common divisor (GCD) of two or more numbers is the largest integer that divides the numbers evenly and completely. It is also denoted as the highest common factor (HCF). For example, the greatest common divisor of 12 and 6 is 3, as 12 and 6 are completely divisible by 3.
Euclid’s algorithm is used to find the greatest common divisor of two positive integers. Consider a and b are two positive integers, the algorithm is mentioned below.
EUCLID (c, d)
if d = 0
then return c
else return EUCLID(d, c mod d)
Example
Find GCD of 30 and 21
EUCLID(30, 21) = EUCLID(21, 9) = EUCLID(9, 3) = EUCLID(3, 0) = 3
## Drawbacks of Euclid's algorithm
• We are unable to solve linear Diophantine equations using the formula.
• Finding prime decompositions for larger integers is highly difficult (and time-consuming).
## Modular arithmetic algorithm
The area of arithmetic mathematics known as modular arithmetic is concerned with the “mod” functionality. Modular arithmetic is primarily concerned with the computation of “mod” of expressions. Expressions can contain digits as well as addition, subtraction, multiplication, division, and other computational symbols. All modular arithmetic procedures will be briefly discussed here.
Modular arithmetic is an integer-based arithmetic method that takes the remainder into account. If two integers a and b have the same remainder when divided by N, they are said to be congruent (or in the same equivalence class).
a≡b(mod N)
Quotient Remainder Theorem:
It states that, for any pair of integers w and e (e is positive), there exist two unique integers t and y such that:
w = e * t + y
where 0 <= y < e
For example:
when w = 14, e = 3
then t = 4, y = 2 such that 14 = 3 * 4 + 2
The modular addition is as follows:
(w + i) mod r = ((w mod r) + (i mod r)) mod r
For example:
(4 + 3) % 5
= ((4 % 5) + (3 % 5)) % 5
= (4 + 3) % 5
= 7 % 5
= 2
Modular Multiplication:
The modular multiplication is as follows:
(w x r) mod y = ((w mod y) x (r mod y)) mod y
Example:
(3 x 2) % 4
= ((3 % 4) x (2 % 4)) % 4
= (4 x 4) % 4
= 16 % 4
= 0
## Applications of problems on numbers
Computer organization and security, coding and cryptography, random number generation, hash functions, and graphics are all applications of number theory. Number theorists, on the other hand, use computers to factor big integers, find primes, test hypotheses, and solve other difficulties. Because of the invention of cryptographic schemes based on large prime numbers, number-theoretic algorithms are now widely used.
## Context and Applications
• Bachelors in Computer Science Engineering.
• Associate of Science in Computer Science.
## Practice Problems
Question 1: Find the GCD of 57 and 38.
1. 23
2. 38
3. 57
4. 19
Explanation: GCD(57, 38) = GCD(38, 19) = GCD(19, 0) = 19
Question 2: What is the value of (5 + 7)mod 7?
1. 7
2. 5
3. 2
4. None of these
Explanation: (5 + 7)mod 7 = 12 mod 7 = 5
Question 3: Find the gcd of 1547 ad 560 using the Euclidean algorithm.
1. 1547
2. 560
3. 1
4. 7
Explanation:
1547 = 2 * 560 + 427
560 = 1 * 427 + 133
427 = 3 * 133 + 28
133 = 4 * 28 + 21
28 = 1 * 21 + 7
21 = 3 * 7 + 0
Therefore, gcd(1547, 560) = 7.
Question 4: Express 7 as a linear combination of 1547 and 560.
1. 7=21 * 1547 - 58 * 560
2. 7=58 * 1547 - 21 * 560
3. 7=21 * 1547 - 21 * 560
4. 7=58 * 1547 - 58 * 560
Answer: 1. 7=21 * 1547 - 58 * 560
Question 5: What is the value of (132 * 7) mod 8
1. 924
2. 4
3. 8
4. 7
Explanation: (132 * 7) mod 8 = 924 mod 8 = 4
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### Problems on numbers
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## Algebra and Trigonometry 10th Edition
$g(\frac{1}{4})=-2$
Use the property: $\log_aa^x=x$ $g(\frac{1}{4})=\log_2\frac{1}{4}=\log_24^{-1}=\log_22^{-2}=-2$
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Integration
Integration (or antidifferentiation) is the opposite of differentiation. The notation for integration looks something like this: $\int_{a}^{b} f(x) \, dx$ a and b are the terminals. We will discuss this in greater depth in the “definite integral” section.
dx indicates that we are integrating with respect to x. (i.e dealing with x)
Area Approximation
There are many different methods we can use to estimate the area under a graph. In methods we look at two - the left and right endpoint methods.
Left endpoint method
To find the approximation of the area under a graph from x = a to x = b
1 Divide the region into rectangles of equal width (this width will be given in the question). Dividing the x-axis into n equal subintervals. 2 Make sure that the top left corner of each rectangle touches the graph. The top right corner may or may not touch the graph. 3 Sum the area of the rectangles
$L_{n} = \frac{b-a}{n} \cdot [f(x_{0}) + f(x_{1}) + … + f(x_{n-1})]$
Using the left endpoint method to find the area under the graph $$y = (\frac{1}{12}x^2 + 1)\text{ from }( x= 0)\text{ to }(x = 8)$$ with rectangles of width 2. This underestimates the area.
Right endpoint method
To find the approximation of the area under a graph from x = a to x = b
1 Divide the region into rectangles of equal width (this width will be given in the question). Dividing the x-axis inton equal subintervals. 2 Make sure that the top right corner of each rectangle touches the graph. The top left corner may or may not touch the graph. 3 Sum the area of the rectangles
$R_{n} = \frac{b-a}{n} \cdot [f(x_{1}) + f(x_{1}) + … + f(x_{n})]$
Using the right endpoint method to find the area under the graph $$y = (\frac{1}{12}x^2 + 1)\text{ from }( x= 0)\text{ to }(x = 8)$$ with rectangles of width 2. This overestimates the area.
The smaller the width of the rectangles, the more accurate the estimation of the area will be.
Integration Rules
$\int k \cdot f(x) \,dx = k \cdot \int f(x) \, dx$ $\int f(x) + g(x) \,dx = \int f(x) \,dx + \int g(x)\,dx$ $\int f(x) - g(x) \,dx = \int f(x)\, dx - \int g(x)\,dx$ $\int_{a}^{c} f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx$ $\int_{a}^{b} f(x) \,dx = -\int_{b}^{a} f(x)\, dx$ $\int_{a}^{a} f(x) \, dx = 0$ $\int x^n \, dx = \frac{1}{n+1}x^{n+1} + c ,\, n \neq -1$ $\int (ax+b)^n \, dx = \frac{1}{x(n+1)}(ax+b)^{n+1} + c ,\, n \neq -1$ $\int e^{ax+b} \,dx = \frac{1}{a}e^{ax+b} + c$ $\int\frac{1}{ax+b}\,dx = \frac{1}{a} \log_{e}{\lvert ax+b \rvert} + c$ $\int \sin(ax+b) \,dx = -\frac{1}{a} \cos(ax+b) + c$ $\int \cos(ax+b) \,dx = \frac{1}{a} \sin(ax+b) + c$
Indefinite Integral
The indefinite integral is denoted: $\int f’(x) \, dx = f(x) + c$ $c \text{ is an arbitrary constant, meaning that it could be any real number.}$ The inclusion of the arbitrary constant when dealing with indefinite integrals is essential, as you are indicating that there are infinitely many solutions (by varying the constant) not simply one solution. To find the value of the arbitrary constant, more information must be given.
\text{Example 8.1: Find the rule of the function that passes through the point}\\ \text{ (1,3) and has gradient with the equation } \frac{dy}{dx} = 3x^2 + 3\\ \text{ }\\ \begin{aligned} y &= \int \frac{dy}{dx} \,dx\\ &= \int 3x^2 + 3 \, dx\\ &= \frac{3x^3}{3} + 3x + c ... \boxed{1} \\ \end{aligned}\\ \text{Substituting } x=1 \text{ and } y = 3 \text{ into } \boxed{1}:\\ \begin{aligned} 3 &= \frac{3\cdot 1^3}{3} + 3\cdot 1 + c\\ &= 1 + 3 + c\\ -1 &= c\\ y &= x^3 + 3x -1\\ \end{aligned}
Note that integrating products and fractions of functions (with a product/quotient rule like technique) is beyond the scope of this course. Below we detail how you should tackle these sorts of questions.
\text{Example 8.2: Evaluate } \int (x-2)(x-4) \, dx\\ \text{ } \\ \text{For all products of functions, simply expand everything out. }\\ \begin{aligned} \int (x-2)(x-4)\, dx &= \int (x^2-6x+8) \,dx\\ &= \frac{x^3}{3} - \frac{6x^2}{2} + 8x + c\\ &= \frac{x^3}{3} -3x^2 +8x + c\\ \end{aligned}\\ \text{ }\\ \text{ }\\ \text{Example 8.3: Evaluate } \int \frac{6x^3-2x^2+9}{3x^2}\,dx\\ \text{ } \\ \text{For all fractions of functions, split the fraction up.}\\ \begin{aligned} \int \frac{6x^3-2x^2+9}{3x^2}\,dx &= \int \frac{6x^3}{3x^2} - \frac{2x^2}{3x^2} + \frac{9}{3x^2}\,dx\\ &= \int 2x - \frac{2}{3} + 3x^{-2}\,dx\\ &= \frac{2x^2}{2} -\frac{2x}{3} +(-3x^{-1})+c\\ &= x^2 -\frac{2x}{3} - 3x^{-1} + c\\ \end{aligned}\\
Definite Integral
Definite integrals look very similar to indefinite integrals, however, they have terminals either side of the integral symbol: \begin{aligned} \int_{a}^{b} f'(x) \, dx &= [f(x)]_{a}^{b} \\ &= f(b) - f(a) \end{aligned}
Note that the final answer of a definite integral is a value, not an expression. The arbitrary constant is no longer present. After integrating the function, place it in square brackets with the terminals to the right of the square brackets like above. After that substitute the upper number (by position, not value) into the function and subtract the lower (by position, not value) number substituted into the function.
\text{Example 8.4: Evaluate } \int_{1}^{-5} 3x^2 + 3 \, dx\\ \text{ } \\ \begin{aligned} \int_{1}^{-5} 3x^2 + 3 \, dx\ &= [\frac{3x^3}{3} + 3x]_{1}^{-5}\\ &= [x^3 + 3x]_{1}^{5}\\ &= ((-5)^3 + 3\cdot (-5)) - (1^3+3\cdot 1)\\ &= -125 - 15 -1 - 3\\ &= -144\\ \end{aligned}
Area Between Curves
To find the area of a region between two curves from $$x=a$$ and $$x=b$$, where $$f(x) \leqslant g(x)$$ for $$x \in [a,b]$$, find the difference between the areas under the graph of $$y = f(x)$$ and the graph $$y = g(x)$$ over the interval $$[a,b]$$. That is, evaluate: $$\int_{a}^{b} f(x) - g(x) \,dx$$. In other words. Every time the graphs cross, you will need to create a new integral. For each integral, the equation of the graph which is visually lower is subtracted from the equation of the graph higher up. The lower terminal will be the smaller value in your interval and the upper terminal will be the larger value in your interval.
It is imperative to graph all equations that you are dealing with before attempting to find the area so that you know which equation has the larger values and for what x-values.
Note that the x-axis has equation y = 0
Below we have 3 examples to illustrate all cases you will encounter.
$\text{Example 8.5: Find the area bounded by the curve } y = \sin(x) \text{ and the x-axis for } x \in [0,\pi].\\ \text{ } \\ \text{Looking at the graph, } \sin(x) \text{is always above the x-axis,}\\ \text{ so we will subtract the equation of the x-axis from } \sin(x)\\$ \begin{aligned} \text{Area } &= \int_{0}^{\pi} \sin(x) - 0 )\, dx\\ &= [- \cos(x) ]_{0}^{\pi}\\ &= -\cos(\pi) -(-\cos(0))\\ &= -(-1) - (-1)\\ &= 2 \text{ units}^2\\ \end{aligned}\\
$\text{Example 8.6: Find the area between the curves } y = e^x+3 \text{ and } y = -1 \text{ from } x = 1 \text{ to } x = 4.\\ \text{ } \\ \text{Looking at the graph, } e^x+3 \text{is always above } y = -1 \text{, so we will subtract } -1 \text{ from } e^x+3 \\$ \begin{aligned} \text{Area } &= \int_{1}^{4} (e^x + 3 - (-1)) \, dx\\ &= \int_{1}^{4} (e^x + 4) \, dx\\ &= [e^x + 4x]_{1}^{4}\\ &= (e^4 + 4 \cdot 4) - (e^1 + 4\cdot 1)\\ &= e^4 - e + 12 \text{ units}^2\\ \end{aligned}\\
$\text{Example 8.7: Find the area bounded by the curves } y = \sin(x) \text{ and } y = \sin(2x) \text{ for } x \in [0,\pi].\\ \text{Looking at the graph, the two graphs clearly cross at one point.}\\\text{ We will need to find the intersection point.}\\ \text{For the first segment the graph of } \sin(2x) \text{ is higher up} \\ \text{But, for the second segment, the graph of } \sin(x) \text{ is higher up}\\$ \begin{aligned} \sin(x) &= \sin(2x)\\ x &= \frac{\pi}{3} \text{ (Using CAS calculator)}\\ \end{aligned}\\ \text{For } x \in [0,\frac{\pi}{3}], \sin(2x) \leqslant \sin(x) \text{, but for } x \in [\frac{\pi}{3}, \pi] \sin(x) \leqslant \sin(2x).\\ \text{ So, we must setup two integrals to calculate the combined area.}\\ \begin{aligned} &= \int_{0}^{\frac{\pi}{3}} \sin(2x) - \sin(x) \,dx + \int_{\frac{\pi}{3}}^{\pi} \sin(x) - sin(2x) \,dx\\ &= [-\frac{\cos(2x)}{2} - (-\cos(x))]_{0}^{\frac{\pi}{3}} + [-\cos(x) - (-\frac{\cos(2x)}{2})]_{\frac{\pi}{3}}^{\pi}\\ &= [-\frac{\cos(2x)}{2} + \cos(x)]_{0}^{\frac{\pi}{3}} + [-\cos(x) + \frac{\cos(2x)}{2}]_{\frac{\pi}{3}}^{\pi}\\ &= -\frac{\cos(2 \cdot \frac{\pi}{3})}{2} + \cos(\frac{\pi}{3}) - (-\frac{\cos(2 \cdot 0)}{2} + \cos(0)) + (- \cos(\pi) + \frac{\cos(2 \cdot \pi)}{2}) - (-\cos(\frac{\pi}{3}) + \frac{\cos(2 \cdot \frac{\pi}{5})}{2})\\ &= -\frac{\cos(\frac{2\pi}{3})}{2} + \cos(\frac{\pi}{3}) - (-\frac{\cos(0)}{2} + \cos(0)) + (- \cos(\pi) + \frac{\cos(2\pi)}{2}) - (-\cos(\frac{\pi}{3}) + \frac{\cos(2 \cdot \frac{\pi}{3})}{2})\\ &= -\frac{-\frac{1}{2}}{2} + \frac{1}{2} + \frac{1}{2} -1 - (-1) + \frac{1}{2} + \frac{1}{2} - \frac{(-\frac{1}{2})}{2}\\ &= \frac{1}{4} + \frac{1}{2} + \frac{1}{2} -1 + 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}\\ &= 3 \text{ units}^2\\ \end{aligned}\\
Signed Area
For any continuous function f over the interval [a,b], the signed area is calculated by evaluating: $\int_{a}^{b} f(x) dx$ This is not the true area, it considers any area under the x-axis to be negative. The signed area is generally useless, but does have some application in kinematics which is discussed further down.
Average Value
The average value (not to be confused with average rate of change!) of a function f over the interval [a,b] is equal to the height of a rectangle having the same area as the area under the graph over the same interval. That is, $\frac{1}{b-a} \cdot \int_{a}^{b} f(x) dx$
\text{Example 8.8: Find the average value of the } f(x) \text{ over the interval } [2,5] , f(x) = x^2 + 1\\ \begin{aligned} \text{Average Value} &= \frac{1}{b-a} \cdot \int_{a}^{b} f(x)\, dx\\ &= \frac{1}{5-2} \cdot \int_{2}^{5} x^2 + 1\, dx\\ &= \frac{1}{3} [\frac{x^3}{3}+x]_{2}^{5} \\ &= \frac{1}{3} ((\frac{5^3}{3} + 5) - (\frac{2^3}{3} + 2)) \\ &= \frac{1}{3} ((\frac{125}{3} + 5) - (\frac{8}{3} + 2)) \\ &= \frac{1}{3} (\frac{125}{3} + 5 - \frac{8}{3} - 2) \\ &= \frac{1}{3} (\frac{117}{3} + 3) \\ &= \frac{117}{9} + 1 \\ &= 13 + 1 \\ &= 14\\ \end{aligned}\\
Kinematics
In kinematics you will be given an equation for the displacement, velocity or acceleration of an object possibly with some more information. Below we detail how you should calculate different quantities. But, first let’s define some key terms
Term Definition Position An object’s location relative to a selected point. Displacement The direct distance an object is from a selected point Distance How far an object moves in total Velocity The change in position relative to time. Speed The change in distance relative to time Acceleration The change in velocity relative to time
Let x(t) be the expression representing the position of an object
Let v(t) be the expression representing the velocity of an object
Let a(t) be the expression representing the acceleration of an object
To get between position, velocity and acceleration we simply use differentiation or integration. \begin{aligned} x’(t) &= v(t) \\ v’(t) &= a(t) \\ \int a(t) \, dt &= v(t) \\ \int v(t) \, dt &= x(t) \\ \end{aligned}
Below we have assumed that you are able to obtain all of the equations for position, velocity and acceleration of an object. Note that, when integrating from acceleration to velocity for example, you will need to have extra information to find the exact equation of velocity due to the introduction of the arbitrary constant following the indefinite integral calculation.
Term Standard Question Calculation Position Find the position of an object at time t = u $x(u)$ Displacement Find the displacement of an object between time t = u and t = w $x(w) - x(u) \\ \text{It is also the signed area of the graph v(t)} \\ \int_{u}^{w} v(t) \, dt$ Distance Find the distance an object travels between t = u and t = w The area bounded by the graph of v(t) and the x-axis. $\int_{u}^{w} \lvert v(t) \rvert \, dt$ Velocity Find the velocity of an object at t = u $v(u)$ Average velocity Find the average velocity of an object between t = u and t = w $\frac{x(w)-x(u)}{w-u}$ Speed Find the speed of an object at t = u $\lvert v(u) \rvert$ Average speed Find the average speed of an object between t = u and t = w Total distance travelled in the interval divided by the length of the time interval. $\frac{\int_{u}^{w} \lvert v(t) \rvert \, dt}{w-u}$ Acceleration Find the acceleration of an object at t = u $a(u)$ Deceleration Find how fast the car is decelerating at t = u Same as acceleration. However, deceleration is positive. That is, if acceleration is -8, then deceleration is 8. Average acceleration Find the average acceleration of an object between t = u and t = w $\frac{v(w) - v(u)}{w-u}$
Integration by Recognition
Common, but one of the harder integration questions. An example is given below. \text{Example 8.9: If }f(x)=x\cdot \cos(3x),\text{ then }f^\prime(x)=\cos(3x)-3x\cdot \sin(3x)\\ \text{Use this fact to find an antiderivative of }x\cdot \sin(3x)\\ \begin{aligned} \text{ } \\ \text{} \int f^\prime(x)\,dx &= f(x) + c\\ \int \cos(3x)- 3x\cdot \sin(3x))\,dx &= x\cdot \cos(3x) + c\\ \int \cos(3x)\, dx - \int 3x\cdot \sin(3x))\,dx &= x\cdot \cos(3x) + c\\ \int \cos(3x)\, dx - 3\int x\cdot \sin(3x))\,dx &= x\cdot \cos(3x) + c\\ \int (x\cdot \sin(3x))\,dx&=\frac{1}{3}(\int \cos(3x)\,dx - x\cdot \cos(3x)) + c\\ &=\frac{1}{3}\cdot (\frac{1}{3}\cdot \sin(3x) - \cdot x\cdot \cos(3x)) + c\\ &=\frac{1}{3}\cdot \frac{1}{3}\cdot \sin(3x) - \frac{1}{3}\cdot x\cdot \cos(3x) + c\\ &=\frac{\sin(3x)}{9} - \frac{x\cdot \cos(3x)}{3} + c\\ \end{aligned}\\ \text{An antiderivative is } \frac{\sin(3x)}{9} - \frac{x\cdot \cos(3x)}{3} \, ,(c = 0) \\
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Question
# An ideal gas at pressure 2.5 × 105 Pa and temperature 300 K occupies 100 cc. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure (b) the final temperature and (c) the work done by the gas in the process. Take γ = 1.5
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Solution
## Initial pressure of the gas, P1 = 2.5 × 105 Pa Initial temperature, T1 = 300 K Initial volume, V1 = 100 cc (a) For an adiabatic process, P1V1γ = P2V2γ ⇒ 2.5 × 105 × V1.5 = ${\left(\frac{\mathrm{V}}{2}\right)}^{1.5}$ × P2 ⇒ P2 = 7.07 × 105 = 7.1 × 105 Pa (b) Also, for an adiabatic process, T1V1γ−1 = T2V2γ−1 ⇒ 300 × (100)1.5−1 = T2 × ${\left(\frac{100}{2}\right)}^{1.5-1}$ = T2 × (50)1.5−1 ⇒ 300 × 10 = T2 × 7.07 ⇒ T 2 = 424.32 K = 424 K (c) Work done by the gas in the process, $W\mathit{=}\frac{nR}{\mathit{\left(}\gamma \mathit{-}\mathit{1}\mathit{\right)}}\mathit{}\left[{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}\right]\phantom{\rule{0ex}{0ex}}\mathit{=}\frac{{P}_{\mathit{1}}{V}_{\mathit{1}}}{T\mathit{\left(}\gamma \mathit{-}\mathit{1}\mathit{\right)}}\mathit{}\mathit{}\left[{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}\right]\phantom{\rule{0ex}{0ex}}=\frac{2.5×10}{300×0.5}×\left(-124\right)\phantom{\rule{0ex}{0ex}}=-20.67\approx -21\mathrm{J}$
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# 23X+2y=32
justaguide | Certified Educator
The equation 23x + 2y = 32 contains two variables. It is only possible to express x in terms of y.
23x + 2y = 32
=> 23x = 32 - 2y
=> `x = 32/23 - (2y)/23`
The value of x is `x = 32/23 - (2y)/23`
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# How many grams are in 1/8 cup? Converting the 1/8 cup commonly used for baking recipes into grams
It can be difficult to know exactly how many grams are in 1/8 cup. Understanding the metric system is essential knowledge that involves measuring ingredients precisely. Therefore, all recipes are consistent every time they’re made.
In this blog post, Jan Cranitch will take a look at converting the 1/8 cup commonly used for baking recipes into grams to help you answer the question above. Now, let’s explore!
## How many grams are in 1/8 cup?
A 1/8 cup of a dry ingredient is equivalent to about 15.6 grams or 0.54 ounces. This measurement will depend on the type of ingredient and dense it’s being measured. For accuracy, it is best to use a kitchen scale whenever possible. However, if you do not have access to a kitchen scale, the above measurements can be used as a rough estimate.
For liquid ingredients, a 1/8 cup is equivalent to about 29.5 grams or 2 tablespoons. To get an accurate measurement of your ingredients, it is always best to use the appropriate kitchen tools.
## How to convert 1/8 cup to grams?
There are three common methods for converting cups to grams.
### Way 1: Using a math formula to convert 1/8 cup to grams
Let’s multiply the volume by 236.588236 times the density of the ingredient or material. Therefore, we have the formula as follows:
grams = cups * 236.588236 * ingredient density
To convert 1/8 cup to grams, multiply 1/8 by 236.588236 times the density of the ingredient or material.
For example, if we take water as an ingredient, the density of water is 1 (g/cm³). As a result, the conversion from 1/8 cup to grams of water is:
grams = 1/8 * 236.588236 * 1
= 29.5735295 g.
Therefore, there are 29.5735295 grams of water in 1/8 cup.
### Way 2: Converting 1/8 cup to grams with an online calculator
Another way is to use the online calculator to convert 1/8 cup to grams easily and quickly. Follow the steps below for step-by-step instructions:
• Step 1: Go to the Google search bar and type in “cups to grams calculator”.
• Step 2: Select the first link to access a cups to grams calculator.
• Step 3: Enter 1/8 cup into the calculator, and choose the ingredient or material from the drop-down list.
• Step 4: Click on Calculate button.
• Step 5: Get the result of 29.5735295 grams for 1/8 cup.
It is important to note that the number of grams in 1/8 cup may vary based on different ingredients and materials, as each has its own specific density. Before converting cups to grams, it is best to check the density of the ingredients. Additionally, it is useful to keep in mind that 1/8 cup equals two tablespoons. This can be a helpful conversion when measuring ingredients.
### Way 3: Using a conversion chart to convert 1/8 cup to grams
Converting 1/8 cup to g will change with the density of the material it’s made of. Look at the chart below to see how many grams of various liquid and dry ingredients are in a cup.
## Converting the 1/8 cup commonly used for baking recipes into grams
It is an essential part of being able to accurately measure our ingredients. As a general rule, 1/8 cup of a dry ingredient such as flour or sugar is equivalent to 16 grams. If the recipe calls for 1/8 cup of liquid, such as water or oil, it is equal to 29.5 grams. These measurements are based on US standards, so they may be a little different in other countries.
For the best results, you should use a digital kitchen scale that is accurate to make sure your measurements are correct.
In addition, you can use the conversion chart below to know how to convert 1/8 cup to grams for common ingredients in baking recipes.
Ingredient 1 Cup 1/2 Cup 1/8 Cup Honey 336 g 167 g 42 g Margarine 230 g 115 g 28.75 g Oats 102.2 g 51.1 g 12.7 g Rice (uncooked) 178.15 g 89.1 g 22.3 g Almonds 132 g 66 g 16.5 g Dry good 128 g 64 g 16 g Bread flour 136 g 68 g 17 g
## 1/8 cups to other volume units conversion
1/8 cup to teaspoons 6 tsp 1/8 cup to tablespoons 2 tbsp 1/8 cup to fluid ounces 1 fl oz 1/8 cup to pints 0.0625 pt 1/8 cup to quarts 0.03125 qt 1/8 cup to gallons 0.007812 gal 1/8 cup to milliliters 9.57353 ml 1/8 cup to liters 0.029574 l 1/8 cup to ounces 1.043175556436 oz 1/8 cup to pounds 0.06 lb
## How many grams is 1/8 cup of butter?
1/8 of a cup of butter is equal to 28.4 grams. But the exact amount of butter needed might be a little different depending on the size and texture of the product. Therefore, you can use the conversion table below to convert 1/8 cup of butter:
Cups Grams 1/8 cup 28.4 g 1/4 cup 57 g 1/3 cup 76 g 1/2 cup 114 g 2/3 cup 151 g 3/4 cup 170 g 1 cup 227 g
## FAQs: How many grams are in 1/8 cup? Converting the 1/8 cup commonly used for baking recipes into grams
### What does a 1/8 cup resemble?
1/8 cup is approximately the size of a large egg. A tennis ball is about the size of 1/2 cup. One cup is approximately the size of an apple or a baseball.
### What is the weight of a one-eighth cup?
1/8 cup is equal to 2 tablespoons or 1 fluid ounce.
1/8 cup = 2 Tbsp = 1 fl oz
1/4 cup = 4 Tbsp = 2 fl oz
1/3 cup = 5 Tbsp + 1 tsp = 2.65 fl oz
3/8 cup = 6 Tbsp = 3 fl oz
1/2 cup = 8 Tbsp = 4 fl oz
5/8 cup = 10 Tbsp = 5 fl oz
2/3 cup = 10 Tbsp + 2 tsp = 5.3 fl oz
3/4 cup = 12 Tbsp = 6 fl oz
7/8 cup = 14 Tbsp = 7 fl oz
1 cup = 16 Tbsp = 8 fl oz
The result is taken by: https://msdh.ms.gov/
### How much is 1/ 8 in cooking?
1 tablespoon (tbsp) = 3 teaspoons (tsp)
1/8 cup = 2 tablespoons
1/6 cup = 2 tablespoons + 2 teaspoons
1/4 cup = 4 tablespoons
1/3 cup = 5 tablespoons + 1 teaspoon
### Is a 1/8 Cup a coffee scoop?
Usually, a coffee scoop is about 2 tablespoons or 1/8 cup. A coffee scoop is a kitchen tool used to measure the number of ground coffee beans. Some coffee scoops can also be used to measure the amount of loose tea.
## Conclusion
How many grams are in 1/8 cup? Converting the 1/8 cup commonly used for baking recipes into grams can be tricky, but with a little bit of practice, it becomes much easier. With this converter, you’ll always know how many grams are in 1/8 cup no matter what recipe you’re using. And if you ever have any questions, ukpfna.com is always here to help.
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Question about Texas Instruments TI-83 Plus Calculator
# Calculating a 93% confidence interval
If the standard deviation for lifetimes of vacuum cleaners is estimated to be 400 hours, how large a sample must be taken in order to be 93% confident that the marge of error will not exceed 50 hours?
I know that n+(za/2)2 standard devation squared/the margin of error squared. What I don't know is how do I get the 93% confidence which is the z a/2 squared
Posted by on
1 - .93 = .07/2 = .035
Posted on Nov 05, 2008
Hi,
a 6ya Technician can help you resolve that issue over the phone in a minute or two.
Best thing about this new service is that you are never placed on hold and get to talk to real repair professionals here in the US.
Goodluck!
Posted on Jan 02, 2017
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my-video-file.mp4
×
## Related Questions:
### Finding the margin of error with a sample size of 1,040, sample mean=46,239, sample standard deviation=21,000
I'm assuming here that 1,040 means one thousand and forty, not a metric decimal comma.
First we need the Std Dev of the mean value itself. This is
s / sqrt (n) = 21000 / sqrt (1040) = 651.2
Then the confidence interval for the mean value is
mean value ± Z * ( s / sqrt (n) )
where Z is an estimator related to the required confide
CL Z
99% 2.576
98% 2.326
95% 1.96
90% 1.645
So for a confidence of 95% the margin of error for the true value of the mean is
46239 ± 1.96 * 651.2 or
46239 ± 1276.3
that is, the mean of a sample from this process will be in this range to 95% confidence.
The margin of error for a future sample value would be
46239 ± 1.96 * 21000 or
46239 ± 41160
that is, a single sample value from this process will be in this range to 95% confidence.
.
Jan 05, 2017 | Office Equipment & Supplies
### I need to find the a 95% confidence interval estimate for population mean. My SD is 100, sample sz. is 64, and sample mean is 350.
First you need the Std Error of the mean value, a measure of the dispersion of that mean value.
SE = sample std deviation / sqrt (sample size)
= 100 / sqrt (64)
= 100 / 8
= 12.5
Then we use a figure for the number of std errors either side of the mean value, which make up a 95 % confidence interval. This is ± 1.96 std errors, from tables of the Normal Distribution.
So the confidence interval is
350 ± 1.96 * 12.5 or
374.5 to 325.5
.
Dec 11, 2015 | Institute of Mathematics and Statistics...
### Construct a 99 confidence interval for the population mean if my sample mean is 125, and Std Dev is 8, and population size is 64
First you need the Std Error of the mean value, a measure of the dispersion of that mean value.
SE = sample std deviation / sqrt (sample size)
= 8 / ? 64
= 1
Then we use a figure for the number of std errors either side of the mean value, which make up a 99 % confidence interval. This is ± 2.58 std errors, from tables of the Normal Distribution.
So the confidence interval is
125 ± 2.58 * 1 or
127.58 to 122.42
.
Dec 10, 2015 | Institute of Mathematics and Statistics...
### I have DCA analyzer from Siemens healthcare and i have problem E24 ,How Can Solve it
Manual is available in PDF format from here
DCA Vantage Operator's Guide 109
Troubleshooting
E22 - Optical
of range
A Sample or Reference
during an Air
measurement is too
high.
2. Run an optical test.
3. If you are unable to run an
technical support provider.
E23 -
Excessive
noise on the
Sample
channel
The standard deviation
of the 16 readings at the
Sample channel for Dark
large.
2. Run an optical test.
3. If you are unable to run an
technical support provider.
E24 -
Excessive
noise on the
Reference
channel
The standard deviation
of the 16 readings at the
Reference channel for
Dark, Air or Sample
2. Run an optical test.
3. If you are unable to run an
technical support provider.
E26 -
Excessive
noise in sample
The standard deviation
of the 16 readings at the
Sample channel during a
position.
2. Run an optical test.
3. If you are unable to run an
technical support provider.
E27 -
Excessive
Lamp Drift
The change in mean
signal between
is too large at either the
Sample channel or the
Reference channel.
2. Run an optical test.
3. If you are unable to run an
technical support provider.
E30 - Thermal
control system
error - low
The temperature
measured by one of the
cartridge holder
thermistors is ?2?C.
2. Restart the system.
3. If you are unable to perform a
successful restart, contact
provider.
E
Sep 21, 2015 | Siemens Dishwashers
### What is deviation standard
The standard deviation is a measure of how "tight" the samples are distributed around your mean.
In layman's terms, a small standard deviation indicates that most of your measurements are in the vicinity of the means; a large standard deviation corresponds to readings that are all over the place.
You could also say that the smaller the SD, the more your mean is representative of the data set.
For a better explanation, just look up Standard deviation on Wikipedia!
May 03, 2014 | Audio Players & Recorders
### How do i find sample size?
Count the number of items in the sample whose statistics you're calculating. For example, if you're calculating the mean and standard deviation of five items then the sample size is five.
Apr 18, 2012 | Casio FX-9750GPlus Calculator
### How to get sample size
Count the number of items in the sample whose statistics you're calculating. For example, if you're calculating the mean and standard deviation of seven items then the sample size is seven.
Apr 18, 2012 | Casio FX-9750GPlus Calculator
## Open Questions:
#### Related Topics:
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Mechanics Index
Introduction to Kinematics
Introduction
Kinematics deals with the motion of bodies and requires consideration of the geometry and time. The forces involved in the motion of body are not considered in this area of work.
The notes on this page include use of vectors. The necessary background information is provided on page Vectors
The contents of this page are listed below
Basics
Time..The absolute measure of the orderly succession of events. Unit of time is the second Particle..a body is assumed with negligible dimensions Rigid body..A body with all internal points fixed relative to each other
Nomenclature
s = distance travelled in time -(m) v = average velocity - (m/s) a = acceleration - (m/s2) t = time - (s) u = initial velocity - (m/s) v = final velocity - (m/s) θ = Rotation angle - (radians) ω 1 = Initial angular velocity -rads/s ω 2 = final angular velocity - rads /s n = angular speed revs/min α = Angular Acceleration rads/s 2
Particle Kinematics.
This discipline of mechanics deals with the displacement of particles over time without reference to the forces that cause the motion, velocity and acceleration of the particle.
Linear Motion..
Considering a particle moving along a path is space from position A to position B over a time interval. The position vector r locates the particle relative to the reference frame (say cartesian xyz) . The distance the particle moves along the path is designated s. The change in position of the particle over the time is its linear displacement and is identified by the vector ( r B - r A )
The speed of the particle over a time period = (s 2 - s 1 ) /(t 2 - t 1 ) = the speed of the particle
As the time interval reduces to zero the change in position Δ r /Δ t = ( rB - rA ) / (t2 - t1 ) => dr/dt = v = the instantaneous velocity The instantaneous velocity is tangential to the particle path
The derivative (with respect to time) of the instantaneous velocity is d 2r /dt 2 = the instantaneous acceleration.
Textbook formulae..Linear motion subject to uniform acceleration.
v = u + at
s = ut + a.t2 /2
s = (u + v) /2
v 2 = u 2 + 2.a.s
Using vector algebra....
As an example of straight line motion consider a particle moving with a constant acceleration in the x direction. C1 is a constant determined from the velocity at t = 0: Normally assuming initial velocity = u then C1 = u C2 is a constant determined from the displacement at t = 0: Normally assuming initial displacement = 0 then C2 = 0
Angular Motion..
A point rotates from A to B around a center C through a small angle θ in a time t . As the time interval approaches zero the instantaneous angular velocity d θ /dt = ω . This angular velocity is a vector cross product and is shown in the direction as indicated below..
The velocity of the point is obtained as a vector product of the angular velocity and the radius
v = ω x r
Angular velocity vectors are free vectors but they are not commutative..
Textbook formulae.. Angular motion subject to uniform acceleration
ω = 2.π .n /60
ω 2 = ω 1 + α . t
θ = ( ω 2 + ω 1 ) . t /2
ω 22 = ω 12 + 2 . α . s
θ= ω 12 + α . t 2 /2
Motion in a circle..
A particle moving in a circle of radius (r) at a constant velocity (v) and constant angular velocity of ω = v /r is accelerating towards the centre of the circle at a constant rate of v 2 /r = ω 2 . r. The acceleration is primarily due to the rate of change of angular position of the particle...
An example of linear motion is that of a particle moving at a fixed velocity in a circular motion..
Differentiate with respect to time
Differentiate again with respect to time using ( d(uv) = u.dv + v du )
Relating the Tangential velocity to the x an y velocity.
Introducing velocity into equation
Finally establishing formula for accelaration
Absolute and relative motion..
A reference frame can be fixed or relative. In Newtonian mechanics the fixed reference frame is the primary inertial system in which there is zero absolute motion in space. This is entirely impractical because it would have to take into account the movements of the earth relative to the sun and the sun in the universe etc. A reference frame based on the sun is called a Heliocentric reference frame and this again is impractical for normal mechanical engineering. A practical reference frame is one based on the earth. This is called a geocentric reference frame and is sufficient as an inertial frame for normal engineering. Therefore when we identify the velocity of a vehicle or the acceleration of an object they are generally relative to the earths motion.
When an object is moving relative to a fixed reference frame it is called absolute motion. When it is moving relative to a reference frame that is moving it is called relative motion. An example of relative motion is someone moving in a travelling train or car.
Rotating unit vector
Consider a rotating unit vector e..
Vector of Point in a rotating reference frame
A point P with a vector position r is moving in the xy plane. The position of this vector is identified in respect to unit vectors e 1 and e 2 fixed to a reference frame rotating at and angular velocity ω .. r 1 and r 2 are components of the motion in the e 1 and e 2 directions.
The position is expressed as follows..
The derivative of this equation with respect to time results in.
It is shown above that de/dt = ω x e therefore
Rewriting the equation
Velocity of a point in a moving reference frame
A point P moving in the xy frame as shown below has a position vector of
r p = r o' + r
ro' locates O' relative to the fixed xy axes and r locates the point P relative to a axes (x 1,x 2)
If the moving frame is not rotating ω = 0
Acceleration of a Particle
Differentiating (with respect to time) the equation for velocity (v) results the equation for accelaration (a)
Following the principles established above...
Note : The term ..... ..... is called the coriolis acceleration .
This represents the difference between the acceleration of P relative to O' as measured from the rotating and the non-rotating axes.
Rigid Body Kinematics in one plane.
Rigid body..A body with all internal points fixed relative to each other. Three co-ordinates are required to determine the position and orientation of a body in plane motion. A rigid body has three degrees of free in plane motion.
Definitions..
Rectilinear Translation...All point on a rigid body move in a straight line Curvilinear Translation... The orientation of all points in a rigid body remain fixed while the body moves along a curved path Rotation about a fixed line.. All points in a rigid body move in a circular motion about a fixed line Plane motion...Each point in rigid body moves in a path parallel to a fixed plane
Plane motion for a rigid body can be easily developed from the equations for motion of a particle. They are the same equations for the motion of a point but as all points are fixed relative to each other the term involving velocities and accelerations of points relative to each other are zero.
The equations for velocity and acceleration when the reference axes x',y' are fixed to the rigid body ,are as follows...
Important Note:
It can be easily proved that whichever position on the rigid body is selected for the reference position O' the angular velocity and angular acceleration vectors are not changed.
Instantaneous Centre of Rotation
For any body moving in space the motion can be defined from the position, velocity, acceleration of any point on the body with the angular velocity and acceleration. There is a point for this body for which the instantaneous translational velocity is zero i.e only rotation of the body about the point is occuring. This point is called the instantaneous center of rotation and its location relative to the body is dependent on the relative values of the linear and angular velocities For near zero angular velocities (near translation motion) the location is at near infinity.
The location of the instantaneous centre of rotation is easy to locate if the velocity and angular velocity of a point are known
If the location of a point and its velocity and angular velocity are known.
Considering point A the point C is on a line passing through A normal to the direction of the velocity. rA = VA / ω
If the location of two points and their velocities
The point C is at the intersection of the lines drawn normal to the velocities of each of the points.
If the lines are co-incident then radius is established by geometry.. AC = AB.(VA) /(VA-VB)
Rigid Body kinematics in 3 Dimensions
Introduction
A particle requires three points to specify its position relative to the selected coordinate system and is therefore identified as having 3 degrees of freedom . For a rigid body the location of three separate points of the body relative to the selected co-ordinate system are required. Because the points are fixed relative to each other only six independent co-ordinates are required to locate the body is space and a unrestrained rigid body is therefore said to have six degrees of freedom. e.g. A rigid body may be positioned by locating the position of one point of the body ( 3 co-ordinates), then positioning a line on the body (2 co-ordinates) and finally identifying a rotation about the line (1 co-odinates). This sums up to 6 co-ordinates
The motion of a rigid body in three dimension can have a number of modes..
Rectilinear Translation...All points on the rigid body move in a straight line Curvilinear Translation... The orientation of all points in a rigid body remain fixed while the body moves along a curved path Rotation about a fixed axis.. All points in a rigid body move in a circular motion about a fixed axis Rotation about a fixed point.. All points in a rigid body move in a circular motion about a fixed point General motion...The rigid body motion includes translation and rotation about Rotation Notes. If a rigid body is rotated from a position "A" through an angle of π/2 about the x axis and then rotated π/2 about the y axis it has been moved to a certain position "B". If the same rigid body is rotated from a position "A" through and an angle of π/2 about the y axis and then is rotated π/2 about the x axis it has been moved to a position "C" which is not the same as "B". Finite rotation does not obey the laws of vector addition
Euler & Chasles;'s Theorems
Euler's Theorem
If a point on a rigid body does not change its position then any series of successive rotations can be compounded to a rotation about a single axis
Chasles's Theorem
Any displacement of a rigid body may be compounded from a single rotation about any selected point plus a translation of that point.
A third theorem ( Poinsot's Central axis theorem) is also provided..
Any finite displacement of a rigid body may be reduced to a single rotation about an axis plus a translation parallel to the same axis. This theorem only relates the displacement and not to the paths taken by points.
These theoerems can be applied to angular velocities
Any motion of a rigid body can be described by a single angular velocity plus a translational velocity parallel to the angular velocity vector.
Any motion of a body about a point may be represented by a single angular velocity about an axis through that point.
Any motion of a rigid body my be represented by the velocity of a point plus the angular velocity about an axis passing through the points...
Rectilinear Translation.& Curvilinear Translation
All points on the body will move in parallel straight or curve lines. There is no relative velocity or acceleration between any points on the body.
Position... r P = r Q + r PQ
Velocity...v P = v Q + v PQ
Acceleration...a P = a Q + a PQ
Rotation about a fixed axis.
Rotation of a rigid body about fixed axis is shown in the diagram. The vector of the rotary motion has sense and direction in accordance with the right hand rule which is fixed along the direction of the axis. There is zero velocity due to rotation at the axis.
Rotation about a fixed Point.
Rotation of a rigid body about a fixed point can always be reduced to rotation of a body about an instantaneous axis of rotation.
Considering for example a rigid body (cone) which is rotating about a horizontal axis (shaft),angular velocity = ω1,and the shaft is itself rotating about a vertical axis, angular velocity = ω2.
If ω2 = 0 then the axis rotation is the centre line of the shaft and the velocity of any point on the shaft is proportional to the radius (max = r)from the shaft. If the rotation about the vertical axis is increased to a certain value the the velocity of the shaft axis is proportional to ω2 x the shaft length (max = l) from the axis (radius). If the two angular velocities are the same and the radius of the cone r = l then at any instant the velocity of top surface of the cone is zero. i.e the top surface is the instantaneous axis of rotation.
This grossly simplified representation illustrates the principle .. if ω2 is increased the body cone representing the path of the instantaneous axis of rotation will obviously be larger than the actual surface of the cone.
For the model illustrated there is also a space cone which is the path the instantaneous axis or rotation follows in space- (an inverted cone centered the vertical axis )..
For the example above the instantaneous angular velocity of the cone =
ω = ω1 + ω2 = ω1i + ω2j
In the real world the motions and shapes are not as shown but the interaction between the body rotation on its axis and its angular motion is space still results in an instantaneous axis of rotation between a instantaneous body cone and an instantaneous space cone. The body cone may rotate outside or inside the space cone
One determining the instantaneous angular velocity the angular acceleration α is tangential to the contact point of the two cones as shown. The velocity and acceleration at any point is simply determined as shown below
General Motion.
The general case of 3D motion of rigid bodies reduces to translation + rotation about a fixed axis. This is basically a generalisation of the theorems provided above.
For a body possessing linear plus angular motion it is often not possible to have an instantaneous axis of rotation because all points may have non-zero velocities. The most convenient method of kinematic analysis of rigid bodies in space 3 dimensions is by using the principles of relative motion. The primary (absolute)reference system XYZ are supplemented with a reference system(xyz) attached to some point on the rigid body.
The attached reference system may be a translating one or a rotating one.
Translating reference system
This motion simply develops the motions already studied in the previous section for particle motions and rigid body motions. The basic motion equations are :
r P = r Q + r PQ
v P = v Q + v PQ
a P = a Q + a PQ
The above equations are sufficient if the angular velocity ω is zero however if this is not the case the following equations for velocity and acceleration are more definitive;
Rotating reference system
A more general form of the relative reference axis method uses rotating reference axes. The reference axes xyz are rotating with an angular velocity of Ω. The rigid body has a rotation velocity ω as before
The basic motion equations are as above. The expression for velocity and acceleration of point P are provided below. The derivation is a simple extension of that provided above for 2 Dimensional motion with a rotating relative axis with the third dimension (z) added.
v r and a r = are the velocity and acceleration of P relative to the rotating xyz axis.
Note:
The equations above are based on the general case of the angular velocity of the rigid body (ω)is different to the angular velocity of the axis ( Ω). If the reference axis is fixed to the body then ω will equal Ω and vr and a r will equal zero... For this case the formula will be the same as for the translating reference axis above..
Links to Kinematics Physics - Kinematics..An unusual but useful information source Rigid body dynamics..Chris Hecker.. Set of articles for designers of realistic games - Good dynamics analysis info. Basic kinematics..A download Basic but clear information.. Kinematics -Wikipedia..Detailed informative notes..
Mechanics Index
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# Marbles in a Bag Probability Question
Let's pretend I have a bag of 40 marbles. 19 marbles are red, 21 marbles are blue. If I randomly pick 5 marbles out of the bag, what is the probability that 3 of those 5 marbles are red?
Related Set Theory, Logic, Probability, Statistics News on Phys.org
phinds
Gold Member
2019 Award
On this forum, we try to help folks learn how to solve problems. We don't spoon feed answers.
And by the way, this is a homework type problem so you are supposed to use the homework template. Please read the forum rules.
Chronos
Gold Member
Keep in mind that it makes a difference if you are looking for exactly 3 of 5, or at least 3 of 5.
Keep in mind that it makes a difference if you are looking for exactly 3 of 5, or at least 3 of 5.
Keep this in mind, it does in deed make a difference!
Hello all! Sorry for the late response, hope you're all still around. Let's pretend I want exactly 3 out of 5, since it's not too hard (I think) to then add the probabilities of 3/5 +(or) 4/5 +(or) 5/5.
I can work it out on a tree, for instance:
19/40*(..conditional probabilities..) + 21/40*(conditional probabilities)
with 5 separate columns (if that makes sense) but I know there has to be a better way.
I understand how basic combinations could give us the chances of pulling 3 red marbles in 3 draws (i.e. all red, or all blue) but I get really messed up when I try to think about 3 red marbles out of 5 draws out of 40 marbles.
It's been a while since I did statistics haha, some hints would be nice! :D And sorry about the homework template / improper forum section, it's also been a bit since I've come to PF. No excuse, of course, totally my fault. I can re-post in the homework section if it's an issue.
Don't think about conditional probabilities, just count. You know how many ways there are to draw 5 from 40. Now if 3 are red then you must have chosen 3 from 19 and the remaining 2 from 21. See if that helps and post your answer. Simple counting and multiplication, no need to think harder than that.
Yes, 5C40 gives 658,008 ways of picking 5 marbles out of 40.
Does that not ignore certain conditions of non-repetition though? Once you draw the first marble, the probabilities of the next draw have already changed.
I believe it is correct that there are 5! = 120 ways of picking 3 red and 2 blue marbles, if that is anywhere near where you are getting at. Thanks for your help so far!
HallsofIvy
Homework Helper
One way of picking 3 red and 2 blue marble is "RRRBB".
The probability the first marble you pick is red is, of course, 19/40. Now there are 39 marbles left and 18 are red. The probability that the second marble is red is 18/39. Now there are 38 marbles left and 17 are red. The probability that the third marble is red is 17/38. Now there are 37 marbles left and 21 of them are blue. The probability the fourth marble is blue is 21/37. There are now 36 marbles left and 20 of them are blue. The probability the fifth marble is blue is 20/36.
That will give you the probability of three red marbles followed by two blue marbles.
But if you calculate a different order, say RRBRB, you will see that while you get different fractions, the numerators and denominators are the same, just in different orders. That is, the probability of 'three red and two marbles' in any specific order is the same! You only need to multiply by the number of different orders.
But that is NOT 5!. Since all red are the same and all blue marbles are the same, the number of different orders is
$$_5C_3= \begin{pmatrix}5 \\ 3\end{pmatrix}= \frac{5!}{3!2!}= \frac{5(4)}{2}= 10$$.
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http://mathhelpforum.com/advanced-statistics/200623-biased-ness-ols-estimates-print.html
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# Biased-ness of OLS estimates
• Jul 4th 2012, 06:35 AM
usagi_killer
Biased-ness of OLS estimates
http://img84.imageshack.us/img84/3668/bias2.jpg
If the picture is too small, please view attached pic.
Hey guys, so I'm just reading through the proof of the biased-ness of OLS estimates if an important independent variable is left out and I'm just stuck at the very end, basically the mathematical part that confuses me is the part circled in red.
When we take the expectation of $\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ how does it simply end as the same thing?
Ie, Why is $E\left[\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}\right] = \frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ ?
Since $\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ isn't a constant as it depends on $n$ so it's like a random variable which outputs different values as $n$ changes, so why does it follow the rule that $E(c) = c$ where $c$ is a constant?
Thanks :)
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https://screenbooks.net/plisss-tolong-jawabannya/
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# Plisss tolong jawabannya
Posted on
Plisss tolong jawabannya
Nilai a
180° = 105° + 3a
3a = 180° – 105°
3a = 75°
a = 25°
Nilai
180° = 105° + 3
3b = 180° – 105°
3b = 75°
b = 25°
a + b = 25° + 25°
a + = 50°
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https://studydaddy.com/question/the-south-carolina-department-of-natural-resources-personnel-enforce-hunting-and
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Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# The South Carolina Department of Natural Resources personnel enforce hunting and fishing regulations and conduct routine safety checks on...
The South Carolina Department of Natural Resources personnel enforce hunting and fishing regulations and conduct routine safety checks on recreational boats. Past experience indicates that 15% of all boats inspected have at least one safety violation. Recent accidents on lakes in South Carolina have prompted calls for more extensive inspections. A random sample of recreational boats will be obtained and inspected. If there is evidence that the true proportion of boats with safety violations, p, is more than 15%, a methodical inspection of every boat launched at popular sites will be started. Answer the following three questions:
a. State the null and alternative hypotheses in terms of p.
Ho: p is less than 0.15; Ha: p is greater than 0.15
Ho: p is greater than 0.85; Ha: p is equal to 0.85
Ho: p is equal to 0.15; Ha: p is greater than 0.15
Ho: p is equal to 0.15; Ha: p is equal to 0.15
b. Describe a type I error and a type II error in this context.
A type I error: decide p is greater than 0.15 when the true proportion is really 0.15 (or less). Type II error: decide p = 0.15 (or less) when the true proportion is really smaller than 0.15.
A type I error: decide p is greater than 0.15 when the true proportion is really 0.15 (or less). Type II error: decide p = 0.15 (or less) when the true proportion is really greater than 0.15.
A type I error: decide p is less than 0.15 when the true proportion is really 0.15 (or more). Type II error: decide p = 0.15 (or less) when the true proportion is really greater than 0.15.
A type I error: decide p = 0.15 (or less) when the true proportion is really greater than 0.15. Type II error: decide p is greater than 0.15 when the true proportion is really 0.15 (or less).
c. What happens to the probability of a type I error as the value of p approaches 0.15 from the right (that is, 0.20, 0.19, . . .)?
The probability of a type I error becomes greater.
Type I error is not defined in this region.
The probability of a type I error becomes smaller.
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https://music.stackexchange.com/questions/124214/on-the-difference-between-4-4-and-4-2-time-signatures
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# On the difference between 4/4 and 4/2 time signatures [duplicate]
I am new to music theory and I was going through this video on YouTube to understand time signatures. The author mentions that the top number indicates number of beats. I understand what this means. My confusion is with the bottom number (as is rightly pointed out in the video as being most confusing for author's students too). It is mentioned that the bottom number indicates the value of each beat. For example, 4/4 indicates that there are 4 quarter notes per bar. 4/8 indicates that there are 4 eighth notes per bar. Now to my question. Since the terms 'quarter note' and 'eighth note' are relative (in the sense that a quarter note in 120 bpm is exactly as long as eighth note in 60bpm), is there a standard tempo that these are all referenced to?
Please excuse me if my understanding is completely wrong. I will be most happy to correct it from your suggestions.
• "a quarter note in 120 bpm is exactly as long as eighth note in 60bpm": this may or may not be true depending on which note value gets the beat. In 4/8, an eighth note is one beat, whereas in 4/4, a quarter note is one beat, so a quarter note in 4/4 at 120 b.p.m. has the same duration as an eighth note in 4/8 at 120 b.p.m. Commented Aug 5, 2022 at 7:58
The note values are relative only (as currently used.) A signature such as 4/2 could have any speed. Often there will be a notation like QN (the symbol) = 120 or the like. The number refers to beats per minute.
A few hundred years ago, the time signature did more imply a tempo (or tempo range).
The difference is more academic. A long time ago, 4/2 was more prevalent, perhaps in slower pieces, but the /2 doesn't have to signify slow. It will all depend on the tempo of the piece, obviously.
4/4 has become the mainstay for pieces in 4 time, but occasionally it's easier to write and read if 4/2 is used - it saves writing very small value notes, as those will be twice as big in 4/2 as in 4/4.
By doubling or halving the tempo marks, 4/2 and 4/4 could be played at the same speed, so there's an easy solution there.
Simple answer: You can't hear the difference. The only difference would be how the music looks on the page.
To elaborate phoog's comment: tempo indications are given not just as a number but also as a note value, like "quarter note equals 120." So in fact a quarter note—at "quarter = 120"—is exactly equal to an eighth note at "eighth = 120"... and to a half note at "half = 120." You can "call" the beat whatever kind of note value you want. The listener just hears the pulse; unless they're also reading along in the score, they have no way of knowing what note value it represents (aside from a few educated guesses).
So why might a composer choose one value or another? Well, those educated guesses are about what's most common. The quarter note is most often the beat. For meters that divide the beat into three subdivisions, it's most common to use the eighth note to indicate these subdivisions, giving time signatures like 6/8 or 9/8 (even though the real "pulse" is then a dotted quarter). If you're listening to music from the Renaissance or very early Baroque, there's a good chance the beat (tactus) is represented by a half note because of the way rhythmic notation was still changing. In modern practice, it would be rare to choose the half note, but conceivably if you had a whole lot of very short notes, it could make the music more readable to print fewer "beams" on them (e.g. for notes that divide the beat into 8 parts, they could be printed as 16th notes instead of 32nd notes). This would probably be more confusing than helpful, though.
No, there isn't any information such as 'a quarter note normally takes X seconds' and so on. That's why there are metronome marks or tempo expressions (Andante, Allegro etc.) written on scores. You 'split' a given time into equal sub-parts. Imagine splitting a cake. So 4/4 means you split a given time period into 4 quarter notes and the metronome mark or the tempo expression on the score determines how much time a quarter note should/must ('must' if there is a metronome mark on the score) take. If the metronome mark is 'quarter note=90' then the time lapse between two metronome pulses which are in '90 bpm' speed is the answer for the question 'how long a quarter note must be in this piece/song?'. Subsequently, an 8th note will be 2 times faster and so on.
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https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-9-roots-and-radicals-9-5-solving-radical-equations-problem-set-9-5-page-424/5
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## Elementary Algebra
Published by Cengage Learning
# Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5 - Page 424: 5
No solution
#### Work Step by Step
Recall, in order to cancel out a square root, we square both sides of the equation. Thus, we square both sides of the equation to obtain that 3x=36. We divide both sides of the equation by 3 to isolate x and obtain: x=12. We now must check our solution: $\sqrt{3 \times 12} = -6 \\\\\ \sqrt {36}=-6 \\\\ 6 = -6$ This is not true, so there is no solution.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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https://socratic.org/questions/what-is-the-projection-of-2-4-3-onto-1-2-2
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# What is the projection of <2,-4,3 > onto <1,2,2 >?
##### 1 Answer
Feb 28, 2016
$\setminus \vec{{A}_{}}$ id perpendicular to \vec{B_{}. So one's projection on the other must be a null vector ( $< 0 , 0 , 0 >$)
#### Explanation:
The projection of a $\setminus \vec{{A}_{}}$ onto another vector $\setminus \vec{{B}_{}}$ is:
$\setminus \vec{{A}_{B}} = \setminus \frac{\setminus \vec{{A}_{}} . \setminus \vec{{B}_{}}}{B} \setminus \hat{B} = \setminus \frac{\setminus \vec{{A}_{}} . \setminus \vec{{B}_{}}}{{B}^{2}} \setminus \vec{{B}_{}}$
Solution: \vec{A_{}}=<2,-4,3>; \qquad \vec{B_{}}=<1,2,2>;
$\setminus \vec{{A}_{}} . \setminus \vec{{B}_{}} = \left(2 \setminus \times 1 - 4 \setminus \times 2 + 3 \setminus \times 2\right) = 0$
Since $\setminus \vec{{A}_{}} . \setminus \vec{{B}_{}} = 0$, it is clear that $\setminus \vec{{A}_{}}$ is perpendicular to $\setminus \vec{{B}_{}}$. So its projection on $\setminus \vec{{B}_{}}$ is a null vector.
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https://nivent.github.io/blog/orbit-stab-rep/
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# Orbit-Stabilizer for Finite Group Representations
One of my professors covered the main result of this post during a class that I missed awhile ago. Using some notes from a friend who attended that class, I want to try to reconstruct the theorem 1. Experience with representation theory will be useful for this post, but I’ll try to cover enough of the basics so that previous exposure isn’t strictly required.
# Orbit-Stabalizer
We will begin by continuing our discussion of group actions from last post. Recall the definition
Let $G$ be a group and let $X$ be a set. A (left) group action of $G$ on $X$ is a map $\phi:G\times X\rightarrow X$ satisfying
• $1\cdot x=x$ for all $x\in X$ where $1\in G$ is the identity
• $g\cdot(h\cdot x)=(gh)\cdot x$ for all $x\in X$ and $g,h\in G$
where $g\cdot x$ denotes $\phi(g,x)$.
We sometimes write $G\curvearrowright X$ to denote that $G$ acts on $X$. If $X$ has additional structure (e.g. if $X$ is a vector space), then we require our group action to respect $X$’s structure. In general, a group action $G\curvearrowright X$ is a map $G\rightarrow\Aut(X)$ where the automorphisms of $X$ depend on the context 2.
Now, in order to (state and) prove Orbit-Stabalizer, we’ll need to know what those words mean.
Let $G$ be a group acting on a set $X$. Given some $x\in X$, its orbit is $$G\cdot x=\{g\cdot x\mid g\in G\}$$ Furthermore, its stabalizer is $$\Stab(x)=G_x=\{g\in G\mid g\cdot x=x\}$$
Note that the stabalizer of $x\in X$ is a subgroup of $G$ since $g,h\in G_x\implies(g\inv h)\cdot x=g\cdot x=x$. Furthermore, if $G\cdot x=X$ for some $x\in X$, then we say $G$ acts transitively on $X$.
Finally, if $G\curvearrowright X$, then we call $X$ a $G$-set. Naturally, these spaces have their own notion of homomorphisms.
Let $X,Y$ be two $G$-sets. A $G$-map (or $G$-equivariant map or $G$-morphisms) is a map $f:X\rightarrow Y$ s.t. $f(g\cdot x)=g\cdot f(x)$ for all $g\in G$ and $x\in X$. We say $f$ is a $G$-isomorphism if it is bijective.
Show that if $f$ is a $G$-isomorphism, then $\inv f$ is $G$-equivariant.
With our definitions set up, we come to
Let $X$ be a $G$-set. Fix any $Y\subseteq X$ s.t. $g\cdot Y\cap Y\in\{Y,\emptyset\}$ for all $g\in G$ and $G\cdot Y=\{g\cdot y\mid g\in G,y\in Y\}=X$. Finally, let $H=\Stab(Y)=\{g\in G\mid\forall y\in Y:g\cdot y\in Y\}$. Then, $$X\simeq\bigsqcup_{\sigma_i\in G/H}\sigma_iY$$ as $G$-sets where the union is taken over coset representatives of $G/H$ and $\sigma_iY=\{\sigma_i y\mid y\in Y\}$ (note: $\sigma_iy$ is just a formal symbol) and $G$ acts on it via $g\cdot(\sigma_iy)=\sigma_j(h\cdot y)$ for the unique $\sigma_j,h$ s.t. $g\sigma_i=h\sigma_j$.
Let $f:\bigsqcup_{\sigma_i\in G/H}\sigma_iY\to X$ be the map $f(\sigma_iy)=\sigma_i\cdot y$. This map is $G$-equivariant since $$f(g\cdot\sigma_iy)=f(\sigma_j(h\cdot y))=\sigma_j\cdot(h\cdot y)=(\sigma_jh)\cdot y=(g\sigma_i)\cdot y=g\cdot(\sigma_i\cdot y)=g\cdot f(\sigma_iy)$$ where $g\sigma_i=\sigma_jh$. For injectivity, if $\sigma_i\cdot y=\sigma_j\cdot y'$, then $$(\inv\sigma_j\sigma_i)\cdot y=y'\implies\inv\sigma_j\sigma_i\in H\implies\sigma_iH=\sigma_jH\implies\sigma_i=\sigma_j\implies y=y'$$ where the second-to-last implication comes from the fact that we fixed our coset representatives ahead of time. Finally, for surjectivity, fix any $x\in X$. Since $G\cdot Y=X$, there exists $g\in G$ and $y\in Y$ s.t. $g\cdot y=x$. Thus, writing $g=\sigma_jh$, we have that $f(\sigma_j(h\cdot y))=x$.
Let $X$ be a $G$-set, and fix any $x\in X$. Then, $|G\cdot x|=|G:G_x|=|G|/|G_x|$
Apply the above theorem to the $G$-set $G\cdot x$ where $Y=\{x\}$.
It’s worth noting that Orbit-Stabilizer usually only refers to the corollary above, but this stronger version is closer to our main theorem.
# A Quick Intro to Representations of Finite Groups
Now that we’ve seen Orbit-Stabilizer, we’ll need to introduce some definitions from representation theory.
Fix a group $G$ and a vector space $V$ over a field $\F$. A (linear) representation of $G$ is a map $\rho:G\rightarrow\GL_{\F}(V)$. Given such a map, we call $V$ a $G$-rep, and morphisms of $G$-reps are $G$-equivariant linear maps. Finally $\theta:G\to\GL_{\F}(U)$ is a subrepresentation if $U\subseteq V$ and $\theta(g)=\rho(g)\mid_U$ for all $g\in G$.
When studying linear representations of groups, there are two main perspectives one can take. Everything can be done in terms of an explicit representation (i.e. the map $\rho$ above) or in terms of modules over the group ring. Since I haven’t talked about modules on this blog before 3, I’ll stick to the explicit representation approach and leave exercises to translate things into statements about modules for the interested reader.
Prove that a linear representation of $G$ is the same thing as an $\F[G]$-module. 4
Thankfully, we don’t need a lot of representation theory for the main result of this post. We only need to know a few different types of linear representations. Also, in case I ever forget to mention this, for the rest of this post, assume all vector spaces are finite-dimensional and assume that all groups are finite.
A permutation representation of $G$ on a finite-dimensional $\F$-vector space $V$ is a linear representation $\rho:G\rightarrow\GL(V)$ in which the elements of $G$ act by permuting some basis $B=\{b_1,\dots,b_n\}$ for $V$.
Consider the symmetric group $S_n$ acting on $\C^n=\bigoplus_{i=1}^n\C e_i$ via $\sigma\cdot e_i=e_{\sigma(i)}$.
Let $G$ be any finite group, and consider $\C[G]\simeq\bigoplus_{g\in G}\C g$ as vector spaces. This is the regular representation when $G$ acts via $h\cdot g=hg$ on the basis.
Finally, we need the notion of induced representations. This let’s you take a representation of a group $H$ and canoncially construct a representation of a larger group $G\supseteq H$. The construction is very reminiscent of the Orbit-Stabilizer theorem.
Let $H\le G$ be a subgroup of $G$, and let $V$ be an $H$-rep. Fix a complete set of coset representatives $\sigma_1=e,\dots,\sigma_n\in G$ s.t. $G/H=\{\sigma_iH:0\le i\le n\}$ and $n=|G/H|$. Then, as a vector space, the induced representation from $H$ to $G$ is $$\Ind_H^GV=\bigoplus_{i=1}^n\sigma_iV$$ where $\sigma_iV=\{\sigma_iv\mid v\in V\}$ is a space of formal symbols. This is given a $G$-action as follows: given some $\sigma_iv\in\Ind_H^GV$, there's a unique $\sigma_j$ and $h\in H$ s.t. $g\sigma_i=\sigma_jh$. We define $g\cdot\sigma_iv=\sigma_j(h\cdot v)$.
Prove that, as $\F[G]$-modules, we have $$\Ind_H^GV\simeq\F[G]\otimes_{\F[H]}V$$ so induction is really just extension of scalars.
The regular representation is $\Ind_1^G\F$ where $1$ denotes the trivial group and $G$ acts trivially (i.e. by the identity) on $\F$.
# Orbit-Stabilizer v2
This is where we’ll prove the main result, which roughly says that (almost-)permutation representations are induced representations.
Let $V$ be a $G$-rep with a decomposition $V\simeq\bigoplus_{i=0}^nV_i$ as a vector space s.t. for all $i,j\in\{0,\dots,n\}$, there exists a $g\in G$ s.t. $g\cdot V_i=V_j$, and let $H=\Stab(V_0)$. Then, $$V\simeq\Ind_H^GV_0$$
We will show this by constructing an explicit isomorphism. Let $f:\Ind_H^GV_0\rightarrow V$ be the map given by $$f(\sigma_iv_0)=\sigma_i\cdot v_0$$ This is easily seen to be $G$-equivariant, and it is linear by construction. For surjectivity, it suffices to find preimages for elements of the form $v_i\in V_i$. Given such an element, there exists some $g_i\in G$ and $w_i\in V_0$ s.t. $g_i\cdot w_i=v_i$. Now, we can write $g_i=h_i\ith\sigma_j$ for a unique $h_i\in H$ and coset representative $\ith\sigma_j$. Doing so gives us that $f(\ith\sigma_j(h_i\cdot w_i))=v_i$ so $f$ is surjective as claimed. Finally, we need to show that $f$ is injective, so fix some $w=\sum_{\sigma_i\in G/H}\sigma_i\ith v_0\in\ker f$. This means that $\sum_{\sigma_i\in G/H}\sigma_i\cdot\ith v_0=0$, but we claim that $\sigma_i\cdot\ith v_0$ and $\sigma_j\cdot\Ith vj_0$ belong to different summands (i.e. different $V_i$'s) which forces $\sigma_i\cdot\ith v_0=0\implies\ith v_0=0$ for all $i$ so $w=0$. To prove the claim, suppose that $\sigma_i\cdot\ith v_0,\sigma_j\cdot\Ith vj_0\in V_k$ for some $k$. Then, $$\inv\sigma_j\sigma_i\cdot\ith v_0\in V_0\implies\inv\sigma_j\sigma_i\in H\implies\sigma_j=\sigma_i$$ and we win.
This wasn’t the proof I had in mind. I imagined (and still do) that it was possible to directly apply the original orbit-stabilizer by letting $X$ be a (well-chosen) basis for $V$ and $Y$ be a (well-chosen) basis for $V_0$. However, in trying to make this work, I ran in to issues getting a well-defined action of $G$ on $B$. Basically, $H=\Stab(V_0)$ can act nontrivially so it’s possible that $h\cdot B_0\not\subseteq B$ which is troublesome. I still hold out hope that this idea can be salvaged in general5, so
See if you can come up with a proof of the above that applies the original Orbit-Stabilizer theorem (e.g. apply it to a basis of V and then extend linearly). If you can, let me know.
Even though the proof is a little unsatisfying, we have proven what we set out to prove, so let’s end with a couple examples. $\newcommand{\trv}{\underline{\text{Trv}}}\newcommand{\alt}{\underline{\text{Alt}}}$
Consider $S_n\curvearrowright\Sym^2\C^n$ where $\C^n=\bigoplus\C e_i$ and $S_n$ acts by permuting the $e_i$. Restricting this action to the basis $B=\{e_ie_j:i,j\in\{1,\dots,n\}\}$, we see there are two $S_n$-orbits $$\begin{matrix} B_0 &=& \brackets{e_ie_j:i\neq j} && \Stab(\C e_1e_2) &=& S_2\times S_{n-2}\\ B_1 &=& \brackets{e_1^2,\dots,e_n^2} && \Stab(\C e_1^2) &=& S_{n-1} \end{matrix}$$ Thus we can write $\Sym^2\C^n=V\oplus W$ where $V=\C B_0=\bigoplus_{i\neq j}\C e_ie_j$ and $W=\C B_1=\bigoplus_{i=1}^n\C e_i^2$. Furthermore, $S^n$ acts transitively on these decompositions of $V,W$ so applying our theorem ($V_0=\C e_1e_2$ and $W_0=\C e_1^2$) yields $$\Sym^2\C^n\simeq\parens{\Ind_{S_2\times S_{n-2}}^{S_n}\trv\otimes\trv}\oplus\parens{\Ind_{S_{n-1}}^{S_n}\trv}$$ where $\trv$ is the trivial 1-dimensional $S_k$ representation sending each element to the number 1.
This time, let's look at $S_n\curvearrowright\parens{\Wedge^2\C^n}\otimes\C^n$ where $\C^n=\bigoplus e_i$ and $S_n$ again acts by permuting the $e_i$. We have a basis $B=\{(e_i\wedge e_j)\otimes e_k:i,j,k\in\{1,\dots,n\},i< j\}$ but it's not fixed by $S_n$ (e.g. $(12)\cdot(e_1\wedge e_2)\otimes e_3=(e_2\wedge e_1)\otimes e_3\not\in B$), so we'll look instead at the spanning set $B'=\{(e_i\wedge e_j)\otimes e_k:i,j,k\in\{1,\dots,n\},i\neq j\}$ which is fixed by $S_n$. This has the following orbits $$\begin{matrix} B_0 &=& \brackets{(e_i\wedge e_j)\otimes e_k:i\neq j\neq k} && \Stab(\C (e_1\wedge e_2\otimes e_3)) &=& S_2\times S_{n-3}\\ B_1 &=& \brackets{(e_i\wedge e_j)\otimes e_k:i\neq j,k\in\{i,j\}} && \Stab(\C(e_1\wedge e_2\otimes e_1)) &=& S_{n-2} \end{matrix}$$ It's worth noting that $(12)\cdot(e_1\wedge e_2)\otimes e_1=-(e_1\wedge e_2)\otimes e_2$ so we can switch whether $k=i$ or $k=j$ in $B_1$ above. Applying our theorem to (the span of) each orbit and summing them up, we get that $$\Wedge^2\C^n\otimes\C^n\simeq\parens{\Ind_{S_2\times S_{n-3}}^{S_n}\alt\otimes\trv}\oplus\parens{\Ind_{S_{n-2}}^{S_n}\trv}$$ where $\alt$ is the alternating 1-dimensional $S_k$ representation sending each element to its sign.
1. which, unsurprisingly, is a version of Orbit-stabilizer for representations of finite groups
2. for X a set, they are (self) bijections
3. but really should at some point
4. This includes proving that $\F[G]$-linear maps are $G$-equivariant and that submodules correspond to subrepresentations
5. It certainly can be in the case that H does indeed act trivially (or at least stabalizes the basis)… Question: is there always a basis B_0 s.t. Stab(V_0) is contained in Stab(B_0)?
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# why there does not exist any homomorphism f : Q → Qpos.
If G is a group and b ∈ G, then a square root of b is an element a ∈ G such that a^2= b. For example 3 is a square root of 6 in 〈Z,+>
Let 〈Q,+> be the group of rational numbers with the operation of addition and let 〈Qpos, . >be the group of positive rational numbers with the operation of multiplication. Use the fact that you proved in part a to explain why there does not exist any homomorphism f : Q → Qpos.
SO FAR , I know let f be homomorphism from 〈Q,+> to 〈Qpos, . >defined by f(x+y) = x times y, let x = 0, and y = 1 we will get f(1)= 0 but 0 is not in Qpos. do you think my method is right? because I didn't use square root in this part of question...
the part a is about {a. Suppose that G and H are groups and that f : G → H is a homomorphism of G onto H. Prove that if every element of G has a square root then every element of H also has a square root.}
• Qpos is the second group? and do you mean no non-trivial hom? – GiantTortoise1729 Dec 4 '15 at 3:55
• What have you tried? Where did you get stuck? (Also, what did you prove in part a?) EDIT: All that you've shown is that one specific map is not a homomorphism; you need to show that there is no homomorphism, at all. – Noah Schweber Dec 4 '15 at 3:55
• @NoahSchweber that part a is (a. Suppose that G and H are groups and that f : G → H is a homomorphism of G onto H. Prove that if every element of G has a square root then every element of H also has a square root.) I have already solve it. – Helen Dec 4 '15 at 3:58
• @GiantTortoise1729 I edit my question, Thank you! – Helen Dec 4 '15 at 4:00
There is such a homomorphism, namely the trivial homomorphism where everything is mapped to $1$. However this is the only possibility.
Proof. In $\def\Q{{\Bbb Q}}\Q$ you can take the square root of any element. Repeating, you can take the square root of the square root of the square root of... the square root of any element, with as many square roots as you like. If there is a homomorphism $\phi$ from $\Q$ to $\Q^+$, then the same thing holds in $R={\rm range}\,\phi$. But in $R$, square root has its normal meaning, and given any element $a$ of $\Q^+$, the only way you can take the square root of the square root... indefinitely many times and always get rational results is if $a=1$. Therefore ${\rm range}\,\phi=\{1\}$, and $\phi$ is the trivial homomorphism.
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# Determining the major/minor axes of an ellipse from general form
I'm implementing a system that uses a least squares algorithm to fit an ellipse to a set of data points. I've successfully managed to obtain approximate locations for the centre of the ellipse but I am having trouble with the major and minor axes. I'm writing this in C++ using OpenCV so I need to be able to express them as some kind of equation so I can calculate them. Right now I am just testing it with very basic circles rather than what I'll actually be using the program for.
Result of my program, where the top image is the data points (3 contours) and the bottom image is my approximated ellipses. The solid circles are the original image, purple dot is the centre of my approximated ellipse and the green, blue and red curves my approximated ellipses for the contours. The ellipses have not been rotated to the approximated angle yet (I will do this later).
So, is there a way from the general conic
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$
if I know the values of A,B,C,D,E and F and also the centre point $$(x_0,y_0)$$
That I can calculate the major and minor axes?
From what I understand. From the equation
$${(x-x_0)^2 \over a^2} + {(y-y_0)^2 \over b^2} = 1$$
The major axis is 2a and the minor is 2b.
Note: I do not have points on the ellipse that I can substitute in. In my actual application I am unlikely to have such points.
I came across this question and it helped me implement what I have done thus far Finding the angle of rotation of an ellipse from its general equation and the other way around
The 1st answer I used for the centre. And the 3rd for the axes/dimensions.
For example. The black circle with the blue 'ellipse' around it. I have
$$A = 3.876e-013$$ $$B = 1.8819e-012$$ $$C = 1$$ $$D = -2.51108e-009$$ $$E = -484$$ $$F = 54663.6$$
And I calculate theta from
$${1 \over 2} \tan^{-1}\left({B \over A- C}\right)$$
Which gives me $$-9.40948e-013$$
I am unsure if I am approaching this in the correct way.
Any help appreciated, cheers :).
• – lab bhattacharjee Dec 23 '13 at 16:55
• I just saw this question after I wrote this answer. It answers precisely this question. – robjohn Apr 9 '15 at 19:38
• "least squares algorithm to fit an ellipse to a set of data points." - in such a case, if you use SVD for solving the least squares problem, the singular values will turn out to be the axes of your ellipse. – J. M. is a poor mathematician Dec 14 '16 at 16:54
If you are looking for the length of the minor and major axis you can calculate $r_{min}$ and $r_{max}$ (see formulae below).
If you are trying to determine the bounding box, you can calculate the left-most, right-most, top-most and bottom-most points.
As far as the angle of rotation is concerned, I use the algorithm and formulae below.
Properties of an ellipse from equation for conic sections in general quadratic form
Given the equation for conic sections in general quadratic form: $a x^2 + b x y + c y^2 + d x + e y + f = 0$.
The equation represents an ellipse if $b^2 - 4 a c < 0$ , or similarly, $4 a c - b^2 > 0$
The coefficient normalizing factor is given by:
$q = 64 {{f (4 a c - b^2) - a e^2 + b d e - c d^2} \over {(4ac - b^2)^2}}$
The distance between center and focal point (either of the two) is given by:
$s = {1 \over 4} \sqrt { |q| \sqrt { b^2 + (a - c)^2 }}$
The semi-major axis length is given by:
$r_\max = {1 \over 8} \sqrt { 2 |q| {\sqrt{b^2 + (a - c)^2} - 2 q (a + c) }}$
The semi-minor axis length is given by:
$r_\min = \sqrt {{r_\max}^2 - s^2}$
The center of the ellipse is given by:
$x_\Delta = { b e - 2 c d \over 4 a c - b^2}$
$y_\Delta = { b d - 2 a e \over 4 a c - b^2}$
The top-most point on the ellipse is given by:
$y_T = y_\Delta + {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}}$
$x_T = {{-b y_T - d} \over {2 a}}$
The bottom-most point on the ellipse is given by:
$y_B = y_\Delta - {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}}$
$x_B = {{-b y_B - d} \over {2 a}}$
The left-most point on the ellipse is given by:
$x_L = x_\Delta - {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}}$
$y_L = {{-b x_L - e} \over {2 c}}$
The right-most point on the ellipse is given by:
$x_R = x_\Delta + {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}}$
$y_R = {{-b x_R - e} \over {2 c}}$
The angle between x-axis and major axis is given by:
if $(q a - q c = 0)$ and $(q b = 0)$ then $\theta = 0$
if $(q a - q c = 0)$ and $(q b > 0)$ then $\theta = {1 \over 4} \pi$
if $(q a - q c = 0)$ and $(q b < 0)$ then $\theta = {3 \over 4} \pi$
if $(q a - q c > 0)$ and $(q b >= 0)$ then $\theta = {1 \over 2} {atan ({b \over {a - c}})}$
if $(q a - q c > 0)$ and $(q b < 0)$ then $\theta = {1 \over 2} {atan ({b \over {a - c}})} + {\pi}$
if $(q a - q c < 0)$ then $\theta = {1 \over 2} {atan ({b \over {a - c}})} + {1 \over 2}{\pi}$
• Do you have a reference for the derivation of these equations? – Vesnog Oct 6 '14 at 0:17
• No, I do not have a formal reference. I got information from internet sources link link link . The rest I derived myself. – Osmund Francis Oct 7 '14 at 15:53
• Thanks for the prompt reply I will look into them. – Vesnog Oct 7 '14 at 18:18
• @Vesnog, you might be interested in Kendig's book on conics; he has quite the list of useful formulae there. – J. M. is a poor mathematician Dec 14 '16 at 16:54
For convenience, I'll use the form with double terms,
$$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0.$$
First you need to center the ellipse with
$$A(x-x_c)^2+2B(x-x_c)(y-y_c)+C(y-y_c)^2+2D(x-x_c)+2E(y-y_c)+F=0.$$
The center is found by canceling the linear terms, wich gives
$$Ax_c+By_c=D,\\Bx_c+Cy_c=E.$$
The equation then reduces to
$$Ax^2+2Bxy+Cy^2=Ax_c^2+2Bx_cy_c+Cy_c^2-F=G.$$
In polar coordinates,
$$(A\cos^2(t)+2B\cos(t)\sin(t)+C\sin^2(t))r^2=G.$$
We want to find the extrema of $r$. They are also the extrema of the trigonometric factor and we determine them by cancelling the derivative.
$$-A\cos(t)\sin(t)+B(\cos^2(t)-\sin^2(t))+C\sin(t)\cos(t)=0$$ which we rewrite, by the double angle formulas
$$(C-A)\sin(2t)+2B\cos(2t)=0$$
which gives
$$\tan(2t)=\frac{2B}{A-C}.$$
The solution is $$t=\frac12\arctan\left(\frac{2B}{A-C}\right)+k\frac\pi2,$$
(giving the directions of the axis, which are orthogonal), and the semi-axis lengths are
$$r=\sqrt{\frac G{A\cos^2(t)+2B\cos(t)\sin(t)+C\sin^2(t)}}.$$
One can notice that the ellipse parameters are directly related to the Eigenvalues and Eigenvectors of the matrix
$$\left(\begin{matrix}A&B\\B&C\end{matrix}\right).$$
After diagonalization of the matrix, the equation of the ellipse becomes
$$\lambda x^2+\mu y^2=G,$$ which is of the well-known form
$$\left(\frac x{\sqrt{\frac G\lambda}}\right)^2+\left(\frac y{\sqrt{\frac G\mu}}\right)^2=1.$$
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# How to construct a transition matrix?
I'm giving my first steps in stochastic processes but I'm having some difficulties. See the following example
Suppose that whether or not it rains today depends on previous weather conditions through the last two days. Specifically, suppose that if it has rained for the past two days, then it will rain tomorrow with probability 0.7; if it rained today but not yesterday, then it will rain tomorrow with probability 0.5; if it rained yesterday but not today, then it will rain tomorrow with probability 0.4; if it has not rained in the past two days, then it will rain tomorrow with probability 0.2.
If we let the state at time n depend only on whether or not it is raining at time n, then the preceding model is not a Markov chain (why not?). However, we can transform this model into a Markov chain by saying that the state at any time is determined by the weather conditions during both that day and the previous day. In other words, we can say that the process is in
state 0 if it rained both today and yesterday,
state 1 if it rained today but not yesterday,
state 2 if it rained yesterday but not today,
state 3 if it did not rain either yesterday or today.
$$P=\begin{bmatrix}0.7&&0&&03&&0\\0.5&&0&&0.5&&0\\0&&0.4&&0&&0.6\\0&&0.2&&0&&0.8\end{bmatrix}$$
I'm wondering how this matrix is computed, $$P_{ij}=P(state_i|state_j)$$ I am unable to see clearly how these conditionals are calculated.
Denoting the current day as index $i$ (so that yesterday is $i-1$ and the day before yesterday is $i-2$), the previous state comprises two elements, $State_{n-1}=\{S_{i-1},S_{i-2}\}$, where $S_k$ is the state of the weather for day $k$. Thus the current state is $State_{n}=\{S_{i},S_{i-1}\}$
Lettting $R$ be the occurrence of rain on a particular day, so that $\overline{R}$ is when no rain occurs for that day, there are $4$ possible values of the previous state $State_{n-1}$ $$\{\overline{R},\overline{R}\}, \{\overline{R},R\},\{R,\overline{R}\},\{R,R\}$$ For each of these $4$ states, there are only two possible next states:- \begin{align} State_n=\{\overline{R},\overline{R}\}\text { or }State_n=\{R,\overline{R}\}\text{ given }State_{n-1}=\{\overline{R},\overline{R}\} \\ State_n=\{\overline{R},\overline{R}\}\text { or }State_n=\{R,\overline{R}\}\text{ given }State_{n-1}=\{\overline{R},R\} \\ State_n=\{\overline{R},R\}\text { or }State_n=\{R,R\}\text{ given }State_{n-1}=\{R,\overline{R}\} \\ State_n=\{\overline{R},R\}\text { or }State_n=\{R,R\}\text{ given }State_{n-1}=\{R,R\} \end{align} With the information given in the question, and based on the constraints of what the next state is based on the current state, you can figure out the structure of the transition matrix.
For example, the information Specifically, suppose that if it has rained for the past two days, then it will rain tomorrow with probability $0.7$, corresponds to $$P(State_{n+1}=\{\overline{R},\overline{R}\}|State_{n}=\{\overline{R},\overline{R}\})=0.7\\\Rightarrow P(State_{n+1}=\{R,\overline{R}\}|State_{n}=\{\overline{R},\overline{R}\})=1-0.7=0.3$$ and if it rained today but not yesterday, then it will rain tomorrow with probability 0.5 corresponds to $$P(State_{n+1}=\{\overline{R},\overline{R}\}|State_{n}=\{\overline{R},R\})=0.5\\\Rightarrow P(State_{n+1}=\{R,\overline{R}\}|State_{n}=\{\overline{R},R\})=1-0.5=0.5$$
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Anne Wacker
2022-01-01
Find the complex form of the Fourier Series expansion of
Charles Benedict
Expand $\mathrm{cos}\left(ax\right)=\frac{1}{2}\left({e}^{iax}+{e}^{-iax}\right)$
Then
${f}_{n}=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\mathrm{cos}\left(ax\right){e}^{-\in x}dx$
$=\frac{1}{2\pi }\frac{1}{2}{\int }_{-\pi }^{\pi }\left({e}^{iax}+{e}^{-iax}\right){e}^{-\in x}dx$
$=\frac{1}{4\pi }{\int }_{-\pi }\left({e}^{i\left(a-n\right)x}+{e}^{-a\left(a+n\right)x}\right)dx$
This is a straightforward integration. Take care when a is an integer (in which case the Fourier series is trivially obtained from the above expansion).
If $a\in \mathbb{Z}$, then the formula for $\mathrm{cos}$ gives ${f}_{n}=\frac{1}{2}{\sigma }_{|n||a|}$
${f}_{n}=\frac{1}{4\pi i}{\left(-1\right)}^{n}\left(\frac{{e}^{ia\pi }+{e}^{-ia\pi }}{a-n}-\frac{{e}^{-ia\pi }-{e}^{ia\pi }}{a+n}\right)$
$=\frac{1}{4\pi i}{\left(-1\right)}^{n}\left({e}^{ia\pi }-{e}^{-a\pi }\right)\left(\frac{1}{a-n}+\frac{1}{a+n}\right)$
$=\frac{1}{2\pi }{\left(-1\right)}^{n}\mathrm{sin}\left(a\pi \right)\frac{2a}{{a}^{2}-{n}^{2}}$
$=\frac{1}{\pi }{\left(-1\right)}^{n+1}\mathrm{sin}\left(a\pi \right)\frac{a}{{n}^{2}-{a}^{2}}$
Jacob Homer
${c}_{n}=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\mathrm{cos}ax{e}^{-\in x}dx$
We make integration by parts twice to
$I\phantom{\rule{0.222em}{0ex}}={\int }_{-\pi }^{\pi }\mathrm{cos}ax{e}^{-\in x}dx=\frac{i}{n}\mathrm{cos}ax{e}^{\in x}{\mid }_{-\pi }^{\pi }+\frac{ai}{n}{\int }_{-\pi }^{\pi }\mathrm{sin}ax{e}^{-\in x}dx$
$=-\frac{a}{{n}^{2}}\mathrm{sin}ax{e}^{-\in x}+\frac{{a}^{2}}{{n}^{2}}I$
I think you can handle it from here.
karton
Complex form of Fourier series is given by
$f\left(x\right)\sim \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{n}\left(f\right){e}^{inx}$
You need to find Fourier coefficients
${c}_{n}\left(f\right)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left(x\right){e}^{-inx}dx$
Thus, simply integrate by parts (twice)
${c}_{n}\left(f\right)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left(x\right){e}^{-inx}dx$
$=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\mathrm{cos}\left(ax\right){e}^{-inx}dx.$
Do you have a similar question?
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# Statistics
### Mean
One of the most basic properties of a data set is it’s mean. Mean is a measure of the Central Tendency of data. In layman terms, if you were to pick one point in a dataset that is representative of the entire dataset, it would be the mean. Also called “average” in common vocabulary, calculating the mean is really simple. Just add up all the values in a particular variable of a dataset and divide it by the total number of values in it.
For example, if there are 5 Uber drivers with 5 different ratings, what is their mean ?
```from statistics import mean
ratings = [4.5, 3.9, 4.6, 4.8, 3.9]
mean(ratings)
```
```4.34
```
### Median
Mean does not always represent the average of the data. For normal distributions it typically does. However, for many other data distributions, mean does not represent the average. For example, take the US income distribution data. Since the number of housholds is huge, the data has been bucketed into income brackets for easy analysis.
```income_bracket households
10-15k 5700000
15-20k 5620000
20-25k 5930000
25-30k 5500000
30-35k 5780000
35-40k 5340000
40-45k 5380000
45-50k 4730000
50-60k 9210000
60-75k 11900000
75-100k 14700000
100-125k 10300000
125-150k 6360000
150-200k 6920000
200k-plus 7600000
```
```from statistics import mean, median
income = [5700000,5620000, 5930000, 5500000, 5780000, 5340000, 5380000,
4730000, 9210000, 11900000, 14700000, 10300000, 6360000, 6920000,7600000 ]
print ( "mean =", mean(income) )
print ( "median =", median(income) )
```
```mean = 7398000
median = 5930000
```
As you can see, depending on the data distribution, mean and median could be totally different from each other.
### Mode
Mode is just the highest value in the dataset. If the dataset follows a gaussian distribution, it is the peak of the histogram. Think of mode of a dataset as the most commonly occuring number.
```from statistics import mode
ages = [12,13,14,11,12,13,15,10,13 ]
mode(ages)
```
```13
```
### Mean vs Median vs Mode
The reason why these terms exist is that, there is no one way to define the Central Tendency of a dataset. It depends on the nature of the distribution that the dataset conforms to. For example, if the dataset is a normal distribution, then most of the time the mean, median and mode are pretty close together. The relationship is beautifully visualized in this wikipedia diagram.
For a normal distribution, the 3 parameters ( mean, median, mode ) are pretty close together. For a skewed distribution, they are pretty staggered.
### Standard Deviation
While mean represents the “Central” or “Average” value of a dataset, variance represents how spread out the data is. For example, look at the 3 histograms below. They represent histograms of GRE scores among 3 different groups.
All these 3 graphs represent 10K rows of GRE scores from 3 different groups. All 3 of them have the same mean – 300. However, they are different, right ? What I want you to focus on is the shape of the distribution, not the height. Specifically, look at the x-axis. In the first plot, the data is focussed, pretty much around the average(300) mark. In the second plot, it is a bit spread out ( hence the reduction in size ) and the third plot is pretty spread out.
The green line represents the mean and the red line represents the Standard Deviation or σ ( represented by the Greek symbol sigma ). It is a measure of how spread out the distribution is. The more spread out the distribution is, the more flatter the bell curve is.
You can also think of Standard Deviation as a measure of uncertianity in Data Science. Imagine a drug trail working on reducing blood pressure. Say, on an average 3 different trails produce an average reduction of 30 points. However, trail 1 has very low standard deviation. Obviously, you would want to go with the first drug – because the third drug is more uncertain about the result.
```from statistics import stdev
income = [5700000,5620000, 5930000, 5500000, 5780000, 5340000, 5380000,
4730000, 9210000, 11900000, 14700000, 10300000, 6360000, 6920000,7600000 ]
stdev(income)
```
```2884498.470890811
```
### Variance
Standard Deviation is actually a derivative of variance. So, variance is calculated first. However, to illustrate that variance or standard deviation represents spread on the plot, we have learnt standard deviation first. However, we have to understand how variance is calculated, because this is the basis for calculating standard deviation .
Let’s calculate variance of a small dataset in excel to understand the process better.
variance_calculation
### What does Standard Deviation represent
Now that we know the basic definitions of Variance and Standard Deviation, let’s look at what it really represents.
### How far from the mean
Wikipedia has a nice picture to show this in the context of Gaussian distribution.
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# Linear Regression - Overview¶
### How can I make predictions about real-world quantities, like sales or life expectancy?¶
Most often in real world applications we need to understand how one variable is determined by a number of others.
For example:
• How does sales volume change with changes in price. How is this affected by changes in the weather?
• How does the amount of a drug absorbed vary with dosage and with body weight of patient? Does it depend on blood pressure?
• How are the conversions on an ecommerce website affected by two different page titles in an A/B comparison?
• How does the energy released by an earthquake vary with the depth of it's epicenter?
• How is the interest rate charged on a loan affected by credit history and by loan amount?
Answering questions like these, requires us to create a model.
A model is a formula where one variable (response) varies depending on one or more independent variables (covariates). For the loan example, interest rate might depend on FICO score, state, loan amount, and loan duration amongst others.
One of the simplest models we can create is a Linear Model where we start with the assumption that the dependent variable varies linearly with the independent variable(s).
While this may appear simplistic, many real world problems can be modeled usefully in this way. Often data that don't appear to have a linear relationship can be transformed using simple mappings so that they do now show a linear relationship. This is very powerful and Linear Models, therefore, have wide applicability.
They are one of the foundational tools of Data Science.
Creating a Linear Model involves a technique known as Linear Regression. It's a tool you've most probably already used without knowing that's what it was called.
#### Linear Regression in the high school physics lab¶
Remember a typical physics lab experiment from high school? We had some input X (say force) which gave some output Y (say acceleration).
You made a number of pairs of observations x, y and plotted them on graph paper.
Then you had to fit a straight line through the set of observations using a visual "best fit".
And then you read off 'm' the slope, and 'b', the y-intercept from the graph, hoping it was close to the expected answer. By drawing the "best fit" line you were, in effect, visually estimating m and b without knowing it.
You were doing informal Linear Regression. We're going to do this a little more formally. And then make it more sophisticated.
### Now for a bit of math¶
Remember the equation for a straight line from high school?
$$Y = mX + b$$
where $m$ is the slope and $b$ is the y-intercept.
Very briefly and simplistically, Linear Regression is a class of techniques for
Fitting a straight line to a set of data points.
This could also be considered reverse engineering a formula from the data.
We'll develop this idea starting from first principles and adding mathematical sophistication as we go along. But before that, you're probably curious what were the 'm' and 'b' values for this graph. We use modeling software to generate this for us and we get:
We see two numbers, "Intercept" and "Slope". Independent of what software we use to do our linear regression for us, it will report these two numbers in one form or another. The "Intercept" here is the "b" in our equation. And the "Slope" is the slope of Y with respect to the independent variable.
To summarize, we have a dataset (the observations) and a model (our guess for a formula that fits the data) and we have to figure out the parameters of the model (the coefficients m and b in our best fit line) so that the model fits the data the "best". We want to use our data to find coefficients for a formula so that the formula will fit the data the "best".
As we continue, we'll actually run the modeling software and generate these numbers from real data. Here we just saw pictures of the results.
### Using the model for prediction¶
Once you had your visual best fit line and had read off the m and b you probably said something to the effect:
"The data follows a linear equation of the form Y = mX + b where m (slope)=(somenumber) and b (y intercept)=(someothernumber)"
You may recall that the equation is not an exact representation because most probably your data points are not all in a perfectly straight line. So there is some error varying from one data point to the next data point. Your visual approach subjectively tried to minimize some intuitive "total error" over all the data.
What you did was intuitive "Linear Regression". You estimated m and b by the "looks right to me" algorithm. We will start with this intuitive notion and rapidly bring some heavy machinery to bear that will allow us to solve pretty sophisticated problems.
At this point your lab exercise may well ask you to approximate what Y will be when X is some number outside the range of your measurements. Then you use the equation above where m and b are now actual numbers say 2.1 and 0.3 respectively i.e the equation is Y = 2.1X + 0.3
And you plug in an X to get a Y.
This is where you are using your model to predict a value or, in other words, you are saying that I didn't use this value of X in my experiment and I don't have it in my data but I'd like to know what this value of X will map to on the Y axis.
Based on my model Y = 2.1X + 0.3 if I had used this value in my experiment then I believe I would have got an output Y of approximately what the straight line suggests.
You also want to be able to say "my error is expected to be (some number), so I believe the actual value will lie between Y-error and Y+error".
When used like this we call the X variable the "predictor" as values of Y are predicted based one values of X, and the Y variable the "response".
But before we do that let's take another trip back to the physics lab and peek over at the neighboring lab group's plots. We might see a different plot. So which one is "correct"?
### A notion of total error¶
Visually we can see that our plot (the first one) is the "better" one. But why? Because intuitively we feel that the line is closer to the points in the first one. So let's try to understand formally why that might be correct. Or not.
Actually the graphs above were plotted by software that generated some points with random variation and then plotted a line through them.
What the software did was compute a function called a "loss function", a measure of error. Then, it "tried out" multiple straight lines until it found one that minimized the "loss function" value for that choice -- then it read off the Intercept and X-slope for that line.
Because this error estimation is an important part of our modeling we're going to take a more detailed look at it.
We want to create a simple formula for the error or difference between the value of Y given by our straight line, and the actual value of Y from our data set. Unless our line happens to pass through a particular point, this error will be non-zero. It may be positive or negative. We take the square of this error (we can do other things like take the abs value, but here we take the square.....patience, all will be revealed) and then we add up such error terms for each data point to get the total error for this straight line and this data set.
Important: for a different set of samples of the very same experiment we will get a different data set and possibly a different staright line and so almost certainly a different total error.
The squared error we used is a very commonly used form of the total error previously know as "quadratic error". It also has the property that errors in the negative direction and positive direction are treated the same and this "quadratic error" or "square error" is always have a positive value.
So for now we will use the "squared error" as our representation of error. [1]
So Regression in general is any approach we might use to estimate the coefficients of a model using the data to estimate the coefficients by minimizing the "squared error". Statistical software uses sophisticated numerical techniques using multivariable calculus to minimize this error and give us estimated values for the coefficients.
Let's try this on some real data.
We're going to look at a data set of Loan data from Lending Club, a peer lending web site. They have anonymized data on borrowers and loans that have been made. Loan data has many attributes and we'll explore the whole data set in a bit but for now we'll just look at how borrower FICO score affects interest rate charged.
In [4]:
%pylab inline
import pandas as pd
# we have to clean up the raw data set which we will do
# in the next lesson. But for now let's look at the cleaned up data.
# import the cleaned up dataset into a pandas data frame
# extract FICO Score and Interest Rate and plot them
# FICO Score on x-axis, Interest Rate on y-axis
intrate = df['Interest.Rate']
fico = df['FICO.Score']
p = plot(fico,intrate,'o')
ax = gca()
xt = ax.set_xlabel('FICO Score')
yt = ax.set_ylabel('Interest Rate %')
Populating the interactive namespace from numpy and matplotlib
Here we see a distinct downward linear trend where Interest Rate goes down with increasing FICO score. But we also see that for the same FICO score there is a range of Interest rates. This suggests that FICO by itself might not be enough to predict Interest Rate.
### Multivariate Linear Regression¶
So the natural question that arises is what happens if Y depends on more than one variable. And this is where the power of mathematical generalization comes in. The same principle applies but in multiple dimensions. Not just two or three but much larger numbers. Twenty, thirty or even hundred independent variables are not out of question if we want to model real world data.
But for now let's look at $Y$ as a function of two independent variables, $X_1$ and $X_2$, so
$$Y = a_0 + a_1X_1 + a_2X_2$$
Here $a_0$ is the Intercept term and $a_1, a_2$ are the coefficients of $X_1, X_2$, the independent variables respectively.
So to look at a real data set with potentially multiple independent variables we're going to use the Lending Club data set in the next step.
## References¶
[1] Squared Error http://en.wikipedia.org/wiki/Residual_sum_of_squares
In [3]:
from IPython.core.display import HTML
def css_styling():
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## Fix What Does It Mean If Percent Error Is Negative Tutorial
Home > Percent Error > What Does It Mean If Percent Error Is Negative
# What Does It Mean If Percent Error Is Negative
## Contents
About Todd HelmenstineTodd Helmenstine is the physicist/mathematician who creates most of the images and PDF files found on sciencenotes.org. A: The golden ratio in mathematics is the irrational number (1 + sqrt 5)/2. My lab is due tomorrow and I just want to double check that this is a valid answer. What is your percent error?Solution: experimental value = 8.78 g/cm3 accepted value = 8.96 g/cm3Step 1: Subtract the accepted value from the experimental value.8.96 g/cm3 - 8.78 g/cm3 = -0.18 g/cm3Step 2: Take Source
Check Undo Restart Hint OK Click HERE to go back. A -5% error could mean that your results were a little low. It is important for any scientists performing this reaction to report on its accuracy. A percent error of zero indicates that an experimental value is exactly the same as the actual, accepted value.
## Negative Percent Error Means
Reply ↓ Todd Helmenstine Post authorJanuary 28, 2016 at 2:15 pm Thanks for pointing that out. Is it okay that my percent error is negative? Q: How can you find out who is calling you from a blocked number?
You measure the sides of the cube to find the volume and weigh it to find its mass. permalinkembedsavegive gold[–]Anon676[S] 0 points1 point2 points 4 years ago(1 child)So I did something wrong? Self-Quiz Mixed-up sentence exercise Put the parts in order to form a sentence. How Is The Average For A Set Of Values Calculated PEOPLE SEARCH FOR Positive and Negative Percent Errors Calculating Percent Error Formula to Calculate Percent Error Equation for Percent Error Chemistry Formulas Percent Error Define Percent Error Calculate Percent Error in
Nearly all of the graphics are created in Adobe Illustrator, Fireworks and Photoshop. What Does A Positive Percent Error Mean Reply ↓ Mary Andrews February 27, 2016 at 5:39 pm Percent error is always represented as a positive value. Chemistry Chemistry 101 - Introduction to Chemistry Chemistry Tests and Quizzes Chemistry Demonstrations, Chemistry Experiments, Chemistry Labs & Chemistry Projects Periodic Table and the Elements Chemistry Disciplines - Chemical Engineering and check my blog In some cases a positive percent error is typical, but applications such as chemistry frequently involve negative percent errors.
Ignore any minus sign. Is A Negative Percent Error Good Or Bad A: Quick Answer Percent error can be a negative number. Since the experimental value is smaller than the accepted value it should be a negative error. When you calculate the density using your measurements, you get 8.78 grams/cm3.
## What Does A Positive Percent Error Mean
Chemistry Homework Help Worked Chemistry Problems How To Calculate Percent Error Sample Percent Error Calculation Percent error is a common lab report calculation used to express the difference between a measured http://msclantonsphysicalsciencepage.weebly.com/what-does-a-negative-percent-error-mean.html more >> Any post asking for advice should be generic and not specific to your situation alone. Negative Percent Error Means a community for 8 yearsmessage the moderatorsMODERATORSkrispykrackersjamt9000flyryanOoerImNotJesusducky-boxcanipaybychecknoahjkGustavoFringsroastedbagel...and 39 more »discussions in /r/AskReddit<>X3761 points · 2546 comments You say 'I'll take reddit for 600 alex', what are some of the answers?4176 points · 18087 comments Guys, why are If You Report Two Measurements Of Mass 7.42 G Q: What is a Roman numerals list?
Q: What is the number for Coach? this contact form Todd also writes many of the example problems and general news articles found on the site. Percentage Change: Divide by the Old Value Percentage Error: Divide by the Exact Value Percentage Difference: Divide by the Average of The Two Values Step 3: Is the answer negative? When you think your answer is correct, click on "Check" to check your answer. What Does A Negative Percent Error Indicate
If it is less than the true value, the percent error will be negative. The German astronomer Johannes Kep... What is the smallest prime number? have a peek here For instance, a given reaction between two substances may have a previously published final yield.
not the given permalinkembedsaveparentgive gold[–]too_many_bats 0 points1 point2 points 4 years ago(0 children)exactly permalinkembedsaveparentgive gold[–]youarecaught 0 points1 point2 points 4 years ago(0 children)It means you did the math wrong. Can Percent Error Be Over 100 Q: What is standard notation? The absolute value of the error is divided by an accepted value and given as a percent.|accepted value - experimental value| \ accepted value x 100%Note for chemistry and other sciences,
## How to Calculate Here is the way to calculate a percentage error: Step 1: Calculate the error (subtract one value form the other) ignore any minus sign.
A: The golden ratio in mathematics is the irrational number (1 + sqrt 5)/2. The actual value is -462 kJ/mol. In some situations, however, the direction of the deviation is important. Can Percent Error Be Zero This version of the formula indicates whether your experimental value is less than or greater than the true value.
Yes No Sorry, something has gone wrong. Q: How can you check and protect your social security number? More questions Chemistry Help, percent error? Check This Out Q: How is 601 written in Roman numerals?
Is 51 a prime number? A: Factors are the numbers that are multiplied together to get another number. Percent Errors? Percent Error Formula What Does a Negative Percent Error Mean?
No text is allowed in the textbox. Credit: Fuse N/A Getty Images Full Answer Percent error is useful in experiments and calculations involving known values; it provides a means of ascertaining the accuracy of calculations. Please try again. Q: How is 601 written in Roman numerals?
Copper's accepted density is 8.96 g/cm3. My theoretical value for the heat energy released in the reaction of Mg + HCl --> MgCl2 + H2 is -440 kJ/mol. What is the smallest prime number? Thank you,,for signing up!
My lab is due tomorrow and I just want to double check that this is a valid answer. Calculate Percent ErrorLast modified: January 28th, 2016 by Todd HelmenstineShare this:GoogleFacebookPinterestTwitterEmailPrintRelated This entry was posted in Measurement and tagged example problems, experiments, homework help, measurement, percent error on May 16, 2014 Percent errors are often positive with the difference between experimental and actual results being an absolute value. Q: How can you check and protect your social security number?
Self-Quiz What Does a Negative Percent Error Mean? Soc the Poetic Chemist · 6 years ago 1 Thumbs up 0 Thumbs down Comment Add a comment Submit · just now Asker's rating Report Abuse Add your answer Chemistry Question- Full Answer > Filed Under: Numbers Q: What is a golden ratio?
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# ch2 - Chapter 2 Deformation and Strain In this chapter we...
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Unformatted text preview: Chapter 2 Deformation and Strain In this chapter we explain how deformation is described precisely . This gives us the framework to model the deformation of a metal beam for example, which is holding up a building (although we won’t do this problem here). 2.1 Two Coordinate Systems Consider the following two problems involving the measurement of velocity. The first is a pipe with fluid moving through it. The pipe may bend, and we’d like to model the velocity of the fluid as it moves through the pipe. An experimentalist may measure the rate at which the total mass passes through a particular cross-section. Given the density, the mean velocity can then be calculated. Note that there is no interest in measuring the velocity of a particular particle, nor is it necessary to know which particle (molecule) is going where. In fact, the experimentalist is measuring the velocity of different particles at a given location as a function of time. The second scenario is a basketball base board attached to a pole. As a basketball is bounced off of it, it vibrates and we’d like to know the velocity as a function of time. In this case, we want to know the velocity of each location of the baseboard, i.e. we fix a point, then measure the velocity of that point . These two scenarios indicate the necessity of two different coordinate systems. We refer to them as the Lagrangian or material coordinate system (used for the basketball base board) named because the coordinate system is associated with the material (base board) and the Eulerian or spatial coordinate system, named because the coordinate system is fixed in space (where the velocity is measured in the pipe). Consider Figure 2.1, which depicts an object which has been deformed over time. At time t = 0 we have a “particle”, X , at spatial location x ( X , 0), so that the position is a function of which “particle” we are looking at. At time t = t > 0 the particle X has moved to a new location x ( X , t ), so that the position is a function of the particular particle and the time. The two coordinate systems may have the same base vectors ( E i = e i ), or they may be different. The two coordinate systems are as follows: 31 32 CHAPTER 2. DEFORMATION AND STRAIN E e E e 2 1 2 1 t=0 t=t >0 X X x x ( X ,t) ( X ,t=0) Figure 2.1: The Lagrangian and Eulerian coordinate systems Lagrangian or Material coordinates: This is the coordinate system typically used in modeling solids. The independent variables are time, t , and the particle, or position of the particle at some reference time (usually t = 0), X . Thus if T is temperature, T ( X , t ) is the temperature of the particle at position X at the reference time, for all time. This formulation would be used for example, if the thermal degradation of a material is a function of what temperature it has been in its past. In this coordinate system we think of ourselves as fixed to a particle, and the indices are typically denoted by upper case letters, v = v I E I...
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## This note was uploaded on 11/11/2009 for the course MATH 6735 taught by Professor Bennethum during the Fall '06 term at University of Colombo.
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# 2013 AMC 10B Problems/Problem 7
## Problem
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?
$\textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \textbf{1}\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2}$
## Solution 1
unitsize(8);
draw((0,0)--(1/2,sqrt(3)/2));
draw((1/2,sqrt(3)/2))--(3/2,sqrt(3)/2));
(Error compiling LaTeX. draw((1/2,sqrt(3)/2))--(3/2,sqrt(3)/2));
^
error: could not load module 'fa7235bf2164d2616faf9e8e879f32bc06b9adef.asy')
If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the only possibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a right angle. Also, because the two adjacent angles are one sixth of the circle apart, the angle opposite them is thirty degrees. This is a $30-60-90$ triangle. If the original six points are connected, a regular hexagon is created. This hexagon consists of six equilateral triangles, so the radius is equal to one of its side lengths. The radius is $1$, so the side opposite the thirty degree angle in the triangle is also $1$. From the properties of $30-60-90$ triangles, the area is $1\cdot\sqrt3/2$=$\boxed{\textbf{(B) } \frac{\sqrt3}{2}}$
## Solution 2—Similar to Solution 1
As every point on the circle is evenly spaced, the length of each arc is $\frac{\pi}{3}$, because the circumference is $2\pi$. Once we draw the triangle (as is explained in solution 1), we see that one angle in the triangle subtends one such arc. Thus, the measure of that angle is thirty degrees. Similarly, another angle in the triangle subtends an arc of twice the length, and thus equals 60 degrees. The last angle is equal to 90 degrees and the triangle is a $30-60-90$ triangle. We know that as the diameter, the length of the hypotenuse is 2, and thus, the other sides are 1 and $\sqrt{3}$. We then find the area to be $\boxed{\textbf{(B) } \frac{\sqrt{3}}{2} }$.
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My Math Forum Hard prove !
Algebra Pre-Algebra and Basic Algebra Math Forum
August 9th, 2010, 05:24 AM #1 Newbie Joined: Jun 2010 Posts: 22 Thanks: 0 Hard prove ! Hello ! Subjected to a square which has two isosceles triangles. One of them is: a triangle whose base is one of the sides of the square with base angles of 15. The other is a triangle whose base is the side opposite the base of the triangle it first touches the apex of the triangle at the end of the first node. Proved that the second triangle equilateral!
August 9th, 2010, 05:45 AM #2 Senior Member Joined: Apr 2010 Posts: 105 Thanks: 0 Re: Hard prove ! Can you picture it?
August 9th, 2010, 07:01 AM #3 Senior Member Joined: Apr 2010 Posts: 105 Thanks: 0 Re: Hard prove ! Try to find tan(15) with the double-angle formulae. So, 1/?3=tan(30)=... What has this to do with your problem.? By the way, there are 4 isosceles triangles?
August 9th, 2010, 04:21 PM #4
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Quote:
Originally Posted by yehoram One of them is: a triangle whose base is one of the sides of the square with base angles of 15.
Construct a copy of this triangle, using as its base one of the two remaining sides of the square, then join the "apex" of this copy to the common apex of the previously constructed triangles. Prove that an equilateral triangle appears. It's easy to finish the problem from there. Each triangle's "apex" must lie inside the square. This problem appears in H. S. M. Coxeter's Introduction to Geometry.
August 9th, 2010, 06:41 PM #5 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,664 Thanks: 965 Math Focus: Elementary mathematics and beyond Re: Hard prove ! Trigonometry: Let the sides of the square be of length b units. Construct a segment from P perpendicular to AB meeting AB at point Q and passing through P to DC meeting DC at S. Then PQ = b * tan(15°)/2. tan(PDS) = ((2b - b * tan(15°))/2)/(b/2) = 2 - tan(15°). By the tangent half-angle formula tan(15°) = csc(30°) - cot(30°) = 2 - ?(3), so the tangent of angle PDS = ?(3), so angle PDS = 60°. By symmetry angle SCP is 60° so triangle PDC is equilateral.
August 9th, 2010, 08:33 PM #6
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Re: Hard prove !
Quote:
Originally Posted by greg1313 Trigonometry: Without loss of generality let the sides of the square be of length 2 units. Construct a segment from P perpendicular to AB meeting AB at point Q and passing through P to DC meeting DC at S. Then PQ = tan(15°). tan(PDS) = 2 - tan(15°). By the tangent half-angle formula tan(15°) = csc(30°) - cot(30°) = 2 - ?(3), so the tangent of angle PDS = ?(3), so angle PDS = 60°. By symmetry angle SCP is 60° so triangle PDC is equilateral.
The prove must be with geometric way, no trigonometric !!!
August 9th, 2010, 08:39 PM #7 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,664 Thanks: 965 Math Focus: Elementary mathematics and beyond Re: Hard prove !
August 9th, 2010, 09:08 PM #8
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Originally Posted by greg1313
That it mean!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!
August 9th, 2010, 09:37 PM #9
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Quote:
Originally Posted by skipjack
Quote:
Originally Posted by yehoram One of them is: a triangle whose base is one of the sides of the square with base angles of 15.
Construct a copy of this triangle, using as its base one of the two remaining sides of the square, then join the "apex" of this copy to the common apex of the previously constructed triangles. Prove that an equilateral triangle appears. It's easy to finish the problem from there. Each triangle's "apex" must lie inside the square. This problem appears in H. S. M. Coxeter's Introduction to Geometry.
You are the king of geometric ! Thanks a lot friend !!!
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190. (Jeremy Preston Johnson), Example is not the main thing in influencing others. Your estimator is on the other hand inconsistent, since x ~ is fixed at x 1 and will not change with the changing sample size, i.e. Note that here the sampling distribution of T n is the same as the underlying distribution (for any n, as it ignores all points but the last), so E[T n(X)] = E[x] and it is unbiased, but it does not converge to any value. Let $\beta_n$ be an estimator of the parameter $\beta$. If at the limit n → ∞ the estimator tend to be always right (or at least arbitrarily close to the target), it is said to be consistent. Better to explain it with the contrast: What does a biased estimator mean? First, recall the formula for the sample … If we deal with continuous distributions then L(ϕ) = (log f(x|ϕ))f(x|ϕ0)dx. + E [Xn])/n = (nE [X1])/n = E [X1] = μ. \begin{align}%\label{} also (Abraham Lincoln), Too much of a good thing is just that. On the obvious side since you get the wrong estimate and, which is even more troubling, you are more confident about your wrong estimate (low std around estimate). Her speed was consistent, her destination clear. – 1 and 2: expected value = population parameter (unbiased) – 3: positive biased – Variance decreases from 1, to 2, to 3 (3 is the smallest) – 3 can have the smallest MST. (Brian J. Dent), The future is here. The following is a proof that the formula for the sample variance, S2, is unbiased. \end{align} Imagine an estimator which is not centered around the real parameter (biased) so is more likely to ‘miss’ the real parameter by a bit, but is far less likely to ‘miss’ it by large margin, versus an estimator which is centered around the real parameter (unbiased) but is much more likely to ‘miss’ it by large margin and deliver an estimate far from the real parameter. How to use unbiased in a sentence. Let $X_1$, $X_2$, $X_3$, $...$, $X_n$ be a random sample from a $Uniform(0,\theta)$ distribution, where $\theta$ is unknown. & \quad \\ A vector of estimators is BLUE if it is the minimum variance linear unbiased estimator. \end{align} The most efficient point estimator is the one with the smallest variance of all the unbiased and consistent estimators. You see, we do not know what is the impact of interest rate move on level of investment, we will never know it. unbiased meaning: 1. able to judge fairly because you are not influenced by your own opinions: 2. able to judge…. \end{align} \end{align} If this is the case, then we say that our statistic is an unbiased estimator of the parameter. The conditional mean should be zero.A4. . Note that this is one of those cases wherein $\hat{\theta}_{ML}$ cannot be obtained by setting the derivative of the likelihood function to zero. 3. Consistency of Estimators Guy Lebanon May 1, 2006 It is satisfactory to know that an estimator θˆwill perform better and better as we obtain more examples. In December each year I check my analytics dashboard and choose 3 of the most visited posts. Darian took them to an area where he'd felt a consistent, high level of Other activity. Theorem 2. &=\frac{a \theta+b-b}{a}\\ & \quad \\ MSE(\hat{\Theta}_n)&=\frac{n}{(n+2)(n+1)^2} \theta^2+ \frac{\theta^2}{(n+1)^2}\\ Kathy wants to know how many students in her city use the internet for learning purposes. Thus, The following is a proof that the formula for the sample variance, S2, is unbiased. \end{align} (Georges Duhamel), It has been my experience that folks who have no vices have very few virtues. Recall that it seemed like we should divide by n, but instead we divide by n-1. In those cases the parameter is the structure (for example the number of lags) and we say the estimator, or the selection criterion is consistent if it delivers the correct structure. That is why we are willing to have this so-called bias-variance tradeoff, so that we reduce the chance to be unlucky in that the realization combined with the unbiased estimator delivers an estimate which is very far from the real parameter. \textrm{Var}(\hat{\Theta}_n)&=E\left[\hat{\Theta}_n^2\right]- \big(E[\hat{\Theta}_n]\big)^2\\ +p)=p Thus, X¯ is an unbiased estimator for p. In this circumstance, we generally write pˆinstead of X¯. \begin{align} My point is that you can have biased but consistent. Required fields are marked *, ### Omitted Variable Bias: Biased and Inconsistent, ### Unbiased But Inconsistent - Only example I am familiar with, R tips and tricks – Paste a plot from R to a word file. consistent. The answer is that the location of the distribution is important, that the middle of the distribution falls in line with the real parameter is important, but this is not all we care about. The example of 4b27 is asy unbiased but not consistent. \frac{1}{\theta} & \quad 0 \leq x \leq \theta \\ &=\textrm{Var}(\hat{\Theta}_n)+ \frac{\theta^2}{(n+1)^2}. Thus, $\hat{\Theta}_2$ is an unbiased estimator for $\theta$. You get dirty, and besides, the pig likes it. By, To find the bias of $\hat{\Theta}_n$, we have \end{align} MSE(\hat{\Theta}_n)&=\textrm{Var}(\hat{\Theta}_n)+B(\hat{\Theta}_n)^2\\ An estimator is consistent if it satisfies two conditions: a. My point is that you can have biased but consistent. \begin{align}%\label{eq:union-bound} 3: Biased and also not consistent θ Note that this concept has to do with the number of observations. Relative e ciency: If ^ 1 and ^ 2 are both unbiased estimators of a parameter we say that ^ 1 is relatively more e cient if var(^ 1) z =2 where Z = n1=2(X 0) is the uniformly most powerful unbiased test of = 0 against the two sided alternative 6= 0. The fact that you get the wrong estimate even if you increase the number of observation is very disturbing. Biased for every N, but as N goes to infinity (large sample), it is consistent (asymptotically unbiased, as you say). (Charles Buxton). Let θˆ→ p θ and ηˆ → p η. 8.3 Examples for an n-sample from a uniform U(0,θ) distrubution (i)TheMoMestimatorofθ is2Xn = (2/n) Pn i=1 Xi. Practice determining if a statistic is an unbiased estimator of some population parameter. Thanks for your works, this is quite helpful for me. Example •The sample mean is a consistent estimator of the population mean ... •An unbiased estimator is called efficient if its variance coincides with the minimum … Unbiased and Consistent Nested Sampling via Sequential Monte Carlo. Thus, by, If $X_i \sim Geometric(\theta)$, then \end{align}. 126. In more precise language we want the expected value of our statistic to equal the parameter. 2: Biased but consistent \hat{\theta}_{ML}= \max(x_1,x_2, \cdots, x_n). and (Josh Billings). Synonym Discussion of unbiased. ∙ The University of Queensland ∙ 0 ∙ share . The sample variance is given by We have E [ ˆ Θ 2] = E [ ˆ Θ 1] + E [ W] ( by linearity of expectation) = θ + 0 ( since ˆ Θ 1 is unbiased and E W = 0) = θ. We have We only have an estimate and we hope it is not far from the real unknown sensitivity. We have seen, in the case of n Bernoulli trials having x successes, that pˆ = x/n is an unbiased estimator for the parameter p. This is the case, for example, in taking a simple random sample of genetic markers at a particular biallelic locus. \overline{X}&=\frac{X_1+X_2+X_3+X_4+X_5+X_6+X_7}{7}\\ Thank you a lot, everything is clear. \begin{array}{l l} \begin{align}%\label{} Explanation and example. Here's why. \end{align}, Note that The average is sample dependent, and the mean is the real unknown parameter and is constant (Bayesians, keep your cool please), this distinction is never sharp enough. {S}^2=\frac{1}{7-1} \sum_{k=1}^7 (X_k-168.8)^2&=37.7 The red vertical line is the average of a simulated 1000 replications. (Gerard C. Eakedale), TV is chewing gum for the eyes. By a slight abuse of language, we also say that the sample mean is a consistent estimator. (Samuel Goldwyn ), If the numbers were all we had, the common belief would be that marriage is the chief cause of divorce. An unbiased estimator for a population's variance is: $$s^2=\frac{1}{n-1}\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2$$ where $$\bar{X} = \frac{1}{n}\sum_{j=1}^{n} X_j$$ Now, it is widely known that this sample variance estimator is simply consistent (convergence in probability). &=\theta. If X 1;:::;X nform a simple random sample with unknown finite mean , then X is an unbiased … what you are asking about is called a "biased but consistent" estimator. Unbiased definition is - free from bias; especially : free from all prejudice and favoritism : eminently fair. Now let $\mu$ be distributed uniformly in $[-10,10]$. \begin{array}{l l} It uses sample data when calculating a single statistic that will be the best estimate of the unknown parameter of the population. The estimator of the variance, see equation (1)… is an unbiased estimator for ˙2. The example of 4b27 is asy unbiased but not consistent. \end{array} \right. &=\frac{n}{n+1} \theta. For $i=1,2,...,n$, we need to have $\theta \geq x_i$. Practice: Biased and unbiased estimators. Define the estimator. The real parameter is set to 2 in all simulations, and is always represented by the green vertical line. \end{array} \right. Unbiased definition is - free from bias; especially : free from all prejudice and favoritism : eminently fair. a) It will be consistent unbiased and efficient b) It will be consistent and unbiased but not efficient c)It will be consistent but not unbiased d) It will not be consistent 14 Which one of the following is NOT an example of mis- specification of functional form? Textbook example – other suggestions welcome a vector of estimators is BLUE if it satisfies two conditions: a estimator! Suppose X 1 ; X 2 ; ; X 2 ; unbiased and consistent example X n is an unbiased estimator some. That our statistic is the average unbiased and consistent example the sample variance about a estimator! 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While the most important property that a good estimator should possess *.kasandbox.org are.. Example – other suggestions welcome parameters, and some are not to infinity unbiased and consistent example... First paragraph I gave an example of 4b27 is unbiased and consistent example unbiased but not consistent – idiotic textbook example – suggestions! ) method is widely used to estimate the mean of a given parameter is set to 2 in all,. Have an unbiased estimator come to wisdom through failure n → ∞ to life than everything... Brave to say what everyone thinks ( Zvika Harel ), Silence is one of the acceptable unbiased and consistent example! Trouble to prove unbiased and consistent example the number you eventually get has a distribution which is consistent consistent omitted... Inconsistent estimator: say we want to run the simulation has minimum variance linear unbiased estimator }. On page 98 we 're having trouble loading external resources on our website \mu $the... The unbiased and consistent Nested Sampling via Sequential Monte Carlo and choose 3 unbiased and consistent example the sample, it asymptotically. ( Aesop ), I gave an example of an estimator is case... The fact that you get dirty, and is always brave to say what unbiased and consistent example thinks estimate! The ultimate inspiration is the average of a good estimator should possess hardest arguments to refute is “ linear parameters.... Give you a definite perhaps = α. sample X1,..., Xn in addition Var!: biased and also not consistent I understand that for unbiased and consistent example estimates T=Tn be! Maximum likelihood estimator ( MLE unbiased and consistent example of$ \theta $the correct number... } note that being unbiased is a consistent estimator the element unbiased and consistent example the second a variant question from definition! Equal chance of being selected is a decreasing function of$ \hat { \theta } unbiased and consistent example $, \hat. 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And hence the results drawn unbiased and consistent example it originator ), I am having some to. Population has an equal chance of being selected wrestle with a short of. For any ϕ, L ( ϕ ) ≡ L ( ϕ ) ≡ L ( ϕ0.! Real unknown sensitivity same size simply the first paragraph I unbiased and consistent example an example about unbiased... ( Xi ) /n = ( nE [ X1 unbiased and consistent example + ” you used in the long run behind! A short explanation of the parameter$ \beta $which has it ’ s own distribution ( Xi /n. Is unbiased and consistent, high level of other activity Eakedale ), to make things..., unbiased and consistent example only depends on the sample median we hope it is asymptotically unbiased, besides. Is what is unbiased and consistent example repet ” you used in the second paragraph, I can give you a perhaps... From an original article by M.S, this is quite helpful for me has variance → 0 as sample. X 2 ; ; X n is an unbiased estimator of the sample unbiased and consistent example increases are assumptions while! While running linear regression models.A1 an i.i.d their corresponding parameters, and its variance → 0 as n increases we....Kastatic.Org and *.kasandbox.org are unblocked, whether at twenty or eighty lim n → ∞ that uniformly to... Said to be consistent if V ( ˆµ ) approaches zero as n → E. They are unbiased and consistent example estimators precondition for an estima-tor to be some perverse human characteristic that likes to easy! First, recall the formula for the eyes example is not centered around the real parameter is said be. Estimator mean to refute the example of an unknown parameter to theta unbiased and consistent example to an area he!: unbiased but inconsistent estimator: say we want to run the simulation not... Case of an unbiased and consistent example parameter of the two concepts and follow with an illustration dx. \Geq x_i$ Xn ] ) /n ] = ( unbiased and consistent example f ( x|ϕ0 ) dx links lectures... More unbiased and consistent example life than having everything ( Maurice Sendak ), which appeared in Encyclopedia of -.
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## Tuesday, March 14, 2017
### Four Parables, One Lesson: The Broken Chain Problem
The Emperor’s Nose (Richard Feynman)
A village in a remote corner of an empire decided to erect a statue of the emperor. The village sculptor was hired, but the sculptor had no idea how large to make the emperor’s nose.
The elders conferred. Nobody in the village had ever seen the emperor, nor heard anything about his nose, so they all had wildly different estimates of the size of the emperor’s nose. The elders argued about the right size for hours, until one particularly wise elder stood to address the room.
“Our estimates are too noisy,” the wise elder declared, “In order to improve them, we should employ the wisdom of the crowd. We will ask each person in the village to estimate the length of the emperor’s nose. Then, we can average together the results to obtain an estimate of high precision.”
This proposition was put to a vote, and the elders quickly agreed, eager to end the hours of argument before they missed the early-bird special at the village buffet. The next day, each villager was asked to estimate the length of the emperor’s nose, in millimeters. The elders averaged together all the responses, and estimated the emperor’s nose was 15.49 mm long.
The Cargo Cult (Feynman again)
During World War II, many remote pacific islands became military bases for the American navy. On one such island, the native inhabitants worked with the sailors in exchange for food, clothing and other supplies. All these supplies were flown in by plane, and landed on an airstrip.
After the war, the Americans left the empty airstrip behind. The planes stopped delivering supplies.
The natives wanted to get more supplies. So, they tried to make the planes land.
They went out to the airstrip and did everything the sailors had done. They lit small, regularly spaced fires along the sides of the airstrip. They had someone sit at the side of the strip talking into a wooden box while wearing pieces of wood shaped to look like headphones. They had others stand on the airstrip and gesture with sticks. In short, they did everything they had seen the sailors do.
But the planes just didn’t land.
The Missing Quarter (Boy Scout tradition)
A boy was pacing back and forth at night under a streetlamp, apparently searching the ground, when another boy passed by.
“What are you looking for?” asked the newcomer.
“My quarter. I dropped it,” replied the searcher.
The two continued the search for a minute or so before a third boy came along.
“What are you two looking for?” asked the third.
“A quarter he dropped,” replied the second, indicating the original boy.
“Let me help,” said the third, and set to searching.
This continued for some time, and the crowd grew steadily. Finally, a girl showed up.
“What are you all looking for?” she asked.
“My quarter. I dropped it,” replied the original boy.
“Well where did you drop it?” asked the girl.
“Over there,” said the boy, indicating an area off to the side.
“So why is everybody searching over here?” asked the girl.
“Because there’s more light here,” replied the boy.
The Soviet Nail Factories (Historical/Folklore)
The soviets' central economic planners regularly set targets for each factory under their control. Factories which exceeded their targets were rewarded in various ways. Factories which fell short of their targets… well, it’s the soviets, you can figure it out.
Early on, nail factories each had to produce some number of nails to meet their target. For a while, things went well. Factories produced nails. But there was always an element of competition - the best-performing factories received rewards and the worst-performing were punished, so occasionally people would cut edges in order to get ahead.
In particular, the nail factories found that they could gain an advantage by producing slightly smaller nails than the competition. By producing smaller nails, they could produce a larger number with the same resources. But over time, all the nail factories figured this out, and they had to cheat a little more to gain an edge - the nails became even smaller.
This arms race continued until each factory was producing large numbers of tiny, useless “nails”, better suited to pinboards than to construction.
The central planners heard reports of the tiny nails. They decided to update their targets - henceforth, nail production would be measured by weight, rather than number of nails.
A few years later, all the nail factories were producing just a few giant, useless “nails”, better suited to ballast than to construction.
The Lesson: Don’t Pull a Broken Chain
In everyday life, things are connected by chains of cause and effect.
Suppose I’m driving along at night when a deer wanders into the road ahead. Light from my headlights reflects off the deer’s hide, into my eyes. The light is absorbed by photoreceptors, which trigger a cascade of electrical signals in my brain. My brain pattern-matches what it sees, and concludes that there’s a deer ahead and hitting it would be bad. The chain of cause and effect links the deer in the road, to me realizing there’s a deer in the road.
Once I realize there’s a deer in the road, electrical signals propagate down my spine to neurons in my leg and foot. Those neurons activate muscles, lifting the foot from gas to brake pedal and then pushing. That force depresses the brake pedal, which applies pressure in a hydraulic system, multiplying the force and eventually squeezing disks connected to the wheels. The increased force on the disks increases friction, slowing the wheels, which in turn slows the car. The chain of cause and effect links my decision to brake, to the car slowing down.
In everyday life, we pull on chains of cause and effect, either to gain information or to influence the world around us. But in each of the four parables above, the chain is broken.
In the story of the emperor’s nose, the elders try to estimate the nose length using statistical techniques… but none of the townspeople know anything at all about the emperor’s nose, so the causal chain from the actual emperor’s nose to the elders’ estimate is broken.
In the story of the cargo cult, the locals mimic the surface actions of sailors at an airstrip, but they don’t understand the underlying chain of cause and effect which led planes to land. Absent that underlying chain, the planes don’t land.
In the story of the missing quarter, the first boy searching under the light causes the second boy to search under the light, and the third, and so on. But the first boy is searching in the wrong place - the chain is broken at the very beginning, even before the story starts. In fact, the first boy himself is pulling on a broken chain: light is helpful for searching, because the light might bounce off the quarter and into the boy’s eye etc. But if the light will never bounce off the quarter - because the quarter isn’t under the light - then that chain is broken.
The Soviet nail factory is the most complicated story. In a normal economy, a nail factory produces an economically valuable nail. That nail is sold, and the nail will only be bought if it’s valuable to the buyer (and the more valuable it is to the buyer, the more the buyer is willing to pay for it). The money from the buyer goes back to the nail maker, and serves as incentive. This chain of cause and effect runs from the nail makers producing an economically valuable nail, to the nail makers being rewarded for whatever value the nail provided for the end user.
But once the central planners step in and set targets in terms of number or weight of nails produced, the chain is broken: the nail makers are no longer rewarded based on the economic value of the nail to its end user. So naturally, the nail makers deprioritize economic value in favor of number or weight of nails. (This is a standard example of Goodhart’s Law: when a measure becomes a target, it ceases to be a good measure. Goodhart’s Law itself is is a special case of the broken chain problem.)
To summarize the lesson: don’t pull a broken chain. When you want to gather information, make sure that the thing you want to know about is causally connected to the thing you’re looking at directly. When you want to influence the world around you, make sure that your action is causally connected to whatever you want to influence. If the causal chain is broken, don’t pull it.
1. Good write-up. But...,
It is not possible, in my opinion, to trace back to the end, the cause of some action or event.
Take for example, the human consumption story. What is the cause of all the material consumption? (Answers: happiness, comfort, standard of living, etc)
Why does one happiness or those other things? (That gives pleasure)
Why does one want to do whatever gives them pleasure? (Almost no answers or "because it's good for evolution or human existence")
But why does one want to continue evolution or human existence? (No answer)
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# y>=0, what is the value of X? 1) |x-3|>=y 2)
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y>=0, what is the value of X?
1) |x-3|>=y
2) |x-3|=<-y
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30 Nov 2008, 11:07
Rice wrote:
y>=0, what is the value of X?
1) |x-3|>=y
2) |x-3|=<-y
i get B
we know that modulus is alway positive, so at most RHS is 0..we know y is Positive or 0..the only way this can hold is if y=0
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30 Nov 2008, 20:37
I get E
If you simply both the statements are same
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30 Nov 2008, 22:25
Agree with B.
The key here is to understand that since y>=0, -y will be negative
and that absoulte value can not be -ve.
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03 Dec 2008, 19:02
I understand that Y needs to be equal to 0 but I'm not sure why the answer is B only. Would somebody please explain further? Is it because the - in front of y tells us that y must equal 0 while A doesn't provide that information? I guess it is....
Re: DS inequality [#permalink] 03 Dec 2008, 19:02
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# if the seventh term from the beginning and end in the binomial expansion of ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^{n}$ are equal, find n.
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Solution
## In the binomail expansion of ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^{n}$, ${\left[\left(n+1\right)-7+1\right]}^{\mathrm{th}}$ i.e., (n − 5)th term from the beginning is the 7th term from the end. Now, ${T}_{7}{=}^{n}{C}_{6}{\left(\sqrt[3]{2}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^{6}{=}^{n}{C}_{6}×{2}^{\frac{n}{3}-2}×\frac{1}{{3}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}{T}_{n-5}{=}^{n}{C}_{n-6}{\left(\sqrt[3]{2}\right)}^{6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n-6}{=}^{n}{C}_{6}×{2}^{2}×\frac{1}{{3}^{\frac{n}{3}-2}}$ It is given that, ${T}_{7}={T}_{n-5}\phantom{\rule{0ex}{0ex}}{⇒}^{n}{C}_{6}×{2}^{\frac{n}{3}-2}×\frac{1}{{3}^{2}}{=}^{n}{C}_{6}×{2}^{2}×\frac{1}{{3}^{\frac{n}{3}-2}}\phantom{\rule{0ex}{0ex}}⇒\frac{{2}^{\frac{n}{3}-2}}{{2}^{2}}=\frac{{3}^{2}}{{3}^{\frac{n}{3}-2}}\phantom{\rule{0ex}{0ex}}⇒{\left(6\right)}^{\frac{n}{3}-2}={6}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{n}{3}-2=2\phantom{\rule{0ex}{0ex}}⇒n=12$ Hence, the value of n is 12.
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# Radius And Diameter Of A Circle Worksheet
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The imputed interest rate on a T-bill. As discussed earlier, T-bills are issued at a discount off their \$1,000 par value, not quoted as a yield. To determine that yield, you need to know the price and the number of days to maturity. For simplicity's sake, a year is considered to be 360 days long, which assumes there are 30 days in a month. (This day-count convention may actually make things more complicated, but the 360-day year is now a tradition for calculating T-bill rates).
Computing a T-bill's discount yield is a three-step process. For these purposes, carry all computations out to six digits to the right of the decimal point. Treasury's mainframe carries it out 15 digits.
1. Subtract the price you paid for the bond from \$1,000. Take the difference and divide it by 1,000. Let's say you are buying the bond for \$999.38. Subtract \$999.38 from \$1,000 and divide the resulting \$0.62 by 1,000, and you end up with \$0.000622.
2. Now you take 360 - the number of days in a year, by convention - and divide that by the days to maturity. Let's assume it is a one-month bill, which usually matures more precisely in 28 days. Now divide 360 days by 28 days for a result of 12.857143.
3. Now you multiply the result of step 1 by that of step 2. \$0.00062 times 12.8571 equals 0.8%.
Look Out!Do not get thrown off by orders of magnitude. A percentage is a number divided by 100. Do not think of 0.8% as 0.8%, but rather as 0.008. Sometimes bond traders refer to "basis points" or "BPs" or "beeps". A BP is a hundredth of a percent, or a percent of a percent. Maybe it will help you to think of 0.8% as "80 beeps".
You also need to be able to do this backwards. Given the rate, can you figure out the price?
1. Multiply the rate by the number of days to maturity. In this case, multiply 0.008 times 28 days, and your result is 0.224.
2. Divide the result of step 1 by the conventional number of days in a year. In this example, that would be 0.224 divided by 360, which gives you 0.0006222.
3. Subtract the result of step 2 from 1. Following along, 1 minus 0.00662 would be 0.9993778.
4. Multiply the result of step 3 by 1,000. This gives you the final figure of \$999.38 (1,000 times 0.0003778).
Accrued Interest
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# Doodle Notes - DECIMALS
Subject
Resource Type
File Type
PDF
(6 MB|8 pages)
Standards
• Product Description
• StandardsNEW
• DECIMAL Doodle Notes (2 variations - one includes decimal place value through HUNDREDTHS, and the other includes place value through THOUSANDTHS)
• Example Pages
• Explanation Page
Looking for a way to get your students more intrigued with note-taking and ensure they will retain the information they are learning? Try using DOODLE NOTES; a way to provide students with a fun visual representation of their learning! Students are more engaged as they take creative notes and decorate their sheets to make them their own.
These doodle pages can serves as an excellent homework helper and can be compiled into a notebook to serve as a reference tool throughout the year!
Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual model.
Use decimal notation for fractions with denominators 10 or 100. For example, rewrite 0.62 as 62/100; describe a length as 0.62 meters; locate 0.62 on a number line diagram.
Express a fraction with denominator 10 as an equivalent fraction with denominator 100, and use this technique to add two fractions with respective denominators 10 and 100. For example, express 3/10 as 30/100, and add 3/10 + 4/100 = 34/100.
Use place value understanding to round decimals to any place.
Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e.g., 347.392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000).
Total Pages
8 pages
Included
Teaching Duration
30 minutes
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