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Author Archive
## How to simulate multilevel/longitudinal data
I was recently talking with my friend Rebecca about simulating multilevel data, and she asked me if I would show her some examples. It occurred to me that many of you might also like to see some examples, so I decided to post them to the Stata Blog.
## Introduction
We simulate data all the time at StataCorp and for a variety of reasons.
One reason is that real datasets that include the features we would like are often difficult to find. We prefer to use real datasets in the manual examples, but sometimes that isn’t feasible and so we create simulated datasets.
We also simulate data to check the coverage probabilities of new estimators in Stata. Sometimes the formulae published in books and papers contain typographical errors. Sometimes the asymptotic properties of estimators don’t hold under certain conditions. And every once in a while, we make coding mistakes. We run simulations during development to verify that a 95% confidence interval really is a 95% confidence interval.
Simulated data can also come in handy for presentations, teaching purposes, and calculating statistical power using simulations for complex study designs.
And, simulating data is just plain fun once you get the hang of it.
Some of you will recall Vince Wiggins’s blog entry from 2011 entitled “Multilevel random effects in xtmixed and sem — the long and wide of it” in which he simulated a three-level dataset. I’m going to elaborate on how Vince simulated multilevel data, and then I’ll show you some useful variations. Specifically, I’m going to talk about:
1. How to simulate single-level data
2. How to simulate two- and three-level data
3. How to simulate three-level data with covariates
4. How to simulate longitudinal data with random slopes
5. How to simulate longitudinal data with structured errors
## How to simulate single-level data
Let’s begin by simulating a trivially simple, single-level dataset that has the form
$y_i = 70 + e_i$
We will assume that e is normally distributed with mean zero and variance $$\sigma^2$$.
We’d want to simulate 500 observations, so let’s begin by clearing Stata’s memory and setting the number of observations to 500.
. clear
. set obs 500
Next, let’s create a variable named e that contains pseudorandom normally distributed data with mean zero and standard deviation 5:
. generate e = rnormal(0,5)
The variable e is our error term, so we can create an outcome variable y by typing
. generate y = 70 + e
. list y e in 1/5
+----------------------+
| y e |
|----------------------|
1. | 78.83927 8.83927 |
2. | 69.97774 -.0222647 |
3. | 69.80065 -.1993514 |
4. | 68.11398 -1.88602 |
5. | 63.08952 -6.910483 |
+----------------------+
We can fit a linear regression for the variable y to determine whether our parameter estimates are reasonably close to the parameters we specified when we simulated our dataset:
. regress y
Source | SS df MS Number of obs = 500
-------------+------------------------------ F( 0, 499) = 0.00
Model | 0 0 . Prob > F = .
Residual | 12188.8118 499 24.4264766 R-squared = 0.0000
Total | 12188.8118 499 24.4264766 Root MSE = 4.9423
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons | 69.89768 .221027 316.24 0.000 69.46342 70.33194
------------------------------------------------------------------------------
The estimate of _cons is 69.9, which is very close to 70, and the Root MSE of 4.9 is equally close to the error’s standard deviation of 5. The parameter estimates will not be exactly equal to the underlying parameters we specified when we created the data because we introduced randomness with the rnormal() function.
This simple example is just to get us started before we work with multilevel data. For familiarity, let’s fit the same model with the mixed command that we will be using later:
. mixed y, stddev
Mixed-effects ML regression Number of obs = 500
Wald chi2(0) = .
Log likelihood = -1507.8857 Prob > chi2 = .
------------------------------------------------------------------------------
y | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons | 69.89768 .2208059 316.56 0.000 69.46491 70.33045
------------------------------------------------------------------------------
------------------------------------------------------------------------------
Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval]
-----------------------------+------------------------------------------------
sd(Residual) | 4.93737 .1561334 4.640645 5.253068
------------------------------------------------------------------------------
The output is organized with the parameter estimates for the fixed part in the top table and the estimated standard deviations for the random effects in the bottom table. Just as previously, the estimate of _cons is 69.9, and the estimate of the standard deviation of the residuals is 4.9.
Okay. That really was trivial, wasn’t it? Simulating two- and three-level data is almost as easy.
## How to simulate two- and three-level data
I posted a blog entry last year titled “Multilevel linear models in Stata, part 1: Components of variance“. In that posting, I showed a diagram for a residual of a three-level model.
The equation for the variance-components model I fit had the form
$y_{ijk} = mu + u_i.. + u_{ij.} + e_{ijk}$
This model had three residuals, whereas the one-level model we just fit above had only one.
This time, let’s start with a two-level model. Let’s simulate a two-level dataset, a model for children nested within classrooms. We’ll index classrooms by i and children by j. The model is
$y_{ij} = mu + u_{i.} + e_{ij}$
For this toy model, let’s assume two classrooms with two students per classroom, meaning that we want to create a four-observation dataset, where the observations are students.
To create this four-observation dataset, we start by creating a two-observation dataset, where the observations are classrooms. Because there are two classrooms, we type
. clear
. set obs 2
. generate classroom = _n
From now on, we’ll refer to classroom as i. It’s easier to remember what variables mean if they have meaningful names.
Next, we’ll create a variable that contains each classroom’s random effect $$u_i$$, which we’ll assume follows an N(0,3) distribution.
. generate u_i = rnormal(0,3)
. list
+----------------------+
| classr~m u_i |
|----------------------|
1. | 1 .7491351 |
2. | 2 -.0031386 |
+----------------------+
We can now expand our data to include two children per classroom by typing
. expand 2
. list
+----------------------+
| classr~m u_i |
|----------------------|
1. | 1 .7491351 |
2. | 2 -.0031386 |
3. | 1 .7491351 |
4. | 2 -.0031386 |
+----------------------+
Now, we can think of our observations as being students. We can create a child ID (we’ll call it child rather than j), and we can create each child’s residual $$e_{ij}$$, which we will assume has an N(0,5) distribution:
. bysort classroom: generate child = _n
. generate e_ij = rnormal(0,5)
. list
+------------------------------------------+
| classr~m u_i child e_ij |
|------------------------------------------|
1. | 1 .7491351 1 2.832674 |
2. | 1 .7491351 2 1.487452 |
3. | 2 -.0031386 1 6.598946 |
4. | 2 -.0031386 2 -.3605778 |
+------------------------------------------+
We now have nearly all the ingredients to calculate $$y_{ij}$$:
$$y_{ij} = mu + u_{i.} + e_{ij}$$
We’ll assume mu is 70. We type
. generate y = 70 + u_i + e_ij
. list y classroom u_i child e_ij, sepby(classroom)
+-----------------------------------------------------+
| y classr~m u_i child e_ij |
|-----------------------------------------------------|
1. | 73.58181 1 .7491351 1 2.832674 |
2. | 72.23659 1 .7491351 2 1.487452 |
|-----------------------------------------------------|
3. | 76.59581 2 -.0031386 1 6.598946 |
4. | 69.63628 2 -.0031386 2 -.3605778 |
+-----------------------------------------------------+
Note that the random effect u_i is the same within each school, and each child has a different value for e_ij.
Our strategy was simple:
2. Create variables for the level ID and its random effect.
3. Expand the data by the number of observations within that level.
4. Repeat steps 2 and 3 until the bottom level is reached.
Let’s try this recipe for three-level data where children are nested within classrooms which are nested within schools. This time, I will index schools with i, classrooms with j, and children with k so that my model is
$y_{ijk} = mu + u_{i..} + u_{ij.} + e_{ijk}$
where
$$u_{i..}$$ ~ N(0,2)
$$u_{ij.}$$ ~ N(0,3)
$$u_{ijk}$$ ~ N(0,5)
Let’s create data for
(level 3, i) 2 schools
(level 2, j) 2 classrooms in each school
(level 1, k) 2 students in most classrooms; 3 students in i==2 & j==2
Begin by creating the level-three data for the two schools:
. clear
. set obs 2
. generate school = _n
. generate u_i = rnormal(0,2)
. list school u_i
+--------------------+
| school u_i |
|--------------------|
1. | 1 3.677312 |
2. | 2 -3.193004 |
+--------------------+
Next, we expand the data so that we have the three classrooms nested within each of the schools, and we create its random effect:
. expand 2
. bysort school: generate classroom = _n
. generate u_ij = rnormal(0,3)
. list school u_i classroom u_ij, sepby(school)
+-------------------------------------------+
| school u_i classr~m u_ij |
|-------------------------------------------|
1. | 1 3.677312 1 .9811059 |
2. | 1 3.677312 2 -3.482453 |
|-------------------------------------------|
3. | 2 -3.193004 1 -4.107915 |
4. | 2 -3.193004 2 -2.450383 |
+-------------------------------------------+
Finally, we expand the data so that we have three students in school 2′s classroom 2, and two students in all the other classrooms. Sorry for that complication, but I wanted to show you how to create unbalanced data.
In the previous examples, we’ve been typing things like expand 2, meaning double the observations. In this case, we need to do something different for school 2, classroom 2, namely,
. expand 3 if school==2 & classroom==2
and then we can just expand the rest:
. expand 2 if !(school==2 & clasroom==2)
Obviously, in a real simulation, you would probably want 16 to 25 students in each classroom. You could do something like that by typing
. expand 16+int((25-16+1)*runiform())
In any case, we will type
. expand 3 if school==2 & classroom==2
. expand 2 if !(school==2 & classroom==2)
. bysort school classroom: generate child = _n
. generate e_ijk = rnormal(0,5)
. generate y = 70 + u_i + u_ij + e_ijk
. list y school u_i classroom u_ij child e_ijk, sepby(classroom)
+------------------------------------------------------------------------+
| y school u_i classr~m u_ij child e_ijk |
|------------------------------------------------------------------------|
1. | 76.72794 1 3.677312 1 .9811059 1 2.069526 |
2. | 69.81315 1 3.677312 1 .9811059 2 -4.845268 |
|------------------------------------------------------------------------|
3. | 74.09565 1 3.677312 2 -3.482453 1 3.900788 |
4. | 71.50263 1 3.677312 2 -3.482453 2 1.307775 |
|------------------------------------------------------------------------|
5. | 64.86206 2 -3.193004 1 -4.107915 1 2.162977 |
6. | 61.80236 2 -3.193004 1 -4.107915 2 -.8967164 |
|------------------------------------------------------------------------|
7. | 66.65285 2 -3.193004 2 -2.450383 1 2.296242 |
8. | 49.96139 2 -3.193004 2 -2.450383 2 -14.39522 |
9. | 64.41605 2 -3.193004 2 -2.450383 3 .0594433 |
+------------------------------------------------------------------------+
Regardless of how we generate the data, we must ensure that the school-level random effects u_i are the same within school and the classroom-level random effects u_ij are the same within classroom.
Concerning data construction, the example above we concocted to produce a dataset that would be easy to list. Let’s now create a dataset that is more reasonable:
$y_{ijk} = mu + u_{i..} + u_{ij.} + e_{ijk}$
where
$$u_{i..}$$ ~ N(0,2)
$$u_{ij.}$$ ~ N(0,3)
$$u_{ijk}$$ ~ N(0,5)
Let’s create data for
(level 3, i) 6 schools
(level 2, j) 10 classrooms in each school
(level 1, k) 16-25 students
. clear
. set obs 6
. generate school = _n
. generate u_i = rnormal(0,2)
. expand 10
. bysort school: generate classroom = _n
. generate u_ij = rnormal(0,3)
. expand 16+int((25-16+1)*runiform())
. bysort school classroom: generate child = _n
. generate e_ijk = rnormal(0,5)
. generate y = 70 + u_i + u_ij + e_ijk
We can use the mixed command to fit the model with our simulated data.
. mixed y || school: || classroom: , stddev
Mixed-effects ML regression Number of obs = 1217
-----------------------------------------------------------
| No. of Observations per Group
Group Variable | Groups Minimum Average Maximum
----------------+------------------------------------------
school | 6 197 202.8 213
classroom | 60 16 20.3 25
-----------------------------------------------------------
Wald chi2(0) = .
Log likelihood = -3710.0673 Prob > chi2 = .
------------------------------------------------------------------------------
y | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons | 70.25941 .9144719 76.83 0.000 68.46707 72.05174
------------------------------------------------------------------------------
------------------------------------------------------------------------------
Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval]
-----------------------------+------------------------------------------------
school: Identity |
sd(_cons) | 2.027064 .7159027 1.014487 4.050309
-----------------------------+------------------------------------------------
classroom: Identity |
sd(_cons) | 2.814152 .3107647 2.26647 3.494178
-----------------------------+------------------------------------------------
sd(Residual) | 4.828923 .1003814 4.636133 5.02973
------------------------------------------------------------------------------
LR test vs. linear regression: chi2(2) = 379.37 Prob > chi2 = 0.0000
The parameter estimates from our simulated data match the parameters used to create the data pretty well: the estimate for _cons is 70.3, which is near 70; the estimated standard deviation for the school-level random effects is 2.02, which is near 2; the estimated standard deviation for the classroom-level random effects is 2.8, which is near 3; and the estimated standard deviation for the individual-level residuals is 4.8, which is near 5.
We’ve just done one reasonable simulation.
If we wanted to do a full simulation, we would need to do the above 100, 1,000, 10,000, or more times. We would put our code in a loop. And in that loop, we would keep track of whatever parameter interested us.
## How to simulate three-level data with covariates
Usually, we’re more interested in estimating the effects of the covariates than in estimating the variance of the random effects. Covariates are typically binary (such as male/female), categorical (such as race), ordinal (such as education level), or continuous (such as age).
Let’s add some covariates to our simulated data. Our model is
$y_{ijk} = mu + u_{i..} + u_{ij.} + e_{ijk}$
where
$$u_{i..}$$ ~ N(0,2)
$$u_{ij.}$$ ~ N(0,3)
$$u_{ijk}$$ ~ N(0,5)
We create data for
(level 3, i) 6 schools
(level 2, j) 10 classrooms in each school
(level 1, k) 16-25 students
(level 3, school i) whether the school is in an urban environment
(level 2, classroom j) teacher’s experience (years)
(level 1, student k) student’s mother’s education level
We can create a binary covariate called urban at the school level that equals 1 if the school is located in an urban area and equals 0 otherwise.
. clear
. set obs 6
. generate school = _n
. generate u_i = rnormal(0,2)
. generate urban = runiform()<0.50
Here we assigned schools to one of the two groups with equal probability (runiform()<0.50), but we could have assigned 70% of the schools to be urban by typing
. generate urban = runiform()<0.70
At the classroom level, we could add a continuous covariate for the teacher's years of experience. We could generate this variable by using any of Stata's random-number functions (see help random_number_functions. In the example below, I've generated teacher's years of experience with a uniform distribution ranging from 5-20 years.
. expand 10
. bysort school: generate classroom = _n
. generate u_ij = rnormal(0,3)
. bysort school: generate teach_exp = 5+int((20-5+1)*runiform())
When we summarize our data, we see that teaching experience ranges from 6-20 years with an average of 13 years.
. summarize teach_exp
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
teach_exp | 60 13.21667 4.075939 6 20
At the child level, we could add a categorical/ordinal covariate for mother's highest level of education completed. After we expand the data and create the child ID and error variables, we can generate a uniformly distributed random variable, temprand, on the interval [0,1].
. expand 16+int((25-16+1)*runiform())
. bysort school classroom: generate child = _n
. generate e_ijk = rnormal(0,5)
. generate temprand = runiform()
We can assign children to different groups by using the egen command with cutpoints. In the example below, children whose value of temprand is in the interval [0,0.5) will be assigned to mother_educ==0, children whose value of temprand is in the interval [0.5,0.9) will be assigned to mother_educ==1, and children whose value of temprand is in the interval [0.9,1) will be assigned to mother_educ==2.
. egen mother_educ = cut(temprand), at(0,0.5, 0.9, 1) icodes
. label define mother_educ 0 "HighSchool" 1 "College" 2 ">College"
. label values mother_educ mother_educ
The resulting frequencies of each category are very close to the frequencies we specified in our egen command.
. tabulate mother_educ, generate(meduc)
mother_educ | Freq. Percent Cum.
------------+-----------------------------------
HighSchool | 602 50.17 50.17
College | 476 39.67 89.83
>College | 122 10.17 100.00
------------+-----------------------------------
Total | 1,200 100.00
We used the option generate(meduc) in the tabulate command above to create indicator variables for each category of mother_educ. This will allow us to specify an effect size for each category when we create our outcome variable.
. summarize meduc*
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
meduc1 | 1200 .5016667 .5002057 0 1
meduc2 | 1200 .3966667 .4894097 0 1
meduc3 | 1200 .1016667 .3023355 0 1
Now, we can create an outcome variable called score by adding all our fixed and random effects together. We can specify an effect size (regression coefficient) for each fixed effect in our model.
. generate score = 70
+ (-2)*urban
+ 1.5*teach_exp
+ 0*meduc1
+ 2*meduc2
+ 5*meduc3
+ u_i + u_ij + e_ijk
I have specified that the grand mean is 70, urban schools will have scores 2 points lower than nonurban schools, and each year of teacher's experience will add 1.5 points to the students score.
Mothers whose highest level of education was high school (meduc1==1) will serve as the referent category for mother_educ(mother_educ==0). The scores of children whose mother completed college (meduc2==1 and mother_educ==1) will be 2 points higher than the children in the referent group. And the scores of children whose mother completed more than college (meduc3==1 and mother_educ==2) will be 5 points higher than the children in the referent group. Now, we can use the mixed command to fit a model to our simulated data. We used the indicator variables meduc1-meduc3 to create the data, but we will use the factor variable i.mother_educ to fit the model.
. mixed score urban teach_exp i.mother_educ || school: ||
classroom: , stddev baselevel
Mixed-effects ML regression Number of obs = 1259
-----------------------------------------------------------
| No. of Observations per Group
Group Variable | Groups Minimum Average Maximum
----------------+------------------------------------------
school | 6 200 209.8 217
classroom | 60 16 21.0 25
-----------------------------------------------------------
Wald chi2(4) = 387.64
Log likelihood = -3870.5395 Prob > chi2 = 0.0000
------------------------------------------------------------------------------
score | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
urban | -2.606451 2.07896 -1.25 0.210 -6.681138 1.468237
teach_exp | 1.584759 .096492 16.42 0.000 1.395638 1.77388
|
mother_educ |
HighSchool | 0 (base)
College | 2.215281 .3007208 7.37 0.000 1.625879 2.804683
>College | 5.065907 .5237817 9.67 0.000 4.039314 6.0925
|
_cons | 68.95018 2.060273 33.47 0.000 64.91212 72.98824
------------------------------------------------------------------------------
------------------------------------------------------------------------------
Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval]
-----------------------------+------------------------------------------------
school: Identity |
sd(_cons) | 2.168154 .7713944 1.079559 4.354457
-----------------------------+------------------------------------------------
classroom: Identity |
sd(_cons) | 3.06871 .3320171 2.482336 3.793596
-----------------------------+------------------------------------------------
sd(Residual) | 4.947779 .1010263 4.753681 5.149802
------------------------------------------------------------------------------
LR test vs. linear regression: chi2(2) = 441.25 Prob > chi2 = 0.0000
“Close” is in the eye of the beholder, but to my eyes, the parameter estimates look remarkably close to the parameters that were used to simulate the data. The parameter estimates for the fixed part of the model are -2.6 for urban (parameter = -2), 1.6 for teach_exp (parameter = 1.5), 2.2 for the College category of mother_educ (parameter = 2), 5.1 for the >College category of mother_educ (parameter = 5), and 69.0 for the intercept (parameter = 70). The estimated standard deviations for the random effects are also very close to the simulation parameters. The estimated standard deviation is 2.2 (parameter = 2) at the school level, 3.1 (parameter = 3) at the classroom level, and 4.9 (parameter = 5) at the child level.
Some of you may disagree that the parameter estimates are close. My reply is that it doesn’t matter unless you’re simulating a single dataset for demonstration purposes. If you are, simply simulate more datasets until you get one that looks close enough for you. If you are simulating data to check coverage probabilities or to estimate statistical power, you will be averaging over thousands of simulated datasets and the results of any one of those datasets won’t matter.
## How to simulate longitudinal data with random slopes
Longitudinal data are often conceptualized as multilevel data where the repeated observations are nested within individuals. The main difference between ordinary multilevel models and multilevel models for longitudinal data is the inclusion of a random slope. If you are not familiar with random slopes, you can learn more about them in a blog entry I wrote last year (Multilevel linear models in Stata, part 2: Longitudinal data).
Simulating longitudinal data with a random slope is much like simulating two-level data, with a couple of modifications. First, the bottom level will be observations within person. Second, there will be an interaction between time (age) and a person-level random effect. So we will generate data for the following model:
$weight_{ij} = mu + age_{ij} + u_{0i.} + age*u_{1i.} + e_{ij}$
where
$$u_{0i.}$$ ~ N(0,3) $$u_{1i.}$$ ~ N(0,1) $$e_{ij}$$ ~ N(0,2)
Let’s begin by simulating longitudinal data for 300 people.
. clear
. set obs 300
. gen person = _n
For longitudinal data, we must create two person-level random effects: the variable u_0i is analogous to the random effect we created earlier, and the variable u_1i is the random effect for the slope over time.
. generate u_0i = rnormal(0,3)
. generate u_1i = rnormal(0,1)
Let’s expand the data so that there are five observations nested within each person. Rather than create an observation-level identification number, let’s create a variable for age that ranges from 12 to 16 years,
. expand 5
. bysort person: generate age = _n + 11
and create an observation-level error term from an N(0,2) distribution:
. generate e_ij = rnormal(0,2)
. list person u_0i u_1i age e_ij if person==1
+-------------------------------------------------+
| person u_0i u_1i age e_ij |
|-------------------------------------------------|
1. | 1 .9338312 -.3097848 12 1.172153 |
2. | 1 .9338312 -.3097848 13 2.935366 |
3. | 1 .9338312 -.3097848 14 -2.306981 |
4. | 1 .9338312 -.3097848 15 -2.148335 |
5. | 1 .9338312 -.3097848 16 -.4276625 |
+-------------------------------------------------+
The person-level random effects u_0i and u_1i are the same at all ages, and the observation-level random effects e_ij are different at each age. Now we’re ready to generate an outcome variable called weight, measured in kilograms, based on the following model:
$weight_{ij} = 3 + 3.6*age_{ij} + u_{0i} + age*u_{1i} + e_{ij}$
. generate weight = 3 + 3.6*age + u_0i + age*u_1i + e_ij
The random effect u_1i is multiplied by age, which is why it is called a random slope. We could rewrite the model as
$weight_{ij} = 3 + age_{ij}*(3.6 + u_{1i}) + u_{01} + e_{ij}$
Note that for each year of age, a person’s weight will increase by 3.6 kilograms plus some random amount specified by u_1j. In other words,the slope for age will be slightly different for each person.
We can use the mixed command to fit a model to our data:
. mixed weight age || person: age , stddev
Mixed-effects ML regression Number of obs = 1500
Group variable: person Number of groups = 300
Obs per group: min = 5
avg = 5.0
max = 5
Wald chi2(1) = 3035.03
Log likelihood = -3966.3842 Prob > chi2 = 0.0000
------------------------------------------------------------------------------
weight | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
age | 3.708161 .0673096 55.09 0.000 3.576237 3.840085
_cons | 2.147311 .5272368 4.07 0.000 1.113946 3.180676
------------------------------------------------------------------------------
------------------------------------------------------------------------------
Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval]
-----------------------------+------------------------------------------------
person: Independent |
sd(age) | .9979648 .0444139 .9146037 1.088924
sd(_cons) | 3.38705 .8425298 2.080103 5.515161
-----------------------------+------------------------------------------------
sd(Residual) | 1.905885 .0422249 1.824897 1.990468
------------------------------------------------------------------------------
LR test vs. linear regression: chi2(2) = 4366.32 Prob > chi2 = 0.0000
The estimate for the intercept _cons = 2.1 is not very close to the original parameter value of 3, but the estimate of 3.7 for age is very close (parameter = 3.6). The standard deviations of the random effects are also very close to the parameters used to simulate the data. The estimate for the person level _cons is 2.1 (parameter = 2), the person-level slope is 0.997 (parameter = 1), and the observation-level residual is 1.9 (parameter = 2).
## How to simulate longitudinal data with structured errors
Longitudinal data often have an autoregressive pattern to their errors because of the sequential collection of the observations. Measurements taken closer together in time will be more similar than measurements taken further apart in time. There are many patterns that can be used to descibe the correlation among the errors, including autoregressive, moving average, banded, exponential, Toeplitz, and others (see help mixed##rspec).
Let’s simulate a dataset where the errors have a Toeplitz structure, which I will define below.
We begin by creating a sample with 500 people with a person-level random effect having an N(0,2) distribution.
. clear
. set obs 500
. gen person = _n
. generate u_i = rnormal(0,2)
Next, we can use the drawnorm command to create error variables with a Toeplitz pattern.
A Toeplitz 1 correlation matrix has the following structure:
. matrix V = ( 1.0, 0.5, 0.0, 0.0, 0.0 \ ///
0.5, 1.0, 0.5, 0.0, 0.0 \ ///
0.0, 0.5, 1.0, 0.5, 0.0 \ ///
0.0, 0.0, 0.5, 1.0, 0.5 \ ///
0.0, 0.0, 0.0, 0.5, 1.0 )
. matrix list V
symmetric V[5,5]
c1 c2 c3 c4 c5
r1 1
r2 .5 1
r3 0 .5 1
r4 0 0 .5 1
r5 0 0 0 .5 1
The correlation matrix has 1s on the main diagonal, and each pair of contiguous observations will have a correlation of 0.5. Observations more than 1 unit of time away from each other are assumed to be uncorrelated.
We must also define a matrix of means to use the drawnorm command.
. matrix M = (0 \ 0 \ 0 \ 0 \ 0)
. matrix list M
M[5,1]
c1
r1 0
r2 0
r3 0
r4 0
r5 0
Now, we’re ready to use the drawnorm command to create five error variables that have a Toeplitz 1 structure.
. drawnorm e1 e2 e3 e4 e5, means(M) cov(V)
. list in 1/2
+---------------------------------------------------------------------------+
| person u_i e1 e2 e3 e4 e5 |
|---------------------------------------------------------------------------|
1. | 1 5.303562 -1.288265 -1.201399 .353249 .0495944 -1.472762 |
2. | 2 -.0133588 .6949759 2.82179 .7195075 -1.032395 .1995016 |
+---------------------------------------------------------------------------+
Let’s estimate the correlation matrix for our simulated data to verify that our simulation worked as we expected.
. correlate e1-e5
(obs=300)
| e1 e2 e3 e4 e5
-------------+---------------------------------------------
e1 | 1.0000
e2 | 0.5542 1.0000
e3 | -0.0149 0.4791 1.0000
e4 | -0.0508 -0.0364 0.5107 1.0000
e5 | 0.0022 -0.0615 0.0248 0.4857 1.0000
The correlations are 1 along the main diagonal, near 0.5 for the contiguous observations, and near 0 otherwise.
Our data are currently in wide format, and we need them in long format to use the mixed command. We can use the reshape command to convert our data from wide to long format. If you are not familiar with the reshape command, you can learn more about it by typing help reshape.
. reshape long e, i(person) j(time)
(note: j = 1 2 3 4 5)
Data wide -> long
-----------------------------------------------------------------------------
Number of obs. 300 -> 1500
Number of variables 7 -> 4
j variable (5 values) -> time
xij variables:
e1 e2 ... e5 -> e
-----------------------------------------------------------------------------
Now, we are ready to create our age variable and the outcome variable weight.
. bysort person: generate age = _n + 11
. generate weight = 3 + 3.6*age + u_i + e
. list weight person u_i time age e if person==1
+-------------------------------------------------------+
| weight person u_i time age e |
|-------------------------------------------------------|
1. | 50.2153 1 5.303562 1 12 -1.288265 |
2. | 53.90216 1 5.303562 2 13 -1.201399 |
3. | 59.05681 1 5.303562 3 14 .353249 |
4. | 62.35316 1 5.303562 4 15 .0495944 |
5. | 64.4308 1 5.303562 5 16 -1.472762 |
+-------------------------------------------------------+
We can use the mixed command to fit a model to our simulated data.
. mixed weight age || person:, residual(toeplitz 1, t(time)) , stddev
Mixed-effects ML regression Number of obs = 1500
Group variable: person Number of groups = 300
Obs per group: min = 5
avg = 5.0
max = 5
Wald chi2(1) = 33797.58
Log likelihood = -2323.9389 Prob > chi2 = 0.0000
------------------------------------------------------------------------------
weight | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
age | 3.576738 .0194556 183.84 0.000 3.538606 3.61487
_cons | 3.119974 .3244898 9.62 0.000 2.483985 3.755962
------------------------------------------------------------------------------
------------------------------------------------------------------------------
Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval]
-----------------------------+------------------------------------------------
person: Identity |
sd(_cons) | 3.004718 .1268162 2.766166 3.263843
-----------------------------+------------------------------------------------
Residual: Toeplitz(1) |
rho1 | .4977523 .0078807 .4821492 .5130398
sd(e) | .9531284 .0230028 .9090933 .9992964
------------------------------------------------------------------------------
LR test vs. linear regression: chi2(2) = 3063.87 Prob > chi2 = 0.0000
Again, our parameter estimates match the parameters that were used to simulate the data very closely.
The parameter estimate is 3.6 for age (parameter = 3.6) and 3.1 for _cons (parameter = 3). The estimated standard deviations of the person-level random effect is 3.0 (parameter = 3). The estimated standard deviation for the errors is 0.95 (parameter = 1), and the estimated correlation for the Toeplitz structure is 0.5 (parameter = 0.5).
## Conclusion
I hope I’ve convinced you that simulating multilevel/longitudinal data is easy and useful. The next time you find yourself teaching a class or giving a talk that requires multilevel examples, try simulating the data. And if you need to calculate statistical power for a multilevel or longitudinal model, consider simulations.
Categories: Statistics Tags:
## How to create animated graphics using Stata
### Introduction
Today I want to show you how to create animated graphics using Stata. It’s easier than you might expect and you can use animated graphics to illustrate concepts that would be challenging to illustrate with static graphs. In addition to Stata, you will need a video editing program but don’t be concerned if you don’t have one. At the 2012 UK Stata User Group Meeting Robert Grant demonstrated how to create animated graphics from within Stata using a free software program called FFmpeg. I will show you how I create my animated graphs using Camtasia and how Robert creates his using FFmpeg.
I recently recorded a video for the Stata Youtube channel called “Power and sample size calculations in Stata: A conceptual introduction“. I wanted to illustrate two concepts: (1) that statistcal power increases as sample size increases, and (2) as effect size increases. Both of these concepts can be illustrated with a static graph along with the explanation “imagine that …”. Creating animated graphs allowed me to skip the explanation and just show what I meant.
### Creating the graphs
Videos are illusions. All videos — from Charles-Émile Reynaud’s 1877 praxinoscope to modern blu-ray movies — are created by displaying a series of ordered still images for a fraction of a second each. Our brains perceive this series of still images as motion.
To create the illusion of motion with graphs, we make an ordered series of slightly differing graphs. We can use loops to do this. If you are not familiar with loops in Stata, here’s one to count to five:
forvalues i = 1(1)5 {
disp "i = i'"
}
i = 1
i = 2
i = 3
i = 4
i = 5
We could place a graph command inside the loop. If, for each interation, the graph command created a slightly different graph, we would be on our way to creating our first video. The loop below creates a series of graphs of normal densities with means 0 through 1 in increments of 0.1.
forvalues mu = 0(0.1)1 {
twoway function y=normalden(x,mu',1), range(-3 6) title("N(mu',1)")
}
You may have noticed the illusion of motion as Stata created each graph; the normal densities appeared to be moving to the right as each new graph appeared on the screen.
You may have also noticed that some of the values of the mean did not look as you would have wanted. For example, 1.0 was displayed as 0.999999999. That’s not a mistake, it’s because Stata stores numbers and performs calculations in base two and displays them in base ten; for a detailed explanation, see Precision (yet again), Part I.
We can fix that by reformating the means using the string() function.
forvalues mu = 0(0.1)1 {
local mu = string(mu', "%3.1f")
twoway function y=normalden(x,mu',1), range(-3 6) title("N(mu',1)")
}
Next, we need to save our graphs. We can do this by adding graph export inside the loop.
forvalues mu = 0(0.1)1 {
local mu = string(mu', "%3.1f")
twoway function y=normalden(x,mu',1), range(-3 6) title("N(mu',1)")
graph export graph_mu'.png, as(png) width(1280) height(720) replace
}
Note that the name of each graph file includes the value of mu so that we know the order of our files. We can view the contents of the directory to verify that Stata has created a file for each of our graphs.
. ls
<dir> 2/11/14 12:12 .
<dir> 2/11/14 12:12 ..
35.6k 2/11/14 12:11 graph_0.0.png
35.6k 2/11/14 12:11 graph_0.1.png
35.7k 2/11/14 12:11 graph_0.2.png
35.7k 2/11/14 12:11 graph_0.3.png
35.7k 2/11/14 12:11 graph_0.4.png
35.8k 2/11/14 12:11 graph_0.5.png
35.9k 2/11/14 12:12 graph_0.6.png
35.7k 2/11/14 12:12 graph_0.7.png
35.8k 2/11/14 12:12 graph_0.8.png
35.9k 2/11/14 12:12 graph_0.9.png
35.6k 2/11/14 12:12 graph_1.0.png
Now that we have created our graphs, we need to combine them into a video.
There are many commercial, freeware, and free software programs available that we could use. I will outline the basic steps using two of them, one a commerical GUI based product (not free) called Camtasia, and the other a free command-based program called FFmpeg.
### Creating videos with Camtasia
Most commercial video editing programs have similar interfaces. The user imports image, sound and video files, organizes them in tracks on a timeline and then previews the resulting video. Camtasia is a commercial video program that I use to record videos for the Stata Youtube channel and its interface looks like this.
We begin by importing the graph files into Camtasia:
Next we drag the images onto the timeline:
And then we make the display time for each image very short…in this case 0.1 seconds or 10 frames per second.
After previewing the video, we can export it to any of Camtasia’s supported formats. I’ve exported to a “.gif” file because it is easy to view in a web browser.
We just created our first animated graph! All we have to do to make it look as professional as the power-and-sample size examples I showed you earlier is go back into our Stata program and modify the graph command to add the additional elements we want to display!
### Creating videos with FFmpeg
Stata user and medical statistician Robert Grant gave a presentation at the 2012 UK Stata User Group Meeting in London entitled “Producing animated graphs from Stata without having to learn any specialized software“. You can read more about Robert by visiting his blog and clicking on About.
In his presentation, Robert demonstrated how to combine graph images into a video using a free software program called FFmpeg. Robert followed the same basic strategy I demonstrated above, but Robert’s choice of software has two appealing features. First, the software is readily available and free. Second, FFmpeg can be called from within the Stata environment using the winexec command. This means that we can create our graphs and combine them into a video using Stata do files. Combining dozens or hundreds of graphs into a single video with a program is faster and easier than using a drag-and-drop interface.
Let’s return to our previous example and combine the files using FFmpeg. Recall that we inserted the mean into the name of each file (e.g. “graph_0.4.png”) so that we could keep track of the order of the files. In my experience, it can be difficult to combine files with decimals in their names using FFmpeg. To avoid the problem, I have added a line of code between the twoway command and the graph export command that names the files with sequential integers which are padded with zeros.
forvalues mu = 0(0.1)1 {
local mu = string(mu', "%3.1f")
twoway function y=normalden(x,mu',1), range(-3 6) title("N(mu',1)")
local mu = string(mu'*10+1, "%03.0f")
graph export graph_mu'.png, as(png) width(1280) height(720) replace
}
. ls
<dir> 2/12/14 12:21 .
<dir> 2/12/14 12:21 ..
35.6k 2/12/14 12:21 graph_001.png
35.6k 2/12/14 12:21 graph_002.png
35.7k 2/12/14 12:21 graph_003.png
35.7k 2/12/14 12:21 graph_004.png
35.7k 2/12/14 12:21 graph_005.png
35.8k 2/12/14 12:21 graph_006.png
35.9k 2/12/14 12:21 graph_007.png
35.7k 2/12/14 12:21 graph_008.png
35.8k 2/12/14 12:21 graph_009.png
35.9k 2/12/14 12:21 graph_010.png
35.6k 2/12/14 12:21 graph_011.png
We can then combine these files into a video with FFmpeg using the following commands
local GraphPath "C:\Users\jch\AnimatedGraphics\example\"
winexec "C:\Program Files\FFmpeg\bin\ffmpeg.exe" -i GraphPath'graph_%03d.png
-b:v 512k GraphPath'graph.mpg
The local macro GraphPath contains the path for the directory where my graphics files are stored.
The Stata command winexec whatever executes whatever. In our case, whatever is ffmpeg.exe, preceeded by ffmpeg.exe‘s path, and followed by the arguments FFmpeg needs. We specify two options, -i and -b.
The -i option is followed by a path and filename template. In our case, the path is obtained from the Stata local macro GraphPath and the filename template is “graph_%03d.png”. This template tells FFmpeg to look for a three digit sequence of numbers between “graph_” and “.png” in the filenames. The zero that precedes the three in the template tells FFmpeg that the three digit sequence of numbers is padded with zeros.
The -b option specifies the path and filename of the video to be created along with some attributes of the video.
Once we have created our video, we can use FFmpeg to convert our video to other video formats. For example, we could convert “graph.mpg” to “graph.gif” using the following command:
winexec "C:\Program Files\FFmpeg\bin\ffmpeg.exe" -r 10 -i GraphPath'graph.mpg
-t 10 -r 10 GraphPath'graph.gif
which creates this graph:
FFmpeg is a very flexible program and there are far too many options to discuss in this blog entry. If you would like to learn more about FFmpeg you can visit their website at www.ffmpeg.org.
### More Examples
I made the preceding examples as simple as possible so that we could focus on the mechanics of creating videos. We now know that, if we want to make professional looking videos, all the complication comes on the Stata side. We leave our loop alone but change the graph command inside it to be more complicated.
So here’s how I created the two animated-graphics videos that I used to create the overall video “Power and sample size calculations in Stata: A conceptual introduction” on our YouTube channel.
The first demonstrated that increasing the effect size (the difference between the means) results in increased statistical power.
local GraphCounter = 100
local mu_null = 0
local sd = 1
local z_crit = round(-1*invnormal(0.05), 0.01)
local z_crit_label = z_crit' + 0.75
forvalues mu_alt = 1(0.01)3 {
twoway ///
function y=normalden(x,mu_null',sd'), ///
range(-3 z_crit') color(red) dropline(0) || ///
function y=normalden(x,mu_alt',sd'), ///
range(-3 5) color(green) dropline(mu_alt') || ///
function y=normalden(x,mu_alt',sd'), ///
range(z_crit' 6) recast(area) color(green) || ///
function y=normalden(x,mu_null',sd'), ///
range(z_crit' 6) recast(area) color(red) ///
title("Power for {&mu}={&mu}{subscript:0} versus {&mu}={&mu}{subscript:A}") ///
xtitle("{it: z}") xlabel(-3 -2 -1 0 1 2 3 4 5 6) ///
legend(off) ///
ytitle("Density") yscale(range(0 0.6)) ///
ylabel(0(0.1)0.6, angle(horizontal) nogrid) ///
text(0.45 0 "{&mu}{subscript:0}", color(red)) ///
text(0.45 mu_alt' "{&mu}{subscript:A}", color(green))
graph export mu_alt_GraphCounter'.png, as(png) width(1280) height(720) replace
local ++GraphCounter
}
The above Stata code created the *.png files that I then combined using Camtasia to produce this gif:
The second video demonstrated that power increases as the sample size increases.
local GraphCounter = 301
local mu_label = 0.45
local power_label = 2.10
local mu_null = 0
local mu_alt = 2
forvalues sd = 1(-0.01)0.5 {
local z_crit = round(-1*invnormal(0.05)*sd', 0.01)
local z_crit_label = z_crit' + 0.75
twoway ///
function y=normalden(x,mu_null',sd'), ///
range(-3 z_crit') color(red) dropline(0) || ///
function y=normalden(x,mu_alt',sd'), ///
range(-3 5) color(green) dropline(mu_alt') || ///
function y=normalden(x,mu_alt',sd'), ///
range(z_crit' 6) recast(area) color(green) || ///
function y=normalden(x,mu_null',sd'), ///
range(z_crit' 6) recast(area) color(red) ///
title("Power for {&mu}={&mu}{subscript:0} versus {&mu}={&mu}{subscript:A}") ///
xtitle("{it: z}") xlabel(-3 -2 -1 0 1 2 3 4 5 6) ///
legend(off) ///
ytitle("Density") yscale(range(0 0.6)) ///
ylabel(0(0.1)0.6, angle(horizontal) nogrid) ///
text(mu_label' 0 "{&mu}{subscript:0}", color(red)) ///
text(mu_label' mu_alt' "{&mu}{subscript:A}", color(green))
graph export mu_alt_GraphCounter'.png, as(png) width(1280) height(720) replace
local ++GraphCounter
local mu_label = mu_label' + 0.005
local power_label = power_label' + 0.03
}
Just as previously, the above Stata code creates the *.png files that I then combine using Camtasia to produce a gif:
Let me show you some more examples.
The next example demonstrates the basic idea of lowess smoothing.
sysuse auto
local WindowWidth = 500
forvalues WindowUpper = 2200(25)5000 {
local WindowLower = WindowUpper' - WindowWidth'
twoway (scatter mpg weight) ///
(lowess mpg weight if weight < (WindowUpper'-250), lcolor(green)) ///
(lfit mpg weight if weight>WindowLower' & weight<WindowUpper', ///
lwidth(medium) lcolor(red)) ///
, xline(WindowLower' WindowUpper', lwidth(medium) lcolor(black)) ///
legend(on order(1 2 3) cols(3))
graph export lowess_WindowUpper'.png, as(png) width(1280) height(720) replace
}
The result is,
The animated graph I created is not yet a perfect analogy to what lowess actually does, but it comes close. It has two problems. The lowess curve changes outside of the sliding window, which it should not and the animation does not illustrate the weighting of the points within the window, say by using differently sized markers for the points in the sliding window. Even so, the graph does a far better job than the usual explanaton that one should imagine sliding a window across the scatterplot.
As yet another example, we can use animated graphs to demonstrate the concept of convergence. There is a FAQ on the Stata website written by Bill Gould that explains the relationship between the chi-squared and F distributions. The animated graph below shows that F(d1, d2) converges to d1*χ^2 as d2 goes to infinity:
forvalues df = 1(1)100 {
twoway function y=chi2(2,2*x), range(0 6) color(red) || ///
function y=F(2,df',x), range(0 6) color(green) ///
title("Cumulative distributions for {&chi}{sup:2}{sub:df} and {it:F}{subscript:df,df2}") ///
xtitle("{it: denominator df}") xlabel(0 1 2 3 4 5 6) legend(off) ///
text(0.45 4 "df2 = df'", size(huge) color(black)) ///
legend(on order(1 "{&chi}{sup:2}{sub:df}" 2 "{it:F}{subscript:df,df2}") cols(2) position(5) ring(0))
local df = string(df', "%03.0f")
graph export converge2_df'.png, as(png) width(1280) height(720) replace
}
The t distribution has a similar relationship with the normal distribution.
forvalues df = 1(1)100 {
twoway function y=normal(x), range(-3 3) color(red) || ///
function y=t(df',x), range(-3 3) color(green) ///
title("Cumulative distributions for Normal(0,1) and {it:t}{subscript:df}") ///
xtitle("{it: t/z}") xlabel(-3 -2 -1 0 1 2 3) legend(off) ///
text(0.45 -2 "df = df'", size(huge) color(black)) ///
legend(on order(1 "N(0,1)" 2 "{it:t}{subscript:df}") cols(2) position(5) ring(0))
local df = string(df', "%03.0f")
graph export converge_df'.png, as(png) width(1280) height(720) replace
}
The result is
### Final thoughts
I have learned through trial and error two things that improve the quality of my animated graphs. First, note that the axes of the graphs in most of the examples above are explicitly defined in the graph commands. This is often necessary to keep the axes stable from graph to graph. Second, videos have a smoother, higher quality appearance when there are many graphs with very small changes from graph to graph.
I hope I have convinced you that creating animated graphics with Stata is easier than you imagined. If the old saying that “a picture is worth a thousand words” is true, imagine how many words you can save using animated graphs.
### Other resources
FFmpeg
Camtasia
Relationship between chi-squared and F distributions
Robert Grant’s blog and examples
Hans Rosling’s 200 Countries recreated using only Stata
Categories: Graphics Tags:
## Measures of effect size in Stata 13
Today I want to talk about effect sizes such as Cohen’s d, Hedges’s g, Glass’s Δ, η2, and ω2. Effects sizes concern rescaling parameter estimates to make them easier to interpret, especially in terms of practical significance.
Many researchers in psychology and education advocate reporting of effect sizes, professional organizations such as the American Psychological Association (APA) and the American Educational Research Association (AERA) strongly recommend their reporting, and professional journals such as the Journal of Experimental Psychology: Applied and Educational and Psychological Measurement require that they be reported.
Anyway, today I want to show you
1. What effect sizes are.
2. How to calculate effect sizes and their confidence intervals in Stata.
3. How to calculate bootstrap confidence intervals for those effect sizes.
4. How to use Stata’s effect-size calculator.
## 1. What are effect sizes?
The importance of research results is often assessed by statistical significance, usually that the p-value is less than 0.05. P-values and statistical significance, however, don’t tell us anything about practical significance.
What if I told you that I had developed a new weight-loss pill and that the difference between the average weight loss for people who took the pill and the those who took a placebo was statistically significant? Would you buy my new pill? If you were overweight, you might reply, “Of course! I’ll take two bottles and a large order of french fries to go!”. Now let me add that the average difference in weight loss was only one pound over the year. Still interested? My results may be statistically significant but they are not practically significant.
Or what if I told you that the difference in weight loss was not statistically significant — the p-value was “only” 0.06 — but the average difference over the year was 20 pounds? You might very well be interested in that pill.
The size of the effect tells us about the practical significance. P-values do not assess practical significance.
All of which is to say, one should report parameter estimates along with statistical significance.
In my examples above, you knew that 1 pound over the year is small and 20 pounds is large because you are familiar with human weights.
In another context, 1 pound might be large, and in yet another, 20 pounds small.
Formal measures of effects sizes are thus usually presented in unit-free but easy-to-interpret form, such as standardized differences and proportions of variability explained.
### The “d” family
Effect sizes that measure the scaled difference between means belong to the “d” family. The generic formula is
The estimators differ in terms of how sigma is calculated.
Cohen’s d, for instance, uses the pooled sample standard deviation.
Hedges’s g incorporates an adjustment which removes the bias of Cohen’s d.
Glass’s Δ was originally developed in the context of experiments and uses the “control group” standard deviation in the denominator. It has subsequently been generalized to nonexperimental studies. Because there is no control group in observational studies, Kline (2013) recommends reporting Glass’s Δ using the standard deviation for each group. Glass’s Delta_1 uses one group’s standard deviation and Delta_2 uses the other group’s.
Although I have given definitions to Cohen’s d, Hedges’s g, and Glass’s Δ, different authors swap the definitions around! As a result, many authors refer to all of the above as just Delta.
Be careful when using software to know which Delta you are getting. I have used Stata terminology, of course.
Anyway, the use of a standardized scale allows us to assess of practical significance. Delta = 1.5 indicates that the mean of one group is 1.5 standard deviations higher than that of the other. A difference of 1.5 standard deviations is obviously large, and a difference of 0.1 standard deviations is obviously small.
### The “r” family
The r family quantifies the ratio of the variance attributable to an effect to the total variance and is often interpreted as the “proportion of variance explained”. The generic estimator is known as eta-squared,
η2 is equivalent to the R-squared statistic from linear regression.
ω2 is a less biased variation of η2 that is equivalent to the adjusted R-squared.
Both of these measures concern the entire model.
Partial η2 and partial ω2 are like partial R-squareds and concern individual terms in the model. A term might be a variable or a variable and its interaction with another variable.
Both the d and r families allow us to make an apples-to-apples comparison of variables measured on different scales. For example, an intervention could affect both systolic blood pressure and total cholesterol. Comparing the relative effect of the intervention on the two outcomes would be difficult on their original scales.
How does one compare mm/Hg and mg/dL? It is straightforward in terms of Cohen’s d or ω2 because then we are comparing standard deviation changes or proportion of variance explained.
## 2. How to calculate effect sizes and their confidence intervals in Stata
Consider a study where 30 school children are randomly assigned to classrooms that incorporated web-based instruction (treatment) or standard classroom environments (control). At the end of the school year, the children were given tests to measure reading and mathematics skills. The reading test is scored on a 0-15 point scale and, the mathematics test, on a 0-100 point scale.
Let’s download a dataset for our fictitious example from the Stata website by typing:
. use http://www.stata.com/videos13/data/webclass.dta
Contains data from http://www.stata.com/videos13/data/webclass.dta
obs: 30 Fictitious web-based learning
experiment data
vars: 5 5 Sep 2013 11:28
size: 330 (_dta has notes)
-------------------------------------------------------------------------------
storage display value
variable name type format label variable label
-------------------------------------------------------------------------------
id byte %9.0g ID Number
treated byte %9.0g treated Treatment Group
agegroup byte %9.0g agegroup Age Group
math float %9.0g Math Score
-------------------------------------------------------------------------------
. notes
_dta:
1. Variable treated records 0=control, 1=treated.
2. Variable agegroup records 1=7 years old, 2=8 years old, 3=9 years old.
We can compute a t-statistic to test the null hypothesis that the average math scores are the same in the treatment and control groups.
. ttest math, by(treated)
Two-sample t test with equal variances
------------------------------------------------------------------------------
Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]
---------+--------------------------------------------------------------------
Control | 15 69.98866 3.232864 12.52083 63.05485 76.92246
Treated | 15 79.54943 1.812756 7.020772 75.66146 83.4374
---------+--------------------------------------------------------------------
combined | 30 74.76904 2.025821 11.09588 70.62577 78.91231
---------+--------------------------------------------------------------------
diff | -9.560774 3.706412 -17.15301 -1.968533
------------------------------------------------------------------------------
diff = mean(Control) - mean(Treated) t = -2.5795
Ho: diff = 0 degrees of freedom = 28
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Pr(T < t) = 0.0077 Pr(|T| > |t|) = 0.0154 Pr(T > t) = 0.9923
The treated students have a larger mean, yet the difference of -9.56 is reported as negative because -ttest- calculated Control minus Treated. So just remember, negative differences mean Treated > Control in this case.
The t-statistic equals -2.58 and its two-sided p-value of 0.0154 indicates that the difference between the math scores in the two groups is statistically significant.
Next, let’s calculate effect sizes from the d family:
. esize twosample math, by(treated) cohensd hedgesg glassdelta
Effect size based on mean comparison
Obs per group:
Control = 15
Treated = 15
---------------------------------------------------------
Effect Size | Estimate [95% Conf. Interval]
--------------------+------------------------------------
Cohen's d | -.9419085 -1.691029 -.1777553
Hedges's g | -.916413 -1.645256 -.1729438
Glass's Delta 1 | -.7635896 -1.52044 .0167094
Glass's Delta 2 | -1.361784 -2.218342 -.4727376
---------------------------------------------------------
Cohen’s d and Hedges’s g both indicate that the average reading scores differ by approximately -0.93 standard deviations with 95% confidence intervals of (-1.69, -0.18) and (-1.65, -0.17) respectively.
Since this is an experiment, we are interested in Glass’s Delta 1 because it is calculated using the control group standard deviation. Average reading scores differ by -0.76 and the confidence interval is (-1.52, 0.02).
The confidence intervals for Cohen’s d and Hedges’s g do not include the null value of zero but the confidence interval for Glass’s Delta 1 does. Thus we cannot completely rule out the possibility that the treatment had no effect on math scores.
Next we could incorporate the age group of the children into our analysis by using a two-way ANOVA to test the null hypothesis that the mean math scores are equal for all groups.
. anova math treated##agegroup
Number of obs = 30 R-squared = 0.2671
Root MSE = 10.4418 Adj R-squared = 0.1144
Source | Partial SS df MS F Prob > F
-----------------+----------------------------------------------------
Model | 953.697551 5 190.73951 1.75 0.1617
|
treated | 685.562956 1 685.562956 6.29 0.0193
agegroup | 47.7059268 2 23.8529634 0.22 0.8051
treated#agegroup | 220.428668 2 110.214334 1.01 0.3789
|
Residual | 2616.73825 24 109.030761
-----------------+----------------------------------------------------
Total | 3570.4358 29 123.118476
The F-statistic for the entire model is not statistically significant (F=1.75, ndf=5, ddf=24, p=0.1617) but the F-statistic for the main effect of treatment is statistically significant (F=6.29, ndf=1, ddf=24, p=0.0193).
We can compute the η2 and partial η2 estimates for this model using the estat esize command immediately after our anova command (note that estat esize works after the regress command too).
. estat esize
Effect sizes for linear models
---------------------------------------------------------------------
Source | Eta-Squared df [95% Conf. Interval]
----------------------+----------------------------------------------
Model | .2671096 5 0 .4067062
|
treated | .2076016 1 .0039512 .4451877
agegroup | .0179046 2 0 .1458161
treated#agegroup | .0776932 2 0 .271507
---------------------------------------------------------------------
The overall η2 indicates that our model accounts for approximately 26.7% of the variablity in math scores though the 95% confidence interval includes the null value of zero (0.00%, 40.7%). The partial η2 for treatment is 0.21 (21% of the variability explained) and its 95% confidence interval excludes zero (0.3%, 20%).
We could calculate the alternative r-family member ω2 rather than η2 by typing
. estat esize, omega
Effect sizes for linear models
---------------------------------------------------------------------
Source | Omega-Squared df [95% Conf. Interval]
----------------------+----------------------------------------------
Model | .1144241 5 0 .2831033
|
treated | .174585 1 0 .4220705
agegroup | 0 2 0 .0746342
treated#agegroup | .0008343 2 0 .2107992
---------------------------------------------------------------------
The overall ω2 indicates that our model accounts for approximately 11.4% of the variability in math scores and treatment accounts for 17.5%. This perplexing result stems from the way that ω2 and partial ω2 are calculated. See Pierce, Block, & Aguinis (2004) for a thorough explanation.
Except for the η2 for treatment, the confidence intervals include 0 so we cannot rule out the possibility that there is no effect. Whether results are practically significant is generically a matter context and opinion. In some situations, accounting for 5% of the variability in an outcome could be very important and in other situations accounting for 30% may not be.
We could repeat the same analyses for the reading scores using the following commands:
. ttest reading, by(treated)
. esize twosample reading, by(treated) cohensd hedgesg glassdelta
. estat esize
. estat esize, omega
None of the t- or F-statistics for reading scores were statistically significant at the 0.05 level.
Even though the reading and math scores were measured on two different scales, we can directly compare the relative effect of the treatment using effect sizes:
Effect Size | Reading Score Math Score
------------------------------------------------------------
Cohen's d | -0.23 (-0.95 - 0.49) -0.94 (-1.69 - -0.18)
Hedges's g | -0.22 (-0.92 - 0.48) -0.92 (-1.65 - -0.17)
Glass's Delta | -0.21 (-0.93 - 0.51) -0.76 (-1.52 - 0.02)
Eta-squared | 0.02 ( 0.00 - 0.20) 0.21 ( 0.00 - 0.44)
Omega-squared | 0.00 ( 0.00 - 0.17) 0.17 ( 0.00 - 0.42)
The results show that the average reading scores in the treated and control groups differ by approximately 0.22 standard deviations while the average math scores differ by approximately 0.92 standard deviations. Similarly, treatment status accounted for almost none of the variability in reading scores while it accounted for roughly 17% of the variability in math scores. The intervention clearly had a larger effect on math scores than reading scores. We also know that we cannot completely rule out an effect size of zero (no effect) for both reading and math scores because several confidence intervals included zero. Whether or not the effects are practically significant is a matter of interpretation but the effect sizes provide a standardized metric for evaluation.
## 3. How to calculate bootstrap confidence intervals
Simulation studies have shown that bootstrap confidence intervals for the d family may be preferable to confidence intervals based on the noncentral t distribution when the variable of interest does not have a normal distribution (Kelley 2005; Algina, Keselman, and Penfield 2006). We can calculate bootstrap confidence intervals for Cohen’s d and Hedges’s g using Stata’s bootstrap prefix:
. bootstrap r(d) r(g), reps(500) nowarn: esize twosample reading, by(treated)
(running esize on estimation sample)
Bootstrap replications (500)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
.................................................. 100
.................................................. 150
.................................................. 200
.................................................. 250
.................................................. 300
.................................................. 350
.................................................. 400
.................................................. 450
.................................................. 500
Bootstrap results Number of obs = 30
Replications = 500
_bs_1: r(d)
_bs_2: r(g)
------------------------------------------------------------------------------
| Observed Bootstrap Normal-based
| Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_bs_1 | -.228966 .3905644 -0.59 0.558 -.9944582 .5365262
_bs_2 | -.2227684 .3799927 -0.59 0.558 -.9675403 .5220036
------------------------------------------------------------------------------
The bootstrap estimate of the 95% confidence interval for Cohen’s d is -0.99 to 0.54 which is slightly wider than the earlier estimate based on the non-central t distribution (see [R] esize for details). The bootstrap estimate is slightly wider for Hedges’s g as well.
## 4. How to use Stata’s effect-size calculator
You can use Stata’s effect size calculators to estimate them using summary statistics. If we know that the mean, standard deviation and sample size for one group is 70, 12.5 and 15 respectively and 80, 7 and 15 for another group, we can use esizei to estimate effect sizes from the d family:
. esizei 15 70 12.5 15 80 7, cohensd hedgesg glassdelta
Effect size based on mean comparison
Obs per group:
Group 1 = 15
Group 2 = 15
---------------------------------------------------------
Effect Size | Estimate [95% Conf. Interval]
--------------------+------------------------------------
Cohen's d | -.9871279 -1.739873 -.2187839
Hedges's g | -.9604084 -1.692779 -.2128619
Glass's Delta 1 | -.8 -1.561417 -.0143276
Glass's Delta 2 | -1.428571 -2.299112 -.5250285
---------------------------------------------------------
We can estimate effect sizes from the r family using esizei with slightly different syntax. For example, if we know the numerator and denominator degrees of freedom along with the F statistic, we can calculate η2 and ω2 using the following command:
. esizei 1 28 6.65
Effect sizes for linear models
---------------------------------------------------------
Effect Size | Estimate [95% Conf. Interval]
--------------------+------------------------------------
Eta-Squared | .1919192 .0065357 .4167874
Omega-Squared | .1630592 0 .3959584
---------------------------------------------------------
## Video demonstration
Stata has dialog boxes that can assist you in calculating effect sizes. If you would like a brief introduction using the GUI, you can watch a demonstration on Stata’s YouTube Channel:
Tour of effect sizes in Stata
## Final thoughts and further reading
Most older papers and many current papers do not report effect sizes. Nowadays, the general consensus among behavioral scientists, their professional organizations, and their journals is that effect sizes should always be reported in addition to tests of statistical significance. Stata 13 now makes it easy to compute most popular effects sizes.
Some methodologists believe that effect sizes with confidence intervals should always be reported and that statistical hypothesis tests should be abandoned altogether; see Cumming (2012) and Kline (2013). While this may sound like a radical notion, other fields such as epidemiology have been moving in this direction since the 1990s. Cumming and Kline offer compelling arguments for this paradigm shift as well as excellent introductions to effect sizes.
American Psychological Association (2009). Publication Manual of the American Psychological Association, 6th Ed. Washington, DC: American Psychological Association.
Algina, J., H. J. Keselman, and R. D. Penfield. (2006). Confidence interval coverage for Cohen’s effect size statistic. Educational and Psychological Measurement, 66(6): 945–960.
Cumming, G. (2012). Understanding the New Statistics: Effect Sizes, Confidence Intervals, and Meta-Analysis. New York: Taylor & Francis.
Kelley, K. (2005). The effects of nonnormal distributions on confidence intervals around the standardized mean difference: Bootstrap and parametric confidence intervals. Educational and Psychological Measurement 65: 51–69.
Kirk, R. (1996). Practical significance: A concept whose time has come. Educational and Psychological Measurement, 56, 746-759.
Kline, R. B. (2013). Beyond Significance Testing: Statistics Reform in the Behavioral Sciences. 2nd ed. Washington, DC: American Psychological Association.
Pierce, C.A., Block, R. A., and Aguinis, H. (2004). Cautionary note on reporting eta-squared values from multifactor ANOVA designs. Educational and Psychological Measurement, 64(6) 916-924
Thompson, B. (1996) AERA Editorial Policies regarding Statistical Significance Testing: Three Suggested Reforms. Educational Researcher, 25(2) 26-30
Wilkinson, L., & APA Task Force on Statistical Inference. (1999). Statistical methods in psychology journals: Guidelines and explanations. American Psychologist, 54, 594-604
Categories: Statistics Tags:
## Update on the Stata YouTube Channel
What is it about round numbers that compels us to pause and reflect? We celebrate 20-year school reunions, 25-year wedding anniversaries, 50th birthdays and other similar milestones. I don’t know the answer but the Stata YouTube Channel recently passed several milestones – more than 1500 subscribers, over 50,000 video views and it was launched six months ago. We felt the need for a small celebration to mark the occasion, and I thought that I would give you a brief update.
I could tell you about re-recording the original 24 videos with a larger font to make them easier to read. I could tell you about the hardware and software that we use to record them including our experiments with various condenser and dynamic microphones. I could share quotes from some of the nice messages we’ve received. But I think it would be more fun to talk about….you!
YouTube collects data about the number of views each video receives as well as summary data about who, what, when, where, and how you are watching them. There is no need to be concerned about your privacy; there are no personal identifiers of any kind associated with these data. But the summary data are interesting, and I thought it might be fun to share some of the data with you.
## Who’s watching?
Figure 1 shows the age distribution of Stata YouTube Channel viewers. If you have ever attended a Stata Conference, you will not be surprised by this graph…until you notice the age group at the bottom. I would not have guessed that 13-17 year olds are watching our videos. Perhaps they saw Stata in the movie “Moneyball” with Brad Pitt and wanted to learn more. Or maybe they were influenced by the latest fashion craze sweeping the youth of the world.
## What are you watching?
We have posted more than 50 videos over a wide range of topics. Figure 2 shows the total number of views for the ten most popular videos. The more popular of the ten are about broad topics. These broader videos are mostly older and have thus had time to accumulate more views.
Even so, these videos receive more views per day currently than do the special topic videos that have been posted more recently. This supports my belief that Stata YouTube Channel viewers tend to be relatively new Stata users who want to learn about general topics, and that means more generic videos in the future. So you and your two post-docs will just have to read the manual if you want to learn how to fit asymmetric power ARCH models with outer-product gradient standard errors.
## When are you watching?
We usually post new videos on Tuesday mornings which might lead you to believe that the peak viewing day would also be Tuesday. Figure 3, however, shows us that the average number of views per day (vpd) is higher on Wednesdays at 420 vpd and in fact peaks on Thursdays at 430 vpd before declining Friday through Sunday.
Figure 4 also shows us that late September may have been not the best time to launch the Stata YouTube Channel. Our early momentum in September and October slowed during the November and December holiday seasons. We were, however, pleased to see that 49 of you spent New Years Eve watching our videos. Perhaps next year we’ll prepare something more festive just for you!
## Where are you watching?
What do the Czech Republic, Pakistan, Uganda, Madagascar, the United Kingdom, the Bahamas, the United States, Montenegro, and Italy have in common? Correct! They are all countries in which you are watching our videos. They are also locations depicted in one of my favorite action films but I’ll leave that to the trivia buffs. I think the most exciting information that we found in our data is that the Stata YouTube Channel is being viewed in 164 countries!
You might not be surprised to learn that roughly half of the people watching the videos live in the United States, the United Kingdom, or Canada. The results may be unexpected when we consider the “view rate” defined as the number of views per 100,000 residents. Figure 5 shows the top 20 countries ranked by view rate for countries with at least four million residents. Denmark had the highest view rate which was nearly twice the rate of Norway which had the second highest view rate. The view rate in Denmark was more than three times the rate in the US and the UK.
## How are you watching?
You might think that I would have anything to report about “how” you are watching the videos, but it turns out that 5.2% of you are watching on mobile devices. Perhaps this explains the 13-17 year old demographic or the 49 people watching on New Year’s Eve. Or maybe we are helping you pass the time in the dentist office waiting room.
## Final thoughts
Six months isn’t much of a milestone. We Stata folk will use any excuse to break out the cake and ice cream. Even so, the Stata YouTube Channel began as an experiment and often experiments do not work out as we would like. This experiment has exceeded our expectations and, as a result, we have started taking requests for videos on our Facebook page and we’ll be adding more videos every week. So thanks for watching and stay tuned!
Now if you will excuse me, I’m going to get some cake and ice cream.
Categories: Resources Tags:
## Multilevel linear models in Stata, part 2: Longitudinal data
In my last posting, I introduced you to the concepts of hierarchical or “multilevel” data. In today’s post, I’d like to show you how to use multilevel modeling techniques to analyse longitudinal data with Stata’s xtmixed command.
Last time, we noticed that our data had two features. First, we noticed that the means within each level of the hierarchy were different from each other and we incorporated that into our data analysis by fitting a “variance component” model using Stata’s xtmixed command.
The second feature that we noticed is that repeated measurement of GSP showed an upward trend. We’ll pick up where we left off last time and stick to the concepts again and you can refer to the references at the end to learn more about the details.
## The videos
Stata has a very friendly dialog box that can assist you in building multilevel models. If you would like a brief introduction using the GUI, you can watch a demonstration on Stata’s YouTube Channel:
Introduction to multilevel linear models in Stata, part 2: Longitudinal data
## Longitudinal data
I’m often asked by beginning data analysts – “What’s the difference between longitudinal data and time-series data? Aren’t they the same thing?”.
The confusion is understandable — both types of data involve some measurement of time. But the answer is no, they are not the same thing.
Univariate time series data typically arise from the collection of many data points over time from a single source, such as from a person, country, financial instrument, etc.
Longitudinal data typically arise from collecting a few observations over time from many sources, such as a few blood pressure measurements from many people.
There are some multivariate time series that blur this distinction but a rule of thumb for distinguishing between the two is that time series have more repeated observations than subjects while longitudinal data have more subjects than repeated observations.
Because our GSP data from last time involve 17 measurements from 48 states (more sources than measurements), we will treat them as longitudinal data.
## Random intercept models
As I mentioned last time, repeated observations on a group of individuals can be conceptualized as multilevel data and modeled just as any other multilevel data. We left off last time with a variance component model for GSP (Gross State Product, logged) and noted that our model assumed a constant GSP over time while the data showed a clear upward trend.
If we consider a single observation and think about our model, nothing in the fixed or random part of the models is a function of time.
Let’s begin by adding the variable year to the fixed part of our model.
As we expected, our grand mean has become a linear regression which more accurately reflects the change over time in GSP. What might be unexpected is that each state’s and region’s mean has changed as well and now has the same slope as the regression line. This is because none of the random components of our model are a function of time. Let’s fit this model with the xtmixed command:
. xtmixed gsp year, || region: || state:
------------------------------------------------------------------------------
gsp | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
year | .0274903 .0005247 52.39 0.000 .0264618 .0285188
_cons | -43.71617 1.067718 -40.94 0.000 -45.80886 -41.62348
------------------------------------------------------------------------------
------------------------------------------------------------------------------
Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval]
-----------------------------+------------------------------------------------
region: Identity |
sd(_cons) | .6615238 .2038949 .3615664 1.210327
-----------------------------+------------------------------------------------
state: Identity |
sd(_cons) | .7805107 .0885788 .6248525 .9749452
-----------------------------+------------------------------------------------
sd(Residual) | .0734343 .0018737 .0698522 .0772001
------------------------------------------------------------------------------
The fixed part of our model now displays an estimate of the intercept (_cons = -43.7) and the slope (year = 0.027). Let’s graph the model for Region 7 and see if it fits the data better than the variance component model.
predict GrandMean, xb
label var GrandMean "GrandMean"
predict RegionEffect, reffects level(region)
predict StateEffect, reffects level(state)
gen RegionMean = GrandMean + RegionEffect
gen StateMean = GrandMean + RegionEffect + StateEffect
twoway (line GrandMean year, lcolor(black) lwidth(thick)) ///
(line RegionMean year, lcolor(blue) lwidth(medthick)) ///
(line StateMean year, lcolor(green) connect(ascending)) ///
(scatter gsp year, mcolor(red) msize(medsmall)) ///
if region ==7, ///
ytitle(log(Gross State Product), margin(medsmall)) ///
legend(cols(4) size(small)) ///
title("Multilevel Model of GSP for Region 7", size(medsmall))
That looks like a much better fit than our variance-components model from last time. Perhaps I should leave well enough alone, but I can’t help noticing that the slopes of the green lines for each state don’t fit as well as they could. The top green line fits nicely but the second from the top looks like it slopes upward more than is necessary. That’s the best fit we can achieve if the regression lines are forced to be parallel to each other. But what if the lines were not forced to be parallel? What if we could fit a “mini-regression model” for each state within the context of my overall multilevel model. Well, good news — we can!
## Random slope models
By introducing the variable year to the fixed part of the model, we turned our grand mean into a regression line. Next I’d like to incorporate the variable year into the random part of the model. By introducing a fourth random component that is a function of time, I am effectively estimating a separate regression line within each state.
Notice that the size of the new, brown deviation u1ij. is a function of time. If the observation were one year to the left, u1ij. would be smaller and if the observation were one year to the right, u1ij.would be larger.
It is common to “center” the time variable before fitting these kinds of models. Explaining why is for another day. The quick answer is that, at some point during the fitting of the model, Stata will have to compute the equivalent of the inverse of the square of year. For the year 1986 this turns out to be 2.535e-07. That’s a fairly small number and if we multiply it by another small number…well, you get the idea. By centering age (e.g. cyear = year – 1978), we get a more reasonable number for 1986 (0.01). (Hint: If you have problems with your model converging and you have large values for time, try centering them. It won’t always help, but it might).
So let’s center our year variable by subtracting 1978 and fit a model that includes a random slope.
gen cyear = year - 1978
xtmixed gsp cyear, || region: || state: cyear, cov(indep)
I’ve color-coded the output so that we can match each part of the output back to the model and the graph. The fixed part of the model appears in the top table and it looks like any other simple linear regression model. The random part of the model is definitely more complicated. If you get lost, look back at the graphic of the deviations and remind yourself that we have simply partitioned the deviation of each observation into four components. If we did this for every observation, the standard deviations in our output are simply the average of those deviations.
Let’s look at a graph of our new “random slope” model for Region 7 and see how well it fits our data.
predict GrandMean, xb
label var GrandMean "GrandMean"
predict RegionEffect, reffects level(region)
predict StateEffect_year StateEffect_cons, reffects level(state)
gen RegionMean = GrandMean + RegionEffect
gen StateMean_cons = GrandMean + RegionEffect + StateEffect_cons
gen StateMean_year = GrandMean + RegionEffect + StateEffect_cons + ///
(cyear*StateEffect_year)
twoway (line GrandMean cyear, lcolor(black) lwidth(thick)) ///
(line RegionMean cyear, lcolor(blue) lwidth(medthick)) ///
(line StateMean_cons cyear, lcolor(green) connect(ascending)) ///
(line StateMean_year cyear, lcolor(brown) connect(ascending)) ///
(scatter gsp cyear, mcolor(red) msize(medsmall)) ///
if region ==7, ///
ytitle(log(Gross State Product), margin(medsmall)) ///
legend(cols(3) size(small)) ///
title("Multilevel Model of GSP for Region 7", size(medsmall))
The top brown line fits the data slightly better, but the brown line below it (second from the top) is a much better fit. Mission accomplished!
## Where do we go from here?
I hope I have been able to convince you that multilevel modeling is easy using Stata’s xtmixed command and that this is a tool that you will want to add to your kit. I would love to say something like “And that’s all there is to it. Go forth and build models!”, but I would be remiss if I didn’t point out that I have glossed over many critical topics.
In our GSP example, we would still like to consider the impact of other independent variables. I haven’t mentioned choice of estimation methods (ML or REML in the case of xtmixed). I’ve assessed the fit of our models by looking at graphs, an approach important but incomplete. We haven’t thought about hypothesis testing. Oh — and, all the usual residual diagnostics for linear regression such as checking for outliers, influential observations, heteroskedasticity and normality still apply….times four! But now that you understand the concepts and some of the mechanics, it shouldn’t be difficult to fill in the details. If you’d like to learn more, check out the links below.
I hope this was helpful…thanks for stopping by.
Multilevel and Longitudinal Modeling Using Stata, Third Edition
Volume I: Continuous Responses
Volume II: Categorical Responses, Counts, and Survival
by Sophia Rabe-Hesketh and Anders Skrondal
or sign up for our popular public training course Multilevel/Mixed Models Using Stata.
Categories: Statistics Tags:
## Multilevel linear models in Stata, part 1: Components of variance
In the last 15-20 years multilevel modeling has evolved from a specialty area of statistical research into a standard analytical tool used by many applied researchers.
Stata has a lot of multilevel modeling capababilities.
I want to show you how easy it is to fit multilevel models in Stata. Along the way, we’ll unavoidably introduce some of the jargon of multilevel modeling.
I’m going to focus on concepts and ignore many of the details that would be part of a formal data analysis. I’ll give you some suggestions for learning more at the end of the post.
The videos
Stata has a friendly dialog box that can assist you in building multilevel models. If you would like a brief introduction using the GUI, you can watch a demonstration on Stata’s YouTube Channel:
Introduction to multilevel linear models in Stata, part 1: The xtmixed command
Multilevel data
Multilevel data are characterized by a hierarchical structure. A classic example is children nested within classrooms and classrooms nested within schools. The test scores of students within the same classroom may be correlated due to exposure to the same teacher or textbook. Likewise, the average test scores of classes might be correlated within a school due to the similar socioeconomic level of the students.
You may have run across datasets with these kinds of structures in your own work. For our example, I would like to use a dataset that has both longitudinal and classical hierarchical features. You can access this dataset from within Stata by typing the following command:
use http://www.stata-press.com/data/r12/productivity.dta
We are going to build a model of gross state product for 48 states in the USA measured annually from 1970 to 1986. The states have been grouped into nine regions based on their economic similarity. For distributional reasons, we will be modeling the logarithm of annual Gross State Product (GSP) but in the interest of readability, I will simply refer to the dependent variable as GSP.
. describe gsp year state region
storage display value
variable name type format label variable label
-----------------------------------------------------------------------------
gsp float %9.0g log(gross state product)
year int %9.0g years 1970-1986
state byte %9.0g states 1-48
region byte %9.0g regions 1-9
Let’s look at a graph of these data to see what we’re working with.
twoway (line gsp year, connect(ascending)), ///
by(region, title("log(Gross State Product) by Region", size(medsmall)))
Each line represents the trajectory of a state’s (log) GSP over the years 1970 to 1986. The first thing I notice is that the groups of lines are different in each of the nine regions. Some groups of lines seem higher and some groups seem lower. The second thing that I notice is that the slopes of the lines are not the same. I’d like to incorporate those attributes of the data into my model.
Components of variance
Let’s tackle the vertical differences in the groups of lines first. If we think about the hierarchical structure of these data, I have repeated observations nested within states which are in turn nested within regions. I used color to keep track of the data hierarchy.
We could compute the mean GSP within each state and note that the observations within in each state vary about their state mean.
Likewise, we could compute the mean GSP within each region and note that the state means vary about their regional mean.
We could also compute a grand mean and note that the regional means vary about the grand mean.
Next, let’s introduce some notation to help us keep track of our mutlilevel structure. In the jargon of multilevel modelling, the repeated measurements of GSP are described as “level 1″, the states are referred to as “level 2″ and the regions are “level 3″. I can add a three-part subscript to each observation to keep track of its place in the hierarchy.
Now let’s think about our model. The simplest regression model is the intercept-only model which is equivalent to the sample mean. The sample mean is the “fixed” part of the model and the difference between the observation and the mean is the residual or “random” part of the model. Econometricians often prefer the term “disturbance”. I’m going to use the symbol μ to denote the fixed part of the model. μ could represent something as simple as the sample mean or it could represent a collection of independent variables and their parameters.
Each observation can then be described in terms of its deviation from the fixed part of the model.
If we computed this deviation of each observation, we could estimate the variability of those deviations. Let’s try that for our data using Stata’s xtmixed command to fit the model:
. xtmixed gsp
Mixed-effects ML regression Number of obs = 816
Wald chi2(0) = .
Log likelihood = -1174.4175 Prob > chi2 = .
------------------------------------------------------------------------------
gsp | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons | 10.50885 .0357249 294.16 0.000 10.43883 10.57887
------------------------------------------------------------------------------
------------------------------------------------------------------------------
Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval]
-----------------------------+------------------------------------------------
sd(Residual) | 1.020506 .0252613 .9721766 1.071238
------------------------------------------------------------------------------
The top table in the output shows the fixed part of the model which looks like any other regression output from Stata, and the bottom table displays the random part of the model. Let’s look at a graph of our model along with the raw data and interpret our results.
predict GrandMean, xb
label var GrandMean "GrandMean"
twoway (line GrandMean year, lcolor(black) lwidth(thick)) ///
(scatter gsp year, mcolor(red) msize(tiny)), ///
ytitle(log(Gross State Product), margin(medsmall)) ///
legend(cols(4) size(small)) ///
title("GSP for 1970-1986 by Region", size(medsmall))
The thick black line in the center of the graph is the estimate of _cons, which is an estimate of the fixed part of model for GSP. In this simple model, _cons is the sample mean which is equal to 10.51. In “Random-effects Parameters” section of the output, sd(Residual) is the average vertical distance between each observation (the red dots) and fixed part of the model (the black line). In this model, sd(Residual) is the estimate of the sample standard deviation which equals 1.02.
At this point you may be thinking to yourself – “That’s not very interesting – I could have done that with Stata’s summarize command”. And you would be correct.
. summ gsp
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
gsp | 816 10.50885 1.021132 8.37885 13.04882
But here’s where it does become interesting. Let’s make the random part of the model more complex to account for the hierarchical structure of the data. Consider a single observation, yijk and take another look at its residual.
The observation deviates from its state mean by an amount that we will denote eijk. The observation’s state mean deviates from the the regionals mean uij. and the observation’s regional mean deviates from the fixed part of the model, μ, by an amount that we will denote ui... We have partitioned the observation’s residual into three parts, aka “components”, that describe its magnitude relative to the state, region and grand means. If we calculated this set of residuals for each observation, wecould estimate the variability of those residuals and make distributional assumptions about them.
These kinds of models are often called “variance component” models because they estimate the variability accounted for by each level of the hierarchy. We can estimate a variance component model for GSP using Stata’s xtmixed command:
xtmixed gsp, || region: || state:
------------------------------------------------------------------------------
gsp | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons | 10.65961 .2503806 42.57 0.000 10.16887 11.15035
------------------------------------------------------------------------------
------------------------------------------------------------------------------
Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval]
-----------------------------+------------------------------------------------
region: Identity |
sd(_cons) | .6615227 .2038944 .361566 1.210325
-----------------------------+------------------------------------------------
state: Identity |
sd(_cons) | .7797837 .0886614 .6240114 .9744415
-----------------------------+------------------------------------------------
sd(Residual) | .1570457 .0040071 .149385 .1650992
------------------------------------------------------------------------------
The fixed part of the model, _cons, is still the sample mean. But now there are three parameters estimates in the bottom table labeled “Random-effects Parameters”. Each quantifies the average deviation at each level of the hierarchy.
Let’s graph the predictions from our model and see how well they fit the data.
predict GrandMean, xb
label var GrandMean "GrandMean"
predict RegionEffect, reffects level(region)
predict StateEffect, reffects level(state)
gen RegionMean = GrandMean + RegionEffect
gen StateMean = GrandMean + RegionEffect + StateEffect
twoway (line GrandMean year, lcolor(black) lwidth(thick)) ///
(line RegionMean year, lcolor(blue) lwidth(medthick)) ///
(line StateMean year, lcolor(green) connect(ascending)) ///
(scatter gsp year, mcolor(red) msize(tiny)), ///
ytitle(log(Gross State Product), margin(medsmall)) ///
legend(cols(4) size(small)) ///
by(region, title("Multilevel Model of GSP by Region", size(medsmall)))
Wow – that’s a nice graph if I do say so myself. It would be impressive for a report or publication, but it’s a little tough to read with all nine regions displayed at once. Let’s take a closer look at Region 7 instead.
twoway (line GrandMean year, lcolor(black) lwidth(thick)) ///
(line RegionMean year, lcolor(blue) lwidth(medthick)) ///
(line StateMean year, lcolor(green) connect(ascending)) ///
(scatter gsp year, mcolor(red) msize(medsmall)) ///
if region ==7, ///
ytitle(log(Gross State Product), margin(medsmall)) ///
legend(cols(4) size(small)) ///
title("Multilevel Model of GSP for Region 7", size(medsmall))
The red dots are the observations of GSP for each state within Region 7. The green lines are the estimated mean GSP within each State and the blue line is the estimated mean GSP within Region 7. The thick black line in the center is the overall grand mean for all nine regions. The model appears to fit the data fairly well but I can’t help noticing that the red dots seem to have an upward slant to them. Our model predicts that GSP is constant within each state and region from 1970 to 1986 when clearly the data show an upward trend.
So we’ve tackled the first feature of our data. We’ve succesfully incorporated the basic hierarchical structure into our model by fitting a variance componentis using Stata’s xtmixed command. But our graph tells us that we aren’t finished yet.
Next time we’ll tackle the second feature of our data — the longitudinal nature of the observations.
Multilevel and Longitudinal Modeling Using Stata, Third Edition
Volume I: Continuous Responses
Volume II: Categorical Responses, Counts, and Survival
by Sophia Rabe-Hesketh and Anders Skrondal
or sign up for our popular public training course “Multilevel/Mixed Models Using Stata“.
There’s a course coming up in Washington, DC on February 7-8, 2013.
Categories: Statistics Tags:
## Using Stata’s SEM features to model the Beck Depression Inventory
I just got back from the 2012 Stata Conference in San Diego where I gave a talk on Psychometric Analysis Using Stata and from the 2012 American Psychological Association Meeting in Orlando. Stata’s structural equation modeling (SEM) builder was popular at both meetings and I wanted to show you how easy it is to use. If you are not familiar with the basics of SEM, please refer to the references at the end of the post. My goal is simply to show you how to use the SEM builder assuming that you already know something about SEM. If you would like to view a video demonstration of the SEM builder, please click the play button below:
The data used here and for the silly examples in my talk were simulated to resemble one of the most commonly used measures of depression: the Beck Depression Inventory (BDI). If you find these data too silly or not relevant to your own research, you could instead imagine it being a set of questions to measure mathematical ability, the ability to use a statistical package, or whatever you wanted.
The Beck Depression Inventory
Originally published by Aaron Beck and colleagues in 1961, the BDI marked an important change in the conceptualization of depression from a psychoanalytic perspective to a cognitive/behavioral perspective. It was also a landmark in the measurement of depression shifting from lengthy, expensive interviews with a psychiatrist to a brief, inexpensive questionnaire that could be scored and quantified. The original inventory consisted of 21 questions each allowing ordinal responses of increasing symptom severity from 0-3. The sum of the responses could then be used to classify a respondent’s depressive symptoms as none, mild, moderate or severe. Many studies have demonstrated that the BDI has good psychometric properties such as high test-retest reliability and the scores correlate well with the assessments of psychiatrists and psychologists. The 21 questions can also be grouped into two subscales. The affective scale includes questions like “I feel sad” and “I feel like a failure” that quantify emotional symptoms of depression. The somatic or physical scale includes questions like “I have lost my appetite” and “I have trouble sleeping” that quantify physical symptoms of depression. Since its original publication, the BDI has undergone two revisions in response to the American Psychiatric Association’s (APA) Diagnostic and Statistical Manuals (DSM) and the BDI-II remains very popular.
The Stata Depression Inventory
Since the BDI is a copyrighted psychometric instrument, I created a fictitious instrument called the “Stata Depression Inventory”. It consists of 20 questions each beginning with the phrase “My statistical software makes me…”. The individual questions are listed in the variable labels below.
. describe qu1-qu20
variable storage display value
name type format label variable label
------------------------------------------------------------------------------
qu1 byte %16.0g response ...feel sad
qu2 byte %16.0g response ...feel pessimistic about the future
qu3 byte %16.0g response ...feel like a failure
qu4 byte %16.0g response ...feel dissatisfied
qu5 byte %16.0g response ...feel guilty or unworthy
qu6 byte %16.0g response ...feel that I am being punished
qu7 byte %16.0g response ...feel disappointed in myself
qu8 byte %16.0g response ...feel am very critical of myself
qu9 byte %16.0g response ...feel like harming myself
qu10 byte %16.0g response ...feel like crying more than usual
qu11 byte %16.0g response ...become annoyed or irritated easily
qu12 byte %16.0g response ...have lost interest in other people
qu13 byte %16.0g qu13_t1 ...have trouble making decisions
qu14 byte %16.0g qu14_t1 ...feel unattractive
qu15 byte %16.0g qu15_t1 ...feel like not working
qu16 byte %16.0g qu16_t1 ...have trouble sleeping
qu17 byte %16.0g qu17_t1 ...feel tired or fatigued
qu18 byte %16.0g qu18_t1 ...makes my appetite lower than usual
qu19 byte %16.0g qu19_t1 ...concerned about my health
qu20 byte %16.0g qu20_t1 ...experience decreased libido
The responses consist of a 5-point Likert scale ranging from 1 (Strongly Disagree) to 5 (Strongly Agree). Questions 1-10 form the affective scale of the inventory and questions 11-20 form the physical scale. Data were simulated for 1000 imaginary people and included demographic variables such as age, sex and race. The responses can be summarized succinctly in a matrix of bar graphs:
Classical statistical analysis
The beginning of a classical statistical analysis of these data might consist of summing the responses for questions 1-10 and referring to them as the “Affective Depression Score” and summing questions 11-20 and referring to them as the “Physical Depression Score”.
egen Affective = rowtotal(qu1-qu10)
label var Affective "Affective Depression Score"
egen physical = rowtotal(qu11-qu20)
label var physical "Physical Depression Score"
We could be more sophisticated and use principal components to create the affective and physical depression score:
pca qu1-qu20, components(2)
predict Affective Physical
label var Affective "Affective Depression Score"
label var Physical "Physical Depression Score"
We could then ask questions such as “Are there differences in affective and physical depression scores by sex?” and test these hypotheses using multivariate statistics such as Hotelling’s T-squared statistic. The problem with this analysis strategy is that it treats the depression scores as though they were measured without error and can lead to inaccurate p-values for our test statistics.
Structural equation modeling
Structural equation modeling (SEM) is an ideal way to analyze data where the outcome of interest is a scale or scales derived from a set of measured variables. The affective and physical scores are treated as latent variables in the model resulting in accurate p-values and, best of all….these models are very easy to fit using Stata! We begin by selecting the SEM builder from the Statistics menu:
In the SEM builder, we can select the “Add Measurement Component” icon:
which will open the following dialog box:
In the box labeled “Latent Variable Name” we can type “Affective” (red arrow below) and we can select the variables qu1-qu10 in the “Measured variables” box (blue arrow below).
When we click “OK”, the affective measurement component appears in the builder:
We can repeat this process to create a measurement component for our physical depression scale (images not shown). We can also allow for covariance/correlation between our affective and physical depression scales using the “Add Covariance” icon on the toolbar (red arrow below).
I’ll omit the intermediate steps to build the full model shown below but it’s easy to use the “Add Observed Variable” and “Add Path” icons to create the full model:
Now we’re ready to estimate the parameters for our model. To do this, we click the “Estimate” icon on the toolbar (duh!):
And the flowing dialog box appears:
Let’s ignore the estimation options for now and use the default settings. Click “OK” and the parameter estimates will appear in the diagram:
Some of the parameter estimates are difficult to read in this form but it is easy to rearrange the placement and formatting of the estimates to make them easier to read.
If we look at Stata’s output window and scroll up, you’ll notice that the SEM Builder automatically generated the command for our model:
sem (Affective -> qu1) (Affective -> qu2) (Affective -> qu3)
(Affective -> qu4) (Affective -> qu5) (Affective -> qu6)
(Affective -> qu7) (Affective -> qu8) (Affective -> qu9)
(Affective -> qu10) (Physical -> qu11) (Physical -> qu12)
(Physical -> qu13) (Physical -> qu14) (Physical -> qu15)
(Physical -> qu16) (Physical -> qu17) (Physical -> qu18)
(Physical -> qu19) (Physical -> qu20) (sex -> Affective)
(sex -> Physical), latent(Affective Physical) cov(e.Physical*e.Affective)
We can gather terms and abbreviate some things to make the command much easier to read:
sem (Affective -> qu1-qu10) ///
(Physical -> qu11-qu20) ///
(sex -> Affective Physical) ///
, latent(Affective Physical ) ///
cov( e.Physical*e.Affective)
We could then calculate a Wald statistic to test the null hypothesis that there is no association between sex and our affective and physical depression scales.
test sex
( 1) [Affective]sex = 0
( 2) [Physical]sex = 0
chi2( 2) = 2.51
Prob > chi2 = 0.2854
Final thoughts
This is an admittedly oversimplified example – we haven’t considered the fit of the model or considered any alternative models. We have only included one dichotomous independent variable. We might prefer to use a likelihood ratio test or a score test. Those are all very important issues and should not be ignored in a proper data analysis. But my goal was to demonstrate how easy it is to use Stata’s SEM builder to model data such as those arising from the Beck Depression Inventory. Incidentally, if these data were collected using a complex survey design, it would not be difficult to incorporate the sampling structure and sample weights into the analysis. Missing data can be handled easily as well using Full Information Maximum Likelihood (FIML) but those are topics for another day.
If you would like view the slides from my talk, download the data used in this example or view a video demonstration of Stata’s SEM builder using these data, please use the links below. For the dataset, you can also type use followed by the URL for the data to load it directly into Stata.
References
Beck AT, Ward CH, Mendelson M, Mock J, Erbaugh J (June 1961). An inventory for measuring depression. Arch. Gen. Psychiatry 4 (6): 561–71.
Beck AT, Ward C, Mendelson M (1961). Beck Depression Inventory (BDI). Arch Gen Psychiatry 4 (6): 561–571
Beck AT, Steer RA, Ball R, Ranieri W (December 1996). Comparison of Beck Depression Inventories -IA and -II in psychiatric outpatients. Journal of Personality Assessment 67 (3): 588–97
Bollen, KA. (1989). Structural Equations With Latent Variables. New York, NY: John Wiley and Sons
Kline, RB (2011). Principles and Practice of Structural Equation Modeling. New York, NY: Guilford Press
Raykov, T & Marcoulides, GA (2006). A First Course in Structural Equation Modeling. Mahwah, NJ: Lawrence Erlbaum
Schumacker, RE & Lomax, RG (2012) A Beginner’s Guide to Structural Equation Modeling, 3rd Ed. New York, NY: Routledge
Categories: Statistics Tags:
StataCorp now provides free tutorial videos on StataCorp’s YouTube channel,
There are 24 videos providing 1 hour 51 minutes of instructional entertainment:
Stata Quick Tour (5:47)
Stata Quick Help (2:47)
Stata PDF Documentation (6:37)
Stata SEM Builder (8:09)
Stata One-way ANOVA (5:15)
Stata Two-way ANOVA (5:57)
Stata Box Plots (4:04)
Stata Basic Scatterplots (5:19)
Stata Bar Graphs (4:15)
Stata Histograms (4:50)
Stata Pie Charts (5:32)
And more are forthcoming.
The inside story
Alright, that’s the official announcement.
Last Friday, 21 September 2012, was an exciting day here at StataCorp. After a couple of years of “wouldn’t it be cool if”, and a couple of months of “we’re almost there”, Stata’s YouTube channel was finally ready for prime time.
Stata’s YouTube Channel was the brainchild of Karen Strope, StataCorp’s Director of Marketing, but I had something to do with it, too. Well, maybe more than something, but I’m a modest guy. Anyway, I thought it sounded like fun and recorded a few prototype videos. Annette Fett, StataCorp’s Graphic Designer, added the cool splash-screen and after a few experiments, we soon had 24 Blu-ray resolution videos. We’ve kicked off with videos covering topics such as a tour of Stata’s interface, how to create basic graphs, how to conduct many common statistical analyses, and more.
My personal favorite is the video entitled Combining Crosstabs and Descriptives because it’s relevant to nearly all Stata users and works well as a video demonstration.
Stata has over 9,000 pages of documentation included in PDF format, a built-in Help system, and a collection of books on general and special topics published by Stata Press, and an extensive collection of dialog boxes that make even the most complex graphs and analyses easy to perform.
So aren’t the videos, ahh, unnecessary?
The problem is, it’s cumbersome to describe how to use all of Stata’s features, especially dialog boxes, in a manual, even when you have 9,000 pages, and 9,000 pages tries even the most dedicated user’s patience.
In a 3-7 minutes video, we can show you how to create complicated graphs or a sophisticated structural equation model.
We have three audiences in mind.
1. Videos for non-Stata users, whom we call future Stata users; videos intended to provide a loosely guided tour of Stata’s features.
2. Videos for new Stata users, such as the person who might simply want to know “How do I calculate a twoway ANOVA in Stata?” or “How do I create a Pie Chart?”. These videos will get them up and running quickly and painlessly.
3. Videos for experienced Stata users who want to learn new tips and tricks.
There’s actually a fourth group that’s of interest, too; experienced Stata users teaching statistics or data analysis classes, who don’t want to spend valuable class time showing their students how to use Stata. They can refer their students to the relevant videos as homework and thus free class time for the teaching of statistics.
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A334067 a(n) is taken to be the smallest natural number greater than a(n-1) which is consistent with the condition "n is a member of the sequence if and only if a(n) is prime" where indices start from 0. 0
1, 2, 3, 5, 6, 7, 11, 13, 14, 15, 16, 17, 18, 19, 23, 29, 31, 37, 41, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 59, 60, 62, 63, 64, 65, 67, 68, 69, 70, 71, 72, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 132, 133, 134, 135, 137, 139, 140, 149 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS a(n) is the minimal sequence for which the sequence generated by the indices of primes in this sequence is equal to itself, where indices start from 0. So if f is a function on 0-indexed integer sequences with infinitely many primes where f returns the increasing sequence of indices of primes of the input sequence b(n), then a(n) is the lexicographically minimal fixed point of f. a(n) has almost the same definition with A079254, except a(n) starts indices from 0, instead of 1. But the resulting sequences does not seem to have any correlation. LINKS EXAMPLE a(0) cannot be 0, since then 0 should be prime, which is not. a(0) = 1 is valid hence a(1) must be the next prime which is a(1) = 2. Then a(2) should be the next prime, hence a(2) = 3. a(3) should be prime, hence a(3) = 5 Since 4 is not in the sequence so far, a(4) must be the next non-prime, which means a(4) = 6. PROG (Python) # is_prime(n) is a Python function which returns True if n is prime, and returns False otherwise. In the form stated below runs with SageMath. def a_list(length): """Returns the list [a(0), ..., a(length-1)].""" num = 1 b = [1] for i in range(1, length): num += 1 if i in b: while not is_prime(num): num += 1 b.append(num) else: while is_prime(num): num += 1 b.append(num) return b print(a_list(63)) CROSSREFS The same definition with A079254 except indices start from 0 instead of 1. Cf. A079000, A079313. Sequence in context: A288863 A121700 A080980 * A134669 A053328 A333786 Adjacent sequences: A334064 A334065 A334066 * A334068 A334069 A334070 KEYWORD nonn AUTHOR Adnan Baysal, Apr 13 2020 STATUS approved
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https://sonichours.com/how-many-seconds-are-there-in-70-years/
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General
# How Many Seconds Are There In 70 Years
It’s difficult to believe that 70 years are equal to the 2,207,520,000 seconds that make up a single day. However, there are many things you can do to convert a year into seconds. Here are some tips to get you started. If you are unsure of the conversion factor, you can use the following chart. After you know how long a year is in seconds, you can try this trick.
In one year, there are 86400 seconds. Then, two days have 172800 seconds. This means that there are a total of 77,536 seconds in seventy years. In other words, there are ten million minutes and twenty-four hours in a single day. A year is made up of seventy thousand seconds. A single day is the same as two months and twenty-four hours are the same as a year.
The answer will be reported in decimal form or in scientific notation. In the United Kingdom, this is called standard form. The second type of result will be in fractions, which are more accurate. In addition to the decimal form, a calculator with a high number of significant places will report an incorrect result. This method is usually not recommended, because the result will be wrong. So, if you want to be precise, make sure to use a calculator with more significant places.
Besides the fact that a day is the same as two days, it is also important to know that a year contains 86400 seconds. That’s a lot of seconds. So, it’s easy to understand how to calculate the time of a year. You can use the calculator to convert the corresponding numbers to seconds and vice versa. If you’re concerned about accuracy, try using a calculator with fewer significant places.
The calculator will give you the answer in either decimal or scientific form. In the United Kingdom, the result will be in the form of fractions. The result of the calculation will be shown in percentages. The second way will give you the number in absolute terms. The third method will give you the answer in a decimal. It is more accurate than the other two options. Then, you can choose to use a more precise unit of time.
The second unit of time is the second. There are four decimal places. In the United States, a second is one thousandth of a millisecond. The third is a kilosecond. A kilosecond is a thousandth of a second. There are a hundred and sixty-four minutes in a day. The fourth place is the second. The fifth place is the third.
The second unit is one thousandth of a second. The first unit is the minute. This is divided by 60. In addition, each second is one tenth of a millionth of a second. A day consists of 24 hours. The second and the minute are equivalent in time. A day is composed of 86400 seconds. If you have a clock, a week is seven days.
A day consists of 86400 seconds. A second is 1/86400 of a second. So, a day has 24 hours. A minute is a thousandth of a second. A year is a hundredth of a second. Its duration is one thousand times that of a day. Then, a minute is equal to one hundredth of a second. Then, a minute is equal a thousandth of a second.
A day is made up of 86400 seconds. It is a multiple of a second. Therefore, a day consists of sixty-four hours, which is 168400 seconds. A minute is one year. A year is one million days. During a week, there are 365 days. Hence, a week consists of seventy-two months. Then, a day contains 168400 seconds.
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https://lcmgcf.com/factors-of-729/
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Factors of 729 | Find the Factors of 729 by Factoring Calculator
Factoring Calculator calculates the factors and factor pairs of positive integers.
Factors of 729 can be calculated quickly with the help of Factoring Calculator i.e. 1, 3, 9, 27, 81, 243, 729 positive integers that divide 729 without a remainder.
Factors of 729 are 1, 3, 9, 27, 81, 243, 729. There are 7 integers that are factors of 729. The biggest factor of 729 is 729.
Factors of:
Factor Tree of 729 to Calculate the Factors
729 3 243 3 81 3 27 3 9 3 3
Factors of 729 are 1, 3, 9, 27, 81, 243, 729. There are 7 integers that are factors of 729. The biggest factor of 729 is 729.
Positive integers that divides 729 without a remainder are listed below.
• 1
• 3
• 9
• 27
• 81
• 243
• 729
Factors of 729 in pairs
• 1 × 729 = 729
• 3 × 243 = 729
• 9 × 81 = 729
• 27 × 27 = 729
• 81 × 9 = 729
• 243 × 3 = 729
• 729 × 1 = 729
Factors of 729 Table
FactorFactor Number
1one
3three
9nine
27twenty seven
81eighty one
243two hundred forty three
729seven hundred twenty nine
How to find Factors of 729?
As we know factors of 729 are all the numbers that can exactly divide the number 729 simply divide 729 by all the numbers up to 729 to see the ones that result in zero remainders. Numbers that divide without remainder are factors and in this case below are the factors
1, 3, 9, 27, 81, 243, 729 are the factors and all of them can exactly divide number 729.
Frequently Asked Questions on Factors of 729
1. What are the factors of 729?
Answer: Factors of 729 are the numbers that leave a remainder zero. The ones that can divide 729 exactly i.e. factors are 1, 3, 9, 27, 81, 243, 729.
2.What are Factor Pairs of 729?
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# The weight of an object will be minimum when it is placed
This question was previously asked in
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1. at the North Pole
2. at the South Pole
3. at the Center of the Earth
4. None of the above
Option 3 : at the Center of the Earth
## Detailed Solution
The correct answer is at the Center of the Earth.
Key Points
• Earth is not a perfect sphere. Its radius at the equator is greater than the poles. Acceleration due to gravity is inversely proportional to the square of its radius.
• So, the acceleration due to gravity is greatest at poles.
• Hence, from the relation, W = mg, it is clear that weight is highest at the poles.
• We know that the weight of the body is the product of mass and acceleration due to gravity and the acceleration due to gravity increases with latitude.
• Now the latitude is minimum at the equator and maximum at the poles So, acceleration due to gravity and hence weight is maximum at the poles and minimum at the equator.
• Acceleration due to gravity g varies slightly over the surface of Earth, so the weight of an object depends on its location and is not an intrinsic property of the object.
• The weight of an object is minimum when it is placed at the center of the earth because when an object is in the center, experiences gravitational pull from all directions & the weight of an object maximum on earth is on the poles. Hence, Option 3 is correct.
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# Harmonic series proof
1. Oct 5, 2009
### James889
Hai,
The harmonic series is given by: $$H_{n} = \sum_{i=1}^n \frac{1}{i}$$
I need to prove that for all positive integers:
$$\sum_{j=1}^n H_{j} = (n+1)H_{n} -n$$
So i have
$$H_{5} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{137}{60}$$
$$H_{5} \neq (5+1)*\frac{137}{60} -5$$
Have i missed something here?
Please excuse my epic fail math skills...
Last edited: Oct 5, 2009
2. Oct 5, 2009
### Office_Shredder
Staff Emeritus
So for your H5 example what they want you to sum is
H1 + H2 + H3 + H4 + H5
When I did that I got what the problem tells you you will get
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# FIND INDICATED TERM FROM NTH TERM OF THE SEQUENCE WORKSHEET
## About "Find indicated term from nth term of the sequence worksheet"
Find indicated term from nth term of the sequence worksheet :
Here we are going to see some practice questions on finding indicated term of the sequence.
## Find indicated term from nth term of the sequence worksheet - Practice questions
(1) Write the first three terms in the following sequence
an = 2n + 5
(2) Write the first three terms in the following sequence
an = (n - 3)/4
(3) Write the first three terms in the following sequence
an = n (n + 2)
(4) Write the first three terms in the following sequence
an = n / (n + 1)
(5) Write the first three terms in the following sequence
an = (2n - 3)/6
(6) Write the first three terms in the following sequence
an = (-1)n-1 5n+1
## Find indicated term from nth term of the sequence worksheet - Solution
Question 1 :
Write the first three terms in the following sequence
an = 2n + 5
Solution :
In order to get first three terms, instead of n we are going apply 1, 2 and 3.
n = 1a1 = 2(1) + 5 = 2 + 5 = 7 n = 2a2 = 2(2) + 5 = 4 + 5 = 9 n = 3a3 = 2(3) + 5 = 6 + 5 = 11
Hence the 1st three terms of the sequence are 7, 9 and 11 respectively.
Question 2 :
Write the first three terms in the following sequence
an = (n - 3)/4
Solution :
In order to get first three terms, instead of n we are going apply 1, 2 and 3.
n = 1a1 = (1 - 3)/4 = -2/4= - 1/2 n = 2a2 = (2 - 3)/4 = -1/4 n = 3a3 = (3 - 3)/4 = 0/4= 0
Hence the 1st three terms of the sequence are -1/2, -1/4 and 0 respectively.
Question 3 :
Write the first three terms in the following sequence
an = n (n + 2)
Solution :
In order to get first three terms, instead of n we are going apply 1, 2 and 3.
n = 1a1 = 1(1 + 2)= 1 (3)= 3 n = 2a2 = 2 (2 + 2)= 2(4)= 8 n = 3a3 = 3 (3 + 2)= 3(5)= 15
Hence the 1st three terms of the sequence are 3, 8 and 15 respectively.
Question 4 :
Write the first three terms in the following sequence
an = n / (n + 1)
Solution :
In order to get first three terms, instead of n we are going apply 1, 2 and 3.
n = 1a1 = 1/(1 + 1)= 1/2 n = 2a2 = 2/(2 + 1)= 2/3 n = 3a3 = 3/(3 + 1)= 3/4
Hence the 1st three terms of the sequence are 1/2, 2/3, and 3/4 respectively.
Question 5 :
Write the first three terms in the following sequence
an = (2n - 3)/6
Solution :
In order to get first three terms, instead of n we are going apply 1, 2 and 3.
n = 1a1 = (2n - 3)/6= (2(1) - 3)/6= (2 - 3)/6= -1/6 n = 2a2 = (2n - 3)/6= (2(2) - 3)/6= (4 - 3)/6= 1/6 n = 3a3 = 3/(3 + 1)= 3/4
Hence the 1st three terms of the sequence are -1/6, 1/6 and 3/4 respectively.
Question 6 :
Write the first three terms in the following sequence
an = (-1)n-1 5n+1
Solution :
In order to get first three terms, instead of n we are going apply 1, 2 and 3.
n = 1an = (-1)n-1 5n+1= (-1)1-1 51+1= (-1)0 52= 25 n = 2an = (-1)n-1 5n+1= (-1)2-1 52+1= (-1)1 53= -125 n = 3an = (-1)n-1 5n+1= (-1)3-1 53+1= (-1)2 54= 625
Hence the 1st three terms of the sequence are 25, -125 and 625 respectively.
After having gone through the stuff given above, we hope that the students would have understood "Find indicated term from nth term of the sequence worksheet"
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# algebra
posted by .
Assume you work at a company where you are paid hourly. You are paid \$8.60 per hour for regular time (less than or equal to forty hours) and time and a half for overtime hours up to fifty hours in one week. If you are asked to work fifty or more hours in one week you are paid double the regular rate for these overtime hours (those past the first 50 hours). Write this information in the form of a piecewise-defined function and be sure to include the domain for each piece. Also, check your model by calculating the pay for working forty hours and for working forty-one hours. Post these values as well.
• algebra -
p =
} 8.60h if h <= 40
} 344 + 12.9h if 40<h<=50
} 473 + 2h if h>50
• algebra -
473 + 17.2h if h>50
For forty-one hours, h=41, so 40<h≤50,
use the second equation.
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You are Here: Home >< Maths
# C2 Binomial Expansion Question 1 watch
1. I can't figure out where I'm going wrong here...
If (1 + 2x)^5 + (1 - 2x)^5 = a + bx + cx^2,
find the values of the constants a,b and c.
I used binomial expansion and I got:
(1 + 2x)^5 + (1 - 2x)^5 =
1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5 + 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5 =
2 + 80x^2 + 160x^4
2. I agree - question seems to be wrong.
3. (Original post by Narik)
I can't figure out where I'm going wrong here...
If (1 + 2x)^5 + (1 - 2x)^5 = a + bx + cx^2,
find the values of the constants a,b and c.
I used binomial expansion and I got:
(1 + 2x)^5 + (1 - 2x)^5 =
1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5 + 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5 =
2 + 80x^2 + 160x^4
Edit: Your calculation is correct but the question only asks for the constant term and the coefficients of and
4. (Original post by Narik)
I can't figure out where I'm going wrong here...
If (1 + 2x)^5 + (1 - 2x)^5 = a + bx + cx^2,
find the values of the constants a,b and c.
I used binomial expansion and I got:
(1 + 2x)^5 + (1 - 2x)^5 =
1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5 + 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5 =
2 + 80x^2 + 160x^4
your working is correct, question seems wrong
5. Just adding voice to the question being wrong in this instance.
6. Unless you square rooted, then you would have a+bx+cx^2 but I dont remember having to do anything like that in C2.
7. (Original post by mcp2)
Unless you square rooted, then you would have a+bx+cx^2 but I dont remember having to do anything like that in C2.
No, the square root of 2+80x^2+160x^4 isn't even a polynomial, let alone one of the form a+bx+x^2.
8. Make x^2 x, would that work?
9. (Original post by mcp2)
Make x^2 x, would that work?
Not really. You're getting into "2+2 = 6" if we make 2 = 3 territory. Far simpler to assume the question's wrong, to be honest.
10. (Original post by DFranklin)
Not really. You're getting into "2+2 = 6" if we make 2 = 3 territory. Far simpler to assume the question's wrong, to be honest.
Why is the question wrong? It asks for certain coefficients , why does it matter that the question does not ask for the coefficient of
11. It doesn't ask for certain coefficients, it asks for the values of a,b,c such that (1 + 2x)^5 + (1 - 2x)^5 = a + bx + cx^2.
There are no such values, so the question is wrong.
12. (Original post by DFranklin)
It doesn't ask for certain coefficients, it asks for the values of a,b,c such that (1 + 2x)^5 + (1 - 2x)^5 = a + bx + cx^2.
There are no such values, so the question is wrong.
Thanks.
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# Volume of Sphere Formula
The volume of the sphere is the measure of space a sphere occupies within its surface as the boundary. Let’s think of a ball as a perfect sphere; loosely speaking, the amount of space that the air occupies inside the ball is its volume.
Don’t mixed things up between a circle and a sphere. They are two different geometric objects. A circle is a two-dimensional figure which means it can be drawn in a plane. You may think of a circle as flat object. On the other hand, there is no way we can draw a sphere on a plane because it lies on a three-dimensional space. It requires an additional axis z-axis that is perpendicular to the regular x and y axes.
Now, finding the volume of a sphere is easy. In order to calculate the volume of the sphere, all we need is its radius. Even if the radius is not given but just the diameter, we can still find the volume. Since the diameter is twice the radius, we can simply divide the diameter by $2$ to get the radius.
So here is the formula of the volume of the sphere if the radius $r$ is given:
$\boxed{\large{V = {\large{{4 \over 3}}}\pi {{\color{red}{r}}^3}}}$
${\color{red}r} = {\text{radius}}$
$\pi\approx3.14$
Note: Pi ($\pi$) is the ratio of the circumference of the circle to its diameter. The value is an irrational number which is approximately equal to $3.1416$.
## Examples of Finding the Volume of Sphere
Example 1: Find the volume of sphere with a radius of $6$ inches. Express your answer in terms of $\pi$.
In this problem, the radius of the sphere is given to us which is $r=6$. That makes our life very easy! We just need to substitute the value of radius into the formula then simplify.
However, we need to be extra careful when we give our final answer. In the problem, we’re told to express our answer in terms of pi, $\pi$. This means when we leave the answer in terms of $\pi$, we’re giving the exact volume of the sphere. It’s different if we use the decimal approximation of $\pi$, like $3.14$. In that case, we’re estimating the sphere’s volume.
To reiterate, in this problem, we want the exact volume of the sphere that’s why we are leaving the answer in terms of $\pi$.
So now we are going to substitute the value of the radius $r=6$ into the formula and then we simplify. Finally, we must not forget to attach the unit of measure which is cubic inches, $\text{in}^3$.
\begin{align*} V &= {4 \over 3}\pi {r^3} \\ \\ & = {4 \over 3}\pi {\left( {\color{red}6} \right)^3} \\ \\ & = {4 \over 3}\pi {\left( 6 \right)^3} \\ \\ & = {4 \over 3}\pi \left( {216} \right) \\ \\ & = {4 \over 3}\left( {216} \right)\pi \\ \\ V &= 288\pi \end{align*}
Therefore, the volume of the sphere is $288\pi$ cubic inches $\text{in}^3$.
Example 2: Determine the volume of a sphere having a radius of $3$ meters. Round your answer to the nearest tenth. Use $\pi=3.1416$.
This problem is similar to example #1 because the radius of the sphere is given to us. However, the difference lies on the use of $\pi$. In example #1, we leave our answer in terms of $\pi$ but in this problem we will use its approximate value of $\pi = 3.1416$. Since we are using the estimated value of $\pi$, it follows that the volume of sphere here is also an approximation. Remember, we are also instructed to round it off to the nearest tenth. That means we round it to one decimal place. Finally, don’t forget to attach the unit of measure which is cubic meter $\text{m}^3$.
\begin{align*} V &= {4 \over 3}\pi {r^3} \\ \\ & = {4 \over 3}\left( {3.1416} \right){\left( 3 \right)^3} \\ \\ & = {4 \over 3}\left( {3.1416} \right)\left( {27} \right) \\ \\ & = {4 \over 3}\left( {27} \right)\left( {3.1416} \right) \\ \\ & = \left( {36} \right)\left( {3.1416} \right) \\ \\ V &\approx 113.1 \end{align*}
Thus, the volume of the sphere is approximately $113.1$ cubic meters $\text{m}^3$.
Calculator check!
Example 3: Calculate the volume of sphere having a diameter of $18$ centimeters. Round your answer to the nearest whole number. Use $\pi=3.14$
This time the diameter of the sphere is given, not the radius. Just remember the fact that the diameter is twice the radius, or the radius is half the diameter.
After finding the radius using the simple fact above, we can proceed as usual.
Be careful, though. Make sure that we follow the instructions. We are going to use the estimated value of $\pi=3.14$. In addition, we will need to round off our final answer to the whole number.
Let’s go ahead and find the radius of the sphere. As we said, the radius is simply half the diameter.
\begin{align*} radius &= {1 \over 2}\left( {diameter} \right) \\ \\ & = {1 \over 2}\left( {18} \right) \\ \\ & = 9 \\ \\ radius &= 9 \end{align*}
Now that we know what the radius is, we can plug in its value into the formula to compute the volume of the sphere.
\begin{align*} V &= {4 \over 3}\pi {r^3} \\ \\ & = {4 \over 3}\left( {3.14} \right){\left( 9 \right)^3} \\ \\ & = {4 \over 3}\left( {3.14} \right)\left( {729} \right) \\ \\ & = {4 \over 3}\left( {729} \right)\left( {3.14} \right) \\ \\ & = \left( {972} \right)\left( {3.14} \right) \\ \\ V &\approx 3,052 \end{align*}
Therefore, the volume of the sphere is approximately $3,052$ cubic centimeters $\text{cm}^3$.
Calculator check!
Example 4: The volume of the sphere is $5,207$ cubic meters. What is the radius of the sphere? Round your answer to the nearest tenth. Use $\pi=3.14$.
In this problem, we are not looking for the volume of the sphere. In fact, the volume of the sphere has already been provided to us. So what’s the deal then? The goal is to manipulate the formula to find the radius of the sphere.
\begin{align*} V &= {4 \over 3}\pi {r^3} \\ \\ 5,207 &= {4 \over 3}\left( {3.14} \right){r^3} \\ \\ 5,207 &= {4 \over 3}\left( {3.14} \right){r^3} \\ \\ 15,621 &= 4\left( {3.14} \right){r^3} \\ \\ 15,621 &= \left( {12.56} \right){r^3} \\ \\ {{15,621} \over {12.56}} &= {r^3} \\ \\ {\left( {{{15,621} \over {12.56}}} \right)^{{1 \over 3}}} &= {r} \\ \\ 10.8 &\approx r \end{align*}
Therefore, the radius of the sphere is approximately $10.8$ meters $\text{m}$.
Calculator check!
Example 5: If the radius of the sphere is doubled, what happens to the volume of the sphere?
Let’s suppose the radius of the original sphere is $r$ while the radius of the new sphere is ${r}^\prime$ (read as r prime).
Since the radius of the new sphere is twice the radius of the original sphere. We have the situation:
${r}^\prime = \color{red}2r$
Here’s the formula of the volume of the NEW sphere with twice the radius of the original.
\begin{align*} {V}^\prime = {4 \over 3}\pi {\left( {r}^\prime \right)^3} \end{align*}
Now, let’s substitute $\color{red}2r$ for $r^\prime$.
\begin{align*} {V}^\prime &= {4 \over 3}\pi {\left( {{\color{red}2r}} \right)^3} \\ \\ & = {4 \over 3}\pi {\left( {\color{red}2} \right)^3}{\left( {\color{red}r} \right)^3} \\ \\ & = {4 \over 3}\pi \left( 8 \right)\left( {{r^3}} \right) \\ \\ &= 8\left( {{\color{blue}{4 \over 3}\pi {r^3}}} \right) \\ \\ {V}^\prime &=8{\color{blue}{V}} \end{align*}
Therefore, if the radius of a sphere is doubled, the volume of the sphere becomes eight times or eight fold.
YOU TRY!
Example 6: If the radius of the sphere is tripled, what happens to the volume of the sphere?
If the radius is tripled, then the radius of the sphere becomes $3r$.
\begin{align*} {V}^\prime &= {4 \over 3}\pi {\left( {{\color{red}3r}} \right)^3} \\ \\ & = {4 \over 3}\pi {\left( {\color{red}3} \right)^3}{\left( {\color{red}r} \right)^3} \\ \\ & = {4 \over 3}\pi \left( 27 \right)\left( {{r^3}} \right) \\ \\ &= 27\left( {{\color{blue}{4 \over 3}\pi {r^3}}} \right) \\ \\ {V}^\prime &=27{\color{blue}{V}} \end{align*}
Therefore, the volume of the sphere is increased by twenty-seven times.
Let’s do a final example. This one is a bit different but still related to this topic. The surface area of the sphere is given and we will have to figure out the volume of the same sphere.
For your information, below is the formula to find the surface area of a sphere.
$\text{Surface Area of Sphere} = 4\pi {r^2}$
It may sound complex, but it’s really not. The key is to use the formula of the area of the sphere to determine the radius of the sphere. Then use the value of the radius to calculate its volume using the volume formula that we have been using from previous examples.
$\text{Volume of Sphere} = {\large{4 \over 3}}\pi {r^3}$
Here we go!
Example 7: The surface area of a sphere is $616$ $\text{ft}^2$. Find the volume of the sphere. Round you answer to two decimal places. Use $\pi=22/7$ as an approximation.
First, we write the formula for the surface area of the sphere ($SA$). Then, we plug in the value of the area which is $616$ and solve for the radius $r$.
$\boxed{SA = 4\pi {{\color{red}r}^2}}$
Let’s get started!
\begin{align*} SA &= 4\pi {r^2} \\ \\ 616 &= 4\left( {{{22} \over 7}} \right){r^2} \\ \\ 616\left( 7 \right) &= 4\left( {22} \right){r^2} \\ \\ 4,312 &= 88{r^2} \\ \\ {{4,312} \over {88}} &= {r^2} \\ \\ 49 &= {r^2} \\ \\ \sqrt {49} &= r \\ \\ 7 &= r \end{align*}
Now that we found the radius of the sphere, it is time to calculate its volume. We have done this before so it’s no big deal. Here’s again the formula for the volume of sphere.
$\boxed{\text{Volume of Sphere} = {\large{4 \over 3}}\pi {r^3}}$
Let’s substitute the radius that we have found above which is $r=7$.
\begin{align*} V &= {4 \over 3}\pi {r^3} \\ \\ & = {4 \over 3}\left( {{{22} \over 7}} \right){\left( {\color{red}7} \right)^3} \\ \\ & = {4 \over 3}\left( {{{22} \over 7}} \right)\left( {343} \right) \\ \\ & = \left( {{{88} \over {21}}} \right)\left( {343} \right) \\ \\ V &\approx 1,437.33 \end{align*}
Hence, the volume of the sphere with a surface area of $616$ $\text{ft}^2$ is approximately $1,437.33$ $\text{ft}^3$.
Calculator check!
Find the radius of the sphere:
Find the volume of the sphere:
YOU TRY!
Example 8: If the surface area of the sphere is $2,464$ $\text{m}^2$, what is its volume? Round you answer to the nearest hundredth. Use $\pi=22/7$ as an approximation.
$\boxed{SA = 4\pi {{\color{red}r}^2}}$
\begin{align*} SA &= 4\pi {r^2} \\ \\ 2,464 &= 4\left( {{{22} \over 7}} \right){r^2} \\ \\ 2,464\left( 7 \right) &= 4\left( {22} \right){r^2} \\ \\ 17,248 &= 88{r^2} \\ \\ {{17,248} \over {88}} &= {r^2} \\ \\ 196 &= {r^2} \\ \\ \sqrt {196} &= r \\ \\ 14 &= r \end{align*}
Now, let’s find the volume using the radius $r=14$.
\begin{align*} V &= {4 \over 3}\pi {r^3} \\ \\ & = {4 \over 3}\left( {{{22} \over 7}} \right){\left( {14} \right)^3} \\ \\ & = {4 \over 3}\left( {{{22} \over 7}} \right)\left( {2,744} \right) \\ \\ & = \left( {{{88} \over {21}}} \right)\left( {2,744} \right) \\ \\ V &\approx 11,498.67 \end{align*}
Therefore, the volume is approximately $11,498.67$ $\text{m}^3$.
You may also be interested in these related math lessons or tutorials:
Volume of Sphere Practice Problems with Answers
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# Dictionary of Lean Six Sigma Tools/Concepts
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The entire extent of the dataset; the difference between the minimum and maximum data points.
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The logical distinction of potential sub-processes that exist within an overall process. These sub-processes are usually distinguished (or grouped) by a factor or category like by time, location, processes, or people.
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A statistical application to correlated values that involves building an equation or model that describes the nature of the correlation or relationship. It will typically look like "Y-Response = Constant + Coefficient(X-Predictor)". A correlation measures the linear association between variables as represented by r, but the quality of the regression is measured by squaring the correlation coefficient, or r2. It measures the proportion of variation explained by the regression model.
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They represent all the deviations for each data point. In statistical tests that analyze a continuous (a.k.a. variable or numerical) value, they are often used as a supplement to validate the original statistical test. The Analysis of Variance (ANOVA) is an example of test where residuals are used.
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An evaluation of the amount of chance of being wrong or yielding an error. Generally risk represents the what we don't know, don't understand, or can't measure, such as our assumptions. It's the opposite of confidence gained by data or evidence; in that way, it can be said it has an inverse relationship with confidence where as more data or evidence is gathered to build confidence, it should likewise decrease the risk in that situation (or at least make us more informed of the risk and what better actions may be necessary). From a statistical standpoint, risk is defined as Alpha Risk (or Type I Error) and Beta Risk (or Type II Error).
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A project may have various roles, but most LSS projects tend to have some roles in common that are either external or internal resources. External resources include an expert in LSS who is leading the project; common roles include a Green Belt, Black Belt, or Master Black Belt. These are external resources because they may not be required to have strong functional experience in the business area for the project, but they should have strong LSS experience. Internal resources include experts within the business area for the project; common roles include a Sponsor, Champion, Process Owner, and Subject Matter Experts (SMEs). In contrast to the external resources, these internal resources aren't necessarily required to have LSS experience, but they should have strong expertise in the business area for the project.
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A measure of the true yield or capability of a process (generally made up of sub-processes). The calculation is (FTYProcessA x FTYProcessB x FTYProcessC x FTYProcessD x etc.). It can help expose "hidden factories" for processing that could risk defects or delays. The measurement of every sub-process should be rolled up to measure their contribution to the overall process they support. Compare to First Time Yield (FTY).
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# A family consisting of three persons—A, B, and C—goes to a
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## Solution for problem 5E Chapter 1
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Problem 5E
A family consisting of three persons—?A, B, ?and ?C?—goes to a medical clinic that always has a doctor at each of stations 1, 2, and 3. During a certain week, each member of the family visits the clinic once and is assigned at random to a station. The experiment consists of recording the station number for each member. One outcome is (1, 2, 1) for ?A ?to station 1, ?B ?to station 2, and ?C ?to station 1. a?. ?List the 27 outcomes in the sample space. b?. ?List all outcomes in the event that all three members go to the same station. c?. ?List all outcomes in the event that all members go to different stations. d?.? ist all outcomes in the event that no one goes to station 2.
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Answer: Step1: a). No, the relevant conceptual population is all scores of all students who participate in the SI in the conjunction with this particular statistics course. b). Randomization protects against various biases and helps ensure that those in the SI group are as similar as possible to the students in the control group. c). There would not be any basis for comparison otherwise.
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##### ISBN: 9780321629111
The full step-by-step solution to problem: 5E from chapter: 1 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM. This full solution covers the following key subjects: station, Outcomes, Event, list, members. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. Since the solution to 5E from 1 chapter was answered, more than 444 students have viewed the full step-by-step answer. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. The answer to “A family consisting of three persons—?A, B, ?and ?C?—goes to a medical clinic that always has a doctor at each of stations 1, 2, and 3. During a certain week, each member of the family visits the clinic once and is assigned at random to a station. The experiment consists of recording the station number for each member. One outcome is (1, 2, 1) for ?A ?to station 1, ?B ?to station 2, and ?C ?to station 1. a?. ?List the 27 outcomes in the sample space. b?. ?List all outcomes in the event that all three members go to the same station. c?. ?List all outcomes in the event that all members go to different stations. d?.? ist all outcomes in the event that no one goes to station 2.” is broken down into a number of easy to follow steps, and 131 words. This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9.
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# Find the expected value of a dice sum
If fair dodecahedron is rolled until at least $k$($k$ is fixed between 2 and 12) is gotten, and $X$ is the sum of all numbers appeared until the last time, what is $E(X)$?
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What is the probability that it will take m rolls? What is the distribution of X over those m rolls? – Yang Apr 13 '12 at 21:27
Once $k$ or greater than that appears, quit the rolling. The sum includes $k$!!! – hkju Apr 13 '12 at 21:35
Doesn't it related with the negative binomial distribution? – hkju Apr 13 '12 at 22:18
When you say "until at least $k$ is gotten", does that mean until a sum of previous rolls is at least $k$ or that an individual roll is at least $k$? – Henry Apr 13 '12 at 23:35
An individual roll is at least $k$. – hkju Apr 15 '12 at 18:54
The probability that any roll is greater than or equal to $k$ is $$\frac{13-k}{12}$$ so the expected number of rolls until a roll of $k$ or greater is $$\frac{12}{13-k}.$$ All but the last one of these rolls is less than $k$, so the sum of those rolls has an expected value of $$\left( \frac{12}{13-k} -1 \right) \frac{1+ (k-1)}{2}.$$ Add to this the expected value of the final roll $$\frac{k+12}{2}$$ and so the expectation of the sum is $$\left( \frac{12}{13-k} -1 \right) \frac{1+ (k-1)}{2} + \frac{k+12}{2} = \frac{78}{13-k}.$$
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I am wondering that the value we want is simply the sum of both expectations? – hkju Apr 13 '12 at 22:21
Yes, we can use the additivity of expectation here. – Matthew Conroy Apr 13 '12 at 22:22
Isn't it a little strange that you both get the same answer even though you claim to be answering different questions? – Michael Lugo Apr 13 '12 at 22:29
Didier edited later. – hkju Apr 13 '12 at 22:39
And now I've edited my answer since Didier is now answering the right question. – Matthew Conroy Apr 13 '12 at 23:03
Let $n=12$ denote the number of faces. If the first roll is $i\geqslant k$, $X=i$. If the first roll is $i\lt k$, $X=i+X'$ where $X'$ is distributed like $X$. Hence, $$\mathrm E(X)=\frac1n\sum_{i\geqslant k}i+\frac1n\sum_{i\lt k}\left(i+\mathrm E(X)\right)=\frac1n\sum_{i=1}^ni+\frac1n\mathrm E(X)\sum_{i=1}^{k-1}1,$$ that is, $$n\mathrm E(X)=\frac{n(n+1)}2+(k-1)\mathrm E(X),$$ hence $$\mathrm E(X)=\frac{n(n+1)}{2(n-k+1)}=\frac{78}{13-k}.$$
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the sum includes last rolling. – hkju Apr 13 '12 at 21:42
Doesn't it depend on $k$? – hkju Apr 13 '12 at 22:04
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# Money Saving Riddle: When Will the Man Save \$ 60?
Are you good at saving money or calculating money matters?
Try this riddle where you will get some tips about money saving.
On 1st of February a man earns \$ 20.
Out of which, he spends \$ 15 on 2nd day.
He again earns \$ 20 on 3rd day and spends \$ 15 on 4th day.
If he saves like this,
On which day will he have \$ 60 in hand?
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By Earning & Spending he would have Rs.5 every 2 days.
So by the 16th day he would have saved Rs.40.
Then on 17th day he would earn Rs. 20 more making it Rs. 60 on the 17th day.
So answer = The man will have Rs. 60 in his hand for the first time on the 17th day.
=======================
## Very interesting maths riddle! Answer in the description
Here is our interesting maths riddle for you. Joe was asked how old he was. He replied: In two years I will be twice as old as I was five years ago. So how old is he? Don’t forget to share your answers in the comment’s section!
Did we make your brain work too much? GOOD! that’s our intention, help keep you thinking. Let’s see if you got it right.
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# 1.4: Bell Numbers
[ "article:topic", "partition", "Bell numbers", "authorname:guichard" ]
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We begin with a definition:
Definition: Partition
partition of a set $$S$$ is a collection of non-empty subsets $$A_i\subseteq S$$, $$1\le i\le k$$ (the parts of the partition), such that $$\bigcup_{i=1}^k A_i=S$$ and for every $$i\not=j$$, $$A_i\cap A_j=\emptyset$$.
Example $$\PageIndex{1}$$:
The partitions of the set $$\{a,b,c\}$$ are $$\{\{a\},\{b\},\{c\}\}$$, $$\{\{a,b\},\{c\}\}$$, $$\{\{a,c\},\{b\}\}$$, $$\{\{b,c\},\{a\}\}$$, and $$\{\{a,b,c\}\}$$.
Partitions arise in a number of areas of mathematics. For example, if $$\equiv$$ is an equivalence relation on a set $$S$$, the equivalence classes of $\equiv$ form a partition of $S$. Here we consider the number of partitions of a finite set $S$, which we might as well take to be $$[n]=\{1,2,3,\ldots,n\}$$, unless some other set is of interest. We denote the number of partitions of an $n$-element set by $B_n$; this numbers are the Bell numbers. From the example above, we see that $B_3=5$. For convenience we let $B_0=1$. It is quite easy to see that $B_1=1$ and $B_2=2$.
There are no known simple formulas for $$B_n$$, so we content ourselves with a recurrence relation.
Theorem 1.4.3: Bell numbers
The Bell numbers satisfy
$$B_{n+1} = \sum_{k=0}^n {n\choose k} B_k.$$
Proof
Consider a partition of $S=\{1,2,\ldots,n+1\}$, $A_1$,…,$A_m$. We may suppose that $n+1$ is in $A_1$, and that $|A_1|=k+1$, for some $k$, $0\le k\le n$. Then $A_2$,…,$A_{m}$ form a partition of the remaining $n-k$ elements of $S$, that is, of $S\backslash A_1$. There are $B_{n-k}$ partitions of this set, so there are $B_{n-k}$ partitions of $S$ in which one part is the set $A_1$. There are ${n\choose k}$ sets of size $k+1$ containing $n+1$, so the total number of partitions of $S$ in which $n+1$ is in a set of size $k+1$ is ${n\choose k}B_{n-k}$. Adding up over all possible values of $k$, this means \eqalignno{ B_{n+1} &= \sum_{k=0}^n {n\choose k} B_{n-k}.& (1.4.1)\cr } We may rewrite this, using theorem 1.3.3, as $$B_{n+1} = \sum_{k=0}^n {n\choose n-k} B_{n-k},$$ and then notice that this is the same as the sum $$B_{n+1} = \sum_{k=0}^n {n\choose k} B_k,$$ written backwards.
$$\square$$
Example $$\PageIndex{1}$$:
We apply the recurrence to compute the first few Bell numbers:
\eqalign{ B_1&=\sum_{k=0}^0 {0\choose 0}B_0 = 1\cdot 1 = 1\cr B_2&=\sum_{k=0}^1 {1\choose k}B_k = {1\choose 0}B_0 + {1\choose 1}B_1 = 1\cdot 1+1\cdot 1 =1+1 =2\cr B_3&=\sum_{k=0}^2 {2\choose k}B_k = 1\cdot 1 + 2\cdot 1 + 1\cdot 2 = 5\cr B_4&=\sum_{k=0}^3 {3\choose k}B_k = 1\cdot 1 + 3\cdot 1 + 3\cdot 2 + 1\cdot 5 = 15\cr }
The Bell numbers grow exponentially fast; the first few are 1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437.
The Bell numbers turn up in many other problems; here is an interesting example. A common need in some computer programs is to generate a random permutation of $1,2,3,\ldots,n$, which we may think of as a shuffle of the numbers, visualized as numbered cards in a deck. Here is an attractive method that is easy to program: Start with the numbers in order, then at each step, remove one number at random (this is easy in most programming languages) and put it at the front of the list of numbers. (Viewed as a shuffle of a deck of cards, this corresponds to removing a card and putting it on the top of the deck.) How many times should we do this? There is no magic number, but it certainly should not be small relative to the size of $n$. Let's choose $n$ as the number of steps.
We can write such a shuffle as a list of $n$ integers, $(m_1,m_2,\ldots,m_n)$. This indicates that at step $i$, the number $m_i$ is moved to the front.
Example $$\PageIndex{1}$$:
Let's follow the shuffle $(3,2,2,4,1)$: \eqalign{ (3)&:\quad 3 1 2 4 5\cr (2)&:\quad 2 3 1 4 5\cr (2)&:\quad 2 3 1 4 5\cr (4)&:\quad 4 2 3 1 5\cr (1)&:\quad 1 4 2 3 5\cr }
Note that we allow "do nothing'' moves, removing the top card and then placing it on top. The number of possible shuffles is then easy to count: there are $n$ choices for the card to remove, and this is repeated $n$ times, so the total number is $n^n$. (If we continue a shuffle for $m$ steps, the number of shuffles is $n^m$.) Since there are only $n!$ different permutations of $1,2,\ldots,n$, this means that many shuffles give the same final order.
Here's our question: how many shuffles result in the original order?
Example 1.4.6 These shuffles return to the original order: $(1,1,1,1,1)$, $(5,4,3,2,1)$, $(4,1,3,2,1)$.
Theorem 1.4.7
The number of shuffles of $[n]$ that result in the original sorted order is $$B_n$$.
Proof
Since we know that $B_n$ counts the number of partitions of $\{1,2,3,\ldots,n\}$, we can prove the theorem by establishing a 1–1 correspondence between the shuffles that leave the deck sorted and the partitions. Given a shuffle $(m_1,m_2,\ldots,m_n)$, we put into a single set all $i$ such that $m_i$ has a single value. For example, using the shuffle $(4,1,3,2,1)$, Since $m_2=m_5$, one set is $\{2,5\}$. All the other values are distinct, so the other sets in the partition are $\{1\}$, $\{3\}$, and $\{4\}$.
Note that every shuffle, no matter what the final order, produces a partition by this method. We are only interested in the shuffles that leave the deck sorted. What we now need is to show that each partition results from exactly one such shuffle.
Suppose we have a partition with $k$ parts. If a shuffle leaves the deck sorted, the last entry, $m_n$, must be 1. If the part containing $n$ is $A_1$, then it must be that $m_i=1$ if and only if $i\in A_1$. If $k=1$, then the only part contains all of $\{1,2,\ldots,n\}$, and the shuffle must be $(1,1,1,\ldots,1)$.
If $k>1$, the last move that is not 1 must be 2, since 2 must end up immediately after 1. Thus, if $j_2$ is the largest index such that $j_2\notin A_1$, let $A_2$ be the part containing $j_2$, and it must be that $m_i=2$ if and only if $i\in A_2$. We continue in this way: Once we have discovered which of the $m_i$ must have values $1,2,\ldots,p$, let $j_{p+1}$ be the largest index such that $j_{p+1}\notin A_1\cup\cdots\cup A_p$, let $A_{p+1}$ be the part containing $j_{p+1}$, and then $m_i=p+1$ if and only if $i\in A_{p+1}$. When $p=k$, all values $m_i$ have been determined, and this is the unique shuffle that corresponds to the partition.
$\square$
Example 1.4.8 Consider the partition $\{\{1,5\},\{2,3,6\},\{4,7\}\}$. We must have $m_7=m_4=1$, $m_6=m_3=m_2=2$, and $m_5=m_1=3$, so the shuffle is $(3,2,2,1,3,2,1)$.
Returning to the problem of writing a computer program to generate a partition: is this a good method? When we say we want a random permutation, we mean that we want each permutation to occur with equal probability, namely, $1/n!$. Since the original order is one of the permutations, we want the number of shuffles that produce it to be exactly $n^n/n!$, but $n!$ does not divide $n^n$, so this is impossible. The probability of getting the original permutation is $B_n/n^n$, and this turns out to be quite a bit larger than $1/n!$. Thus, this is not a suitable method for generating random permutations.
The recurrence relation above is a somewhat cumbersome way to compute the Bell numbers. Another way to compute them is with a different recurrence, expressed in the Bell triangle, whose construction is similar to that of Pascal's triangle: $$\matrix{ A_{1,1}\cr A_{2,1}&A_{2,2}\cr A_{3,1}&A_{3,2}&A_{3,3}\cr A_{4,1}&A_{4,2}&A_{4,3}&A_{4,4}\cr} \qquad\matrix{ 1\cr 1&2\cr 2&3&5\cr 5&7&10&15\cr}$$ The rule for constructing this triangle is: $A_{1,1}=1$; the first entry in each row is the last entry in the previous row; other entries are $A_{n,k}=A_{n,k-1}+A_{n-1,k-1}$; row $n$ has $n$ entries. Both the first column and the diagonal consist of the Bell numbers, with $A_{n,1}=B_{n-1}$ and $A_{n,n}=B_n$.
$A_{n,k}$ may be interpreted as the number of partitions of $\{1,2,\ldots,n+1\}$ in which $\{k+1\}$ is the singleton set with the largest entry in the partition. For example, $A_{3,2}=3$; the partitions of $3+1=4$ in which $2+1=3$ is the largest number appearing in a singleton set are $\{\{1\},\{2,4\},\{3\}\}$, $\{\{2\},\{1,4\},\{3\}\}$, and $\{\{1,2,4\},\{3\}\}$.
To see that this indeed works as advertised, we need to confirm a few things. First, consider $A_{n,n}$, the number of partitions of $\{1,2,\ldots,n+1\}$ in which $\{n+1\}$ is the singleton set with the largest entry in the partition. Since $n+1$ is the largest element of the set, all partitions containing the singleton $\{n+1\}$ satisfy the requirement, and so the $B_n$ partitions of $\{1,2,\ldots,n\}$ together with $\{n+1\}$ are exactly the partitions of interest, that is, $A_{n,n}=B_n$.
Next, we verify that under the desired interpretation, it is indeed true that $A_{n,k}=A_{n,k-1}+A_{n-1,k-1}$ for $k>1$. Consider a partition counted by $A_{n,k-1}$. This contains the singleton $\{k\}$, and the element $k+1$ is not in a singleton. If we interchange $k$ and $k+1$, we get the singleton $\{k+1\}$, and no larger element is in a singleton. This gives us partitions in which $\{k+1\}$ is a singleton and $\{k\}$ is not. Now consider a partition of $\{1,2,\ldots,n\}$ counted by $A_{n-1,k-1}$. Replace all numbers $j>k$ by $j+1$, and add the singleton $\{k+1\}$. This produces a partition in which both $\{k+1\}$ and $\{k\}$ appear. In fact, we have described how to produce each partition counted by $A_{n,k}$ exactly once, and so $A_{n,k}=A_{n,k-1}+A_{n-1,k-1}$.
Finally, we need to verify that $A_{n,1}=B_{n-1}$. We know that $A_{1,1}=1=B_0$. Now we claim that for $n>1$, $$A_{n,1}=\sum_{k=0}^{n-2}{n-2\choose k}A_{k+1,1}.$$ In a partition counted by $A_{n,1}$, 2 is the largest element in a singleton, so $\{n+1\}$ is not in the partition. Choose any $k\ge 1$ elements of $\{3,4,\ldots,n\}$ to form the set containing $n+1$. There are $A_{n-k-1,1}$ partitions of the remaining $n-k$ elements in which 2 is the largest element in a singleton. This accounts for ${n-2\choose k}A_{n-k-1,1}$ partitions of $\{1,2,\ldots,n+1\}$, or over all $k$: $$\sum_{k=1}^{n-2}{n-2\choose k}A_{n-k-1,1}= \sum_{k=1}^{n-2}{n-2\choose n-k-2}A_{n-k-1,1}= \sum_{k=0}^{n-3} {n-2\choose k}A_{k+1,1}.$$ We are missing those partitions in which 1 is in the part containing $n+1$. We may produce all such partitions by starting with a partition counted by $A_{n-1,1}$ and adding $n+1$ to the part containing 1. Now we have $$A_{n,1} = A_{n-1,1}+\sum_{k=0}^{n-3} {n-2\choose k}A_{k+1,1}= \sum_{k=0}^{n-2} {n-2\choose k}A_{k+1,1}.$$ Although slightly disguised by the shifted indexing of the $A_{n,1}$, this is the same as the recurrence relation for the $B_n$, and so $A_{n,1}=B_{n-1}$ as desired.
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# Find the Winning Strategy for a Coin Flip Game
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Your opponent specifies 3 successives results of tosses of a coin, e.g. HHT. You then specify another such result, e.g. THT. The winner is the person whose sequence appears first when a fair coin is tossed successively and independently. Find the strategy which will allow you, the second player, to win at least 2/3 of the time. More specifically, for each possible choice of your opponent, find the choice that gives you the maximum probability of winning and calculate that probability.
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Solution:
Let us consider a few cases for the successive and independent tosses. Assume that you opponent choose a sequence HHT.
First Case: If your opponent wins at the 3rd successive toss, then it is obvious that there is only one way to win, which is by having a pattern of HHT.
Second Case: If your opponent wins at the 4th successive toss, then there will be two possible patterns of HHHT and THHT.
Third Case: If your opponent wins at the 5th successive toss, then there will be four possible patterns of HHHHT, HTHHT, THHHT, and TTHHT.
Fourth Case: If your opponent wins at the 6th successive toss, then there will be seven possible patterns of HHHHHT, HTHHHT, HTTHHT, THHHHT, THTHHT, TTHHHT, and TTTHHT.
Consider closely to second, third, and fourth cases, if you chose a sequence THH, then you will reduce the winning chance of your opponent. More specifically, let begin with the ...
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\$2.49
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Playground Swings A playground swing was one of Tesla's favorite examples of a resonant system. It's easy to measure its natural frequency. Each time the swing moves forward and then returns to its starting position counts as one cycle. Using a stop watch determine the length of time a swing needs to complete say 20 cycles. Divide 20 cycles by the time and you have the swings frequency in cycles per second or Hertz (Hz). Since a swing is basically a pendulum it's possible to calculate its resonant or natural frequency using pendulum equations as follows: f = 1/2π (g/L)0.5 where: g = gravity constant = 9.8 m/s/s for Earth L = Length Note that the natural frequency of the swing is not influenced by the mass of the person in it. In other words' it makes no difference whether a swing has a large adult or a small child in it. It will have the about the same natural frequency. Slight differences can be caused by slightly different locations of the person's center of mass. This is located about two inches below the navel. When people are sitting the center of mass is in about the same place relative to the seat of the swing regardless of whether the person is an adult or a child. If a forcing function is applied to a swing at the natural frequency of the swing it will resonate. The amplitude of the swing will increase during each back and forth cycle. The forcing function can be provided by a second person pushing on the swing. In this case even a small child can make a large adult swing by pushing in sync with the swing's back and forth cycle. The forcing function can also be provided by the person in the swing. In this case the person in the swing shifts her center of mass very slightly by changing the position of her legs or torso. This creates a slight pushing force which makes the swing go higher and higher. It takes a very small force but it has to be timed perfectly. The big question is what keeps the swing from flying apart or spinning over the top of the swing's frame and subsequently killing its rider? After all, if it is a resonating system then it should be very dangerous to keep applying force in time with the swing's frequency. The answer is fairly simple. The equation given above is only good for small angles. When the swing goes beyond a certain height it is no longer possible for the person in it to apply the necessary small force in sync with the natural frequency because the natural frequency changes. In other words the motion of the system is naturally limited. Suggested Activities Visit a playground and measure the natural frequency of a swing. It should make very little difference whether the person in it is large or small. Attach a flimsy piece of thread to the person in the swing. Instruct them not to assist in making the swing move and then attempt to make the swing resonate by pulling on the thread without breaking it. If the the force is applied in time with the natural frequency of the swing it will make the swing resonate. Back to Contents
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# How do you calculate income elasticity of demand in calculus?
## How do you calculate income elasticity of demand in calculus?
How to Calculate Price Elasticity of Demand with Calculus
1. Take the partial derivative of Q with respect to P, ∂Q/∂P. For your demand equation, this equals –4,000.
2. Determine P0 divided by Q0. Because P is \$1.50, and Q is 2,000, P0/Q0 equals 0.00075.
3. Multiply the partial derivative, –4,000, by P0/Q0, 0.00075.
What is income elasticity of demand with example?
Income Elasticity of Demand (YED) is defined as the responsiveness of demand when a consumer’s income changes. For example, if a person experiences a 20% increase in income, the quantity demanded for a good increased by 20%, then the income elasticity of demand would be 20%/20% = 1. This would make it a normal good.
### Is demand elastic or inelastic calculus?
If \(E \lt 1\), we say demand is inelastic. In this case, raising prices increases revenue. If \(E \gt 1\), we say demand is elastic. In this case, raising prices decreases revenue.
How do you find the elasticity of a function?
In economics, the price elasticity of demand refers to the elasticity of a demand function Q(P), and can be expressed as (dQ/dP)/(Q(P)/P) or the ratio of the value of the marginal function (dQ/dP) to the value of the average function (Q(P)/P).
#### What is income elasticity of demand answer?
Income elasticity of demand is an economic measure of how responsive the quantity demand for a good or service is to a change in income. The formula for calculating income elasticity of demand is the percent change in quantity demanded divided by the percent change in income.
Is a demand of inelastic?
Inelastic demand is when a buyer’s demand for a product does not change as much as its change in price. When the price increases, people will still purchase roughly the same amount of goods or services as they did before the increase because their needs stay the same.
## What are the factors affecting income elasticity of demand?
5 Factors Affecting the Price Elasticity of Demand
• Nature or type of Good. The Elasticity of Demand for a good is affected by its nature.
• Availability of Substitutes. The Price Elasticity of Demand for a good, with a large number of substitutes available, is very high.
• Price Level.
• Income Levels.
• Time Period.
What is the formula for income elasticity of demand?
Income elasticity of demand refers to the sensitivity of the quantity demanded for a certain good to a change in real income of consumers who buy this good, keeping all other things constant. The formula for calculating income elasticity of demand is the percent change in quantity demanded divided by the percent change in income.
### How to calculate the elasticity of demand for cheap garments?
Therefore, the income elasticity of demand for cheap garments is -0.92, i.e. it is an inferior good. The formula for income elasticity of demand can be derived by using the following steps: Step 1: Firstly, determine the initial real income and the quantity demanded at that income level that are denoted by I 0 and D 0 respectively.
How is Julie’s elasticity of demand inelastic?
Julie’s elasticity of demand is inelastic, since it is less than 1. Problem : If Neil’s elasticity of demand for hot dogs is constantly 0.9, and he buys 4 hot dogs when the price is \$1.50 per hot dog, how many will he buy when the price is \$1.00 per hot dog?
#### How is the elasticity of demand related to other variables?
In economics, demand elasticity refers to how sensitive the demand for a good is to changes in other economic variables. The cross elasticity of demand measures the responsiveness in the quantity demanded of one good when the price changes for another good.
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# How do you write an equation of a line with points (-2,-4), (2,4)?
Feb 12, 2017
$\left(y + \textcolor{red}{4}\right) = \textcolor{b l u e}{2} \left(x + \textcolor{red}{2}\right)$
Or
$\left(y - \textcolor{red}{4}\right) = \textcolor{b l u e}{2} \left(x - \textcolor{red}{2}\right)$
Or
$y = 2 x$
#### Explanation:
he point-slope formula can be used to find the equation. First we must determine the slope. The slope can be found by using the formula: $m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$
Where $m$ is the slope and ($\textcolor{b l u e}{{x}_{1} , {y}_{1}}$) and ($\textcolor{red}{{x}_{2} , {y}_{2}}$) are the two points on the line.
Substituting the values from the points in the problem gives:
$m = \frac{\textcolor{red}{4} - \textcolor{b l u e}{- 4}}{\textcolor{red}{2} - \textcolor{b l u e}{- 2}}$
$m = \frac{\textcolor{red}{4} + \textcolor{b l u e}{4}}{\textcolor{red}{2} + \textcolor{b l u e}{2}} = \frac{8}{4} = 2$
The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.
Substituting the slope we calculated and the first point gives:
$\left(y - \textcolor{red}{- 4}\right) = \textcolor{b l u e}{2} \left(x - \textcolor{red}{- 2}\right)$
$\left(y + \textcolor{red}{4}\right) = \textcolor{b l u e}{2} \left(x + \textcolor{red}{2}\right)$
We can also substitute the slope we calculated and the second point giving:
$\left(y - \textcolor{red}{4}\right) = \textcolor{b l u e}{2} \left(x - \textcolor{red}{2}\right)$
We can also solve this equation for $y$ to put the equation in the familiar slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$
Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.
$y - \textcolor{red}{4} = \left(\textcolor{b l u e}{2} \times x\right) - \left(\textcolor{b l u e}{2} \times \textcolor{red}{2}\right)$
$y - \textcolor{red}{4} = 2 x - 4$
$y - \textcolor{red}{4} + 4 = 2 x - 4 + 4$
$y - 0 = 2 x - 0$
$y = 2 x$
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https://math.stackexchange.com/questions/24521/how-do-i-come-up-with-a-function-to-count-a-pyramid-of-apples?noredirect=1
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How do I come up with a function to count a pyramid of apples?
My algebra book has a quick practical example at the beginning of the chapter on polynomials and their functions. Unfortunately it just says "this is why polynomial functions are important" and moves on. I'd like to know how to come up with the function (the teacher considers this out of scope). Even suggesting google search terms to find out more information on this sort of thing would be helpful.
The example
Consider a stack of apples, in a pyramid shape. It starts with one apple at the top. The next layer is 2x2, then 3x3, and so on. How many apples are there, given x number of layers?
The polynomial function
$$f(x) = \frac{2x^3 + 3x^2 + x}{6}$$
What I do and don't understand
Thanks to @DJC, I now know this is a standard function to generate Square Pyramidal Numbers, which is part of Faulhaber's formula. Faulhaber's formula appears to be about quickly adding sequential coefficients which all have the same exponent. Very cool. But how does one get from:
$$\sum_{k=1}^{n} k^p$$
to the spiffy function above? If I'm sounding stupid, how do I make the question better?
Fwiw, I'm in intermediate algebra in the USA. The next course would be Trigonometry or Calculus. (to help people place my current knowledge level)
The first layer has $1^2$ elements; the second has $2^2$ elements; the next has $3^2$ elements, and so on.
So this is a matter of finding the formula for $$1^2 + 2^2 + 3^2 + \cdots + n^2$$ in terms of $n$.
For squares, this is a well known formula: $1^2+2^2+\cdots +n^2 = \frac{n(n+1)(2n+1)}{6}$.
Note that $$\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} = \frac{2n^3 + 3n^2 + n}{6} = \frac{n(2n^2+3n+1)}{6} = \frac{n(2n+1)(n+1)}{6}$$ same formula.
Of course, you might be more interested in how to figure out the formulas in the first place. How does one express $1^k + 2^k + 3^k + \cdots + n^k$?
This is known as Faulhaber's formula. For the general case, $$1^p + 2^p + \cdots + n^p = \frac{1}{p+1}\sum_{j=0}^p(-1)^jB_j\binom{p+1}{j} n^{p+1-j}$$ where $B_m$ is the $m$th Bernoulli number.
For $p=2$, you get the spiffy formula with: \begin{align*} 1^2+\cdots+n^2 &= \frac{1}{3}\left(\binom{3}{0}B_0n^3 - \binom{3}{1}B_1n^2 + \binom{3}{2}B_2n - \binom{3}{3}B_3\right)\\ &= \frac{1}{3}\left(B_0n^3 - 3B_1n^2 + 3B_2n - B_3\right)\\ &= \frac{1}{3}\left(n^3 - 3\left(-\frac{1}{2}\right)n^2 + 3\left(\frac{1}{6}\right)n - 0\right)\\ &= \frac{1}{3}n^3 +\frac{1}{2}n^2 + \frac{1}{6}n\\ &= \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}, \end{align*} using the fact that $B_0 = 1$ $B_1 = -\frac{1}{2}$, $B_2 = \frac{1}{6}$, and $B_3=0$ (in fact, $B_n = 0$ for all odd $n\gt 1$).
• This answer provides a lot of things for me to research. I'm trying to understand Bernoulli's numbers and Faulhaber's formula, and binomial coefficients (first encounter with those..). While your and Paul's answers I think are good, I'd feel wrong marking one of them as The Answer until I actually understand it, even though it has more to do with my lack of knowledge. – djeikyb Mar 3 '11 at 10:40
You can prove the formula works using induction. I'm not sure what level you're at mathematically, so don't take this the wrong way if you have heard it... but if you haven't heard it, it's the following fact. If $P(n)$ is some true/false statement based on a natural number $n$, and $P(1)$ is true, and whenever $P(k)$ is true then $P(k+1)$ is also true, then $P(n)$ is true for all natural numbers $n$.
In this example, $P(1)$ is the statement that $1=\frac{1}{3}+\frac{1}{2}+\frac{1}{6}$, obviously true. Now suppose that $P(k)$ is true, that is, that $f(k)$ is known to be $\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$. Then $f(k+1)$ is just the $k$th plus $(k+1)^2=k^2+2k+1$. This is
$\displaystyle \frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}+k^2+2k+1=\frac{2k^3+3k^2+k+6k^2+12k+6}{6}$ $\displaystyle =\frac{2k^3+9k^2+13k+6}{6}=\frac{(2k^3+6k^2+6k+2)+3k^2+6k+3)+(k+1)}{6}$ $\displaystyle =\frac{2(k+1)^3+3(k+1)^2+(k+1)}{6}=\frac{(k+1)^3}{3}+\frac{(k+1)^2}{2}+\frac{(k+1)}{6}.$
How you actually come up with a formula like this in the first place is a different matter. "Watch for patterns" is really what it boils down to, but there are a couple shortcuts. The rule is simple enough that you can guess it's a polynomial. The numbers you're adding to get from one stage to the next are squares, so you can guess that it's a cubic. One way to check this is the "method of common differences." Your sequence is
$\displaystyle 0,1,5,14,30,55,\dotsc$
so the differences between successive terms are
$\displaystyle 1,4,9,16,25,\dotsc$
and the differences between successive terms of that sequence are
$\displaystyle 3,5,7,9,\dotsc$
and finally
$\displaystyle 2,2,2,\dotsc$
The final sequence is constant, the one above that is linear, the one above that is quadratic, and the one above that is cubic. So we have $f(n)=an^3+bn^2+cn+d$. Then by substituting different values for $n$ and solving the system of equations, we can find out what $a,b,c,$ and $d$ are. (It was obvious in this case what the first difference sequence was, but in other cases it may not be so obvious.)
I was going to conclude by noting that the $n$th triangular number is $\frac{(n)(n+1)}{1\cdot 2}$, and the $n$th pyramidal number is $\frac{(n)(n+1)(n+2)}{1\cdot 2\cdot 3}$, and guessing that things worked the same in higher dimensions. But as it turns out, they don't. Take that how you will.
• Hmm. I think this answers my question, but it's taking me a while to understand the p(n) and p(k+1) stuff. Although I've some appreciation now for why it's considered outside the class's scope. – djeikyb Mar 3 '11 at 10:32
• Hmm, perhaps I stated it too formally. The basic principle is this. You can easily see that your formula holds for $n=1$. What I proved above is that, if we know it holds for a specific value of $n$, it also holds for the next value. So since it holds for $1$, it holds for $2$... and since it holds for $2$, it holds for $3$... and so on, until it "proves itself" for every value of $n$. (This is intuitive, but be warned that it isn't very formal.) – Paul VanKoughnett Mar 3 '11 at 18:04
This formula is derived in many different ways in the book Concrete Mathematics. For example, one method which hasn't been mentioned yet here is to use "discrete calculus", which easily lets you compute sums of "falling factorial powers" $k^{\underline{m}} = k(k-1)(k-2) \dots (k-m+1)$ almost as when you integrate ordinary powers $x^m$. Upon writing $k^2 = k(k-1)+k = k^{\underline{2}} + k^{\underline{1}}$, your sum becomes $$\sum_{0 \le k < n+1} (k^{\underline{2}} + k^{\underline{1}}) = \left[ \frac{k^{\underline{3}}}{3} + \frac{k^{\underline{2}}}{2} \right]_{k=0}^{n+1} = \frac{(n+1)n(n-1)}{3} + \frac{(n+1)n}{2} = \frac{(n+1)n(2n+1)}{6},$$ which is your polynomial $f(n)$. (See the book for explanation of why it this works.)
• Intriguing. I don't expect you to dispense a course's worth of info, but what is the thing that looks like a minus sign next to the k, underneath the elevated two (and one, and three)? – djeikyb Mar 3 '11 at 10:49
• @djeikyb: The exponent is underlined, to show that it's not an ordinary power like $17^4 = 17 \cdot 17 \cdot 17 \cdot 17$, but instead a "falling power" like $17^{\underline{4}} = 17 \cdot 16 \cdot 15 \cdot 14$. Overlined exponents denote "rising powers" like $17^{\overline{4}} = 17 \cdot 18 \cdot 19 \cdot 20$. There are also other notations in use for this. See for example this Wikipedia article: en.wikipedia.org/wiki/Pochhammer_symbol. – Hans Lundmark Mar 3 '11 at 15:37
You can prove it works by induction. It works for $x=1$, as $f(x)=1$. So assume it works for $x: f(x) = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{6}$ and we want to show it works for $x+1$: \begin{align} f(x+1)&=f(x)+(x+1)^2 \\ &=\frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{6} + (x+1)^2 \\ &=\frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{6} +\frac{3x^2+3x+1}{3}+\frac{2x+1}{2}+\frac{1}{6} \\& =\frac{(x+1)^3}{3} + \frac{(x+1)^2}{2} + \frac{x+1}{6}\end{align}
Ok so let me so something similar to Carsten.
Consider expanding $(k+1)^3 - k^3 = 3k^2 + 3k + 1$.
Now if we were to sum both sides over $k=1,2,...,n$ then we would find that on the left most of the terms cancel, e.g. $(5^3 - 4^3) + (4^3 - 3^3) + ... + (2^3 - 1^3) = 5^3 - 1^3$. On the right you have sums you already have a formula for along with the one you are after.
So:
$(n+1)^3 - 1 = 3\sum_{k=1}^{n} k^2 + 3\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1$
i.e. $\sum_{k=1}^{n} k^2 = \frac{1}{3}((n+1)^3 - 3(\sum_{k=1}^{n} k) - n - 1) = \frac{1}{3}((n+1)^3 - 3\frac{(n)(n+1)}{2} - n-1) = ... = \frac{n(n+1)(2n+1)}{6}$
We can also solve this problem by finding a recurrence formula. First assume your solution is going to be a polynomial in the following form $$An^4 +Bn^3+Cn^2+Dn+En^0$$ where $A,B,C,D,E$ are the coefficients for which we are solving. We will solve it using linear algebra by setting up the matrix below.
\begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\[0.3em] 16 & 8 & 4&2&1&5 \\[0.3em] 81&27&9&3&1&14 \\[0.3em] 256&64&16&4&1&30 \\[0.3em] 625&125&25&5&1&55 \end{bmatrix}
Label each row as $a_1, a_2, a_3, a_4,a_5$ where each subscript corresponds to $n$ in the polynomial and each column, except for column six, corresponds to the exponents in the polynomial. So you can see that the first column is $1^4, 2^4, 3^4, 4^4, 5^4$. The second column is $1^3, 2^3, 3^3, 4^3, 5^3$, The third column is $1^2, 2^2, 3^2, 4^2, 5^2$. The fourth column is $1^1, 2^1, 3^1, 4^1, 5^1$. The fifth column is $1^0, 2^0, 3^0, 4^0, 5^0$. Column six represents the cumulative amounts in each row through five rows of the square pyramid. Solving the matrix will result in the following matrix
\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\[0.3em] 0 & 1 & 0&0&0&\frac{1}{3} \\[0.3em] 0&0&1&0&0&\frac{1}{2} \\[0.3em] 0&0&0&1&0&\frac{1}{6} \\[0.3em] 0&0&0&0&1&0 \end{bmatrix}
Based on the solution to the matrix our final polynomial is $$a(n) = 0n^4 + \frac{1}{3}n^3 + \frac{1}{2}n^2 +\frac{1}{6}n +0n^0$$ where $A=0, B =\frac{1}{3}, C=\frac{1}{2}, D=\frac{1}{6}, E=0$
It is evident from the result that we did not need to start with the polynomial we started with an a $5$ x $6$ matrix. We could have started with $An^3 +Bn^3 +Cn^2 +Dn^1+ En^0$ and a $4$ x $5$ matrix. The final result let us know that when the coefficient $A$ turned out to be $0$.
If you wanted the formula for the tetrahedral pyramid of apples just replace the last column in the original matrix with $1, 4, 10, 20, 35$, the cumulative number of apples for row one through row five and solve.
You have gotten lots of good answers here. I'd say that you'd profit very much from trying to understand Hans' answer, and maybe even check out the book he mentions from the library to see if there are things in it that you can understand at your level. Let me just add a very brief summary of Paul's answer.
First convince yourself that claiming that $$\sum_{k=1}^n k^2=f(n)\quad\text{for all n\in\mathbb N}$$ is actually the same as claiming $$f(0)=0\qquad\text{and}\qquad f(n+1)-f(n)=(n+1)^2\quad\text{for all n\in\mathbb N.}$$ I hope that this makes sense after you have thought about it for a while. Now as soon as it does, you do at least know how to check the formula given by your book: Just compute $f(0)$ and $f(n+1)-f(n)$.
Now how do you come up with such a formula? Well, if you already know or suspect that $f(n)$ is a polynomial of degree three (which is not unreasonable, since it is related to the volume of the pyramid, you can set $$f(x)=ax^3+bx^2+cx+d$$ and then figure out what $a,b,c,d$ have to be for $f(0)=0$ and $f(n+1)-f(n)=(n+1)^2$ to hold. If you know systems of linear equations then it should not be to difficult. Otherwise, here is another approach. Instead set $$f(x)=rx(x-1)(x-2)+sx(x-1)+tx+u.$$ (Compare with Hans' answer.) We know that we must have $f(0)=1$, $f(1)=1$, $f(2)=5$, $f(3)=14$. Now we get \begin{align*} f(0)=0\quad&\Longrightarrow\quad u=0\\ f(1)=1\quad&\Longrightarrow\quad t+u=1\quad\Longrightarrow\quad t=1\\ f(2)=5\quad&\Longrightarrow\quad 2s+2t+u=5\quad\Longrightarrow\quad s=\frac32\\ f(3)=14\quad&\Longrightarrow\quad 6r+6s+3t+u=14\quad\Longrightarrow\quad r=\frac13\\ \end{align*} (I hope my calculations are correct.) Of course, if you do want to take on faith that a polynomial of degree $3$ works, you will still have to verify that $f(n+1)-f(n)=(n+1)^2$.
I hope that this was at a level accessible to you.
The binomial coefficient identities $$\sum_{k=0}^n\binom k1=\binom{n+1}2,$$ $$\sum_{k=0}^n\binom k2=\binom{n+1}3,$$ $$\sum_{k=0}^n\binom k3=\binom{n+1}4,$$ etc. are simple and natural. To derive the sum-of-squares formula you asked about, observe that $$\binom{n+1}3=\sum_{k=0}^n\binom k2=\sum_{k=0}^n\frac{k^2-k}2=\frac12\sum_{k=0}^nk^2-\frac12\sum_{k=0}^nk=\frac12\sum_{k=0}^nk^2-\frac12\binom{n+1}2,$$ whence $$\sum_{k=0}^nk^2=2\binom{n+1}3+\binom{n+1}2=\frac{n(n+1)(2n+1)}6.$$
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1. ## Trying to Understand
I have searched far and wide for a suitable explaination without much success so hoping I can find the answer here.
With conditional formatting assuming my work area is B11 to R41, when creating a formula I would write \$B11=2 then in the box to cover where the formula is valid it would be =\$B\$11:\$R\$41
What I would like to know is if I enter a "2" in for example B33 then the formatting works for that row, which part of the formula is doing this? How does B33 relate to just that row for example and why is it not defined in the formula box ( it just refers to =\$B11=2?
Thanks
Alan
2. Allen,
The behavior you see is cause by the use of the MIXED reference in your formula, e.g. \$B11.
The B part is FIXED by the \$ however the 11 (row) part is not so it will change to operate for each row.
cf1.JPG
Notice how it operates for row 12 with a value of 2 in B12.
If you change the formula to =\$B\$11=2
cf2.JPG
Notice how now since the reference is fixed on both parts (or an Absolute Reference in Excel speak) nothing is formatted because \$b\$11 is NOT equal to 2. However if \$B\$11 is set to 2 all cells will be formatted.
cf3.JPG
HTH
3. ## The Following User Says Thank You to RetiredGeek For This Useful Post:
4. To expand on RG's comments. It is about Absolute vs relative references in cells. One way to look at it would be to imagine that you would copy the formula:
=\$B11=2
into the range
=\$B\$11:\$R\$41
and imagine what the formulas would be [ignoring that they would be circular references]
Since the Column B is Absolute [=preceded by a dollar sign(\$B)] but the row is relative [not preceded by a dollar sign] in the cells in row 11 B11:R11] the formula would amount to:
=B11=2
So for any cell in row11, if B would be 2 they would all be "TRUE" and meet the condition.
That is the same for each row. Row 40 for example in Cells B40:R40 would all have:
=B40=2
So again they are all comparing to the value in col B for that row.
Each cell can have 1 of 4 abs/rel combinations [Tip: when entering formulas the <F4> key toggles through the 4 possibilities:]
=B11
=\$B\$11
=B\$11
=\$B11
If you had used B11 and copied the [=B11=2] into the B11:R41 cells, the cond format would be true when a particular cell had the value of 2
If you had used [=\$B\$11=2] into the B11:R41 cells, the cond format would be true for all the cells when B11 equals 2
If you had used [=B\$11=2] into the B11:R41 cells, the cond format would be true for any cells in the column when the value in row 11 equals 2
And to reiterate your example, when you had used [=\$B11=2] into the B11:R41 cells, the cond format would be true for any cells in the row when the value in Col B equals 2
Hope this helps,
Steve
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In linear algebra, the transpose of a linear map between two vector spaces, defined over the same field, is an induced map between the dual spaces of the two vector spaces. The transpose or algebraic adjoint of a linear map is often used to study the original linear map. This concept is generalised by adjoint functors.
## Definition
Let ${\displaystyle X^{\#))$ denote the algebraic dual space of a vector space ${\displaystyle X.}$ Let ${\displaystyle X}$ and ${\displaystyle Y}$ be vector spaces over the same field ${\displaystyle {\mathcal {K)).}$ If ${\displaystyle u:X\to Y}$ is a linear map, then its algebraic adjoint or dual,[1] is the map ${\displaystyle {}^{\#}u:Y^{\#}\to X^{\#))$ defined by ${\displaystyle f\mapsto f\circ u.}$ The resulting functional ${\displaystyle {}^{\#}u(f):=f\circ u}$ is called the pullback of ${\displaystyle f}$ by ${\displaystyle u.}$
The continuous dual space of a topological vector space (TVS) ${\displaystyle X}$ is denoted by ${\displaystyle X^{\prime }.}$ If ${\displaystyle X}$ and ${\displaystyle Y}$ are TVSs then a linear map ${\displaystyle u:X\to Y}$ is weakly continuous if and only if ${\displaystyle {}^{\#}u\left(Y^{\prime }\right)\subseteq X^{\prime },}$ in which case we let ${\displaystyle {}^{t}u:Y^{\prime }\to X^{\prime ))$ denote the restriction of ${\displaystyle {}^{\#}u}$ to ${\displaystyle Y^{\prime }.}$ The map ${\displaystyle {}^{t}u}$ is called the transpose[2] or algebraic adjoint of ${\displaystyle u.}$ The following identity characterizes the transpose of ${\displaystyle u}$:[3] ${\displaystyle \left\langle {}^{t}u(f),x\right\rangle =\left\langle f,u(x)\right\rangle \quad {\text{ for all ))f\in Y^{\prime }{\text{ and ))x\in X,}$ where ${\displaystyle \left\langle \cdot ,\cdot \right\rangle }$ is the natural pairing defined by ${\displaystyle \left\langle z,h\right\rangle :=z(h).}$
## Properties
The assignment ${\displaystyle u\mapsto {}^{t}u}$ produces an injective linear map between the space of linear operators from ${\displaystyle X}$ to ${\displaystyle Y}$ and the space of linear operators from ${\displaystyle Y^{\#))$ to ${\displaystyle X^{\#}.}$ If ${\displaystyle X=Y}$ then the space of linear maps is an algebra under composition of maps, and the assignment is then an antihomomorphism of algebras, meaning that ${\displaystyle {}^{t}(uv)={}^{t}v{}^{t}u.}$ In the language of category theory, taking the dual of vector spaces and the transpose of linear maps is therefore a contravariant functor from the category of vector spaces over ${\displaystyle {\mathcal {K))}$ to itself. One can identify ${\displaystyle {}^{t}\left({}^{t}u\right)}$ with ${\displaystyle u}$ using the natural injection into the double dual.
• If ${\displaystyle u:X\to Y}$ and ${\displaystyle v:Y\to Z}$ are linear maps then ${\displaystyle {}^{t}(v\circ u)={}^{t}u\circ {}^{t}v}$[4]
• If ${\displaystyle u:X\to Y}$ is a (surjective) vector space isomorphism then so is the transpose ${\displaystyle {}^{t}u:Y^{\prime }\to X^{\prime }.}$
• If ${\displaystyle X}$ and ${\displaystyle Y}$ are normed spaces then
${\displaystyle \|x\|=\sup _{\|x^{\prime }\|\leq 1}\left|x^{\prime }(x)\right|\quad {\text{ for each ))x\in X}$ and if the linear operator ${\displaystyle u:X\to Y}$ is bounded then the operator norm of ${\displaystyle {}^{t}u}$ is equal to the norm of ${\displaystyle u}$; that is[5][6] ${\displaystyle \|u\|=\left\|{}^{t}u\right\|,}$ and moreover, ${\displaystyle \|u\|=\sup \left\{\left|y^{\prime }(ux)\right|:\|x\|\leq 1,\left\|y^{*}\right\|\leq 1{\text{ where ))x\in X,y^{\prime }\in Y^{\prime }\right\}.}$
### Polars
Suppose now that ${\displaystyle u:X\to Y}$ is a weakly continuous linear operator between topological vector spaces ${\displaystyle X}$ and ${\displaystyle Y}$ with continuous dual spaces ${\displaystyle X^{\prime ))$ and ${\displaystyle Y^{\prime },}$ respectively. Let ${\displaystyle \langle \cdot ,\cdot \rangle :X\times X^{\prime }\to \mathbb {C} }$ denote the canonical dual system, defined by ${\displaystyle \left\langle x,x^{\prime }\right\rangle =x^{\prime }x}$ where ${\displaystyle x}$ and ${\displaystyle x^{\prime ))$ are said to be orthogonal if ${\displaystyle \left\langle x,x^{\prime }\right\rangle =x^{\prime }x=0.}$ For any subsets ${\displaystyle A\subseteq X}$ and ${\displaystyle S^{\prime }\subseteq X^{\prime },}$ let ${\displaystyle A^{\circ }=\left\{x^{\prime }\in X^{\prime }:\sup _{a\in A}\left|x^{\prime }(a)\right|\leq 1\right\}\qquad {\text{ and ))\qquad S^{\circ }=\left\{x\in X:\sup _{s^{\prime }\in S^{\prime ))\left|s^{\prime }(x)\right|\leq 1\right\))$ denote the (absolute) polar of ${\displaystyle A}$ in ${\displaystyle X^{\prime ))$ (resp. of ${\displaystyle S^{\prime ))$ in ${\displaystyle X}$).
• If ${\displaystyle A\subseteq X}$ and ${\displaystyle B\subseteq Y}$ are convex, weakly closed sets containing the origin then ${\displaystyle {}^{t}u\left(B^{\circ }\right)\subseteq A^{\circ ))$ implies ${\displaystyle u(A)\subseteq B.}$[7]
• If ${\displaystyle A\subseteq X}$ and ${\displaystyle B\subseteq Y}$ then[4]
${\displaystyle [u(A)]^{\circ }=\left({}^{t}u\right)^{-1}\left(A^{\circ }\right)}$ and ${\displaystyle u(A)\subseteq B\quad {\text{ implies ))\quad {}^{t}u\left(B^{\circ }\right)\subseteq A^{\circ }.}$
• If ${\displaystyle X}$ and ${\displaystyle Y}$ are locally convex then[5]
${\displaystyle \operatorname {ker} {}^{t}u=\left(\operatorname {Im} u\right)^{\circ }.}$
### Annihilators
Suppose ${\displaystyle X}$ and ${\displaystyle Y}$ are topological vector spaces and ${\displaystyle u:X\to Y}$ is a weakly continuous linear operator (so ${\displaystyle \left({}^{t}u\right)\left(Y^{\prime }\right)\subseteq X^{\prime ))$). Given subsets ${\displaystyle M\subseteq X}$ and ${\displaystyle N\subseteq X^{\prime },}$ define their annihilators (with respect to the canonical dual system) by[6]
{\displaystyle {\begin{alignedat}{4}M^{\bot }:&=\left\{x^{\prime }\in X^{\prime }:\left\langle m,x^{\prime }\right\rangle =0{\text{ for all ))m\in M\right\}\\&=\left\{x^{\prime }\in X^{\prime }:x^{\prime }(M)=\{0\}\right\}\qquad {\text{ where ))x^{\prime }(M):=\left\{x^{\prime }(m):m\in M\right\}\end{alignedat))}
and
{\displaystyle {\begin{alignedat}{4}{}^{\bot }N:&=\left\{x\in X:\left\langle x,n^{\prime }\right\rangle =0{\text{ for all ))n^{\prime }\in N\right\}\\&=\left\{x\in X:N(x)=\{0\}\right\}\qquad {\text{ where ))N(x):=\left\{n^{\prime }(x):n^{\prime }\in N\right\}\\\end{alignedat))}
• The kernel of ${\displaystyle {}^{t}u}$ is the subspace of ${\displaystyle Y^{\prime ))$ orthogonal to the image of ${\displaystyle u}$:[7]
${\displaystyle \ker {}^{t}u=(\operatorname {Im} u)^{\bot ))$
• The linear map ${\displaystyle u}$ is injective if and only if its image is a weakly dense subset of ${\displaystyle Y}$ (that is, the image of ${\displaystyle u}$ is dense in ${\displaystyle Y}$ when ${\displaystyle Y}$ is given the weak topology induced by ${\displaystyle \operatorname {ker} {}^{t}u}$).[7]
• The transpose ${\displaystyle {}^{t}u:Y^{\prime }\to X^{\prime ))$ is continuous when both ${\displaystyle X^{\prime ))$ and ${\displaystyle Y^{\prime ))$ are endowed with the weak-* topology (resp. both endowed with the strong dual topology, both endowed with the topology of uniform convergence on compact convex subsets, both endowed with the topology of uniform convergence on compact subsets).[8]
• (Surjection of Fréchet spaces): If ${\displaystyle X}$ and ${\displaystyle Y}$ are Fréchet spaces then the continuous linear operator ${\displaystyle u:X\to Y}$ is surjective if and only if (1) the transpose ${\displaystyle {}^{t}u:Y^{\prime }\to X^{\prime ))$ is injective, and (2) the image of the transpose of ${\displaystyle u}$ is a weakly closed (i.e. weak-* closed) subset of ${\displaystyle X^{\prime }.}$[9]
### Duals of quotient spaces
Let ${\displaystyle M}$ be a closed vector subspace of a Hausdorff locally convex space ${\displaystyle X}$ and denote the canonical quotient map by ${\displaystyle \pi :X\to X/M\quad {\text{ where ))\quad \pi (x):=x+M.}$ Assume ${\displaystyle X/M}$ is endowed with the quotient topology induced by the quotient map ${\displaystyle \pi :X\to X/M.}$ Then the transpose of the quotient map is valued in ${\displaystyle M^{\bot ))$ and ${\displaystyle {}^{t}\pi :(X/M)^{\prime }\to M^{\bot }\subseteq X^{\prime ))$ is a TVS-isomorphism onto ${\displaystyle M^{\bot }.}$ If ${\displaystyle X}$ is a Banach space then ${\displaystyle {}^{t}\pi :(X/M)^{\prime }\to M^{\bot ))$ is also an isometry.[6] Using this transpose, every continuous linear functional on the quotient space ${\displaystyle X/M}$ is canonically identified with a continuous linear functional in the annihilator ${\displaystyle M^{\bot ))$ of ${\displaystyle M.}$
### Duals of vector subspaces
Let ${\displaystyle M}$ be a closed vector subspace of a Hausdorff locally convex space ${\displaystyle X.}$ If ${\displaystyle m^{\prime }\in M^{\prime ))$ and if ${\displaystyle x^{\prime }\in X^{\prime ))$ is a continuous linear extension of ${\displaystyle m^{\prime ))$ to ${\displaystyle X}$ then the assignment ${\displaystyle m^{\prime }\mapsto x^{\prime }+M^{\bot ))$ induces a vector space isomorphism ${\displaystyle M^{\prime }\to X^{\prime }/\left(M^{\bot }\right),}$ which is an isometry if ${\displaystyle X}$ is a Banach space.[6]
Denote the inclusion map by ${\displaystyle \operatorname {In} :M\to X\quad {\text{ where ))\quad \operatorname {In} (m):=m\quad {\text{ for all ))m\in M.}$ The transpose of the inclusion map is ${\displaystyle {}^{t}\operatorname {In} :X^{\prime }\to M^{\prime ))$ whose kernel is the annihilator ${\displaystyle M^{\bot }=\left\{x^{\prime }\in X^{\prime }:\left\langle m,x^{\prime }\right\rangle =0{\text{ for all ))m\in M\right\))$ and which is surjective by the Hahn–Banach theorem. This map induces an isomorphism of vector spaces ${\displaystyle X^{\prime }/\left(M^{\bot }\right)\to M^{\prime }.}$
## Representation as a matrix
If the linear map ${\displaystyle u}$ is represented by the matrix ${\displaystyle A}$ with respect to two bases of ${\displaystyle X}$ and ${\displaystyle Y,}$ then ${\displaystyle {}^{t}u}$ is represented by the transpose matrix ${\displaystyle A^{T))$ with respect to the dual bases of ${\displaystyle Y^{\prime ))$ and ${\displaystyle X^{\prime },}$ hence the name. Alternatively, as ${\displaystyle u}$ is represented by ${\displaystyle A}$ acting to the right on column vectors, ${\displaystyle {}^{t}u}$ is represented by the same matrix acting to the left on row vectors. These points of view are related by the canonical inner product on ${\displaystyle \mathbb {R} ^{n},}$ which identifies the space of column vectors with the dual space of row vectors.
## Relation to the Hermitian adjoint
The identity that characterizes the transpose, that is, ${\displaystyle \left[u^{*}(f),x\right]=[f,u(x)],}$ is formally similar to the definition of the Hermitian adjoint, however, the transpose and the Hermitian adjoint are not the same map. The transpose is a map ${\displaystyle Y^{\prime }\to X^{\prime ))$ and is defined for linear maps between any vector spaces ${\displaystyle X}$ and ${\displaystyle Y,}$ without requiring any additional structure. The Hermitian adjoint maps ${\displaystyle Y\to X}$ and is only defined for linear maps between Hilbert spaces, as it is defined in terms of the inner product on the Hilbert space. The Hermitian adjoint therefore requires more mathematical structure than the transpose.
However, the transpose is often used in contexts where the vector spaces are both equipped with a nondegenerate bilinear form such as the Euclidean dot product or another real inner product. In this case, the nondegenerate bilinear form is often used implicitly to map between the vector spaces and their duals, to express the transposed map as a map ${\displaystyle Y\to X.}$ For a complex Hilbert space, the inner product is sesquilinear and not bilinear, and these conversions change the transpose into the adjoint map.
More precisely: if ${\displaystyle X}$ and ${\displaystyle Y}$ are Hilbert spaces and ${\displaystyle u:X\to Y}$ is a linear map then the transpose of ${\displaystyle u}$ and the Hermitian adjoint of ${\displaystyle u,}$ which we will denote respectively by ${\displaystyle {}^{t}u}$ and ${\displaystyle u^{*},}$ are related. Denote by ${\displaystyle I:X\to X^{*))$ and ${\displaystyle J:Y\to Y^{*))$ the canonical antilinear isometries of the Hilbert spaces ${\displaystyle X}$ and ${\displaystyle Y}$ onto their duals. Then ${\displaystyle u^{*))$ is the following composition of maps:[10]
${\displaystyle Y{\overset {J}{\longrightarrow ))Y^{*}{\overset ((}^{\text{t))u}{\longrightarrow ))X^{*}{\overset {I^{-1)){\longrightarrow ))X}$
## Applications to functional analysis
Suppose that ${\displaystyle X}$ and ${\displaystyle Y}$ are topological vector spaces and that ${\displaystyle u:X\to Y}$ is a linear map, then many of ${\displaystyle u}$'s properties are reflected in ${\displaystyle {}^{t}u.}$
• If ${\displaystyle A\subseteq X}$ and ${\displaystyle B\subseteq Y}$ are weakly closed, convex sets containing the origin, then ${\displaystyle {}^{t}u\left(B^{\circ }\right)\subseteq A^{\circ ))$ implies ${\displaystyle u(A)\subseteq B.}$[4]
• The null space of ${\displaystyle {}^{t}u}$ is the subspace of ${\displaystyle Y^{\prime ))$ orthogonal to the range ${\displaystyle u(X)}$ of ${\displaystyle u.}$[4]
• ${\displaystyle {}^{t}u}$ is injective if and only if the range ${\displaystyle u(X)}$ of ${\displaystyle u}$ is weakly closed.[4]
## References
1. ^ Schaefer & Wolff 1999, p. 128.
2. ^ Trèves 2006, p. 240.
3. ^ Halmos (1974, §44)
4. Schaefer & Wolff 1999, pp. 129–130
5. ^ a b Trèves 2006, pp. 240–252.
6. ^ a b c d Rudin 1991, pp. 92–115.
7. ^ a b c Schaefer & Wolff 1999, pp. 128–130.
8. ^ Trèves 2006, pp. 199–200.
9. ^ Trèves 2006, pp. 382–383.
10. ^ Trèves 2006, p. 488.
## Bibliography
• Halmos, Paul (1974), Finite-dimensional Vector Spaces, Springer, ISBN 0-387-90093-4
• Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277.
• Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
• Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.
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# Factors of 70
Factors are numbers that can divide 70 without leaving a remainder. To find the factors of 70, you can divide 70 by different numbers and see which ones result in a whole number.
Enter Number
## What is the Factors of 70?
Factors of 70: 1, 2, 5, 7, 10, 14, 35, 70.
Explanation:
1 x 70 = 70, So 1 and 70 are factors.
2 x 35 = 70: So, 2 and 35 are factors.
5 x 14 = 70: So, 5 and 14 are factors.
7 x 10 = 70: So, 7 and 10 are factors.
### Factors Pairs of 70
Factor pairs are pairs of numbers that, when multiplied together, result in 70.
Factors Pairs of 70
(1, 70)
(2, 35)
(5, 14)
(7, 10)
Explanation:
Each pair, when multiplied together, equals 70.
For example, 2 x 35 = 70, so (2, 35) is a factor pair.
### Prime Factorization of 70
Prime factorization is breaking down a number into its prime number factors. Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves.
Prime Factorization of 70:
70 = 2 x 5 x 7
Explanation:
2, 5, and 7 are all prime numbers.
When multiplied together, they give 70: 2 x 5 x 7 = 70.
### Factor Tree of 70
`````` 70
/ \
2 35
/ \
5 7 ``````
More Factors:
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# How to Solve Linear Equations in Two Variables?
A linear equation in two variables is a linear equation in which there are two variables. In the following guide, you will learn more about linear equations in two variables and solving them.
We can recognize a linear equation in two variables that are expressed as $$ax+by+c=0$$, which consists of two variables $$x$$ and $$y$$ and the highest degree of the given equation is $$1$$.
## Step-by-step guide tolinear equations in two variables
The standard form of a two-variable linear equation is $$ax+ by+ c= 0$$ where $$x$$ and $$y$$ are the two variables. Graphical representation of linear equations in two variables consists of two straight lines that can be intersecting lines, parallel lines, or coincident lines.
### Forms of linear equations in two variables
A linear equation in two variables can be in different forms such as standard form, intercept form, and point-slope form.
### Methods for solving linear equations in two variables
There are four methods to solve a system of linear equations in two variables:
• Graphical Method
• Substitution Method
• Cross Multiplication Method
• Elimination Method
#### Graphical method
The steps for solving linear equations in two variables graphically are presented below:
• Step 1: To solve a system of two equations in two variables graphically, we graph each equation.
• Step 2: To graph an equation manually, first convert it to the form $$y=mx+b$$ by solving the equation for $$y$$.
• Step 3: Start putting the values of $$x$$ as $$0, 1, 2,$$ and so on and find the corresponding values of $$y$$, or vice-versa.
• Step 4: Identify the point where both lines meet.
• Step 5: The point of intersection is the solution of the given system.
But the two lines may not always intersect. Sometimes they may be parallel. In that case, the system of linear equations in the two variables has no answer. In some other cases, both lines coincide with each other. In that case, each point on that line is a solution to the given system and hence the given system has an infinite number of solutions. If the system has a solution, then it is said to be consistent; otherwise, it is said to be inconsistent.
#### Method of substitution
To solve a system of two linear equations in two variables using the substitution method, we must use the following steps:
• Step 1: Solve one of the equations for one variable.
• Step 2: Substitute this in another equation to get an equation based on a single variable.
• Step 3: Solve it for the variable.
• Step 4: Substitute it in any of the equations to get the value of another variable.
#### Method of elimination
To solve a system of linear equations in two variables using the elimination method, we use the following steps:
• Step 1: Arrange the equations in the standard form: $$ax+by+c=0$$ or $$ax+by=c$$.
• Step 2: Check if adding or subtracting the equations would result in the cancellation of a variable.
• Step 3: If not, multiply one or both equations by either the coefficient of $$x$$ or $$y$$ so that their addition or subtraction would result in the cancellation of any one of the variables.
• Step 4: Solve the resulting single variable equation.
• Step 5: Substitute it in any of the equations to get the value of another variable.
Note:
When solving equations using the substitution method or the elimination method:
• If we get an equation that is true (i.e., something like $$0 = 0$$, $$-1 = -1$$, etc), then it means that the system has an infinite number of solutions.
• If we get an equation that is false (i.e., something like $$0 = 2$$, $$3= -1$$, etc), then it means that the system has no solution.
### Linear Equations in Two Variables – Example 1:
Solve the following system of equations using the substitution method.
$$x+2y-7=0$$
$$2x-5y+13=0$$
Solution:
First, solve the equation, $$x+2y-7=0$$ for $$y$$:
$$x+2y-7=0$$
$$2y=7-x$$
$$y=\frac{(7-x)}{2}$$
Now, substitute this in the equation, $$2x-5y+13=0$$:
$$2x-5y+13=0$$
$$2x-5(\frac{(7-x)}{2})+13=0$$
$$2x-(\frac{35}{2})+(\frac{5x}{2})+13=0$$
$$2x + (\frac{5x}{2}) = \frac{35}{2} – 13$$
$$\frac{9x}{2} = \frac{9}{2}$$
$$x=1$$
Substitute $$x=1$$ this in the equation $$y=\frac{(7-x)}{2}$$:
$$y=\frac{(7-1)}{2} = 3$$
Therefore, the solution of the given system is $$x=1$$ and $$y=3$$.
## Exercises forLinear Equations in Two Variables
1. A boat running downstream covers a distance of $$20 km$$ in $$2$$ hours while covering the same distance upstream, takes $$5$$ hours. What is the speed of the boat in still water?
2. Solve the system of equations using the elimination method. $$2x+3y-9=0,\:3x+2y-11=0$$
3. Find the solution of the following system of equations graphically. $$x−y=0,\:x+y−6=0$$
1. $$\color{blue}{7 \frac{km}{h}}$$
2. $$\color{blue}{x=3 , y=1}$$
3. $$\color{blue}{x=3 , y=3}$$
### What people say about "How to Solve Linear Equations in Two Variables? - Effortless Math: We Help Students Learn to LOVE Mathematics"?
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## Sound Is Frequencies
• Most sounds have high periodicity
• Fourier's Theorem (FOO-ree-YAY or thereabouts) says that an infinitely repeating sound can be represented as a sum of sinusoids
• The ear hears/decomposes a sum of sinusoids
• Yet PCM is a sequence of samples over time: frequency is not represented
• The Nyquist Limit is hard to think of as a signal change over time thing
## Time and Frequency
• We have: a continuous waveform, a function $$f(t)$$ representing sound pressure
• We want: a continuous spectrum, showing the amplitude and phase of sine waves at every frequency $$\hat{f}(\omega)$$
• Wait, amplitude and phase from a single function? Yes, representing a frequency as a complex number with the usual geometric interpretation
$$f(\omega) = a + b i$$
$$|f(\omega)| = \sqrt{a^2 - b^2}$$
$$\theta(f(\omega)) = tan^{-1}(a, b)$$
• Note: you will see both i and j for $$\sqrt{-1}$$ in different contexts
• Note: we freely mix between angular frequency $$\omega$$ and "normal" frequency f (dammit — we'll be using f as a symbol for both frequency and a generic function) via
$$\omega = 2 \pi f$$
because once around the circle is one cycle
## The Euler Formula
• Euler's Formula says complex exponential is a sinusoid:
$$e^{i (\omega t + \theta)} = cos(\omega t + \theta) + i~sin(\omega t + \theta)$$
$$= e^{i \omega t} e^{i \theta}$$
• Starting point for "phasor analysis"
• Now our sum of sinusoids can be represented as a sum of exponentials, making things easier (?)
## The Time Domain and the Frequency Domain
• Reminder: Fourier claims that every $$f(t)$$ can be represented as some $$\hat{f}(\omega)$$ (more or less)
• We think of the first kind of thing as "in the time domain", the second as "in the frequency domain"
• Converting from a single frequency to its time domain representation is "easy":
$$f(t) = e^{-i \omega t}$$
• Even for a single sinusoid, converting the other way isn't immediately obvious
## Continuous Fourier Decomposition: The Fourier Transform
• Let's just get the Fourier Transform out there:
$$\hat{f}(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt$$
$$f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \hat{f}(\omega) e^{i \omega t} d\omega$$
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# Resistors in Series and Parallel
In the previous two tutorials we have learnt how to connect individual resistors together to form either a Series Resistor Network or a Parallel Resistor Network and we used Ohms Law to find the various currents flowing in and voltages across each resistor combination.
But what if we want to connect various resistors together in “BOTH” parallel and series combinations within the same circuit to produce more complex resistive networks, how do we calculate the combined or total circuit resistance, currents and voltages for these resistive combinations.
Resistor circuits that combine series and parallel resistors networks together are generally known as Resistor Combination or mixed resistor circuits. The method of calculating the circuits equivalent resistance is the same as that for any individual series or parallel circuit and hopefully we now know that resistors in series carry exactly the same current and that resistors in parallel have exactly the same voltage across them.
For example, in the following circuit calculate the total current ( IT ) taken from the 12v supply.
At first glance this may seem a difficult task, but if we look a little closer we can see that the two resistors, R2 and R3 are actually both connected together in a “SERIES” combination so we can add them together to produce an equivalent resistance the same as we did in the series resistor tutorial. The resultant resistance for this combination would therefore be:
R2 + R3 = 8Ω + 4 Ω = 12 Ω
So we can replace both resistor R2 and R3 above with a single resistor of resistance value 12 Ω
So our circuit now has a single resistor RA in “PARALLEL” with the resistor R4. Using our resistors in parallel equation we can reduce this parallel combination to a single equivalent resistor value of R(combination) using the formula for two parallel connected resistors as follows.
The resultant resistive circuit now looks something like this:
We can see that the two remaining resistances, R1 and R(comb) are connected together in a “SERIES” combination and again they can be added together (resistors in series) so that the total circuit resistance between points A and B is therefore given as:
R( A B ) = Rcomb + R1 = 6 Ω + 6 Ω = 12 Ω.
and a single resistance of just 12 Ω can be used to replace the original four resistors connected together in the original circuit.
Now by using Ohm´s Law, the value of the circuit current ( I ) is simply calculated as:
So any complicated resistive circuit consisting of several resistors can be reduced to a simple single circuit with only one equivalent resistor by replacing all the resistors connected together in series or in parallel using the steps above.
Related Products: Resistor Fixed Single-Through Hole
It is sometimes easier with complex resistor combinations and resistive networks to sketch or redraw the new circuit after these changes have been made, as this helps as a visual aid to the maths. Then continue to replace any series or parallel combinations until one equivalent resistance, REQ is found. Lets try another more complex resistor combination circuit.
## Resistors in Series and Parallel Example No2
Find the equivalent resistance, REQ for the following resistor combination circuit.
Again, at first glance this resistor ladder network may seem a complicated task, but as before it is just a combination of series and parallel resistors connected together. Starting from the right hand side and using the simplified equation for two parallel resistors, we can find the equivalent resistance of the R8 to R10 combination and call it RA.
RA is in series with R7 therefore the total resistance will be RA + R7 = 4 + 8 = 12Ω as shown.
This resistive value of 12Ω is now in parallel with R6 and can be calculated as RB.
RB is in series with R5 therefore the total resistance will be RB + R5 = 4 + 4 = 8Ω as shown.
This resistive value of is now in parallel with R4 and can be calculated as RC as shown.
RC is in series with R3 therefore the total resistance will be RC + R3 = 8Ω as shown.
This resistive value of is now in parallel with R2 from which we can calculated RD as:
RD is in series with R1 therefore the total resistance will be RD + R1 = 4 + 6 = 10Ω as shown.
Then the complex combinational resistive network above comprising of ten individual resistors connected together in series and parallel combinations can be replaced with just one single equivalent resistance ( REQ ) of value 10Ω.
Related Products: Resistor Networks and Arrays
When solving any combinational resistor circuit that is made up of resistors in series and parallel branches, the first step we need to take is to identify the simple series and parallel resistor branches and replace them with equivalent resistors. This step will allow us to reduce the complexity of the circuit and help us transform a complex combinational resistive circuit into a single equivalent resistance remembering that series circuits are voltage dividers and parallel circuits are current dividers.
However, calculations of more complex T-pad Attenuator and resistive bridge networks which cannot be reduced to a simple parallel or series circuit using equivalent resistances require a different approach. These more complex circuits need to be solved using Kirchoff’s Current Law, and Kirchoff’s Voltage Law which will be dealt with in another tutorial.
In the next tutorial about Resistors, we will look at the electrical potential difference (voltage) across two points including a resistor.
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### JEE Mains Previous Years Questions with Solutions
4.5
star star star star star
1
### AIEEE 2003
Let $\overrightarrow F$ be the force acting on a particle having position vector $\overrightarrow r ,$ and $\overrightarrow \tau$ be the torque of this force about the origin. Then
A
$\overrightarrow {r.} \overrightarrow \tau = 0\,\,$ and $\overrightarrow {F.} \overrightarrow \tau \ne 0\,\,$
B
$\overrightarrow {r.} \vec \tau \ne 0{\mkern 1mu} {\mkern 1mu}$ and $\overrightarrow {F.} \overrightarrow \tau = 0\,\,$
C
$\overrightarrow {r.} \vec \tau \ne 0{\mkern 1mu}$ and $\overrightarrow {F.} \overrightarrow \tau \ne 0$
D
$\overrightarrow {r.} \vec \tau = 0{\mkern 1mu}$ and $\overrightarrow {F.} \overrightarrow \tau = 0\,\,$
## Explanation
As we know $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F$
So the angle between $\overrightarrow \tau$ and $\overrightarrow r$ is ${90^ \circ }$ and the angle between $\overrightarrow t$ and $\overrightarrow F$ is also ${90^ \circ }.$
We also know that the dot product of two vectors which have an angle of ${90^ \circ }$ between them is zero.
$\therefore$ $\overrightarrow {r.} \vec \tau = 0{\mkern 1mu}$ and $\overrightarrow {F.} \overrightarrow \tau = 0\,\,$
Therefore $(d)$ is the correct option.
2
### AIEEE 2003
A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is
A
${L \over 4}$
B
$2L$
C
$4L$
D
${L \over 2}$
## Explanation
We know Rotational Kinetic Energy$={1 \over 2}I{\omega ^2},$
Angular Momentum $L = I\omega \Rightarrow I = {L \over \omega }$
$\therefore$ Initial $K.E. = {1 \over 2}{L \over \omega } \times {\omega ^2} = {1 \over 2}L\omega$
Final $K.E'$ = ${{K.E} \over 2}$ = ${1 \over 2}{L'} \times 2\omega$
$\therefore$ ${{K.E} \over {K.E'}} = {{L \times \omega } \over {L' \times \omega '}}$
$\Rightarrow {{K.E} \over {{{K.E} \over 2}}} = {{L \times \omega } \over {L' \times 2\omega }}$
$\therefore$ $L' = {L \over 4}$
3
### AIEEE 2002
Moment of inertia of a circular wire of mass $M$ and radius $R$ about its diameter is
A
${{M{R^2}} \over 2}$
B
$M{R^2}$
C
$2M{R^2}$
D
${{M{R^2}} \over 4}$
## Explanation
Moment of Inertia of a circular wire about an axis $nn'$ passing through the centre of the circle and perpendicular to the plane of the circle $= M{R^2}$
As shown in the figure, $X$-axis and $Y$-axis lies in the plane of the ring. Then by perpendicular axis theorem
${I_X} + {I_Y} = {I_Z}$
$\Rightarrow 2{I_X} = M{R^2}\,$ $\left[ \, \right.$ as ${I_X} - {I_Y}$ (by symmetry) and ${I_Z} = M{R^2}$ $\left. \, \right]$
$\therefore$ ${I_X} = {1 \over 2}M{R^2}$
4
### AIEEE 2002
Two identical particles move towards each other with velocity $2v$ and $v$ respectively. The velocity of center of mass is
A
$v$
B
$v/3$
C
$v/2$
D
zero
## Explanation
The velocity of center of mass of two particle system is
${v_c} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$
$= {{m\left( {2v} \right) + m\left( { - v} \right)} \over {m + m}}$
$= {v \over 2}$
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# What Is Variance?
DEFINITION
Variance is the difference between actual and budgeted income and expenses.
## What Is Variance?
Variance is a measure of the difference between actual and expected results. In personal budgeting and management accounting, it's used to determine whether an individual or organization has exceeded or fallen short of its budgeted income and expenses.
• Alternate definition: In statistics, variance represents the spread of a set of numbers and is calculated as the average squared deviation from the mean.
## How Variance Works
At the end of the budgeting or accounting period, an individual or business may calculate the variance between their actual and expected income and expenses to determine whether they went over or fell under budget. By assessing variance, a person or business can take the corrective steps needed to bring actual and budgeted amounts into alignment during the next accounting period and thereby more efficiently allocate dollars (as well as staff and other resources at a business) and negotiate better financial arrangements.
For example, let's say that Bob is a college student who pays for college expenses with a combination of wages from a job and a student loan. He budgets for \$2,100 in income and \$2,000 in expenses for the month, which would leave him with a budget surplus of \$100. During the month, he brings in \$2,100 in income but incurs \$2,075 in expenses thanks to an unplanned parking ticket, resulting in an actual budget surplus of only \$25. At the end of the month, he calculates that the variance between his expected and actual income is \$0 (\$2,100 less \$2,100). But the variance between his expected and actual expenses is \$75 (\$2,000 less \$2,075).
Finally, he calculates the variance between his expected and actual budget surplus, which is \$75 (\$100 less \$25). In other words, he has gone \$75 over budget for the month. If he wants to keep the same expected budgetary figures next month but wants them to align with his actual results, he needs to cut unplanned expenses or increase his income.
Businesses typically calculate variance as part of a variance analysis that breaks down variance according to type (for example, cost variance or profit variance).
## Types of Variance
Variance may be measured at multiple levels, including:
• Income variance: This is the difference between actual and expected income. If the actual figure is higher than you expected, the variance is said to be favorable. If it's less than the expected figure, you have an unfavorable variance in income. In a budget, you may choose to further break down income into categories (salary and freelance income, for example) and assess the variance at the income category level.
• Expense or cost variance: This represents the difference between actual and budgeted expenditures. If you spend less than you budgeted for, the difference is said to be a favorable variance; if you spent more than you budgeted for, you have an unfavorable variance. You may also calculate variance for a specific expense category (for example, the variance between your expected and actual food and transportation costs. Businesses often track the variance in different types of costs, such as direct material or labor costs and overhead.
• Surplus/deficit or profit/loss variance: An individual may also choose to calculate the variance between the actual and expected surplus, or in the case of an income shortfall, the variance between the actual or expected deficit. Similarly, a business may choose to measure the variance between actual and expected profit or loss.
## How to Correct for Variance
Your goal is not to avoid variance completely—that's almost impossible, as you likely have both fixed and variable expenses. Instead, your goal should be to minimize variance. How you go about minimizing variance depends on the specific cause of variance in your budget, so you'll first need to assess that cause.
If, for example, your expenses put you over budget because you tend to spend more on food when you dine out with friends, look for ways to curb spending in the eating-out category ("go dutch" or throw potlucks at home, for example) or compensate for added expenses in that category by decreasing expenses in another category like clothing. If your income rather than spending is the problem, look for a job that pays more, get a second job, or create a passive income stream (such as a money-making blog).
Sometimes, variance is artificially created (for example, your accounting software might divide the cost of annual insurance premiums over 12 months). As a result, you might notice favorable variance in certain months and unfavorable variance in others, but you generally don't have to take any specific corrective action for this sort of variance.
Similarly, if you run a business and face unfavorable expense variance, it may be because the price of raw materials or labor has increased. A possible solution is to work with a different supplier to secure cheaper raw materials, use less raw materials, or reduce overhead or other expenses. If you can't reduce your expenses, you may be able to compensate for the higher expenses by increasing the sales volume or sales price. If you're not bringing in enough revenue, you may need to lower your product price or change the product mix by innovating.
### Key Takeaways
• Variance is a term used in personal and business budgeting for the difference between actual and expected results and can tell you how much you went over or under the budget.
• It can be measured at multiple levels, including income, expenses, and the budget surplus or deficit.
• You can correct for variance by examining the cause of it in your budget and then cutting expenses or increasing income as needed.
### Article Sources
1. UMN Libraries. "14.3 The Financial Planning Process." Accessed Nov. 1, 2020.
2. BYU Idaho. "What Causes Variance?" Accessed Nov. 1, 2020.
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# Thermodynamics 6: The Ideal Gas Law and Kinetic Theory of
Gases
18.6, 20.1
## PHYSICS 120 Advancing Physics I
Friday 10 May, 2019
Learning objectives
## By the end of this class, you should be able to:
• Apply the ideal gas law to situations involving the pressure, volume,
temperature, and the number of molecules of a gas.
• Use the unit of moles in relation to numbers of molecules, molecular mass
and macroscopic mass.
• Explain the relations between microscopic and macroscopic quantities in a
gas.
• Solve problems involving the distance and time between a gas molecule’s
collisions.
1
Announcement
## • Individual Learning Profile (Link)
• Reminders:
• Teammates 3 on Monday 13 May
• Assignment 8 by Wednesday 15 May
• Quiz 4 on Wednesday 15 May
• Lab 4 by Friday 17 May
2
## 3. The kinetic theory of gases
3
Atoms and moles
Number density
Number density
In an N-atom system that fills volume V,
N
number density =
V
## Distinguish number density (N/V) from
mass density (M/V).
4
Atomic mass and atomic mass number
5
Moles and molar mass
## NA = 6.02 × 1023 mol −1
Moles of substance
For a substance containing N basic
particles, the number of moles n in the
substance is
Which of these contains more atoms:
N 1 mol of helium gas (A = 4) or 1 mol of
n=
NA oxygen gas (A = 16)?
6
Moles and molar mass
## For a system of mass M consisting of For a system of mass M consisting of
atoms or molecules with atomic or atoms or molecules with molar mass
molecular mass m, the number of atoms Mmol , the number of moles n is
or molecules N is
M
M n=
N= Mmol
m
7
Atoms and moles
## Reference values and geometry:
Example 18.2, p. 516 • Table 18.1 (on the next slide)
How many moles of oxygen (O2 ) are in • Table 18.2 (see Slide 5)
100 g of oxygen gas? • The atomic mass number of gold is
Exercise 18.9, p. 533 197.
Two moles of gold are shaped into a • The surface area of a sphere = 4πr2
sphere. What is the sphere’s diameter? • The volume of a sphere = 43 πr3
8
Atoms and moles
9
Ideal gas law
Gases
## Images: Australian Geographic, 2014. Iceland’s volcanic eruptions.
10
Gases and gas laws
## Image: Figure 2.2 in University Physics, Openstax.
11
Gases and gas laws
12
Boyle’s law
13
Charles’ law
14
Gas laws
## Boyle’s law or Mariotte’s law: At constant temperature, the product of the
pressure and the volume of an ideal gas is constant.
1
p∝ p1 V1 = p2 V2
V
Charles’ law: At constant pressure, the volume and temperature of a gas are
directly proportional.
V1 V2
V∝T =
T1 T2
Amonton’s or Gay-Lussac’s law: The pressure exerted on a container of a
fixed volume by a gas is directly proportional to the temperature of the gas.
p1 p2
p∝T =
T1 T2
15
Ideal gas law
## Ideal gas law
The ideal gas law characterises the
relations of four state variables – the
absolute pressure of a gas p, the volume
the gas occupies V, the number of moles
in the gas n, the absolute temperature of
the gas T – for a gas in thermal
equilibrium. Experimentally, for gases at
low density (such that their molecules
occupy a negligible fraction of the total
volume) and at temperatures well above
the boiling point, these proportionalities
hold to a good approximation. 16
Ideal gas law
## Ideal gas law
pV = nRT
where p is in Pa (= N m−2 ), V is in m −3 , n
is the number of moles in the gas,
R = 8.31 J mol−1 K−1 , T is in K.
pV = NkB T
## where N is the number of molecules in
the gas, kB = NRA = 1.38 × 10−23 J K−1 .
17
Ideal gas law
## Exercise 18.21, p. 533
A rigid container holds 2.0 mol of gas at a pressure of 1.0 atm and a temperature
of 30 ◦C. (1 atm ≈ 101.3 kPa)
## (a) What is the container’s volume?
(b) What is the pressure if the temperature is raised to 130 ◦C?
18
Ideal gas law
## Exercise 18.26, p. 533
A gas at temperature T0 and atmospheric pressure fills a cylinder. The gas is
transferred to a new cylinder with three times the volume, after which the
pressure is half the original pressure. What is the new temperature of the gas?
19
Ideal gas law
## Adapted from Example 18.6, p. 524
‘Standard temperature and pressure’, abbreviated STP, are T = 0 ◦C and p = 1 atm.
Consider a gas that behaves like an ideal gas at STP:
## (a) Determine the volume of 1.00 mol of such gas at STP.
(b) Use your answer for (a) to estimate how many molecules you breathe in with
a 1.0-L breath of air at STP.
(c) Determine the average volume per gas molecule at STP.
(d) Estimate the average distance between such gas molecules at STP.
20
The kinetic theory of gases
Micro-macro connection
## If matter really consists of atoms and molecules, then the macroscopic
properties of matter, such as temperature, pressure, specific heat capacity,
should be related to the microscopic motion of those atoms and molecules.
21
Molecular speeds and collisions
22
Molecular speeds and collisions
## Mean free path
If a molecule has Ncoll collisions as it
travels distance L, the average distance
between collisions, is called the mean
free path λ.
L
λ=
Ncoll
23
Molecular speeds and collisions
1
λ= √
4 2π(N/V)r2
24
Molecular speeds and collisions
## Example 20.1, p. 572
What is the mean free path of a nitrogen molecule at 1.0 atm pressure and room
temperature (20 ◦C)?
Although laboratory measurements are necessary to determine atomic and
molecular radii, it is reasonable to approximate atoms in a monatomic gas have
r ≈ 0.5 × 10−10 m and diatomic molecules have r ≈ 1.0 × 10−10 m.
Try this
Determine the mean free path of a nitrogen molecule at STP (1.0 atm and 0 ◦C)
and compare it with your calculation for the average separation between gas
molecules at STP in the previous exercise.
25
Molecular speeds and collisions
26
## Example 18.2 3.13 mol of oxygen molecules
Exercise 18.9 diameter = 3.39 cm
Exercise 18.21
(a) V = 0.050 m3
(b) p = 1.3 atm
Exercise 18.26 Tnew = 32 T0
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Home > Library > Statistics > Interval Data: Definition, Characteristics and Collection
# Interval Data: Definition, Characteristics and Collection
Published by at August 31st, 2021 , Revised On July 20, 2023
Whether it is market research or any other form of economic, educational, or social research, data types are a great demand that supports most statistical transformations and tests. This demand is made in order to make better the analysis process so that a proper conclusion can be drawn.
When conducting market research and studying data types, interval data is quantitative data measured along a scale, really crucial, and must be taken into consideration. We discover that interval data is perfectly compatible with most of these statistical tests. It has distinct attributes, making it the most in-demand type compared to its counterparts.
Having that said, this blog, therefore, thoroughly discusses what interval data is and how it can be analyzed today.
## What is Interval Data?
Interval data is yet another type of data that can be calculated along a scale where every point is placed at an equal interval from another, just as the name explains itself.
It is one of the two types of discrete data. The data collected on a thermometer is an example of interval data, as its gradation markings are equally distanced from each other. Interval data is always expressed in the form of numbers, unlike ordinal data.
Moreover, arithmetic operations are all performed on this type of data. But there is a but here. These operations are bound to subtraction and addition only.
That was easy, no?
Let us now look at some of the characteristics of interval data.
## Characteristics of Interval Data
Below are some of the essential characteristics of interval data you must know about:
### 1. Arithmetic Operation
As we just mentioned in the previous heading, one can perform arithmetic operations on this kind of data, such as subtractions and additions. You cannot, however, divide or multiply interval data. When calculating interval data, practically every statistical analysis is applicable due to its quantitative nature. This contains the mean, mode, and median but is not restricted to them.
Even if a variable is negative, it can be measured on an interval scale. Whether positive or negative, the interval scale reads both values. For instance, if we take a winter temperature, say -10 degrees Celsius, it can be measured and read with an interval scale. Similarly, those with positive values would also be read by it.
### 3. Interval Difference
You will note that the difference between each interval will be equal. For instance, the difference between 20 degrees Celsius and 40 degrees Celsius will be the same as the difference between 15 degrees Celsius and 25 degrees Celsius. You must already be familiar with these characteristics.
### 4. Quantitativeness
As we said earlier, interval data is one of the two kinds of numerical data; it is either quantitative or parametric. Interval data, sometimes known as integer, shows the quantitative value and performs operations of an integer. That is why it is quite different from categorical data, such as ordinal and nominal data, which takes numerical value but demonstrates qualitative nature.
### 5. Measurement Scale
Interval data is calculated with an interval scale that gives you the order and shows the exact difference in the values. It is again different from ordinal data because ordinal data did not tell us about the order and direction, with no standardized difference in the values of variables.
An example of an internal data measurement scale would be a ruler where the markings between intervals are equal. Thus, it is an interval scale. Like literally!
## Is the Statistics assignment pressure too much to handle?
### How about we handle it for you?
Put in the order right now in order to save yourself time, money, and nerves at the last minute.
## What is The Process of Interval Data Collection?
Since we are pretty gnostic about what interval data is and know some of its characteristics, it is high time we study the interval data collection process now. Do not worry; it is not as difficult as it sounds.
So, there are many ways to collect interval data. We will discuss a few in this guide. Methods or techniques used by researchers depend on the data usage, the person collecting the data, and the audience being targeted.
Following are interval data collection techniques:
1. Observations
2. Surveys and Questionnaires
3. Interviews
### Observation
In this data collection method, researchers make systematic observations via counting. This could be counting the number of individuals present at a particular event at a specific time and location or the number of persons attending an event at a specific location. Naturalistic and standardized observation approaches are the two types of observation methods.
### Surveys and Questionnaires
If you are wondering how daunting would it be to prepare questionnaires and surveys, you do not really have to. There is a lot of online software that will do this job for you in a matter of seconds. For instance, you get your survey created with Formplus, a questionnaire builder. Surveys are used to improve the experience of respondents while working towards gaining trust from them. When it comes to the types or kinds of surveys you can go for; there are two major ones. One is called a web-based questionnaire, and the second is an online questionnaire. You do not really need to know each type as both are exactly how they sound.
### Interviews
This is probably the most fun technique for most people.
Are you an interview person or a survey person?
Respondents in this technique are interviewed so that information could be collected and analyzed. However, these interviews are different from the casual interviews that we see happening around where people are swerving from the topic at hand. Interviews for interval data collection are planned and structured, with researchers asking a set of standardized questions. Even the interviews can be taken through different mediums. For instance, in the current COVID-19 times, an online interview would be more suitable. You can also go for a face-to-face interview, depending on the nature of your study.
Interval data is yet another type of data that can be calculated along a scale where every point is placed at an equal interval from another, just as the name explains itself.
The data collected on a thermometer is an example of interval data, as its gradation markings are equally distanced from each other. Interval data is always expressed in the form of numbers, unlike ordinal data.
Methods or techniques used by researchers depend on the data usage, the person collecting the data, and the audience being targeted.
Following are interval data collection techniques:
• Observations
• Surveys and Questionnaires
• Interviews
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# Physics
posted by on .
I really don't get why I am getting this question wrong? Please can you explain to me how to do this question step by step so that I can understand it please? :( i have been stuck on this question for quite a while now.
Determine the stopping distances for an automobile with an initial speed of 89 km/h and human reaction time of 2.0 s for the following:
(a) an acceleration a = -4.0 m/s2.
(b) a = -8.0 m/s2.
• Physics - ,
Vo=85000m/h = 85000m/3600s = 23.61 m/s.
a. V = Vo + a*t
T = (V-Vo)/a + Tr
T = (0-23.61)/-4 + 2s = 7.9 s = Stopping time.
d = Vo*t + 0.5a*t^2
d = 23.61*7.9 + 0.5*-4*7.9^2 = 61.7m
b. T = (0-23.61)/-8 + 2s = 4.95 s.
d=23.61*4.95 + 0.5*(-8)*4.95^2=18.86 m.
• Physics - ,
CORRECTION:
Vo=89,000m/1h = 89000/3600s = 24.72 m/s
a. V = Vo + va*t
T = (V-V0)/a + Tr
T = (0-24.72)/-4 + 2s.=8.18 s.=Stopping
time.
d = Vo*t + 0.5a*t^2
d=24.72*8.18 + 0.5*(-40*8.18*2=68.38 m.
b. T = (0-24.72)/-8 + 2s = 5.09 s.
d=24.72*5.09 + 0.5*(-8)*5.09^2= 22.19 m.
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# Statistics
Statistics
Statistics
1.Use the following grades and numbers of students achieving those grades. Make a Cumulative Frequency Chart for the above grades.
Grade number of students getting the grade
A 19
B 11
C 9D 5
F 6
2. Find the measures of central tendency: mean, median, mode.
3. Find the measures of dispersion range, standard deviation.
4. What does standard deviation measure?
____________________________________________________________
Probability
1. Find the probability of rolling at least one 3 in 4 rolls of a fair single die.
2. Find the probability of flipping a coin and getting at least one tail in 4 flips.
3. Chart for Heads and Tails Probability for flipping 3 coins at once or one coin 3 times.
HHHHHT
H TH
H T T
THH
THT
TTH
T TT
4 Find the following probabilities below using the chart above. Leave answers as fractions or decimals.
1. P( 3 tails)________________
3. P(more than 1 head)_______
4. P(fewer than 3 tails)_______
5. P(At least two heads)________
5.Use a standard deck of 52 playing cards comprised of 13 hearts, 13 spades, 13 diamonds, and 13 clubs find the following probabilities when selecting one card at a time from the deck.
SPADES BLACK A K Q J 10 9 8 7 6 5 4 3 2
HEARTS RED A K Q J 10 9 8 7 6 5 4 3 2
DIAMONDS RED A K Q J 10 9 8 7 6 5 4 3 2
CLUBS BLACK A K Q J 10 9 8 7 6 5 4 3 2
1. P(Heart)
2. P (Red Card)
3. P(King)
4. P(Ace) or P(diamond)
5. Find the probability of selecting a club and then a spade; you put the first card back into the deck before selecting the second card
6. Find the probability of selecting a Queen and then a spade; you do not put the first card back into the deck before selecting the second card
7. How many ways can 13 third graders line up for the launch line.
8. How many 9-digit zip codes can be formed if there are 10 numbers available with no restrictions.
9. What number is the answer using nCr where n = 12 and r =4
10. What number is the answer using nPr where n =15 and r= 3
_________________________________________________________________
Finance
1.Find the Amount of savings if the principle is \$25,000, APR = 3.5% time 5 years compounded quarterly. Use formula 2
http://www.myarmortizationchart.com/
You are buying a house for \$350,000. Select 30 years mortgage
1. Find the 20% down payment
2.What is the amount of the mortgage?
3. What is the monthly payment?
4. What are the monthly taxes?
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# How to find the covariance matrix of a polygon?
Imagine you have a polygon defined by a set of coordinates $$(x_1,y_1)...(x_n,y_n)$$ and its centre of mass is at $$(0,0)$$. You can treat the polygon as a uniform distribution with a polygonal boundary.
I'm after a method that will find the covariance matrix of a polygon.
I suspect that the covariance matrix of a polygon is closely related to the second moment of area, but whether they are equivalent I'm not sure. The formulas found in the wikipedia article I linked seem (a guess here, it's not especially clear to me from the article) to refer to the rotational inertia around the x, y and z axes rather than the principal axes of the polygon.
(Incidentally, if anyone can point me to how to calculate the principal axes of a polygon, that would also be useful to me)
It is tempting to just perform PCA on the coordinates, but doing so runs into the issue that the coordinates are not necessarily evenly spread around the polygon, and are therefore not representative of the density of the polygon. An extreme example is the outline of North Dakota, whose polygon is defined by a large number of points following the Red river, plus only two more points defining the western edge of the state.
• By "find", I assume simply sampling from the polygon, then calculating the covariance of the samples, is not what you have in mind? Jul 4, 2019 at 6:32
• Also, can you edit your post to include coordinates for your polygon, so people can play around with it? Jul 4, 2019 at 6:33
• @StephanKolassa I mean treating the polygon as a uniform bivariate probability density with polygonal boundary. Sure, you can sample points and the limit would be the same thing, but i'm looking for an a-priori method. The picture is just an illustration from paint that I used. The real world data I intend to use are the outlines of states and regions. Jul 4, 2019 at 9:46
• You are correct that the usual term for "covariance matrix" is moment of inertia or second moment. The principal axes are oriented in its eigendirections. Running PCA on the coordinates is incorrect: it is tantamount to assuming all the mass is located at the vertices. The most direct methods of computation of the barycenter--the first moment--are discussed in my post at gis.stackexchange.com/a/22744/664. The second moments are computed in the same way with minor modifications. Special considerations are needed on the sphere.
– whuber
Jul 7, 2019 at 13:21
• It works the other way: compute the inertial tensor and find its principal axes from that. The technique in your case involves Green's Theorem, which shows that the requisite integrals $$\mu_{k,l}(\mathcal{P})=\iint_{\mathcal{P}}x^ky^l\mathrm{d}x\mathrm{d}y$$ can be computed as contour integrals around $\partial\mathcal{P}$ of the one-form $\omega$ where $\mathrm{d}\omega=x^ky^l\mathrm{d}x\mathrm{d}y.$ Such forms are easy to find because any suitable linear combination of $x^ky^{l+1}\mathrm{d}x$ and $x^{k+1}y^l\mathrm{d}y$ will work. The contour integral is a sum of integrals over the edges.
– whuber
Jul 8, 2019 at 20:22
Let's do some analysis first.
Suppose within the polygon $$\mathcal{P}$$ its probability density is proportional function $$p(x,y).$$ Then the constant of proportionality is the inverse of the integral of $$p$$ over the polygon,
$$\mu_{0,0}(\mathcal{P})=\iint_{\mathcal P} p(x,y) \mathrm{d}x\,\mathrm{d}y.$$
The barycenter of the polygon is the point of average coordinates, computed as their first moments. The first one is
$$\mu_{1,0}(\mathcal{P})=\frac{1}{\mu_{0,0}(\mathcal{P})} \iint_{\mathcal P} x\,p(x,y)\mathrm{d}x\,\mathrm{d}y.$$
The inertial tensor can be represented as the symmetric array of second moments computed after translating the polygon to put its barycenter at the origin: that is, the matrix of central second moments
$$\mu^\prime_{k,l}(\mathcal{P}) = \frac{1}{\mu_{0,0}(\mathcal{P})} \iint_{\mathcal P} \left(x - \mu_{1,0}(\mathcal{P})\right)^k\,\left(y - \mu_{0,1}(\mathcal{P})\right)^l\,p(x,y)\mathrm{d}x\,\mathrm{d}y$$
where $$(k,l)$$ range from $$(2,0)$$ to $$(1,1)$$ to $$(0,2).$$ The tensor itself--aka covariance matrix--is
$$I(\mathcal{P}) = \pmatrix{\mu^\prime_{2,0}(\mathcal{P}) & \mu^\prime_{1,1}(\mathcal{P}) \\ \mu^\prime_{1,1}(\mathcal{P}) & \mu^\prime_{0,2}(\mathcal{P})}.$$
A PCA of $$I(\mathcal{P})$$ yields the principal axes of $$\mathcal{P}:$$ these are the unit eigenvectors scaled by their eigenvalues.
Next, let's work out how to do the calculations. Because the polygon is presented as a sequence of vertices describing its oriented boundary $$\partial\mathcal P,$$ it is natural to invoke
Green's Theorem: $$\iint_{\mathcal{P}} \mathrm{d}\omega = \oint_{\partial\mathcal{P}}\omega$$ where $$\omega = M(x,y)\mathrm{d}x + N(x,y)\mathrm{d}y$$ is a one-form defined in a neighborhood of $$\mathcal{P}$$ and $$\mathrm{d}\omega = \left(\frac{\partial}{\partial x}N(x,y) - \frac{\partial}{\partial y}M(x,y)\right)\mathrm{d}x\,\mathrm{d}y.$$
For instance, with $$\mathrm{d}\omega = x^k y^l \mathrm{d}x\mathrm{d}y$$ and constant (i.e., uniform) density $$p,$$ we may (by inspection) select one of the many solutions, such as $$\omega(x,y) = \frac{-1}{l+1}x^k y^{l+1}\mathrm{d}x.$$
The point of this is that the contour integral follows the line segments determined by the sequence of vertices. Any line segment from vertex $$\mathbf{u}$$ to vertex $$\mathbf{v}$$ can be parameterized by a real variable $$t$$ in the form
$$t \to \mathbf{u} + t\mathbf{w}$$
where $$\mathbf{w} \propto \mathbf{v}-\mathbf{u}$$ is the unit normal direction from $$\mathbf{u}$$ to $$\mathbf{v}.$$ The values of $$t$$ therefore range from $$0$$ to $$|\mathbf{v}-\mathbf{u}|.$$ Under this parameterization $$x$$ and $$y$$ are linear functions of $$t$$ and $$\mathrm{d}x$$ and $$\mathrm{d}y$$ are linear functions of $$\mathrm{d}t.$$ Thus the integrand of the contour integral over each edge becomes a polynomial function of $$t,$$ which is easily evaluated for small $$k$$ and $$l.$$
Implementing this analysis is as straightforward as coding its components. At the lowest level we will need a function to integrate a polynomial one-form over a line segment. Higher level functions will aggregate these to compute the raw and central moments to obtain the barycenter and inertial tensor, and finally we can operate on that tensor to find the principal axes (which are its scaled eigenvectors). The R code below performs this work. It makes no pretensions of efficiency: it is intended only to illustrate the practical application of the foregoing analysis. Each function is straightforward and the naming conventions parallel those of the analysis.
Included in the code is a procedure to generate valid closed, simply connected, non-self-intersecting polygons (by randomly deforming points along a circle and including the starting vertex as its final point in order to create a closed loop). Following this are a few statements to plot the polygon, display its vertices, adjoin the barycenter, and plot the principal axes in red (largest) and blue (smallest), creating a polygon-centric positively-oriented coordinate system.
#
# Integrate a monomial one-form x^k*y^l*dx along the line segment given as an
# origin, unit direction vector, and distance.
#
lintegrate <- function(k, l, origin, normal, distance) {
# Binomial theorem expansion of (u + tw)^k
expand <- function(k, u, w) {
i <- seq_len(k+1)-1
u^i * w^rev(i) * choose(k,i)
}
# Construction of the product of two polynomials times a constant.
omega <- normal[1] * convolve(rev(expand(k, origin[1], normal[1])),
expand(l, origin[2], normal[2]),
type="open")
# Integrate the resulting polynomial from 0 to distance.
sum(omega * distance^seq_along(omega) / seq_along(omega))
}
#
# Integrate monomials along a piecewise linear path given as a sequence of
# (x,y) vertices.
#
cintegrate <- function(xy, k, l) {
n <- dim(xy)[1]-1 # Number of edges
sum(sapply(1:n, function(i) {
dv <- xy[i+1,] - xy[i,] # The direction vector
lambda <- sum(dv * dv)
if (isTRUE(all.equal(lambda, 0.0))) {
0.0
} else {
lambda <- sqrt(lambda) # Length of the direction vector
-lintegrate(k, l+1, xy[i,], dv/lambda, lambda) / (l+1)
}
}))
}
#
# Compute moments of inertia.
#
inertia <- function(xy) {
mass <- cintegrate(xy, 0, 0)
barycenter = c(cintegrate(xy, 1, 0), cintegrate(xy, 0, 1)) / mass
uv <- t(t(xy) - barycenter) # Recenter the polygon to obtain central moments
i <- matrix(0.0, 2, 2)
i[1,1] <- cintegrate(uv, 2, 0)
i[1,2] <- i[2,1] <- cintegrate(uv, 1, 1)
i[2,2] <- cintegrate(uv, 0, 2)
list(Mass=mass,
Barycenter=barycenter,
Inertia=i / mass)
}
#
# Find principal axes of an inertial tensor.
#
principal.axes <- function(i.xy) {
obj <- eigen(i.xy)
t(t(obj$$vectors) * obj$$values)
}
#
# Construct a polygon.
#
circle <- t(sapply(seq(0, 2*pi, length.out=11), function(a) c(cos(a), sin(a))))
set.seed(17)
radii <- (1 + rgamma(dim(circle)[1]-1, 3, 3))
#
# Compute principal axes.
#
i.xy <- inertia(xy)
axes <- principal.axes(i.xy$$Inertia) sign <- sign(det(axes)) # # Plot barycenter and principal axes. # plot(xy, bty="n", xaxt="n", yaxt="n", asp=1, xlab="x", ylab="y", main="A random polygon\nand its principal axes", cex.main=0.75) polygon(xy, col="#e0e0e080") arrows(rep(i.xy$$Barycenter[1], 2),
rep(i.xy$$Barycenter[2], 2), -axes[1,] + i.xy$$Barycenter[1], # The -signs make the first axis ..
-axes[2,]*sign + i.xy$$Barycenter[2],# .. point to the right or down. length=0.1, angle=15, col=c("#e02020", "#4040c0"), lwd=2) points(matrix(i.xy$$Barycenter, 1, 2), pch=21, bg="#404040")
• +1 Wow, this is a great answer! Jul 9, 2019 at 15:47
Edit: Didn't notice that whuber had already answered. I'll leave this up as an example of another (perhaps less elegant) approach to the problem.
### The covariance matrix
Let $$(X,Y)$$ be a random point from the uniform distribution on a polygon $$P$$ with area $$A$$. The covariance matrix is:
$$C = \begin{bmatrix} C_{XX} & C_{XY} \\ C_{XY} & C_{YY} \end{bmatrix}$$
where $$C_{XX} = E[X^2]$$ is the variance of $$X$$, $$C_{YY} = E[Y^2]$$ is the variance of $$Y$$, and $$C_{XY} = E[XY]$$ is the covariance between $$X$$ and $$Y$$. This assumes zero mean, since the polygon's center of mass is located at the origin. The uniform distribution assigns constant probability density $$\frac{1}{A}$$ to every point in $$P$$, so:
$$C_{XX} = \frac{1}{A} \underset{P}{\iint} x^2 dV \quad C_{YY} = \frac{1}{A} \underset{P}{\iint} y^2 dV \quad C_{XY} = \frac{1}{A} \underset{P}{\iint} x y dV \tag{1}$$
### Triangulation
Instead of trying to directly integrate over a complicated region like $$P$$, we can simplify the problem by partitioning $$P$$ into $$n$$ triangular subregions:
$$P = T_1 \cup \cdots \cup T_n$$
In your example, one possible partitioning looks like this:
There are various ways to produce a triangulation (see here). For example, you could compute the Delaunay triangulation of the vertices, then discard edges that fall outside $$P$$ (since it may be nonconvex as in the example).
Integrals over $$P$$ can then be split into sums of integrals over the triangles:
$$C_{XX} = \frac{1}{A} \sum_{i=1}^n \underset{T_i}{\iint} x^2 dV \quad C_{YY} = \frac{1}{A} \sum_{i=1}^n \underset{T_i}{\iint} y^2 dV \quad C_{XY} = \frac{1}{A} \sum_{i=1}^n \underset{T_i}{\iint} x y dV \tag{2}$$
A triangle has nice, simple boundaries so these integrals are easier to evaluate.
### Integrating over triangles
There are various ways to integrate over triangles. In this case, I used a trick that involves mapping a triangle to the unit square. Transforming to barycentric coordintes might be a better option.
Here are solutions for the integrals above, for an arbitrary triangle $$T$$ defined by vertices $$(x_1,y_1), (x_2,y_2), (x_3,y_3)$$. Let:
$$v_x = \left[ \begin{smallmatrix} x_1 \\ x_2 \\ x_3 \end{smallmatrix} \right] \quad v_y = \left[ \begin{smallmatrix} y_1 \\ y_2 \\ y_3 \end{smallmatrix} \right] \quad \vec{1} = \left[ \begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix} \right] \quad L = \left[ \begin{smallmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{smallmatrix} \right]$$
Then:
$$\underset{T}{\iint} x^2 dV = \frac{A}{6} \text{Tr}(v_x v_x^T L) \quad \underset{T}{\iint} y^2 dV = \frac{A}{6} \text{Tr}(v_y v_y^T L) \quad \underset{T}{\iint} x y dV = \frac{A}{12} (\vec{1}^T v_x v_y^T \vec{1} + v_x^T v_y) \tag{3}$$
### Putting everything together
Let $$v_x^i$$ and $$v_y^i$$ contain the x/y coordinates of the vertices for each triangle $$T_i$$, as above. Plug $$(3)$$ into $$(2)$$ for each triangle, noting that the area terms cancel out. This gives the solution:
$$C_{XX} = \frac{1}{6} \sum_{i=1}^n \text{Tr} \big( v_x^i (v_x^i)^T L \big) \quad C_{YY} = \frac{1}{6} \sum_{i=1}^n \text{Tr} \big( v_y^i (v_y^i)^T L \big) \quad C_{XY} = \frac{1}{12} \sum_{i=1}^n \big( \vec{1}^T v_x^i (v_y^i)^T \vec{1} + (v_x^i)^T v_y^i \big) \tag{4}$$
### Principal axes
The principal axes are given by the eigenvectors of the covariance matrix $$C$$, just as in PCA. Unlike PCA, we have an analytic expression for $$C$$, rather than having to estimate it from sampled data points. Note that the vertices themselves are not a representative sample from the uniform distribution on $$P$$, so one can't simply take the sample covariance matrix of the vertices. But, $$C$$ *is* a relatively simple function of the vertices, as seen in $$(4)$$.
• +1 This can be simplified by allowing oriented triangles, thereby eliminating the need for a proper triangulation. Instead, you can just establish an arbitrary center $O$ and sum the (signed) values over the triangles $OP_iP_{i+1}:$ this is how it's often done because it's much less fussy. It's easy to see that such a summation is essentially the same thing as applying Green's Theorem, because each term in the summation ultimately is a function of the edge $P_iP_{i+1}.$ This approach is illustrated in the "Area" section at quantdec.com/SYSEN597/GTKAV/section2/chapter_11.htm.
– whuber
Jul 9, 2019 at 15:46
• @whuber Interesting, thanks for pointing this out Jul 9, 2019 at 16:00
• Both of these answers are good, albeit a bit over my education level. Once I'm sure I fully understand them I'll try to figure out who gets the bounty. Jul 11, 2019 at 2:21
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# Graph coloring problem (possibly related to partitions)
Given an undirected graph I'd like to color each node either black or red such that at most half of every node's neighbors have the same color as the node itself. As a first step, I'd like to show that this is always possible.
I believe this problem is the essence of the math quiz #2 in Communications of the ACM 02/2011 where I found it, so you might consider this a homework-like question. The quiz deals with partitions but I found it more natural to formulate the problem as a graph-coloring problem.
Coming from computer science with some math interest I'm not sure how to approach this and would be glad about some hints. One observation is that any node of degree 1 forces its neighbor to be of the opposite color. This could lead to a constructive proof (or a greedy algorithm) that provides a coloring. However, an existence proof would be also interesting.
-
Hint: Start with a random colouring and try to increase the number of edges which have differently coloured endpoints.
Spoiler:
Pick a node which has more than half of it's neighbour of the same colour as itself and flip it's colour. Now show that, as a result, the number of edges with different coloured endpoints increases by at least 1. Repeat.
-
Perhaps related is the following well-known riddle: At each step, all the nodes (simultaneously) choose their color as the majority color of their neighbors (or black in case of a tie). Show that this process converges to either a fixed point or a $2$-cycle.
This charming riddle is solved by comparing the state at time $t$ to the state at time $t+2$, and using a potential function. This solution was published, and generalized in a sequence of papers.
In your case, you want to take the anti-majority. Perhaps the same techniques work, though I'm not sure how a $2$-cycle will help you.
-
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### Author Topic: Problem 3 (Read 12975 times)
#### James McVittie
• Full Member
• Posts: 20
• Karma: 1
##### Problem 3
« on: October 20, 2012, 10:01:33 PM »
What is implied by the word "weird", is it just something unexpected that comes up or something that hasn't been discussed in the course? Thanks!
#### Victor Ivrii
• Elder Member
• Posts: 2511
• Karma: 0
##### Re: Problem 3
« Reply #1 on: October 21, 2012, 03:18:11 AM »
What is implied by the word "weird", is it just something unexpected that comes up or something that hasn't been discussed in the course? Thanks!
Why you don't try to solve and see by yourself?
« Last Edit: October 21, 2012, 04:19:25 AM by Victor Ivrii »
#### James McVittie
• Full Member
• Posts: 20
• Karma: 1
##### Re: Problem 3
« Reply #2 on: October 22, 2012, 11:58:32 AM »
Are we allowed to assume that solutions are real or must we always assume in greatest generality that the solution could be complex?
Thanks
#### Victor Ivrii
• Elder Member
• Posts: 2511
• Karma: 0
##### Re: Problem 3
« Reply #3 on: October 22, 2012, 12:13:10 PM »
Are we allowed to assume that solutions are real or must we always assume in greatest generality that the solution could be complex?
Thanks
Have you read the preamble of this HA? It says that you may assume that eigenvalues are real (but nothing about eigenfunctions).
#### Shu Wang
• Jr. Member
• Posts: 11
• Karma: 0
##### Re: Problem 3
« Reply #4 on: October 23, 2012, 05:20:03 AM »
uh, say we assume some form of solution for X(x) and T(t), so there will be 3 coefficients. When we setup the matrix for A,B,C it becomes 2x3 matrix since we're given 2 B.C. I was wondering if we were to solve for eigenvalue for the matrix, do we need to take into account for all combinations of 2x2 matrices? (In other words, break the matrix into A&B, B&C, A&C). Or am I just completely off the question? o.O
#### Victor Ivrii
• Elder Member
• Posts: 2511
• Karma: 0
##### Re: Problem 3
« Reply #5 on: October 23, 2012, 08:05:21 AM »
uh, say we assume some form of solution for X(x) and T(t), so there will be 3 coefficients.
Where? and don't mix coefficients for $X$ and $T$!
#### Bowei Xiao
• Full Member
• Posts: 17
• Karma: 2
##### Re: Problem 3
« Reply #6 on: October 23, 2012, 12:42:40 PM »
Is there supposed to be condition at t=0? Because I feel there's no restriction for interior and that can go wildly I guess?
#### Victor Ivrii
• Elder Member
• Posts: 2511
• Karma: 0
##### Re: Problem 3
« Reply #7 on: October 23, 2012, 04:31:06 PM »
We are looking at $u(x,t)=X(x)T(t)$.
Plug it in the equation and separating variables find equations to $X(x)$ and $T(t)$ in the standard way.
Plug into both boundary conditions. $u(0,x)=0$ implies what? (Standard)
Another b.c. implies ODE to $T(t)$. Solve it and use in conjugation with everything else. This is the only non-standard (albeit rather easy) part.
#### Peishan Wang
• Full Member
• Posts: 32
• Karma: 6
##### Re: Problem 3
« Reply #8 on: October 24, 2012, 11:20:15 PM »
Please let me know if there's anything wrong with the answer attached. Thanks
« Last Edit: October 24, 2012, 11:30:22 PM by Peishan Wang »
#### Peishan Wang
• Full Member
• Posts: 32
• Karma: 6
##### Re: Problem 3
« Reply #9 on: October 24, 2012, 11:21:14 PM »
Part 3 (to check if there's 0 or negative eigenvalues).
OK. -- V.I.
« Last Edit: October 25, 2012, 12:19:12 PM by Victor Ivrii »
#### Victor Ivrii
• Elder Member
• Posts: 2511
• Karma: 0
##### Re: Problem 3
« Reply #10 on: October 25, 2012, 03:33:06 PM »
So, as Peishan did: plugging $u=X(x)T(t)$ into equation and boundary conditions we get after separation of variables
\begin{align}
& \frac{T''}{T}=c^2\frac{X''}{X}=-c^2\lambda, \label{eq-1}\\
& X(0)=0,\label{eq-2}\\
& \frac{T'}{T}=-i\alpha \frac{X'(l)}{X(l)}=-i\beta\label{eq-3}
\end{align}
and from(\ref{eq-3}) we conclude that $\beta$ is a constant and $T=e^{-i\beta t}$ (do not care about constant factor) and then $\lambda=c^{-2}\beta^2$
\begin{align}
&X''+c^{-2}\beta^2 X=0 \label{eq-4}\\
& X(0)=0,\label{eq-5}\\
& X'(l)=\alpha^{-1}\beta X(l)\label{eq-6}
\end{align}
and the weirdness of this problem is that spectral parameter $\beta$ is present in both equation and the boundary condition. Then (\ref{eq-4})--(\ref{eq-5}) imply that $X= \sin (c ^{-1}\beta x)$ and (\ref{eq-6}) that $c^{-1}\beta \cos (c^{-1}\beta l)=\alpha{-1}\beta \sin (c^{-1}\beta l)$ which is equivalent to $\tan (c^{-1}\beta l)= c^{-1}\alpha$ i.e. $\beta = cl^{-1}\arctan c^{-1}\alpha +cl^{-1} n$ with $n\in \mathbb{Z}$.
Then exactly like in Peisan HA.
#### Chiara Moraglia
• Jr. Member
• Posts: 9
• Karma: 1
##### Re: Problem 3
« Reply #11 on: November 09, 2012, 05:55:45 PM »
Hi,
I have a question regarding (3) in your above response Professor. Where is beta coming from? Is it just equal to alpha*X′(l)X(l) and this is why beta is constant? I am also unclear as to how you obtain λ.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > plusfeq Structured version Visualization version GIF version
Theorem plusfeq 17170
Description: If the addition operation is already a function, the functionalization of it is equal to the original operation. (Contributed by Mario Carneiro, 14-Aug-2015.)
Hypotheses
Ref Expression
plusffval.1 𝐵 = (Base‘𝐺)
plusffval.2 + = (+g𝐺)
plusffval.3 = (+𝑓𝐺)
Assertion
Ref Expression
plusfeq ( + Fn (𝐵 × 𝐵) → = + )
Proof of Theorem plusfeq
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 fnov 6721 . . 3 ( + Fn (𝐵 × 𝐵) ↔ + = (𝑥𝐵, 𝑦𝐵 ↦ (𝑥 + 𝑦)))
21biimpi 206 . 2 ( + Fn (𝐵 × 𝐵) → + = (𝑥𝐵, 𝑦𝐵 ↦ (𝑥 + 𝑦)))
3 plusffval.1 . . 3 𝐵 = (Base‘𝐺)
4 plusffval.2 . . 3 + = (+g𝐺)
5 plusffval.3 . . 3 = (+𝑓𝐺)
63, 4, 5plusffval 17168 . 2 = (𝑥𝐵, 𝑦𝐵 ↦ (𝑥 + 𝑦))
72, 6syl6reqr 2674 1 ( + Fn (𝐵 × 𝐵) → = + )
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1480 × cxp 5072 Fn wfn 5842 ‘cfv 5847 (class class class)co 6604 ↦ cmpt2 6606 Basecbs 15781 +gcplusg 15862 +𝑓cplusf 17160 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1836 ax-6 1885 ax-7 1932 ax-8 1989 ax-9 1996 ax-10 2016 ax-11 2031 ax-12 2044 ax-13 2245 ax-ext 2601 ax-sep 4741 ax-nul 4749 ax-pow 4803 ax-pr 4867 ax-un 6902 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-3an 1038 df-tru 1483 df-ex 1702 df-nf 1707 df-sb 1878 df-eu 2473 df-mo 2474 df-clab 2608 df-cleq 2614 df-clel 2617 df-nfc 2750 df-ne 2791 df-ral 2912 df-rex 2913 df-rab 2916 df-v 3188 df-sbc 3418 df-csb 3515 df-dif 3558 df-un 3560 df-in 3562 df-ss 3569 df-nul 3892 df-if 4059 df-pw 4132 df-sn 4149 df-pr 4151 df-op 4155 df-uni 4403 df-iun 4487 df-br 4614 df-opab 4674 df-mpt 4675 df-id 4989 df-xp 5080 df-rel 5081 df-cnv 5082 df-co 5083 df-dm 5084 df-rn 5085 df-res 5086 df-ima 5087 df-iota 5810 df-fun 5849 df-fn 5850 df-f 5851 df-fv 5855 df-ov 6607 df-oprab 6608 df-mpt2 6609 df-1st 7113 df-2nd 7114 df-plusf 17162 This theorem is referenced by: mgmb1mgm1 17175 mndfo 17236 cnfldplusf 19692 symgtgp 21815
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# WRITING EQUATIONS IN SLOPE-INTERCEPT FORM - "MATH MUGSHOT"
Subject
Resource Type
Common Core Standards
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PDF (Acrobat) Document File
4 MB|11 pages
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Product Description
Student of ALL ages love to draw and color… especially in Math Class!
Math Mugshots are a GREAT way to review math concepts while providing some “color” to math class. You can use this activity to review for a test, as a homework assignment, as extra credit, or as an assessment.
Mathematical Concepts in THIS activity:
• Write an equation when given the GRAPH of an equation.
• Write an equation when given a TABLE OF VALUES.
• Write an equation when given the SLOPE and the Y-INTERCEPT.
• Write an equation when given the SLOPE and an ORDERED PAIR.
• Write an equation when given the Y-INTERCEPT and an ORDERED PAIR. • Write an equation when given TWO ORDERED PAIRS.
How to use this activity:
1. Hand out student worksheets and instruct students to solve each of the problems.
2. When all of the problems are solved, hand out answer sheets. Tell the
students to find and highlight their answers on this sheet. This will
help them to check their work and make corrections, if necessary.
3. Hand out copies of the blank “WANTED” poster. Each correct answer
corresponds to specific feature of the Mugshot. Students will LOVE
drawing and coloring their suspects!
4. Hint: It looks GREAT if the kids outline their pictures with a black Sharpie!
Common Core Standards:
CCSS.MATH.CONTENT.8.F.B.4
Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.
Click the links below for more "mugshots" and other products your students will enjoy!
Integers and Order of Operations Mugshot!
Linear Equations and Slope-Intercept Form Mugshot!
Solving Multi-Step Equations Mugshot
Scatterplots, Functions, and Slope Mugshot
Polynomials and Factoring Mugshot
7th Grade Common Core Review Mugshot
Total Pages
11 pages
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Teaching Duration
1 hour
Report this Resource
\$2.50
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## Percentage Of Calculator
• Post author:
• Post category:News
Percentage Of Calculator. The answer are calculated automatically as you type! To calculate the percentage of a number, we need to use a different formula such as:
Use alcula's percentage calculator to compute percentages and answer questions such as: 250 is 8 percent of what amount? In other words, when you are purchasing a product for \$8000 with discount of 12%, you have pay \$7040 and you will save \$960.
### Let 10% Of 80 = X.
By substituting the corresponding values, you will get 20% x 550= b. Use alcula's percentage calculator to compute percentages and answer questions such as: (40% * 100 + 80% * 60) / (100 + 60) = 55%.
### The 50 Less 12.5% Value Is 43.75.
The easiest way of doing this is by converting the percent values to decimals and then multiplying the two values by each other, and multiply this result by 100 to get the percent of a percent value. 25% = 50% / 2 (\$12 / \$60) × 100% = 20%.
### P/100 * Number = X.
20% × \$60 = (20/100) × \$60 = 0.2 × \$60 = \$12. \$12 / 20% = (\$12 / 20) × 100 = \$60. In the calculator window, choose the question you need answered and enter the 2 quantities that you already know.
### What Is Percentage Difference Calculator?
If we remove the % sign, then we need to express the above formulas as; Convert the problem to an equation using the percentage formula: Before going into details like how our percent composition calculator works or finding the percent composition of a component, let's recall what percent composition was once!
### Percentage Of Marks =(1156/1200) × 100.
The answer are calculated automatically as you type! Percentage of a value calculation. Understanding how to calculate percentages roots itself in the word.
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# Finding residue of complex function, the result is different when using Laurent series and residue theorem.
Find the residue of $$f(z)=\dfrac{z^3+2z+1}{(z-1)(z+3)}$$ on simple pole $$z=1$$.
If I using residue theorem, I have \begin{align} \underset{z=1} {\operatorname{Res}} f(z) = \lim\limits_{z\to 1} (z-1)\dfrac{z^3+2z+1}{(z-1)(z+3)}=\dfrac{1+2+1}{4}=1. \end{align}
If I using Laurent series method, I have \begin{align} f(z)&=\dfrac{z^3+2z+1}{(z-1)(z+3)}\\ &=(z-2)+\dfrac{9z-5}{(z-1)(z+3)}\\ &= -1+(z-1)+\dfrac{9(z+3)-32}{(z-1)(z+3)}\\ &= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{z-1}\cdot\dfrac{1}{z+3}\\ &= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{z-1}\cdot\dfrac{1}{z-1+4}\\ &= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\cdot\dfrac{1}{1+\dfrac{4}{z-1}}\\ &= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\sum\limits_{n=0}^{\infty} (-1)^n \left(\dfrac{4}{z-1}\right)^n\\ &= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\sum\limits_{n=0}^{\infty} (-4)^n \left(z-1\right)^{-n}\\ &= -1+(z-1)+\dfrac{9}{(z-1)}+\sum\limits_{n=0}^{\infty} (-32)(-4)^n \left(z-1\right)^{-n-2}. \end{align} Now I have coefficient of $$(z-1)^{-1}$$ is $$9$$, so we can conclude \begin{align} \underset{z=1} {\operatorname{Res}} f(z) =9. \end{align} My question
When I using residue theorem and Laurent series method, why the result is distinct? What my mistake in my work?
• I think you can't use Taylor expansion of $\frac{1}{1 +\frac{4}{z - 1}}$, because $|\frac{4}{z-1}| \to \infty$ at $1$, but you need $\frac{4}{z-1}$ to be close to zero to use it. – Nik Pronko Apr 28 '20 at 13:26
The "Laurent series method" had an incorrect conclusion going from the fourth to the fifth lines. There was a still an order $$1$$ term on the far right, it just still had to be pulled out.
We can calculate the Laurent series another way. From partial fraction decomposition we have that
$$\frac{4}{(z-1)(z+3)} = \frac{1}{z-1}-\frac{1}{z+3}$$
which means we can rewrite the function as
$$f(z) = \frac{z^3+2z+1}{4(z-1)} - \frac{z^3+2z+1}{4(z+3)}$$
This time the term on the far right has no singularity at $$z=1$$ so it will not contribute to the residue. We will only focus on the term on the left.
Next, shift the variables in the numerator to the desired center:
$$\frac{z^3+2z+1}{4(z-1)} = \frac{(z-1)^3+3(z-1)^2+5(z-1)+4}{4(z-1)}$$
$$= \frac{1}{4}(z-1)^2+\frac{3}{4}(z-1) + \frac{5}{4} + \frac{1}{z-1}$$
which has a residue of $$1$$.
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https://scribesoftimbuktu.com/solve-for-m-3-20m-7-12/
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# Solve for m -3/20=m-7/12
-320=m-712
Rewrite the equation as m-712=-320.
m-712=-320
Move all terms not containing m to the right side of the equation.
Add 712 to both sides of the equation.
m=-320+712
To write -320 as a fraction with a common denominator, multiply by 33.
m=-320⋅33+712
To write 712 as a fraction with a common denominator, multiply by 55.
m=-320⋅33+712⋅55
Write each expression with a common denominator of 60, by multiplying each by an appropriate factor of 1.
Multiply 320 and 33.
m=-3⋅320⋅3+712⋅55
Multiply 20 by 3.
m=-3⋅360+712⋅55
Multiply 712 and 55.
m=-3⋅360+7⋅512⋅5
Multiply 12 by 5.
m=-3⋅360+7⋅560
m=-3⋅360+7⋅560
Combine the numerators over the common denominator.
m=-3⋅3+7⋅560
Simplify the numerator.
Multiply -3 by 3.
m=-9+7⋅560
Multiply 7 by 5.
m=-9+3560
m=2660
m=2660
Cancel the common factor of 26 and 60.
Factor 2 out of 26.
m=2(13)60
Cancel the common factors.
Factor 2 out of 60.
m=2⋅132⋅30
Cancel the common factor.
m=2⋅132⋅30
Rewrite the expression.
m=1330
m=1330
m=1330
m=1330
The result can be shown in multiple forms.
Exact Form:
m=1330
Decimal Form:
m=0.43‾
Solve for m -3/20=m-7/12
### Solving MATH problems
We can solve all math problems. Get help on the web or with our math app
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CC-MAIN-2022-49
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https://www.clutchprep.com/chemistry/practice-problems/104015/the-following-equilibrium-pressures-were-observed-at-a-certain-temperature-for-t
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The Reaction Quotient Video Lessons
Example
# Problem: The following equilibrium pressures were observed at a certain temperature for the reactionN2(g) + 3H2(g) ⇌ 2NH3(g)PNH3 = 3.1 x 10 -2 atmPN2 = 8.5 x 10 -1 atmPH2 = 3.1 x 10 -3 atmCalculate the value for the equilibrium constant Kp at this temperature.If PN2 = 0.525 atm, PNH3 = 0.0167 atm, and PH2 = 0.00761 atm, does this represent a system at equilibrium?
###### FREE Expert Solution
100% (328 ratings)
###### Problem Details
The following equilibrium pressures were observed at a certain temperature for the reaction
N2(g) + 3H2(g) ⇌ 2NH3(g)
PNH3 = 3.1 x 10 -2 atm
PN2 = 8.5 x 10 -1 atm
PH2 = 3.1 x 10 -3 atm
Calculate the value for the equilibrium constant Kp at this temperature.
If PN2 = 0.525 atm, PNH3 = 0.0167 atm, and PH2 = 0.00761 atm, does this represent a system at equilibrium?
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https://developer.aliyun.com/article/1381734
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# 3D Hough变换点云平面检测算法
1.适用场景分析：
2.算法思路：
2.1“投票”算法
2.2法向量的转换
2.3累加器
2.4需要考虑的问题：
3.针对具体应用的改进：
4.分析以及与类似算法进行对比：
HT算法的优点：
HT的缺点：
5.参考文献：
6.实现代码：
#define PI 3.141592653
void HoughTransform(const std::vector<Point>& input, double& A, double& B, double& C, double& D)
{
int n = input.size();
if (n < 3)
return;
double theta_start=0, theta_end=PI;
double phi_start=0, phi_end=PI;
//double phi_start = 0.25*PI, phi_end = 0.75*PI;
double anglestep=PI/90, disstep=0.1;
boundingbox box;
calcboundbox(input, box);
double d_start = -box.diag() / 2.0, d_end = box.diag() / 2.0;
int thetas = ceil((theta_end - theta_start) / anglestep);
int phis = ceil((phi_end - phi_start) / anglestep);
int dises = ceil( box.diag()/disstep);
int*** cube = new int**[thetas];
for (int i = 0; i < thetas;++i)
{
cube[i] = new int*[phis];
for (int j = 0; j < phis; ++j)
{
cube[i][j] = new int[dises];
memset(cube[i][j], 0, sizeof(int)*dises);
}
}
//cos(theta)sin(phi)X+sin(theta)sin(phi)Y+cos(phi)Z = D
Point ptCenter = box.center();
for (int i = 0; i < n;++i)
{
const Point& ptOrigin = input[i];
Point point = ptOrigin - ptCenter;
double theta = theta_start;
for(int j = 0; j < thetas; ++j)
{
int** row = cube[j];
double phi = phi_start;
for (int k = 0; k < phis; ++k)
{
int* col = row[k];
double sinphi = sin(phi);
double d = cos(theta)*sinphi*point.x + sin(theta)*sinphi*point.y + cos(phi)*point.z;
int d_index = floor((d - d_start) / disstep);
++(col[d_index]);
phi += anglestep;
if (phi > phi_end)
break;
}
theta += anglestep;
if (theta > theta_end)
break;
}
}//all points
int buf = 1;
int maxcount = 0;
int xmax, ymax, zmax;
for (int i = 0; i < thetas;++i)
for (int j = 0; j < phis; ++j)
for (int k = buf; k < dises - buf;++k)
{
int count = 0;
for (int x = i - buf; x <= i + buf; ++x)
for (int y = j - buf; y <= j + buf; ++y)
for (int z = k - buf; z <= k + buf; ++z)
{
count += cube[x<0?x+thetas:x%thetas][y<0?y+phis:y%phis][z];
}
if (count > maxcount)
{
xmax = i;
ymax = j;
zmax = k;
maxcount = count;
}
}
double theta = theta_start + xmax*anglestep;
double phi = phi_start + ymax*anglestep;
double d = d_start + zmax*disstep;
A = cos(theta)*sin(phi);
B = sin(theta)*sin(phi);
C = cos(phi);
D = -d - (A*ptCenter.x + B*ptCenter.y+C*ptCenter.z);
//std::cout << A << " , " << B << " , " << C << " , "<< D << std::endl;
//释放cube
for (int i = 0; i < thetas; ++i)
{
int** row = cube[i];
for (int j = 0; j < phis;++j)
{
int* col = row[j];
delete[] col;
}
delete[] row;
}
delete[] cube;
}
class Point
{
public:
double x, y, z;
Point(double ix,double iy,double iz) :
x(ix), y(iy), z(iz){}
Point operator-(const Point& pt) const
{
return Point(x - pt.x, y - pt.y, z - pt.z);
}
};
typedef Point Vector;
class boundingbox
{
public:
double x_min, x_max;
double y_min, y_max;
double z_min, z_max;
public:
double diag() const
{
double dx = x_max - x_min;
double dy = y_max - y_min;
double dz = z_max - z_min;
return sqrt(dx*dx + dy*dy + dz*dz);
}
boundingbox():
x_min(std::numeric_limits<double>::max()),
y_min(std::numeric_limits<double>::max()),
z_min(std::numeric_limits<double>::max()),
x_max(-std::numeric_limits<double>::max()),
y_max(-std::numeric_limits<double>::max()),
z_max(-std::numeric_limits<double>::max())
{}
Point center() const
{
return Point((x_max + x_min) / 2.0,(y_min+y_max) / 2.0, (z_min+z_max) / 2.0);
}
};
void calcboundbox(const std::vector<Point>& input, boundingbox& box)
{
for (int i = 0, n = input.size(); i < n;++i)
{
auto point = input[i];
if (point.x < box.x_min)
box.x_min = point.x;
if (point.y < box.y_min)
box.y_min = point.y;
if (point.z < box.z_min)
box.z_min = point.z;
if (point.x > box.x_max)
box.x_max = point.x;
if (point.y > box.y_max)
box.y_max = point.y;
if (point.z > box.z_max)
box.z_max = point.z;
}
}
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CC-MAIN-2024-38
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https://im.kendallhunt.com/k5_es/teachers/grade-3/unit-7/lesson-5/lesson.html
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# Lesson 5
## Warm-up: Conversación numérica: Dividamos entre 7 (10 minutes)
### Narrative
This Number Talk prompts students to rely on properties of operations and the relationship between multiplication and division to divide within 100. The reasoning here helps students develop fluency in division.
### Launch
• Display one expression.
• “Hagan una señal cuando tengan una respuesta y puedan explicar cómo la obtuvieron” // “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Encuentra mentalmente el valor de cada expresión.
• $$70\div7$$
• $$77\div7$$
• $$63\div7$$
• $$56\div7$$
### Activity Synthesis
• “¿Cómo usaron hechos que ya se saben para encontrar hechos que no se sabían?” // “How did you use facts you know to find facts you didn’t know?” (I used $$70\div7$$ and thought about one more group to find $$77\div7$$. I used $$70\div7$$ and one less group to find $$63\div7$$.)
## Activity 1: Todas las maneras (20 minutes)
### Narrative
The purpose of this activity is to deepen students’ understanding that a shape can belong to multiple categories because of its attributes. Students analyze shapes and determine all the ways that each one could be named. The names may refer to a broad category such as triangle or quadrilateral, or a narrower subcategory such as rhombus or rectangle. As they name the different categories students need to be precise both about the meaning of the categories and verifying the properties of the different shapes (MP6).
MLR8 Discussion Supports. Synthesis: For each observation that is shared, invite students to turn to a partner and restate what they heard using precise mathematical language.
Engagement: Provide Access by Recruiting Interest. Leverage choice around perceived challenge. Invite students to select at least 4 of the 6 problems.
Supports accessibility for: Organization, Attention, Social-Emotional Skills
### Launch
• Groups of 2
• “Examinen el cuadrilátero del primer problema. Individualmente, marquen todos los nombres que se pueden usar para describir el cuadrilátero. Prepárense para compartir su razonamiento” // “Look at the quadrilateral in the first problem. Work independently to circle all the names that you could use to describe the quadrilateral. Be prepared to share your reasoning.”
• 1 minute: independent work time
• “Discutan sus respuestas y su razonamiento con su compañero” // “Discuss your responses and reasoning with your partner.”
• 2 minutes: partner discussion
• Share responses.
### Activity
• “Terminen de resolver los demás problemas individualmente. Prepárense para explicar su razonamiento” // “Complete the rest of the problems independently. Be prepared to explain your reasoning.”
• 3–5 minutes: independent work time
• “Ahora discutan sus respuestas con su compañero. Para cada manera en la que describieron la figura, asegúrense de explicar cómo razonaron. Asegúrense también de hacerle preguntas a su compañero si tienen dudas sobre cómo razonó” // “Now, discuss your answers with your partner. Be sure to explain your reasoning for each way you described the shape. Also, be sure to ask your partner if you have any questions about their reasoning.”
• 5–7 minutes: partner discussion
• Monitor for students who notice that some shapes can be described using multiple terms.
### Student Facing
Selecciona todas las maneras en las que se puede describir cada figura. Prepárate para explicar tu razonamiento.
1. triángulo
4. rombo
5. rectángulo
1. triángulo
3. hexágono
4. rombo
5. rectángulo
1. triángulo
3. pentágono
4. rombo
5. rectángulo
1. triángulo
3. hexágono
4. rombo
5. rectángulo
1. hexágono
3. triángulo
5. rectángulo
6. rombo
1. hexágono
3. triángulo
4. rombo
5. rectángulo
### Student Response
If students use only one name for a shape that can be named in multiple ways, consider asking:
• “¿Cómo describiste la figura?” // “How did you describe the shape?”
• “¿Hay otros nombres que se pueden usar para describir la figura?” // “Are there any other names that could be used to describe the shape?”
### Activity Synthesis
• Select 1–2 students to share the terms they selected for each of the last four quadrilaterals and their reasoning.
• “¿Alguien puede expresar el razonamiento de _____ de otra forma?” // “Who can restate _____’s reasoning in a different way?”
• “¿Alguien quiere agregar algo al razonamiento de _____?” // “Does anyone want to add on to _____’s reasoning?”
• “¿Están de acuerdo o en desacuerdo? ¿Por qué?” // “Do you agree or disagree? Why?”
• “La última figura se puede describir con 4 de las opciones. ¿Cómo es posible que se pueda describir de tantas maneras?” // “The last shape can be described with 4 of the choices. How is it possible that it can be described in so many ways?” (It is a quadrilateral because it has 4 sides. It is a rhombus because it’s a quadrilateral with 4 sides that are the same length. It is a rectangle because it has 4 right angles and 2 pairs of sides that are the same length. It is a square because it has 4 sides that are the same length and 4 right angles. The last three are more specific descriptions of a quadrilateral.)
## Activity 2: Dibujemos una figura que no sea . . . (15 minutes)
### Narrative
The purpose of this activity is for students to apply what they know about the defining attributes of rectangles, rhombuses, and squares to draw shapes that are not those quadrilaterals. They use geometric attributes to explain why their drawings meet the criteria.
### Launch
• Groups of 2
• “Piensen un minuto en cómo pueden dibujar una figura para cada una de estas descripciones” // “Take a minute and think about how you could draw a shape for each one of these descriptions.”
• 1 minute: quiet think time
### Activity
• “Ahora, con su compañero, dibujen una figura para cada enunciado. Prepárense para explicar cómo saben que cada figura corresponde a la descripción dada” // “Now, work with your partner to draw a shape for each statement. Be ready to explain how you know each shape matches the description given.”
• 7–10 minutes: partner work time
### Student Facing
2. Dibuja un cuadrilátero que no sea un rombo.
3. Dibuja un cuadrilátero que no sea un rectángulo.
### Activity Synthesis
• Select students to share their drawings and explanations for the first three problems.
• Highlight explanations that include the defining attributes of squares, rectangles, and rhombuses.
• Invite students to share as many different quadrilaterals as they can think of for the last problem. Display as many as possible.
## Lesson Synthesis
### Lesson Synthesis
“En las últimas lecciones, ¿cómo ha cambiado su manera de pensar sobre cómo puede verse un cuadrilátero?” // “How has your thinking changed over the last few lessons about what a quadrilateral can look like?” (Before, when I thought of quadrilaterals, I thought of rectangles and squares, but now I know they can look so different. Some have right angles and some don’t. Some have sides with equal length and some don’t. They all look really different even though they have some things in common.)
## Student Section Summary
### Student Facing
En esta sección, aprendimos a clasificar figuras según sus características, como el número de lados, las longitudes de sus lados y si los ángulos eran ángulos rectos. También clasificamos cuadriláteros y triángulos en grupos más específicos.
Aprendimos que a una figura se le puede dar un nombre según sus características. Por ejemplo:
• Si un triángulo tiene un ángulo recto, entonces es un triángulo rectángulo.
• Si un cuadrilátero tiene 2 pares de lados que tienen la misma longitud y 4 ángulos rectos, entonces es un rectángulo.
• Si un cuadrilátero tiene lados que tienen todos la misma longitud, entonces es un rombo.
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https://www.gizmodo.com.au/2016/05/the-mathematical-explanation-for-why-you-cant-catch-a-falling-dollar-bill-with-your-fingers/
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# The Mathematical Explanation For Why You Can’t Catch A Falling Dollar Bill With Your Fingers
Here’s a really interesting mathematical explanation on how the “catch a dollar” trick works. You know the trick: a person holds a bill vertically and says you can keep the bill if you can catch the dollar with your fingers when it drops. You never catch it. It’s really hard! Why?
As Numberphile explains, the reaction time in most humans is about 0.2 seconds. That is, it takes about that much time for our eyes to tell our brains to tell our body to do something (in this case, catch the bill). That doesn’t seem that long though, so why is still so hard to get the dollar?
It can be explained with the free fall formula for distance (1/2 gravity x the square of the time falling). When you fill in the formula (gravity at 9.8 m/s and our 0.2 reaction time), you’ll find what is basically, sort of the human reaction time in physical form: 20 cm. In order to catch something, you’ll need it to travel more than 20 cm. A dollar bill measures less than 20 cm (around 15 cm), so you can’t react in time before the dollar bill falls all the way through.
Obviously, some people have faster reaction times, so this doesn’t apply to everybody — but who knew maths could be so delightful?
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# Probability of periodically happening event occurring at a given time based on previous data
Assume that an event $X$ happens periodically over time with a period $P_X$. When it starts it lasts for a time $T_X$. $P_X$ and $T_X$ may vary slightly. There is no correlation between $P_X$ and $T_X$. For example:
1. Event $X$ starts and lasts for 7 days.
2. 20 days of nothing
3. Event $X$ starts and lasts for 5 days.
4. 22 days of nothingt
5. Event $X$ starts and lasts for 6 days.
6. 19 days of nothing
7. etc.
I have a data set over previous time and now I want to be able to predict the probability, as well as the confidence level, of the event occurring e.g 3 months from now, on a given date.
I hope you can help. And sorry if I'm not using the correct statistical terms. I have a decent math knowledge, but never did much with statistics.
Assume that the time between events is a Poisson random variable $X\sim\mathcal{P(\lambda})$ with rate $\lambda$. This is a reasonable model under the given circumstances if it is acceptable that the length of the time intervals between events are independent of each other. If your $P_X$ is, for instance, $20$ days, then $\lambda$ is $\frac{1}{20}$ per day.
Say you want to find the probability that an event occurs after $n$ periods, i.e., after a time $nP_X=\frac{n}{\lambda}.$ To find this, you simply have to add the rates (see this answer), so $$p(X\;\text{happens after}\;n\;\text{periods})=\mathcal{P}(n\lambda).$$
It would then be standard to assume that the length of $X$, call it $Y\sim\mathcal{N}(0,\mu)$ where the event after $n$ periods have been placed at the origin, is normally distributed, with some standard deviation $\mu$, which is related to your $T_X$.
• @Timo A comment on the independence: Remember that $X$ is a random variable specifying the time in between events, and not the event itself (which is not "a quantity" that you can measure). I will edit my answer to include this point more clearly. – Bobson Dugnutt Apr 15 '17 at 11:10
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# Principles of Medical Statistics
2
Principles of Medical Statistics
Julie Morris
University of Manchester, Manchester, UK
## Abstract
A basic knowledge of statistics is an essential skill for medical practitioners. Definitions and worked examples of key statistical concepts are provided, including descriptive summary statistics, hypothesis testing, statistical comparison tests, and correlation. Measures of diagnostic accuracy tests and elements of study design, including sample size and randomisation, are also presented.
Keywordssummary statistics; confidence intervals; significance tests; correlation; diagnostic tests; study design; systematic reviews
## 2.1 Introduction
Statistics and statistical methodology have a central role to play in many areas of the medical world, from research to evidence‐based medicine to published papers in medical journals. It is therefore essential that medical practitioners have a basic understanding of statistical concepts, issues, and techniques. This chapter explains basic statistical principles and covers summary statistics, significance tests, and the design of studies. Examples from the field of urology illustrate each statistical point.
## 2.2 Descriptive Statistics
Data can be summarised in various ways. The most appropriate statistics depend on the type of data and the distribution of values.
### 2.2.1 Qualitative or Categorical Data
For categorical data, such as presence or absence of disease or stage of disease (mild, moderate, or severe), data are summarised by the number of subjects in each category and the percentage in each category. Example 2.1 emphasises the need to look at both the actual numbers as well as the percentages when interpreting categorical data.
We define rate as the number of events (e.g. cases of disease) per unit of population during a particular period of time. Table 2.1 includes the definition of two specific rates, ‘incidence’ and ‘prevalence’.
### 2.2.2 Quantitative or Numerical Data
For quantitative data (numerical data such as length of stay or systolic blood pressure), there is a greater selection of appropriate summary statistics.
Example 2.2 shows data from a cohort of patients undergoing a surgical procedure.
There are a number of different summary statistics available, and they are shown in Table 2.1.
You may ask, ‘Why are different summary statistics used? The answer is that the appropriate summary statistics to choose depend on how the data are distributed.
If data are normally distributed, then the appropriate summary statistics are the mean, the standard deviation (SD), or the variance. [Note the variance is the square of the SD].
If data are not normally distributed, then the appropriate summary statistics are the median, the range, or the interquartile range. The latter is the 25th to the 75th percentile (i.e. when the data are ordered from the smallest to the largest, the lower value of the interquartile range is the ‘quarter’ point, and the higher value is the ‘three‐quarter’ point). It contains the middle 50% of the data and is sometimes used instead of the range when the sample size of the data set is very large (n = 100 or more).
Hence, for quantitative data, the distribution of data needs to be assessed before deciding on the appropriate summary statistics.
The easiest way to assess the distribution of data is to look at a histogram of the values (Figure 2.1).
The data are separated into sections (bars) – in this example the bars correspond to 5 ml categories – and the height of the bars correspond to the number of people in that particular section (i.e. blood loss category). Data are said to be normally distributed if the shape of the histogram is a symmetric upturned U‐shape or bell‐shape.
This is true for the blood loss data, and hence, blood loss would be said to be normally distributed.
Figure 2.2 shows a histogram of the operating time, and is an example of not normally distributed data.
The histogram shows a very skewed distribution (not a symmetric bell shape). These data are ‘positively skewed’ (the tail of the distribution is towards the right), which is a common occurrence in medical data where just a few patients have high abnormal values. These values are called ‘outliers’ and would have a large influence on the average (mean) value, which would make it an inappropriate summary statistic. Median values are unaffected by outliers. For ‘negatively skewed’ data, the tail of the distribution is towards the left, and there would be a few small abnormal values (outliers).
There are a number of ways of checking data distributions. Any combination of these four properties of a normal distribution can be used.
• Symmetric histogram
• Mean is approximately the same as the median
• Standard deviation is less than mean
• No outliers
Figure 2.3 shows a flow chart to aid the selection of the most appropriate summary statistics.
## 2.3 Confidence Intervals
The estimated mean change in maximum voiding pressure (MVP) after treatment with a muscle relaxant was 16 cmH 2 0 with 95% confidence interval (8 cmH 2 0, 24 cmH 2 0).’
The main purpose of confidence intervals (CIs) is to indicate the precision with which an estimate is calculated from the study sample. It presents a range of values in which the population value (the ‘true’ value) may lie with a reasonable level of confidence. The (im)precision is indicated by the width of the CI; the narrower the interval is, the better the precision. The width depends on three factors. The width decreases (and precision increases) with a larger sample size, lower variability between subjects, and lower confidence (a 90% CI will be narrower than a 95% CI).
It is important to note that the interpretation of the CI does not directly relate to the actual observations. That is, a 95% CI does not contain 95% of observations. Instead, it relates to the accuracy of the estimated effect size (e.g. the mean difference pre to post for a single group of patients).
CIs can be thought of as bridging the gap between summary statistics and formal significance tests (our next topic).
If the 95% CI in the example is changed to (−4 cmH 2 0, 36 cmH 2 0), then, because it contains zero, it means that a zero‐change in MVP is feasible. Hence, the study has not shown good evidence that there is any increase in MVP with muscle relaxant. This reflects the result that would be obtained by a formal significance test. That is, there is no significant change in MVP.
## 2.4 Significance Tests
When carrying out a clinical study, the aim is to measure the strength of evidence provided by the data for and against a specific proposition.
Suppose we have two types of surgery, X and Y, and wish to find out whether the complication rate after X is lower than after Y.
The results of the study are depicted in Table 2.2:
Do these data show enough evidence that, in general, patients having surgery X have a significant chance of a lower complication rate than those having surgery Y?
To answer this question, we carry out a statistical significance test.
In carrying out such a test, we are looking to see whether the data from the study supports one of two scenarios or hypotheses. One scenario is that the complication rates are the same (the null hypothesis). The other scenario is that the complication rates are different (the alternative hypothesis).
In general terms:
Null hypothesis:
Effects of X and Y are the same
Alternative hypothesis:
Effects of X and Y are different
To find out which scenario is best supported by the data, we calculate a ‘p‐value’.
The p‐value is a probability. Probabilities correspond to the chance of something (e.g. an event/situation) happening or being true. It takes values between 0 and 1.
A probability of 0 means that there is no chance of the event happening or the situation being true. A probability of 1 means that we are certain that the event does happen or that the situation is true.
In the context of significance tests, the p‐value corresponds to the probability of the null scenario/hypothesis being true, given the evidence from the study data. It is derived using a mathematical formula on the study data.
If the p‐value is small (by convention ≤ 0.05), then we say that the null scenario (that X and Y have the same effect) is unlikely to be true. Hence, we say the alternative scenario (that X and Y have different effects) is likely to be true. We state, therefore, that the difference between X and Y is ‘statistically significant’.
If the p‐value is large (by convention > 0.05), then we say that the null scenario (that X and Y have the same effect) could be true. We state, therefore, that the difference between X and Y is ‘not statistically significant’.
In the example, the p‐value for the comparison of complication rates was 0.15.
How should this be interpreted?
It is greater than 0.05, and hence, we conclude that the data have not given us sufficient evidence to determine that there is a difference between X and Y. Thus, patients having surgery X do not have a significant chance of a lower complication rate than those having surgery Y, and we say the difference is not statistically significant.
### 2.4.1 What Statistical Test Should Be Used?
This depends on the type of data and the specific comparison being made.
Figure 2.4 shows the process for selecting the appropriate test for the comparison of different groups of subjects when the outcome is a quantitative or numerical measure (Example 2.3).
Aug 6, 2020 | Posted by in UROLOGY | Comments Off on Principles of Medical Statistics
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# SOLUTION: Can someone help with this please Solve by completing the square. m squared - (25 over 2)m = 5 over 2 The solutions are ? (hint Use redicalsas needed. Use a comma to s
Algebra -> -> SOLUTION: Can someone help with this please Solve by completing the square. m squared - (25 over 2)m = 5 over 2 The solutions are ? (hint Use redicalsas needed. Use a comma to s Log On
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Click here to see ALL problems on Quadratic Equations Question 259080: Can someone help with this please Solve by completing the square. m squared - (25 over 2)m = 5 over 2 The solutions are ? (hint Use redicalsas needed. Use a comma to separate answers.) THank you for your helpAnswer by richwmiller(10283) (Show Source): You can put this solution on YOUR website!"Use redicalsas needed." I am not sure I know how! It has been a long time since I used any redicalsas. Needed or not! lol. Use a comma to separate answers. How generous they are with commas today! m^2-25m/2=5/2 to complete the square we take 1/2 of the coefficient of m and and square it. (25/4)^2 and add it to both sides. m^2-25m/2+625/16=40/16+625/16=665/16 now we take the square root of both sides (m-25/4)^2=sqrt(665/16) m=25/4+\- sqrt(665/16) m=25/4+ sqrt(665/16) m=25/4- sqrt(665/16) and ,
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# Interpolation
```Interpolation
Assoc. Prof. Dr. Siridech Boonsang
Electrical Engineering
xi
x
xi+1
Spline Interpolation
• Polynomials are the most common choice of
interpolants.
• There are cases where polynomials can lead to
erroneous results because of round off error and
overshoot.
• Alternative approach is to apply lower-order
polynomials to subsets of data points. Such
connecting polynomials are called spline functions.
Spline provides a superior approximation of the behavior
of functions that have local, abrupt changes (d).
Linear Spline
The first order splines for a group of ordered data
points can be defined as a set of linear functions:
f ( x) f ( x0 ) m0 ( x x0 )
f ( x) f ( x1 ) m1 ( x x1 )
x0 x x1
x1 x x2
f ( x) f ( xn1 ) mn1 ( x xn1 ) xn1 x xn
f ( xi 1 ) f ( xi )
mi
xi 1 xi
Linear spline - Example
Fit the following data with first order splines. Evaluate the
function at x = 5.
x
f(x)
3.0
4.5
7.0
9.0
2.5
1.0
2.5
0.5
2.5 1
m
0.6
7 4.5
f (5) f (4.5) m(5 4.5)
1.0 0.6 0.5
1.3
Parametric Continuity Conditions
•
Zero order parametric (C0): Simply the curves meet,
C1(1) = C2(0) .
•
First order parametric (C1): The first parametric
derivations for two successive curve sections are equal at
their joining point, C´1(1)= C´2(0)
Parametric Continuity Conditions
•
Second order parametric (C2): Both the first and second
parametric derivatives of the two curve sections are the
same at the intersection, C˝1(1)= C˝2(0)
Linear Spline
• The main disadvantage of linear spline is that they are not
smooth. The data points where 2 splines meets called (a knot),
the changes abruptly.
• The first derivative of the function is discontinuous at these
points.
• Using higher order polynomial splines ensure smoothness at
the knots by equating derivatives at these points.
• Objective: to derive a second order polynomial for each
2
interval between data points. f i ( x) ai x bi x ci
• Terms: Interior knots and end points
For n+1 data points:
• i = (0, 1, 2, …n),
• n intervals,
• 3n unknown
constants (a’s, b’s and
c’s)
• The function values of adjacent polynomial must be
equal at the interior knots 2(n-1).
ai 1 xi 1 bi 1 xi 1 ci 1 f i ( xi 1 ) i 2, 3, 4,..., n
2
ai xi 1 bi xi 1 ci f i ( xi 1 )
2
i 2, 3, 4,..., n
• The first and last functions must pass through the end
points (2).
a1 x0 b1 x0 c1 f ( x0 )
2
an xn bn xn cn f ( xn )
2
• The first derivatives at the interior knots must be equal
(n-1).
f i ( x) 2ai x bi
'
2ai 1 xi 1 bi 1 2ai xi 1 bi
• Assume that the second derivate is zero at the first point
(1)
a1 0
(The first two points will be connected by a straight line)
Fit the following data with quadratic splines. Estimate
the value at x = 5.
x
f(x)
3.0
4.5
7.0
9.0
2.5
1.0
2.5
0.5
Solutions:
There are 3 intervals (n=3), 9 unknowns.
1. Equal interior points:
For first interior point (4.5, 1.0)
The 1st equation:
x12 a1 x1b1 c1 f ( x1 )
(4.5) 2 a1 4.5b1 c1 f (4.5)
The 2nd equation:
20.25 a1 4.5 b1 c1 1.0
x12 a2 x1b2 c2 f ( x1 )
(4.5) 2 a2 4.5b2 c2 f (4.5)
20.25a2 4.5b2 c2 1.0
For
second interior point (7.0, 2.5)
The 3rd equation:
x22 a2 x2b2 c2 f ( x2 )
(7) 2 a2 7b2 c2 f (7)
49a2 7b2 c2 2.5
The 4th equation:
x22 a3 x2b3 c3 f ( x2 )
(7) a3 7b3 c3 f (7)
2
49a3 7b3 c3 2.5
First and last functions pass the end points
For the start point (3.0, 2.5)
x a1 x0 b1 c1 f ( x0 )
2
0
9a1 3b1 c1 2.5
For the end point (9, 0.5)
x a x3b3 c3 f ( x3 )
2
3 1
81a3 9b3 c3 0.5
Equal derivatives at the interior knots.
For first interior point (4.5, 1.0)
2 x1 a1 b1 2 x1 a2 b2
9a1 b1 9a2 b2
For second interior point (7.0, 2.5)
2 x2 a2 b2 2 x3 a3 b3
14a2 b2 14a3 b3
Second derivative at the first point is 0
f '' ( x0 ) a1 0
4.5 1
0 0
0 0
0 0
3 1
0 0
1 0
0 0
0
0
0
0
0
20.25 4.5 1
0
0
49
7
1
0
0
0
0
0
49
7
0
0
0
0
0
0
0
0
81
9
0
0
9 1
0
14
0
1
14 1
0 b1 1
c
0 1 1
0 a 2 2 .5
1 b2 2.5
2 .5
0 c2
1 a3 0.5
0 b3 0
0 c3 0
Solving these 8 equations with 8 unknowns
a1 0,
b1 1,
c1 5.5
a2 0.64, b2 6.76, c2 18.46
a3 1.6, b3 24.6,
c3 91.3
f1 ( x) x 5.5,
3.0 x 4.5
f 2 ( x) 0.46 x 2 6.76 x 18.46,
4.5 x 7.0
f3 ( x) 1.6 x 2 24.6 x 91.3,
7.0 x 9.0
Cubic Splines
Objective: to derive a third order polynomial for
each interval between data points.
Terms: Interior knots and end points
f i ( x) ai x bi x ci x d i
3
2
For n+1 data points:
• i = (0, 1, 2, …n),
• n intervals,
• 4n unknown constants (a’s, b’s ,c’s and d’s)
Cubic Splines
• The function values must be equal at the interior knots (2n-2).
• The first and last functions must pass through the end points (2).
• The first derivatives at the interior knots must be equal (n-1).
• The second derivatives at the interior knots must be equal (n-1).
• The second derivatives at the end knots are zero (2), (the 2nd derivative
function becomes a straight line at the end points)
Alternative technique to get Cubic Splines
• The second derivative within each interval [xi-1, xi ] is a straight line. (the 2nd
derivatives can be represented by first order Lagrange interpolating polynomials.
x xi
x xi 1
''
f i ( x) f i ( xi 1 )
f i ( xi )
xi 1 xi
xi xi 1
''
''
A straight line
connecting the first
knot f’’(xi-1) and the
second knot f’’(xi)
The second derivative at any point x within the interval
Cubic Splines
• The last equation can be integrated twice
2 unknown constants of integration can be evaluated by
applying the boundary conditions:
1. f(x) = f (xi-1) at xi-1
2. f(x) = f (xi) at xi
''
f i ( x)
''
f i ( xi 1 )
xi x 3 f i ( xi ) x xi 1 3
6xi xi 1
6xi xi 1
f i ( xi 1 ) f i ( xi 1 )xi xi 1
xi x
6
xi xi 1
''
''
f i ( xi )
f i ( xi )xi xi 1
x xi 1
6
xi xi 1
Unknowns:
f ' ' ( xi )
f ' ' ( xi 1 )
i = 0, 1,…, n
Cubic Splines
•For each interior point xi (n-1):
f i1 ( xi ) f i ( xi )
'
'
( xi xi 1 ) f '' ( xi 1 ) 2( xi 1 xi 1 ) f '' ( xi )
( xi 1 xi ) f '' ( xi 1 )
6
f ( xi 1 ) f ( xi )
xi 1 xi
6
f ( xi 1 ) f ( xi )
xi xi 1
This equation result with n-1 unknown second
derivatives where, for boundary points:
f˝(xo) = f˝(xn) = 0
Cubic Splines - Example
Fit the following data with cubic splines
Use the results to estimate the value at x=5.
x
f(x)
3.0
2.5
4.5
1.0
7.0
2.5
9.0
0.5
Solution:
Natural Spline:
f '' ( x0 ) f '' (3) 0,
f '' ( x3 ) f '' (9) 0
Cubic Splines - Example
For 1st interior point (x1 = 4.5)
x
f(x)
3.0
2.5
4.5
1.0
7.0
2.5
9.0
0.5
- xi xi 1 x1 x0 4.5 3.0 1.5
- xi 1 xi 1 x2 x0 7 3.0 4
- xi 1 xi x2 x1 7 4.5 2.5
Apply the following equation:
( xi xi 1 ) f '' ( xi 1 ) 2( xi 1 xi 1 ) f '' ( xi ) ( xi 1 xi ) f '' ( xi 1 )
6
6
f ( xi 1 ) f ( xi )
f ( xi 1 ) f ( xi )
xi 1 xi
xi xi 1
Cubic Splines - Example
6
6
1.5 f (3) 2 4 f (4.5) 2.5 f (7)
(2.5 1)
(2.5 1)
2.5
1.5
''
f
(3) 0
Since
''
''
''
8 f '' (4.5) 2.5 f '' (7) 9.6 .............. (eq.1)
For 2nd interior point (x2 = 7 )
x
3.0
4.5
7.0
9.0
f(x)
2.5
1.0
2.5
0.5
xi xi 1 x2 x1 7 4.5 2.5
xi 1 xi 1 x3 x1 9 4.5 4.5
xi 1 xi x3 x2 9 7 2
Cubic Splines - Example
Apply the following equation:
( xi xi 1 ) f '' ( xi 1 ) 2( xi 1 xi 1 ) f '' ( xi ) ( xi 1 xi ) f '' ( xi 1 )
6
6
f ( xi 1 ) f ( xi )
f ( xi 1 ) f ( xi )
xi 1 xi
xi xi 1
2.5 f '' (4.5) 2 4.5 f '' (7) 2 f '' (9)
Since
6
6
(0.5 2.5)
(1 2.5)
2
2.5
f '' (9) 0
2.5 f '' (4.5) 9 f '' (7) 9.6 ............. (equ 2)
Cubic Splines - Example
Solve the two equations:
8 f i '' (4.5) 2.5 f i '' (7) 9.6
''
''
yeild
f
(
4
.
5
)
1
.
67909
,
f
(7) 1.53308
''
''
2.5 f i (4.5) 9 f i (7) 9.6
The first interval (i=1), apply for the equation:
''
''
f (x )
f i ( xi )
3
x xi 1 3
f i ( x) i i 1 xi x
6 xi xi 1
6 xi xi 1
''
f i ( xi 1 ) f i '' ( xi 1 ) xi xi 1
f i ( xi )
f i ( xi ) xi xi 1
xi x
x xi 1
6
6
xi xi 1
xi xi 1
f1 ( x) 0 ( xi 3)3
1.67909
2.5 0(1.5)
1 1.67909(1.5)
( x 3)3
4
.
5
x
( x 3)
6(1.5)
1
.
5
6
1
.
5
6
f1 ( x) 0.186566 ( x 3)3 1.6667 (4.5 x) 0.24689( x 3)
Cubic Splines - Example
The 2nd interval (i =2), apply for the equation:
f 2 ( x)
1.67909
1.53308
1 1.67909(2.5)
7 x
(7 x ) 3
( x 4.5)3
6(2.5)
6(2.5)
6
2.5
2.5 1.53308(2.5)
( x 4.5)
6
2.5
f 2 ( x) 0.111939(7 x)3 0.102205 ( x 4.5)3 0.29962(7 x) 1.638783 ( x 4.5)
The 3rd interval (i =3),
f 3 ( x) 0.127757(9 x)3 1.761027 (9 x) 0.25 ( x 7)
For x = 5:
f 2 ( x) f 2 (5) 1.102886
```
| 5,363 | 9,093 |
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| 4.15625 | 4 |
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| 413,530,831 | 20,873 |
Just because harmonics is becoming a more prevalent problem, that doesn't mean the subject is getting any easier to understand Harmonics are AC voltages and currents with frequencies that are integer multiples of the fundamental frequency. On a 60-Hz system, this could include 2nd order harmonics (120 Hz), 3rd order harmonics (180 Hz), 4th order harmonics (240 Hz), and so on. Normally, only odd-order
Just because harmonics is becoming a more prevalent problem, that doesn't mean the subject is getting any easier to understand
Harmonics are AC voltages and currents with frequencies that are integer multiples of the fundamental frequency. On a 60-Hz system, this could include 2nd order harmonics (120 Hz), 3rd order harmonics (180 Hz), 4th order harmonics (240 Hz), and so on. Normally, only odd-order harmonics (3rd, 5th, 7th, 9th) occur on a 3-phase power system. If you observe even-order harmonics on a 3-phase system, you more than likely have a defective rectifier in your system.
If you connect an oscilloscope to a 120V receptacle, the image on the screen usually isn't a perfect sine wave. It may be very close, but it will likely be different in one of several ways. It might be slightly flattened or dimpled as the magnitude approaches its positive and negative maximum values (Fig. 1). Or perhaps the sine wave is narrowed near the extreme values, giving the waveform a peaky appearance (Fig. 2 below). More than likely, random deviations from the perfect sinusoid occur at specific locations on the sine wave during every cycle (Fig. 3 below).
The flattened and dimpled sinusoid in Fig. 1 has the mathematical equation, y=sin (x)+0.25 sin (3x). This means a 60-Hz sinusoid (the fundamental frequency) added to a second sinusoid with a frequency three times greater than the fundamental (180 Hz) and an amplitude ¼ (0.25 times) of the fundamental frequency produces a waveform similar to the first part of Fig. 1. The 180-Hz sinusoid is called the third harmonic, since its frequency is three times that of the fundamental frequency.
Similarly, the peaky sinusoid in Fig. 2 has the mathematical equation, y=sin (x) -0.25 sin (3x). This waveform has the same composition as the first waveform, except the third harmonic component is out of phase with the fundamental frequency, as indicated by the negative sign preceding the “0.25 sin (3x)” term. This subtle mathematical difference produces a very different appearance in the waveform.
The waveform in Fig. 3 contains several other harmonics in addition to the third harmonic. Some are in phase with the fundamental frequency and others out of phase. As the harmonic spectrum becomes richer in harmonics, the waveform takes on a more complex appearance, indicating more deviation from the ideal sinusoid. A rich harmonic spectrum may completely obscure the fundamental frequency sinusoid, making a sine wave unrecognizable.
Analyzing harmonics. When the magnitudes and orders of harmonics are known, reconstructing the distorted waveform is simple. Adding the harmonics together, point by point, produces the distorted waveform. The waveform in Fig. 1 is synthesized in Fig. 4 by adding the magnitudes of the two components, the fundamental frequency (red waveform) and the third harmonic (blue waveform), for each value of x, which results in the green waveform.
Decomposing a distorted waveform into its harmonic components is considerably more difficult. This process requires Fourier analysis, which involves a fair amount of calculus. However, electronic equipment has been developed to perform this analysis on a real-time basis. One manufacturer offers a 3-phase power analyzer that can digitally capture 3-phase waveforms and perform a host of analysis functions, including Fourier analysis, to determine harmonic content. Another manufacturer offers similar capabilities for single-phase applications. Easy-to-use analyzers like these can help detect and diagnose harmonic-related problems on most power systems.
What causes harmonics? If harmonic voltages aren't generated intentionally, where do they come from? One common source of harmonics is iron core devices like transformers. The magnetic characteristics of iron are almost linear over a certain range of flux density, but quickly saturate as the flux density increases. This nonlinear magnetic characteristic is described by a hysteresis curve. Because of the nonlinear hysteresis curve, the excitation current waveform isn't sinusoidal. A Fourier analysis of the excitation current waveform reveals a significant third harmonic component, making it similar to the waveform shown in Fig. 2.
Core iron isn't the only source of harmonics. Generators themselves produce some 5th harmonic voltages due to magnetic flux distortions that occur near the stator slots and nonsinusoidal flux distribution across the air gap. Other producers of harmonics include nonlinear loads like rectifiers, inverters, adjustable-speed motor drives, welders, arc furnaces, voltage controllers, and frequency converters.
Semiconductor switching devices produce significant harmonic voltages as they abruptly chop voltage waveforms during their transition between conducting and cutoff states. Inverter circuits are notorious for producing harmonics, and are in widespread use today. An adjustable-speed motor drive is one application that makes use of inverter circuits, often using pulse width modulation (PWM) synthesis to produce the AC output voltage. Various synthesis methods produce different harmonic spectra. Regardless of the method used to produce an AC output voltage from a DC input voltage, harmonics will be present on both sides of the inverter and must often be mitigated.
Effects of harmonics. Besides distorting the shape of the voltage and current sinusoids, what other effects do harmonics cause? Since harmonic voltages produce harmonic currents with frequencies considerably higher than the power system fundamental frequency, these currents encounter much higher impedances as they propagate through the power system than does the fundamental frequency current. This is due to “skin effect,” which is the tendency for higher frequency currents to flow near the surface of the conductor. Since little of the high-frequency current penetrates far beneath the surface of the conductor, less cross-sectional area is used by the current. As the effective cross section of the conductor is reduced, the effective resistance of the conductor is increased. This is expressed in the following equation:
where R is the resistance of the conductor, ρ is the resistivity of the conductor material, L is the length of the conductor, and A is the cross-sectional area of the conductor. The higher resistance encountered by the harmonic currents will produce a significant heating of the conductor, since heat produced — or power lost — in a conductor is I2R, where I is the current flowing through the conductor.
This increased heating effect is often noticed in two particular parts of the power system: neutral conductors and transformer windings. Harmonics with orders that are odd multiples of the number three (3rd, 9th, 15th, and so on) are particularly troublesome, since they behave like zero-sequence currents. These harmonics, called triplen harmonics, are additive due to their zero-sequence-like behavior. They flow in the system neutral and circulate in delta-connected transformer windings, generating excessive conductor heating in their wake.
Reducing the effects of harmonics. Because of the adverse effect of harmonics on power system components, the IEEE developed standard 519-1992 to define recommended practices for harmonic control. This standard also stipulates the maximum allowable harmonic distortion allowed in the voltage and current waveforms on various types of systems.
Two approaches are available for mitigating the effects of excessive heating due to harmonics, and a combination of the two approaches is often implemented. One strategy is to reduce the magnitude of the harmonic waveforms, usually by filtering. The other method is to use system components that can handle the harmonics more effectively, such as finely stranded conductors and k-factor transformers.
Harmonic filters can be constructed by adding an inductance (L) in series with a power factor correction capacitor (C). The series L-C circuit can be tuned for a frequency close to that of the troublesome harmonic, which is often the 5th. By tuning the filter in this way, you can attenuate the unwanted harmonic.
Filtering isn't the only means of reducing harmonics. The switching angles of an inverter can be preselected to eliminate some harmonics in the output voltage. This can be a very cost-effective means of reducing inverter-produced harmonics.
Since skin effect is responsible for the increased heating caused by harmonic currents, using conductors with larger surface areas will lessen the heating effects. This can be done by using finely stranded conductors, since the effective surface area of the conductor is the sum of the surface area of each strand.
Specially designed transformers called k-factor transformers are also advantageous when harmonic currents are prevalent. They parallel small conductors in their windings to reduce skin effect and incorporate special core designs to reduce the saturation effects at the higher flux frequencies produced by the harmonics.
You should also increase the size of neutral conductors to better accommodate triplen harmonics. Per the FPN in 210.4(A) and 220.22 of the 2002 NEC, “A 3-phase, 4-wire wye-connected power system used to supply power to nonlinear loads may necessitate that the power system design allow for the possibility of high harmonic neutral currents.” And per 310.15(B)(4)(c), “On a 4-wire, 3-phase wye circuit where the major portion of the load consists of nonlinear loads, harmonic currents are present on the neutral conductor: the neutral shall therefore be considered a current-carrying conductor.” It's important to note that the duct bank ampacity tables in B.310.5 through B.310.7 are designed for a maximum harmonic loading on the neutral conductor of 50% of the phase currents.
Harmonics will undoubtedly continue to become more of a concern as more equipment that produces them is added to electrical systems. But if adequately considered during the initial design of the system, harmonics can be managed and their detrimental effects avoided.
Fehr is an independent engineering consultant located in Clearwater, Fla.
| 2,123 | 10,560 |
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| 421,009,800 | 7,355 |
# 204123 (number)
204,123 (two hundred four thousand one hundred twenty-three) is an odd six-digits composite number following 204122 and preceding 204124. In scientific notation, it is written as 2.04123 × 105. The sum of its digits is 12. It has a total of 2 prime factors and 4 positive divisors. There are 136,080 positive integers (up to 204123) that are relatively prime to 204123.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 12
• Digital Root 3
## Name
Short name 204 thousand 123 two hundred four thousand one hundred twenty-three
## Notation
Scientific notation 2.04123 × 105 204.123 × 103
## Prime Factorization of 204123
Prime Factorization 3 × 68041
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 204123 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 204,123 is 3 × 68041. Since it has a total of 2 prime factors, 204,123 is a composite number.
## Divisors of 204123
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 272168 Sum of all the positive divisors of n s(n) 68045 Sum of the proper positive divisors of n A(n) 68042 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 451.8 Returns the nth root of the product of n divisors H(n) 2.99996 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 204,123 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 204,123) is 272,168, the average is 68,042.
## Other Arithmetic Functions (n = 204123)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 136080 Total number of positive integers not greater than n that are coprime to n λ(n) 68040 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 18272 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 136,080 positive integers (less than 204,123) that are coprime with 204,123. And there are approximately 18,272 prime numbers less than or equal to 204,123.
## Divisibility of 204123
m n mod m 2 3 4 5 6 7 8 9 1 0 3 3 3 3 3 3
The number 204,123 is divisible by 3.
## Classification of 204123
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (204123)
Base System Value
2 Binary 110001110101011011
3 Ternary 101101000010
4 Quaternary 301311123
5 Quinary 23012443
6 Senary 4213003
8 Octal 616533
10 Decimal 204123
12 Duodecimal 9a163
20 Vigesimal 15a63
36 Base36 4di3
## Basic calculations (n = 204123)
### Multiplication
n×y
n×2 408246 612369 816492 1020615
### Division
n÷y
n÷2 102062 68041 51030.8 40824.6
### Exponentiation
ny
n2 41666199129 8505029564808867 1736072149857480358641 354372255445358463246876843
### Nth Root
y√n
2√n 451.8 58.8795 21.2556 11.534
## 204123 as geometric shapes
### Circle
Diameter 408246 1.28254e+06 1.30898e+11
### Sphere
Volume 3.56258e+16 5.23593e+11 1.28254e+06
### Square
Length = n
Perimeter 816492 4.16662e+10 288674
### Cube
Length = n
Surface area 2.49997e+11 8.50503e+15 353551
### Equilateral Triangle
Length = n
Perimeter 612369 1.8042e+10 176776
### Triangular Pyramid
Length = n
Surface area 7.2168e+10 1.00233e+15 166666
## Cryptographic Hash Functions
md5 166ac1cceef4d74cfb37a873310d7da4 7c0c9358ea38104404603e0467e5094d884a25b2 bba38541f8adcbc60eba90022d05127508f89858a4b297a411f75bcd1a56a084 a1d0118f8bb85813cdfa1439fd65b83784ea92e92fabce517b544c7eed54e4b4ba505c3e2ded11e451b75ac17c85428ed7f1f219696f71f82b031c690a9d9f0b 673db2eceb5e28bcb20beb12fe35a6f34e0a2d75
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# physics180
posted by .
A satellite is in circular orbit at a height R above earth's surface.
a)find orbital period.
b)what height is required for a circular orbit with a period double that found in part (a)?
• physics180 -
A satellite is in circular orbit at a height R above earth's surface.
a)find orbital period.
b)what height is required for a circular orbit with a period double that found in part (a)?
Allow me to modify your terms.
a) The orbit radius R = Re + h where Re = the earth's radius and h = the altitude of the satellite in feet.
The orbital period of a satellite derives from
T = 2(Pi)sqrt(R^3/µ) where T = the orbital period in seconds, R = the radius of the satellite orbit and µ = the earth's gravitational constant = 1.407974x10^16 ft^3/sec^2.
b) For T = 4(Pi)sqrt[(Re + h)^3/µ)
h = R - Re = [(Tµ)/(16Pi^2)]^(1/3) - Re
| 238 | 851 |
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| 636,093,664 | 7,863 |
## Use Newton’s method with the specified initial approximation x1 to find x3.
1. The problem statement, all variables and given/known data
Use Newton’s method with the specified initial approximation to find , the third approximation to the root of the given equation. (Give your answer to four decimal places.)
x^5+2=0, x_1=-1
2. Relevant equations
3. The attempt at a solution
x5+2=0, x1=-1
y'=5x4
x(n+1)=xn-(x5+2)/(5x4 )
For n=1
x2=-1-((-1)5+2)/(5(-1)4 )=-6/5=-1.2
For n=2
x3=-1.2-((-1.2)5+2)/(5(-1.2)4 )= -1.2-0.4883/10.368=-1.2+0.047=-1.1530
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Quote by phillyolly Please verify my answer. 1. The problem statement, all variables and given/known data Use Newton’s method with the specified initial approximation to find , the third approximation to the root of the given equation. (Give your answer to four decimal places.) x^5+2=0, x_1=-1 2. Relevant equations 3. The attempt at a solution x5+2=0, x1=-1 y'=5x4 x(n+1)=xn-(x5+2)/(5x4 )
This should be
x(n+1)=xn-(xn5+2)/(5xn4 )
For n=1 x2=-1-((-1)5+2)/(5(-1)4 )=-6/5=-1.2 For n=2 x3=-1.2-((-1.2)5+2)/(5(-1.2)4 )= -1.2-0.4883/10.368=-1.2+0.047=-1.1530
Looks good, but you aren't at "4 decimal places" yet. Continue until you get two consecutive results that are the same to 4 decimal places (and I would recommend doing the calculations to at least 5 decimal places until then).
Hi! What do you mean by
Quote by HallsofIvy Looks good, but you aren't at "4 decimal places" yet.
?
The task says to do the third approximation, which I found already....
Do you mean I should do the fourth...and the fifth?...
Mentor
## Use Newton’s method with the specified initial approximation x1 to find x3.
I think that HallsOfIvy missed the part about the third approximation, so you're done.
| 636 | 2,008 |
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| 343,214,875 | 42,223 |
# How many square units are in an office?
## How many square units are in an office?
This is defined as the area of the office in square units. The formula for finding the area of a figure is length x width = area. If an office measures 13 units by 9 units, then the office is 13 units long and 9 units wide. The area is 13 x 9 = 117 square units, or 117 units squared.
## Is area square units or units?
A shape with an area of three square metres would have the same area as three such squares. In mathematics, the unit square is defined to have area one, and the area of any other shape or surface is a dimensionless real number….
Area
SI unit Square metre [m2]
In SI base units 1 m2
Dimension
## How many square units are in an office that is 13 unit by 9 units?
Square units 13 by 9 of an office means office of length 13 units and breadth 9 units. Now its area is 13x 9 = 117 square units or units square.
## How many cans of paint are needed to cover an area of 2200 square units if one can of paint covers in area 400 square units?
So, in total you will need 6 cans of paint to cover all 2,200 sq units.
## Why is area calculated in square units?
Answer: Area is measured in square units because The area of a figure is the number of squares required to cover it completely, like tiles on a floor. Area of a square = side times side. Since each side of a square is the same, it can simply be the length of one side squared.
## How many cans of paint are needed?
Paint Calculator Tip 2: Two gallon cans of paint cover up to 800 square feet, which is enough to cover an average size room. This is the most common amount needed, especially when considering second coat coverage. Paint Calculator Tip 3: Three gallon cans of paint cover up to 1200 square feet.
## What is a square that is 1 unit wide and 1 unit long?
A unit square is a square that is one unit long by one unit wide where the unit of length can be millimeters, inches, feet, miles, etc. To find the area of a rectangle, you need to imagine it being cut into its unit squares, and then count the number of unit squares inside that rectangle.
## Is a perimeter square units?
Since all sides of a square are equal, we only need one side to find its perimeter. The perimeter of the given square is: a + a + a + a = 4 a units. Hence, the formula of the perimeter of a square = 4 × (length of any one side).
## How do you calculate area and perimeter?
Divide the perimeter by 4: that gives you the length of one side. Then square that length: that gives you the area. In this example, 14 ÷ 4 = 3.5.
## What area will 1 Litre of paint cover?
As a general rule, 1 litre of paint will cover between 6 and 6.5 metres squared of wall. So, to calculate how many litres of paint you will need, divide the total paintable surface area by 6.5.
## How many square feet will 5 gallons of paint cover?
1,800 square feet
According to our paint estimator, 5 gallons of paint can cover as much as 1,800 square feet. A quart of paint will coat about 90 square feet of space.
## How do you convert between square units?
You can use the formula for the area of a square, A=s2 , to convert square units. Since 1 yard = 3 feet, substitute s by 3 . So, 1 square yard is equal to 9 square feet.
| 810 | 3,261 |
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# What does "isomorphic" mean in linear algebra?
My professor keeps mentioning the word "isomorphic" in class, but has yet to define it... I've asked him and his response is that something that is isomorphic to something else means that they have the same vector structure. I'm not sure what that means, so I was hoping anyone could explain its meaning to me, using knowledge from elementary linear algebra only. He started discussing it in the current section of our textbook: General Vector Spaces.
I've also heard that this is an abstract algebra term, so I'm not sure if isomorphic means the same thing in both subjects, but I know absolutely no abstract algebra, so in your definition if you keep either keep abstract algebra out completely, or use very basic abstract algebra knowledge, that would be appreciated.
• en.wikipedia.org/wiki/Isomorphic Commented Jul 12, 2013 at 3:20
• I've already read it. Wasn't very helpful as my knowledge in whatever topic it was in is very limited. Commented Jul 12, 2013 at 3:22
• Here an isomorphism just a bijective linear map between linear spaces. Two linear spaces are isomorphic if there exists a linear isomorphism between them.
– mdg
Commented Sep 21, 2014 at 21:07
Isomorphisms are defined in many different contexts; but, they all share a common thread.
Given two objects $G$ and $H$ (which are of the same type; maybe groups, or rings, or vector spaces... etc.), an isomorphism from $G$ to $H$ is a bijection $\phi:G\rightarrow H$ which, in some sense, respects the structure of the objects. In other words, they basically identify the two objects as actually being the same object, after renaming of the elements.
In the example that you mention (vector spaces), an isomorphism between $V$ and $W$ is a bijection $\phi:V\rightarrow W$ which respects scalar multiplication, in that $\phi(\alpha\vec{v})=\alpha\phi(\vec{v})$ for all $\vec{v}\in V$ and $\alpha\in K$, and also respects addition in that $\phi(\vec{v}+\vec{u})=\phi(\vec{v})+\phi(\vec{u})$ for all $\vec{v},\vec{u}\in V$. (Here, we've assumed that $V$ and $W$ are both vector spaces over the same base field $K$.)
• Why are Isomorphisms useful? Commented Jul 12, 2013 at 3:33
• Lots of reasons; one of the primary ones, however, is that it allows you to replace an object that you need to deal with with another object which has the same structure, but you are more familar with. For instance, The vector space $P_2(\mathbb{R})=\{ax^2+bx+c\mid a,b,c\in\mathbb{R}\}$ of polynomials of degree at most 2 is isomorphic to $\mathbb{R}^3$; which one do you have more intuition about when it comes to vector space properties? Commented Jul 12, 2013 at 3:37
• @NickPeterson, better to include your comment in the answer.. i almost missed it. It really helped me to understand its usefulness Commented Mar 30, 2020 at 6:42
• @NickPeterson Could you please give an example of a property of $P_2(\mathbb{R})$ which perhaps becomes transparent or nearly transparent once we use this isomorphism but is hard to discover otherwise? Commented Oct 25, 2020 at 19:51
• @user1752323 do you know every polynomial $p(t)$ of degree n can be uniquely identified with n+1 points on its curve. say for 2nd degree polynomials if you use $\{1,x,x^2\}$ as the basis to represent any polynomial in $\mathbb{R}^3$ and choose the points at which you evaluate the function to be t = 1,2,3 then create a linear transformation T that for input p(t) gives out the vector [p(1), p(2), p(3)]. by showing that the kernel of this transformation is 0 you can show that its invertible and hence for any given p(1), p(2) and p(3) there is a unique polynomial of degree 2. Commented Jan 9, 2022 at 17:54
Two vector spaces $V$ and $W$ are said to be isomorphic if there exists an invertible linear transformation (aka an isomorphism) $T$ from $V$ to $W$.
The idea of a homomorphism is a transformation of an algebaric structure (e.g. a vector space) that preserves its algebraic properties. So an homomorphism of a vector space should preserve the basic algebraic properties of the vector space, in the following sense:
$1$. Scalar multiplication and vector addition in $V$ is carried over to scalar multiplication and vector addition in $W$:
For any vectors $x,y$ in $V$ and scalars $a,b$ from the underlying field, $T(ax+by)=aT(x)+bT(y)$.
$2$. The identity element of $V$ is carried over to the identity element of $W$:
If $0_V$ is the identity vector in $V$, then $T(0_V)$ is the identity vector in $W$.
$3$. Vector inversion in $V$ is carried over to vector inversion.
$T(-v)=-T(v)$ for all $v$ in $V$.
$1$ is precisely the property that defines linear transformations, and $2$ and $3$ are redundant (they follow from $1$). So linear transformations are the homomorphisms of vector spaces.
An isomorphism is a homomorphism that can be reversed; that is, an invertible homomorphism. So a vector space isomorphism is an invertible linear transformation. The idea of an invertible transformation is that it transforms spaces of a particular "size" into spaces of the same "size." Since dimension is the analogue for the "size" of a vector space, an isomorphism must preserve the dimension of the vector space.
So this is the idea of the (finite-dimensional) vector space isomorphism: a linear (i.e. structure-preserving) dimension-preserving (i.e. size-preserving, invertible) transformation.
Because isomorphic vector spaces are the same size and have the same algebraic properties, mathematicians think of them as "the same, for all intents and purposes."
Isomorphism is a rather general notion that occurs in lots of contexts.
Essentially, it means "the same."
In linear algebra, we call two vector spaces $V$ and $W$ isomorphic if there exist linear maps $\alpha: V\mapsto W$ and $\beta: W\mapsto V$ such that $\alpha \circ \beta = \text{id}_W$ and $\beta \circ \alpha = \text{id}_V$. When you have these maps, you can then, using $\alpha$, associate to every vector $v\in V$ a vector $\alpha(v) \in W$. The fact that $\alpha$ is linear means that this map respects the structure of $V$ as a vector space (e.g. for any two vectors $v,w\in V$, $\alpha(v) + \alpha(w) = \alpha(v+w))$, and $\beta$, the inverse to $\alpha$, ensures that you can do the same thing in reverse, from $W$ to $V$. This is probably what your professor ment by having "the same vector structure."
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The sum of the weekly salaries of 5 employees is \$3,250. If each of the 5 salaries is to increase by 10 percent, then the average (arithmetic mean) weekly salary per employee will increase by
The sum of the weekly salaries of 5 employees is \$3,250. If each of the 5 salaries is to increase by 10 percent, then the average (arithmetic mean) weekly salary per employee will increase by
A. \$52.50
B. \$55.00
C. \$57.50
D. \$62.50
E. \$65.00
Full explanation coming soon. Send us a note if you’d like this added to the express queue!
You’ll find tons of practice questions, explanations for GMAT Official Guide questions, and strategies on our GMAT Question of the Day page.
Here are a few other extra challenging GMAT questions with in depth explanations:
Here’s a tough function question from the GMAT Prep tests 1 and 2:
For which of the following functions is f(a+b) = f(b) + f(a) for all positive numbers a and b?
And a very challenging word problem from the Official Guide. Almost no-one gets this one on the first try but there is a somewhat simple way through it:
Last Sunday a certain store sold copies of Newspaper A for \$1.00 each and copies of Newspaper B for \$1.25 each, and the store sold no other newspapers that day. If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
Tanya’s letters from the GMAT Prep tests. This one often gets GMAT tutoring students caught up in a tangled net. With combinatorics it’s important to stay practical. We’ll take a look at how to do that in the explanation:
Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
Here’s an exponents puzzle that comes up a lot in GMAT tutoring sessions:
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is
This is one of the most difficult questions in the GMAT universe. That said, there is a simple way to solve it that relies on a fundamental divisibility rule every GMAT studier should know:
For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is?
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# Statistics – Pooled Variance (r)
Pooled Variance/Change is the weighted normal for assessing the fluctuations of two autonomous variables where the mean can differ between tests however the genuine difference continues as before.
### Example
Problem Statement:
Compute the Pooled Variance of the numbers 1, 2, 3, 4 and 5.
Solution:
#### STEP 1
Decide the normal (mean) of the given arrangement of information by including every one of the numbers then gap it by the aggregate include of numbers given the information set.
Mean=1+2+3+4+55=155=3Mean=1+2+3+4+55=155=3
#### STEP 2
At that point, subtract the mean worth with the given numbers in the information set.
⇒(1−3),(2−3),(3−3),(4−3),(5−3)⇒−2,−1,0,1,2⇒(1−3),(2−3),(3−3),(4−3),(5−3)⇒−2,−1,0,1,2
#### STEP 3
Square every period’s deviation to dodge the negative numbers.
⇒(−2)2,(−1)2,(0)2,(1)2,(2)2⇒4,1,0,1,4⇒(−2)2,(−1)2,(0)2,(1)2,(2)2⇒4,1,0,1,4
#### STEP 4
Now discover Standard Deviation utilizing the underneath equation
S=∑X−M2n−1−−−−−−−√S=∑X−M2n−1
Standard Deviation = 1√04√=1.58113104=1.58113
#### STEP 5
Pooled Variance (r) =((aggregate check of numbers −1)×Var)(aggregate tally of numbers−1), (r)=(5−1)×2.5(5−1), =(4×2.5)4=2.5Pooled Variance (r) =((aggregate check of numbers −1)×Var)(aggregate tally of numbers−1), (r)=(5−1)×2.5(5−1), =(4×2.5)4=2.5
Hence, Pooled Variance (r) =2.5
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Algebra Tutorials!
Sunday 15th of September
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
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Number of inequalities to solve: 23456789
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Treat radicals like you would a variable (like x). Only like radicals can be added or subtracted.
Example
1. simplify radical to simplest terms
2. Combine like radicals and reduce
## Multiplying radicals by a binomial
Again, treat the radical like a variable and multiple each term by the multiplier
Example
Take each term in the first bracket and multiply it by each term in the 2nd.
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A radical should never be left as a denominator in a fraction. Multiple the fraction by a radical to eliminate the fraction.
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Notice that if the denominator is a binomial we have to multiple by its conjugate (we keep the values the same, but change the sign in -between the terms). The product of two conjugates is always a rational number.
Steps:
1. multiple the radical by a conjugate
2. reduce
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# Every point in the xy plane satisfying the condition ax + by ≥ c is sa
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Veritas Prep GMAT Instructor
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Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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23 Nov 2010, 19:00
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Let's try this relatively simple question:
Every point in the xy plane satisfying the condition ax + by ≥ c is said to be in region R. If a, b and c are real numbers, does any point of region R lie in the third quadrant?
(1) Slope of the line represented by ax + by – c = 0 is 2.
(2) The line represented by ax + by – c = 0 passes through (-3, 0).
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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24 Nov 2010, 14:25
1
Quote:
2. The line represented by ax + by – c = 0 passes through (-3, 0).
with that information we don't know the slope of the line. It could be posivite, negative or a zero slope. If it has either positive or negative some solutions will lie in the III quadrant, if it has a zero slope-no. That is why I thought it's not sufficient.
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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27 Nov 2010, 13:23
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rockroars: I appreciate the effort for the diagrams. But you made a tiny judgment error. Let me explain the answer in detail.
First of all, notice that ax + by – c = 0 or ax + by = c is the equation of the same line. A line divides the plane into two regions. One of them, where every point (x, y) satisfies ax + by ≥ c, is region R.
Statement 1: Slope of line is 2
Attachment:
Ques1.jpg [ 7.64 KiB | Viewed 5020 times ]
the line will pass through third quadrant and hence both regions will lie in the third quadrant. Sufficient.
Statement 2: The line passes through (-3, 0).
Attachment:
Ques2.jpg [ 7.56 KiB | Viewed 5018 times ]
A line passing through (-3, 0) could be the blue line or the green line. In either case, the line will pass through the third quadrant and hence, will have both regions in the third quadrant. So it is sufficient too? What about the x axis? That is also a line passing through (-3, 0). It does not pass through the third quadrant. We would need the equation of the line to find out whether our region R lies in the third quadrant. The equation of x axis is y = 0. So the required region is y ≥ 0 i.e. the first and second quadrant. Hence using just this information, we cannot say whether a point of region R lies in the third quadrant or not.
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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27 Nov 2010, 19:21
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VeritasPrepKarishma wrote:
Let's try this relatively simple question:
Every point in the xy plane satisfying the condition ax + by ≥ c is said to be in region R. If a, b and c are real numbers, does any point of region R lie in the third quadrant?
1. Slope of the line represented by ax + by – c = 0 is 2.
2. The line represented by ax + by – c = 0 passes through (-3, 0).
Any line with a positive slope will pass from 1st and 3rd quadrant, and either "one of the other two quadrants" or the "origin".
S1: Slope is positive. Sufficient.
S2: Line passes from (-3,0) so the line either passes from 3rd Quadrant or the equation is of X-axis. Not Sufficient.
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PS: Correct me if I am wrong.
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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03 May 2013, 20:17
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Rajkiranmareedu wrote:
Karishma: I didn't understand. Could you explain a bit.
VeritasPrepKarishma wrote:
rockroars: I appreciate the effort for the diagrams. But you made a tiny judgment error. Let me explain the answer in detail.
First of all, notice that ax + by – c = 0 or ax + by = c is the equation of the same line.A line divides the plane into two regions. One of them, where every point (x, y) satisfies ax + by ≥ c, is region R.
Statement 1: Slope of line is 2
Attachment:
Ques1.jpg
the line will pass through third quadrant and hence both regions will lie in the third quadrant. Sufficient.
Statement 2: The line passes through (-3, 0).
Attachment:
Ques2.jpg
A line passing through (-3, 0) could be the blue line or the green line. In either case, the line will pass through the third quadrant and hence, will have both regions in the third quadrant. So it is sufficient too? What about the x axis? That is also a line passing through (-3, 0). It does not pass through the third quadrant. We would need the equation of the line to find out whether our region R lies in the third quadrant. The equation of x axis is y = 0. So the required region is y ≥ 0 i.e. the first and second quadrant. Hence using just this information, we cannot say whether a point of region R lies in the third quadrant or not.
Check out this post. It discusses this question in detail: http://www.veritasprep.com/blog/2010/12 ... s-part-ii/
A line (which by definition, extends infinitely at both sides) given by ax+by - c = 0 splits a region into two sections - one on the left side of the line and the other on the right side of the side. One of these two regions will satisfy ax+by - c < 0 and the other will satisfy ax+by - c > 0.
How do you know which region satisfies which inequality? Put a point from that region in the inequalities and see what it satisfies e.g. if (0, 0) doesn't lie on the line, put x = 0, y = 0
If c is negative, ax+by - c < 0 will be satisfied and the region that contains the point (0, 0) i.e. the origin of the axis will satisfy ax+by - c < 0.
In that case, the other region will satisfy ax+by - c > 0.
Which quadrants will lie in any particular region depends on where the line is located. If it is a vertical line passing through first and fourth quadrant, second and third quadrant will lie to its left and some part of first and fourth quadrants will also lie to its left. Rest of the first and fourth quadrants will lie to its right (make a line and see what i mean)
Similarly, try making a line with a positive slope, say 2. It will pass through first and third quadrants in ALL cases (remember, a line extends indefinitely at both ends). Since it will pass through the third quadrant, both the regions will have a part of the third quadrant.
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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24 Nov 2010, 02:02
Is it A ?
I think A is sufficient becase if the slope is 2, then some solutions lie in the III quadrant.
B seems to me insufficient because I need at least the coordinates of 1 more point to find the slope.
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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24 Nov 2010, 10:38
medanova wrote:
Is it A ?
I think A is sufficient becase if the slope is 2, then some solutions lie in the III quadrant.
B seems to me insufficient because I need at least the coordinates of 1 more point to find the slope.
Your answer is correct. Good! Though, I am not convinced with the reasoning. Think a little more...
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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24 Nov 2010, 14:51
A : Sufficient ... Since slope is 2, hence positive. This means that the line must pass through the third quadrant. The inequality ax+by>=c represents one side of the line. Since the line passes through the 3rd quadrant, either side has points from the 3rd quadrant. Hence the region will always havea bit of third quadrant.
B : Insuffcient ... Any line except the one with slope=0 will pass through 3rd quadrant and the above logic applies. But the line with slope 0 may or may not have the thrid quadrant points in the region included depending on which side we choose (sign of c)
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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27 Nov 2010, 13:04
I think the answer should be C.
We get the details of line and regions completely only after we club 1 and 2. After which we will be in a position to decide if Region R passes through 3rd Quad.
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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27 Nov 2010, 13:35
shrouded1 wrote:
A : Sufficient ... Since slope is 2, hence positive. This means that the line must pass through the third quadrant. The inequality ax+by>=c represents one side of the line. Since the line passes through the 3rd quadrant, either side has points from the 3rd quadrant. Hence the region will always havea bit of third quadrant.
B : Insuffcient ... Any line except the one with slope=0 will pass through 3rd quadrant and the above logic applies. But the line with slope 0 may or may not have the thrid quadrant points in the region included depending on which side we choose (sign of c)
I am used to perfect answers from you shrouded1... But I think you missed out on a point here. c = 0 we know because it has to be x axis since it passes through (-3, 0). Since y >= 0 is the first and second quadrant hence we know that the region R may or not lie in third quadrant. If instead, we had ax+by <=c, statement 2 would be sufficient too.
If I am missing something here, let me know. (I would like to believe that I didn't err in a question I made myself!)
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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27 Nov 2010, 13:55
@VeritasPrepKarishma:
Okay! I realized where I made a mistake, a line with slope 2 is acute to x-axis. I assumed it to be obtuse to X axis.
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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03 May 2013, 11:02
Karishma: I didn't understand. Could you explain a bit.
VeritasPrepKarishma wrote:
rockroars: I appreciate the effort for the diagrams. But you made a tiny judgment error. Let me explain the answer in detail.
First of all, notice that ax + by – c = 0 or ax + by = c is the equation of the same line.A line divides the plane into two regions. One of them, where every point (x, y) satisfies ax + by ≥ c, is region R.
Statement 1: Slope of line is 2
Attachment:
Ques1.jpg
the line will pass through third quadrant and hence both regions will lie in the third quadrant. Sufficient.
Statement 2: The line passes through (-3, 0).
Attachment:
Ques2.jpg
A line passing through (-3, 0) could be the blue line or the green line. In either case, the line will pass through the third quadrant and hence, will have both regions in the third quadrant. So it is sufficient too? What about the x axis? That is also a line passing through (-3, 0). It does not pass through the third quadrant. We would need the equation of the line to find out whether our region R lies in the third quadrant. The equation of x axis is y = 0. So the required region is y ≥ 0 i.e. the first and second quadrant. Hence using just this information, we cannot say whether a point of region R lies in the third quadrant or not.
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Posts: 234
Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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03 May 2013, 12:30
Let's try this relatively simple question:
Every point in the xy plane satisfying the condition ax + by ≥ c is said to be in region R. If a, b and c are real numbers, does any point of region R lie in the third quadrant?
1. Slope of the line represented by ax + by – c = 0 is 2.
2. The line represented by ax + by – c = 0 passes through (-3, 0).
This is bit tricky question.
You have to know that a line with positive slope makes an acute angle with the x axis or it always passes through 1 and 3rd quadrant. This is like a theorm. Helpful in some cases. Since Slope is +. It gives the soln.
2.Now here you have look back at the question and see the value of a,b and c. The points a,b and c will always pass through 3rd quadrant barring one case. In case the line is x - axis itself. ie x = 0 . and c =0 and b = 1.
Hence A gives a clear soln.
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink]
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Re: Every point in the xy plane satisfying the condition ax + by ≥ c is sa [#permalink] 28 Nov 2019, 08:06
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# Gravitational Fields Questions
Q6. Communications satellites are usually placed in a geo-synchronous orbit.
(a) State two features of a geo-synchronous orbit.
It has a period of 24 hours or its period equals the period of the Earth's rotation
It remains in a fixed position relative to surface of Earth
It has an equatorial orbit
It has the same angular speed as the Earth (2 MAX)
(2 marks)
(b) Given that the mass of the Earth is 6.00 × 1024 kg and its mean radius is 6.40 × 106 m,
(i) show that the radius of a geo-synchronous orbit must be 4.23 × 107 m,
The gravitational pull on the satellite becomes the centripetal force holding the satellite in orbit. So,
GMm/r2 = mv2/r
GM/r =v2
v = 2πr/T
GM/r = 4π2r2/T2
GMT2 = 4π2r3
r3 = GMT2/4π2
r3 = (6.67 x 10-11 x 6.00 × 1024 x (24 x 602)2)/4π2
r3 = 7.57 x 1022
r = 4.23 x 107m
OR use GMm/r2 = mω2r and T = 2π/ω - it gives the same outcome
(ii) calculate the increase in potential energy of a satellite of mass 750 kg when it is raised from the Earth's surface into a geo-synchronous orbit.
V = -GM/r
ΔV = -GM x (1/R - 1/r)
ΔV = -GM x ( -1/(6.40 × 106) - 1/(4.23 × 107))
ΔV = -GM x 1.32 × 10-7
ΔV = -6.67 x 10-11 x 6.00 × 1024 x 1.32 × 10-7
ΔV = -5.31 x 107 J/kg
ΔW = mΔV = 750 x -5.31 x 107 = -3.98 x 1010 J
Therefore there is an increase of 3.98 x 1010 J
(6 marks)
(Total 8 marks)
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# Thread: Calculate the velocity of the brick using an assumption & draw the displacement-time graph
1. Calculate the velocity of the brick at the bottom of the inclined plane.
State the assumption you made for your calculation in part above.
& Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)
Workings:
I am unable to think on the first two but I think the displacement and time graph should be something like ,
Many Thanks
2. To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.
So, if we equate the initial potential energy to the final kinetic energy, we have:
$\displaystyle mgh=\frac{1}{2}mv^2$
What do you get upon solving for $v$?
3. Thread Author
Originally Posted by MarkFL
To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.
So, if we equate the initial potential energy to the final kinetic energy, we have:
$\displaystyle mgh=\frac{1}{2}mv^2$
What do you get upon solving for $v$?
Thank you very much MarkFL
$\displaystyle mgh=\frac{1}{2}mv^2$
$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}mv^2$
$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}*2kg * v^2$
$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}*2kg * v^2$
$\displaystyle 100 J= v^2$
$\displaystyle 10 J= v$
Correct ?
4. I would solve the equation first, and then plug in the numbers:
$\displaystyle mgh=\frac{1}{2}mv^2$
$\displaystyle v=\sqrt{2gh}$
Okay, at this point we can plug in:
$\displaystyle g\approx9.81\,\frac{\text{m}}{\text{s}^2},\,h=5\text{ m}$
$\displaystyle v\approx\sqrt{2\left(9.81\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)}=3\sqrt{\frac{109}{10}}\,\frac{\text{m}}{\text{s}}\approx9.9045\,\frac{\text{m}}{\text{s}}$
5. Thread Author
Originally Posted by MarkFL
I would solve the equation first, and then plug in the numbers:
$\displaystyle mgh=\frac{1}{2}mv^2$
$\displaystyle v=\sqrt{2gh}$
Okay, at this point we can plug in:
$\displaystyle g\approx9.81\,\frac{\text{m}}{\text{s}^2},\,h=5\text{ m}$
$\displaystyle v\approx\sqrt{2\left(9.81\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)}=3\sqrt{\frac{109}{10}}\,\frac{\text{m}}{\text{s}}\approx9.9045\,\frac{\text{m}}{\text{s}}$
Thank you very much
speaking about the graph
Originally Posted by mathlearn
Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)
Is the above drawn graph correct?
6. For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?
7. Originally Posted by mathlearn
speaking about the graph
Is the above drawn graph correct?
Hey mathlearn!
Here's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):
On the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.
As a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).
Note that $v=\int_0^t a\,dt$.
However, we're asked for the displacement, which we can call $d$.
We have that $d=\int_0^t v\,dt$.
What would the displacement graph look like?
8. Originally Posted by mathlearn
$\displaystyle 100 J= v^2$
$\displaystyle 10 J= v$
Recheck your units, speed is not measured in J. (And the square root of a J is not J!)
-Dan
9. Thread Author
Originally Posted by MarkFL
For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?
This kind of a curve with constant acceleration and with deceleration.
Originally Posted by I like Serena
Hey mathlearn!
Here's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):
On the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.
As a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).
Note that $v=\int_0^t a\,dt$.
However, we're asked for the displacement, which we can call $d$.
We have that $d=\int_0^t v\,dt$.
What would the displacement graph look like?
The displacement time graph would look like the above graph
Originally Posted by topsquark
Recheck your units, speed is not measured in J. (And the square root of a J is not J!)
-Dan
Thanks for the catch
10. Originally Posted by mathlearn
This kind of a curve with constant acceleration and with deceleration...
Why do you have the brick returning to zero displacement?
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+0
# Help Fast!
0
1
3
+41
Let x,y,z be nonzero real numbers, such that no two are equal, and
$$x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x}.$$
Find all possible numeric values of xyz
I got 1 as one of the answers.
Jul 23, 2024
#1
+1659
+1
Ok, first off, let's split the one equation into a system of equations so that each term is equal to another. We get
$$x + 1/y = y + 1/z\\ x + 1/y = z + 1/x\\ y + 1/z = z + 1/x$$
I won't really show the steps, although I will if needed. Let me know if you want to.
$$(x, y, z) = (-\frac{1}{z-1}, \frac{z-1}{z}, z)\\ (x, y,z) = (-\frac{1}{z+1}, -\frac{z+1}{z}, z)$$
We want to find the value of xyz since we need the product.
Thus, multiplying these together, we get
$$(-\frac{1}{z-1})(\frac{z-1}{z})z=-\frac{1}{z-1}\left(z-1\right)=-1$$
$$(-\frac{1}{z+1} )(-\frac{z+1}{z}) z=\frac{1}{z}z=1$$
So good job! You were correct about 1. The other value is -1.
I didn't go into too much detail, but let me know if you need more assistance.
Thanks! :)
Jul 23, 2024
edited by NotThatSmart Jul 23, 2024
#2
+41
0
Thanks again. I think I can fill in the gaps!
MeldHunter Jul 23, 2024
#3
+1659
+1
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What is the equation of the line with slope m= -1/25 that passes through (7/5, 1/10) ?
1 Answer
Jun 25, 2016
In point slope form:
$y - \frac{1}{10} = - \frac{1}{25} \left(x - \frac{7}{5}\right)$
In slope intercept form:
$y = - \frac{1}{25} x + \frac{39}{250}$
Explanation:
Given a slope $m$ and a point $\left({x}_{1} , {y}_{1}\right)$ through which a line passes, its equation can be written in point slope form:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
In our example, $m = - \frac{1}{25}$ and $\left({x}_{1} , {y}_{1}\right) = \left(\frac{7}{5} , \frac{1}{10}\right)$, so we get the equation:
$y - \frac{1}{10} = - \frac{1}{25} \left(x - \frac{7}{5}\right)$
Expanding and rearranging, this can be expressed as:
$y = - \frac{1}{25} x + \frac{39}{250}$
which is in slope intercept form:
$y = m x + b$
with $m = - \frac{1}{25}$ and $b = \frac{39}{250}$
graph{(y - 1/10 + 1/25(x-7/5))(x^2+(y-39/250)^2-0.0017)((x-7/5)^2+(y-1/10)^2-0.0017)=0 [-1.76, 3.24, -1.17, 1.33]}
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# Search by Topic
#### Resources tagged with Cubes similar to Building with Cubes:
Filter by: Content type:
Stage:
Challenge level:
### There are 22 results
Broad Topics > 3D Geometry, Shape and Space > Cubes
### Cubes Cut Into Four Pieces
##### Stage: 1 Challenge Level:
Eight children each had a cube made from modelling clay. They cut them into four pieces which were all exactly the same shape and size. Whose pieces are the same? Can you decide who made each set?
### Green Cube, Yellow Cube
##### Stage: 2 Challenge Level:
How can you paint the faces of these eight cubes so they can be put together to make a 2 x 2 cube that is green all over AND a 2 x 2 cube that is yellow all over?
### Start Cube Drilling
##### Stage: 1 Challenge Level:
Imagine a 3 by 3 by 3 cube. If you and a friend drill holes in some of the small cubes in the ways described, how many will have holes drilled through them?
### Four Layers
##### Stage: 1 and 2 Challenge Level:
Can you create more models that follow these rules?
### Triple Cubes
##### Stage: 1 and 2 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### A Puzzling Cube
##### Stage: 2 Challenge Level:
Here are the six faces of a cube - in no particular order. Here are three views of the cube. Can you deduce where the faces are in relation to each other and record them on the net of this cube?
### Construct-o-straws
##### Stage: 2 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
### Holes
##### Stage: 1 and 2 Challenge Level:
I've made some cubes and some cubes with holes in. This challenge invites you to explore the difference in the number of small cubes I've used. Can you see any patterns?
### Dicey
##### Stage: 2 Challenge Level:
A game has a special dice with a colour spot on each face. These three pictures show different views of the same dice. What colour is opposite blue?
### Painted Faces
##### Stage: 2 Challenge Level:
Imagine a 3 by 3 by 3 cube made of 9 small cubes. Each face of the large cube is painted a different colour. How many small cubes will have two painted faces? Where are they?
##### Stage: 2 Challenge Level:
Make a cube with three strips of paper. Colour three faces or use the numbers 1 to 6 to make a die.
### Cube Drilling
##### Stage: 2 Challenge Level:
Imagine a 4 by 4 by 4 cube. If you and a friend drill holes in some of the small cubes in the ways described, how many will not have holes drilled through them?
### Three Sets of Cubes, Two Surfaces
##### Stage: 2 Challenge Level:
How many models can you find which obey these rules?
### Next Size Up
##### Stage: 2 Challenge Level:
The challenge for you is to make a string of six (or more!) graded cubes.
### Cubic Conundrum
##### Stage: 2 Challenge Level:
Which of the following cubes can be made from these nets?
### Three Cubed
##### Stage: 2 Challenge Level:
Can you make a 3x3 cube with these shapes made from small cubes?
### Christmas Presents
##### Stage: 2 Challenge Level:
We need to wrap up this cube-shaped present, remembering that we can have no overlaps. What shapes can you find to use?
### The Third Dimension
##### Stage: 1 and 2 Challenge Level:
Here are four cubes joined together. How many other arrangements of four cubes can you find? Can you draw them on dotty paper?
### Dice, Routes and Pathways
##### Stage: 1, 2 and 3
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Cubes
##### Stage: 1 and 2 Challenge Level:
Investigate the number of faces you can see when you arrange three cubes in different ways.
### Thinking 3D
##### Stage: 2 and 3
How can we as teachers begin to introduce 3D ideas to young children? Where do they start? How can we lay the foundations for a later enthusiasm for working in three dimensions?
### Paper Folding - Models of the Platonic Solids
##### Stage: 2, 3 and 4
A description of how to make the five Platonic solids out of paper.
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## Featured resource
### Defining Mathematics Education *REDUCED*
The Seventy-fifth Yearbook is a celebration and reflection of the history of the NCTM yearbooks, and is a great resource on the key issues of mathematics education through the years.
Members: $21.00 inc.GST Others:$ 26.25 inc.GST
Home > Topdrawer > Fractions
# Fractions
Students' first experiences of number are with positive integers — counting numbers and the operations of addition and subtraction. When they begin working with multiplication and division they are also introduced to rational numbers — more specifically, fractions. Many young students find the transition from whole-number thinking to rational-number thinking quite difficult.
It can be unsettling for students to discover that there are many more numbers between zero and one. To then extend that understanding beyond one (to include improper fractions and mixed numerals) can be a challenge.
One source of confusion is the fact that the symbols for fractions involve two numerals rather than just one. Two numerals are needed because a fraction expresses a mathematical relationship between two quantities.
To further complicate learning about fractions, there are various meanings and uses for fractions, such as part-whole, division and ratio. There is also a huge variety in the ways fractions can be represented, such as area diagrams, lengths, volumes and discrete items.
All of these complications make good quality teaching essential. Attention must be given to developing conceptual understanding and 'fraction sense', rather than relying on procedural understanding and practised 'rules'.
A conceptual understanding of fractions is essential for problem solving, proportional reasoning, probability and algebra.
## Big ideas
What is a fraction?
Think beyond your immediate response. Are there other ways fractions are used in mathematics?
## Misunderstandings
Research into student thinking has revealed that many students have misconceptions that hamper understanding and restrict the development of effective strategies for working with fractions.
## Good teaching
Good teaching of fractions involves learning experiences that enable students to build a strong sense of what fractions mean in a variety of contexts and using a variety of representations to develop effective strategies.
## Assessment
Assessment of students' understanding of mathematics serves two main purposes — to inform further teaching, and to provide feedback to students on their own learning.
## Activities
Student activities that appear in other parts of the drawer have been collected here.
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# Help
0
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All vertices of pentagon ABCDE lie on a circle. We know that AB = BC = CD = DE and angle ABC = 128. Find angle ACE in degrees.
Jan 13, 2019
#1
+533
0
i dont know how to solve this, but i think a way to make it easier is to draw all the diagonals
Jan 14, 2019
#2
+111437
+1
Just a rough pic for reference......definitely not exact!!!!
A
E B
D
C
Since ABC = 128°.......then major arc CA is twice 128° = 256°
Therefore.......arc ABC = 360° - 256° = 104°
But.....since AB = BC....then each of these chords must span an arc of 52°
And since AB = BC = CD = DE.....these equal chords each span an arc of 52°
So....arc EA must be 360 - 4(52) = 152°
And angle ACE is an inscrbed angle subtending EA....so it's measure is (1/2) measure arc EA = 76°
Jan 14, 2019
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https://oeis.org/A266152
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A266152 Least positive integer y such that n = x^4 - y^3 + z^2 for some positive integers x and z, or 0 if no such y exists. 14
8, 1, 2, 17, 1, 3, 139, 19, 37, 1, 3, 9, 2, 7, 3, 1411, 1, 2, 2, 1, 5, 4, 387, 3, 1, 1, 4, 7, 9, 2, 35, 1, 33, 2, 6, 5, 1, 4, 3, 11, 1, 6, 2, 429, 2, 5, 11, 179, 73, 1, 15, 1, 4, 3, 11, 3, 5, 2, 3, 15, 5, 6, 7, 3, 1, 6, 4, 6337, 8, 16, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Conjecture: Any integer m can be written as x^4 - y^3 + z^2, where x, y and z are positive integers. This is slightly stronger than the conjecture in A266003. See also A266153 for a related sequence, and A266212 for a stronger conjecture. If n is a positive square, then a(n) = 1. - Altug Alkan, Dec 23 2015 LINKS Zhi-Wei Sun, Table of n, a(n) for n = 0..10000 Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120. EXAMPLE a(0) = 8 since 0 = 4^4 - 8^3 + 16^2. a(6) = 139 since 6 = 36^4 - 139^3 + 1003^2. a(15) = 1411 since 15 = 119^4 - 1411^3 + 51075^2. a(11019) = 71383 since 11019 = 4325^4 - 71383^3 + 3719409^2. MATHEMATICA SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]] Do[y=1; Label[bb]; Do[If[SQ[n+y^3-x^4], Print[n, " ", y]; Goto[aa]], {x, 1, (n+y^3)^(1/4)}]; y=y+1; Goto[bb]; Label[aa]; Continue, {n, 0, 70}] CROSSREFS Cf. A000290, A000578, A000583, A262827, A266003, A266004, A266153, A266212. Sequence in context: A156944 A227424 A248965 * A021127 A010155 A019607 Adjacent sequences: A266149 A266150 A266151 * A266153 A266154 A266155 KEYWORD nonn AUTHOR Zhi-Wei Sun, Dec 22 2015 STATUS approved
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# Tutor profile: Brittany B.
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Brittany B.
UC Berkeley student with years of tutoring experience
Tutor Satisfaction Guarantee
## Questions
### Subject:SAT
TutorMe
Question:
A grocery store must pay \$1500 a month to rent their facilities and makes and average of \$64 per customer who visits. Write an equation which models the store's profits, where S represents the total profit and C represents the number of customers. In addition, determine the minimum number of customers they must have visit in a month and spend the average amount in order for the store to break even.
Inactive
Brittany B.
S = 64C - 1500 (explanation: the store makes an average of \$64 per costumer, so this is accounted for in the equation by 64 times C, the number of customers. However, to find the store's profits, we must also subtract their expenses, which include the rent of \$1500). To find how many customers they need to visit to break even, we take S = 0 and solve for C. 0 = 64C - 1500 1500 = 64C C = 1500/64 C = 23.4375 but since we cant have .4375 of a person, we round up to 24 customers.
### Subject:Literature
TutorMe
Question:
What is one motif that appears in Shakespeare's Hamlet? How is it used/what does it signify?
Inactive
Brittany B.
Hearing and ears are a common motif in Hamlet. References to these things often signify deceit and the power that words have to corrupt and even kill. Notable uses include the fact that King Hamlet was killed by poison poured in his ear, and the following quote spoken to Hamlet by his father's ghost: "So the whole ear of Denmark is by a forgèd process of my death rankly abused" (Act 1 scene 5).
### Subject:Calculus
TutorMe
Question:
What is the fundamental theorem of calculus? What does it tell us and what can we find with this knowledge?
Inactive
Brittany B.
The fundamental theorem of calculus states that the integral of a continuous function f over the closed interval [a, b] is equal to F(a) - F(b), where F is the antiderivative of f. From this, we can find the area under the curve f from point a to b.
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http://www.numbersaplenty.com/2137
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Search a number
2137 is a prime number
BaseRepresentation
bin100001011001
32221011
4201121
532022
613521
76142
oct4131
92834
102137
111673
1212a1
13c85
14ac9
15977
hex859
2137 has 2 divisors, whose sum is σ = 2138. Its totient is φ = 2136.
The previous prime is 2131. The next prime is 2141. The reversal of 2137 is 7312.
Adding to 2137 its reverse (7312), we get a palindrome (9449).
It can be divided in two parts, 21 and 37, that multiplied together give a palindrome (777).
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1296 + 841 = 36^2 + 29^2 .
2137 is a truncatable prime.
It is a cyclic number.
It is not a de Polignac number, because 2137 - 23 = 2129 is a prime.
It is a d-powerful number, because it can be written as 211 + 34 + 1 + 7 .
It is a nialpdrome in base 13 and base 15.
It is not a weakly prime, because it can be changed into another prime (2131) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1068 + 1069.
It is an arithmetic number, because the mean of its divisors is an integer number (1069).
It is an amenable number.
2137 is a deficient number, since it is larger than the sum of its proper divisors (1).
2137 is an equidigital number, since it uses as much as digits as its factorization.
2137 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 42, while the sum is 13.
The square root of 2137 is about 46.2276973253. The cubic root of 2137 is about 12.8805628395.
The spelling of 2137 in words is "two thousand, one hundred thirty-seven".
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https://algorithmist.wordpress.com/2008/12/04/degrafa-introduction-to-splines-part-iii/
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# Degrafa – Introduction to Splines Part III
In part I and part II of this series, we saw that splines are essentially piecewise polynomials designed to fit a set of data points or knots. The spline is called a piecewise function as the definition of the curve depends on the ‘piece’ or segment. From the previous example with five knots, there are four polynomial functions, the definition of which depends on the segment. One polynomial is defined for the segment from knot 0 to knot1. Another is defined for the segment from knot 1 to knot 2, and so on. Another name for this type of curve is a composite curve as it is composed of multiple, independent constituent curves.
Believe it or not, you are probably already familiar with drawing certain types of splines with your previous Flash work. If you have ever drawn line segments between points using moveTo() and lineTo(), you just drew a piecewise linear spline that interpolates the points at each moveTo() and lineTo(). You did not have to define each (linear) polynomial as Flash did the drawing for you. A linear polynomial has two coefficients, i.e. y = c0 + c1*x. Two conditions are required to determine the coefficients. These are uniquely determined by having the line segment pass through each knot. So, one set of coefficients is determined by a line from knot 0 to knot 1. Another line segment is determined by having a line pass through knot 1 and knot 2. The process continues for each new knot.
A quadratic spline would be piecewise quadratic, meaning there would be one quadratic polynomial for each segment; knot 0 to knot1, knot 1 to knot 2, and so on. Three conditions would be required for each segment as there are three coefficients in a quadratic polynomial.
The cubic Bezier spline in Degrafa is piecewise cubic, meaning there is a unique cubic Bezier curve for each segment. If there are five knots, there are four cubic Bezier curves. Since each cubic is ‘well behaved’, this avoids the oscillation problems with high-order interpolating polynomials.
The cubic Bezier curve requires four control points to uniquely determine its coefficients. Two control points are fixed as the spline must interpolate the knots. So, for example, the outermost control points for the first cubic Bezier curve are knot 0 and knot 1. The outermost control points for the second cubic Bezier curve are knot 1 and knot 2.
We are free to choose how the other control points are chosen. I mentioned in part II that to achieve a smooth curve through each knot, certain continuity conditions are applied at the joins. The obvious continuity is that the curve pass through each knot. Otherwise, what’s the point of the spline in the first place? Technically, this is called geometric continuity.
The spline would be pretty worthless if there were cusps or ‘rough spots’ at the joins. Instead, we want the spline to look as if it were smoothly drawn through each point, almost as if by hand. Intuitively, this means we want the slope of the curve going into one join to be the same as the slope of the curve going out of the same join.
If you’ve taken calculus, you recognize this as continuity of first derivative at each join. If you haven’t taken calculus, think about the slope of the line segment at the end of the Bezier control cage going into a join and the slope of the line segment at the beginning of the Bezier control cage going out of the same join.
These control cages are sets of four points that determine a unique cubic Bezier curve. The line segments above are called in-tangents and out-tangents. They are illustrated for the first cubic Bezier segment in the following diagram.
By matching the in- and out-tangent at each join, the spline passes through each join as if it we drew a single smooth curve, even though it is comprised of completely independent Bezier curves.
Matching tangent direction alone does not supply enough information to completely determine a set of control points for each Bezier curve. Fortunately, this gives us some design flexibility in terms of a ‘tension’ parameter. If you are curious about the technical details, you can read this TechNote.
The next section will discuss why so much attention is devoted to cubic splines and why the cubic Bezier spline was the first to be ported to Degrafa.
## 3 thoughts on “Degrafa – Introduction to Splines Part III”
1. JT says:
Cool – how many parts to this series?
2. JT – perhaps six or seven; don’t really know. I want to cover everything from ‘what is a spline’ to how to create/manipulate the Degrafa cubic bezier spline in MXML and script.
regards,
– jim armstrong
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https://meangreenmath.com/2017/06/03/engaging-students-deriving-the-pythagorean-theorem-7/
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# Engaging students: Deriving the Pythagorean theorem
In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.
I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).
This student submission comes from my former student Jillian Greene. Her topic, from Geometry: deriving the Pythagorean theorem.
How can technology be used to effectively engage students with this topic?
Geometers Sketchpad is a fantastic resource to be able to more intuitively explore aspects of geometry without the approximation that often comes from using a graphing calculator or a pencil and paper. There is an exploratory activity that can either allow students to discover the Pythagorean Theorem in a different way, or just to reinforce the relationships between the sides. Have students create of a line segment AB with a length of one unit, whatever the measurement might be. Then create a right isosceles triangle using AB as the two equal sides. Now the students will build off of this triangle, making more right triangle (not necessarily isosceles) using the hypotenuse as one of the legs of the next triangle, and the other leg having the same length as AB. Do this 6 times and find the length of final triangle’s hypotenuse. Now explain what the pattern is, and how the relationships work. The final product should look like this:
The final side should be sqrt(7), and the hypotenuses should go sqrt(2), sqrt(3), sqrt(4)…all the way up to x. Hopefully students will be fascinated by the relationship!
What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?
The Pythagorean Theorem was first theorized by Pythagoras, right? Wrong! There’s a very rich history that comes with this theorem that finds a relation in the sides on right triangles. Actually, there were clay tablets indicating an understanding of this theorem found in Babylonian settlements from more than 1000 years before Pythagoras. The Yale tablet, depicted below, has numbers written out in the Babylonian system that give the number “1.414212963” which is very close to √2 = 1.414213562, indicating an understanding of the 1-1-√2 relationship.
Similarly, there are relics from the Chinese and the Egyptian people having either the relationship between the legs figured out, or the existence of 3-4-5 triangles, or a “Pythagorean triple.” The Egyptians made sure their corners on their buildings were 90 degrees by using a rope with 12 evenly spaced notches to make a 3-4-5 triangle. So where does Pythagoras come in? Pythagoras was the first one to formulate a proof in regards to this theorem. So where are his proofs? Well, Pythagoras felt strongly against allowing anyone to record his teachings in any way, so there is no physical proof left behind. However, from what we know about Pythagoras, it is safe to assume that he approached it geometrically.
How could you as a teacher create an activity or project that involves your topic?
Hello Detective, thank you for coming in to help today. Scar Tellub, 24 year old male, brown hair, green eyes, was found shot early this morning. He was shot for an unknown purpose, but is luckily recovering now. However, we are determined to find this shooter. We know from eye witness testimonies that the gunshot came from overhead, from the top of a nearby building. We know from where the bodies were found, Mr. Tellub was standing perfectly in the center of three buildings, specifically he was 9 feet away from each building. From the entry and exit of the bullet, we can tell the gun was shot from 15 feet away. We have three possible suspects that could be the culprit, but we need your mathematical prowess to help us nail the bad person.
These are the possible shooters:
1. Madison Bloodi: 19 years old, blonde hair/blue eyes, babysitter. Spotted atop the first building, Trump Tower (20 feet tall), at the time of the shooting.
2. Hunter Kilt: 34 years old, brown hair/brown eyes, landscaper. Spotted atop the second building, the Eiffel Tower (6 feet tall), at the time of the shooting.
3. Winston Payne: 26 years old, black hair/green eyes, lawyer. Spotted atop the third building, the Leaning Tower of Pisa (12 feet tall), at the time of the shooting.
Again, thank you for your time, Detective. We know full well that you won’t let us down. Please draw us a photo and show us your work for all three suspects so we can provide them to the judge. Happy mathing!
References:
http://jwilson.coe.uga.edu/emt668/emat6680.f99/challen/pythagorean/lesson4/lesson4.html
http://www.ualr.edu/lasmoller/pythag.html
(I did a similar activity to the murder one with students before, but I cannot find it online again, so I wrote a new one kind of similar to what I remember)
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Purchase Solution
# Power Functions
Not what you're looking for?
We say a critical region C is of size  if = max P{(X1,....,Xn)  C], 0
We define the power function of a critical region to be c()= P[(X1,...Xn) C], 1
1)Let X have a pdf of the from f(x; )= x^(-1), 0<x<1, 0 elsewhere, where  {: =1,2}. To test the simple hypothesis H0: =1 against the alternative simple hypothesis H1: =2, use a random sample X1, X2 of size n=2 and define the critical region to be C= {(x1,x2): ¾ <= x1x2}. Find the power function of the test.
##### Solution Summary
We say a critical region C is of size a if = max P0{(X1,....,Xn) E C], 0
We define the power function of a critical region to be c()= P[(X1,...Xn) C], 1
1)Let X have a pdf of the from f(x; )= x^(-1), 0<x<1, 0 elsewhere, where  {: =1,2}. To test the simple hypothesis H0: =1 against the alternative simple hypothesis H1: =2, use a random sample X1, X2 of size n=2 and define the critical region to be C= {(x1,x2): ¾ <= x1x2}. Find the power function of the test.
Solution provided by:
###### Education
• BSc , Wuhan Univ. China
• MA, Shandong Univ.
###### Recent Feedback
• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
• "excellent work"
• "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
• "Thank you"
• "Thank you very much for your valuable time and assistance!"
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# Thursday September 12, 2024
## Calculating 59 days before Sunday November 10, 2024 by hand
This page helps you figure out the date that is 59 days before Sunday November 10, 2024. We've made a calculator to find the date before a certain number of days before a specific date. If you want to change the question on this page, you have two choices: you can change the URL in your browser's address bar, or go to our days before specific date calculatorto type in a new question or days from a specific date if you want to add 59 days.
But for all you time sickos out there who want to calculate 59 days before Sunday November 10, 2024 - here's how you do it:
1. Start with the Input Date (Sunday November 10, 2024): Write it down! I can't stress this enough
2. Count in Weeks: Recognize that 59 days is approximately 8.428571428571429 weeks. Count forward 8.428571428571429 weeks (11.8 work weeks) from the input date. This takes you to .
3. Add Remaining Days: Since you've counted 8.428571428571429 weeks, you only need to add the remaining days to reach Thursday September 12, 2024
4. Use Mental Math: If Sunday November 10, 2024 is a Thursday, then compare that to if 59 is divisible by 7. That way, you can double-check if September 12 matches that Thursday.
## Thursday September 12, 2024 Stats
• Day of the week: Thursday
• Month: September
• Day of the year: 256
## Counting 59 days backward from Sunday November 10, 2024
Counting backward from today, Thursday September 12, 2024 is 59 before now using our current calendar. 59 days is equivalent to:
59 days is also 1416 hours. Thursday September 12, 2024 is 70% of the year completed.
## Within 59 days there are 1416 hours, 84960 minutes, or 5097600 seconds
Thursday Thursday September 12, 2024 is day number 256 of the year. At that time, we will be 70% through 2024.
## In 59 days, the Average Person Spent...
• 12673.2 hours Sleeping
• 1685.04 hours Eating and drinking
• 2761.2 hours Household activities
• 821.28 hours Housework
• 906.24 hours Food preparation and cleanup
• 283.2 hours Lawn and garden care
• 4956.0 hours Working and work-related activities
• 4559.52 hours Working
• 7462.32 hours Leisure and sports
• 4049.76 hours Watching television
## Famous Sporting and Music Events on September 12
• 1846 Poet and playwright Robert Browning (34) weds fellow poet Elizabeth Barrett (40) at Marylebone Church in London
• 1992 Stefan Edberg beats Michael Chang 6-7, 7-5, 7-6, 5-7, 6-4 in the longest match in US Open history (5 hours, 26 minutes)
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https://justaaa.com/statistics-and-probability/883519-a-machine-that-puts-corn-flakes-into-boxes-is
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Question
# A machine that puts corn flakes into boxes is adjusted to put an average of 15.4...
A machine that puts corn flakes into boxes is adjusted to put an average of 15.4 ounces into each box, with standard deviation of 0.22 ounce. If a random sample of 17 boxes gave a sample standard deviation of 0.33 ounce, do these data support the claim that the variance has increased and the machine needs to be brought back into adjustment? (Use a 0.01 level of significance.) (i) Give the value of the level of significance.
State the null and alternate hypotheses.
H0: σ2 = 0.0484; H1: σ2 > 0.0484
H0: σ2 = 0.0484; H1: σ2 < 0.0484
H0: σ2 = 0.0484; H1: σ2 ≠ 0.0484
H0: σ2 < 0.0484; H1: σ2 = 0.0484
(ii) Find the sample test statistic. (Round your answer to two decimal places.)
(iii) Find or estimate the P-value of the sample test statistic.
P-value > 0.100
0.050 < P-value < 0.100
0.025 < P-value < 0.050
0.010 < P-value < 0.025
0.005 < P-value < 0.010
P-value < 0.005
(iv) Conclude the test.
Since the P-value ≥ α, we fail to reject the null hypothesis.
Since the P-value < α, we reject the null hypothesis.
Since the P-value < α, we fail to reject the null hypothesis.
Since the P-value ≥ α, we reject the null hypothesis.
(v) Interpret the conclusion in the context of the application.
At the 1% level of significance, there is sufficient evidence to conclude that the variance has increased and the machine needs to be adjusted.
At the 1% level of significance, there is insufficient evidence to conclude that the variance has increased and the machine needs to be adjusted.
given,
significance level,
Degree of freedom=17-1=16
Critical Value are given
Test staticstics
Decision making
There is enough evidence to support the calim. that standard the variance has increased and the machine needs to be brought back into adjustmen.
..........................
H0: σ2 = 0.0484; H1: σ2 > 0.0484
test statistics = 36
P-value < 0.005
Since the P-value < α, we reject the null hypothesis.
At the 1% level of significance, there is sufficient evidence to conclude that the variance has increased and the machine needs to be adjusted.
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September 16th, 2022
## High School Resource
### Classroom Idea: Testing an Odometer
• An odometer is a device that measures how far a bicycle (or a car) travels. Sometimes an odometer is not adjusted accurately and gives readings which are consistently too high or too low.
• Paul did an experiment to check his bicycle odometer. He cycled 10 laps around a race track. One lap of the track is 0.4 kilometers long. When he started, his odometer read 1945.68 and after 10 laps, the his odometer read 1949.88.10 laps Compare how far Paul really traveled with what his odometer read. Make a table that shows numbers of laps in multiples of 10 up to 60 laps, the distance Paul really travels, and the distance the odometer would say he traveled.
• Draw a graph to show how the distance shown by the odometer is related to the real distance traveled.
• Find a rule or formula that Paul can use to change his incorrect odometer readings into accurate distances he has gone from the start of his ride.
• An odometer measures how far a bicycle travels by counting the number of times the wheel turns around. It then multiplies this number by the circumference of the wheel. To do this right, the odometer has to be set for the right wheel circumference. If it is set for the wrong circumference, its readings are consistently too high or too low. Before Paul’s experiment, he estimated that his wheel circumference was 210 cm. Then he set his odometer for this circumference. Use the results of his experiment to find a more accurate estimate for the circumference.
Related NY State Academic Standards: N-VM.1, N-VM.2, N-VM.3, and more!
Be sure to check out our Educational Resources, featuring thousands of activities, lesson plans, constructed-response questions, rubrics, teacher resources, multimedia, and more!
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http://mathhelpforum.com/calculus/198220-vectors.html
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1. ## vectors
I was given: a=2i - 3j - k, b=2i - j and c = -3j + k
(a) Find a + b, |a + b| Which I'm sure I'm correct with)
a + b = 4i - 2j - k
|a + b| = *33 (* representing a square root)
Here I found -(a+c), assuming it would be interpreted in the direction c towards a ??
That is a + c = 2i - 2k
I found unit vector then to be: - 4/*40 (2i - 6j)
And therefore unit vector a-c: 4/*40 (-2i + 6j)
I'm confused, because if I did it a-c I'd get 0/*8
thanks for any help
2. ## Re: vectors
Originally Posted by cellae
I was given: a=2i - 3j - k, b=2i - j and c = -3j + k
(a) Find a + b, |a + b| Which I'm sure I'm correct with)
a + b = 4i - 2j - k
|a + b| = *33 (* representing a square root)
Here I found -(a+c), assuming it would be interpreted in the direction c towards a ??
That is a + c = 2i - 2k
I found unit vector then to be: - 4/*40 (2i - 6j)
And therefore unit vector a-c: 4/*40 (-2i + 6j)
I'm confused, because if I did it a-c I'd get 0/*8
thanks for any help
Are you sure your a+b is correct? a=2i-3j-k and b=2i-j gives a+b=4i-4j-k.
a-c=2i-3j-k-(-3j+k)=2i-2k
Unit vector in direction of a-c : $\frac{1}{\sqrt{2}}$(i-k)
3. ## Re: vectors
Oh cool !
I then got asked: Find a unit vector to b and to a-c:
So I got b - b.(a-c)/|a-c|
= b - 4/4 (2j - 2k)
b - (2j - 2k)
thus (2i - j) - (2j - 2k)
= 2i - 3j - 2k
(seems wrong to me :/)
4. ## Re: vectors
just figured i was wrong there.. all good
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http://mathhelpforum.com/calculus/54526-volume-paraboloid-print.html
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# volume of a paraboloid
• Oct 19th 2008, 01:21 PM
calculusgeek
volume of a paraboloid
I need help with determining the limits using cylindrical coordinates
the volume is contained between the surface of a paraboloid z=4-x^2-y^2 and the plane z=0
• Oct 19th 2008, 01:34 PM
calculusgeek
do i use the limits for theta 0 and 2pi
for r 2 and 0 and for z 4 - r^2
• Oct 19th 2008, 01:38 PM
Chris L T521
Quote:
Originally Posted by calculusgeek
do i use the limits for theta 0 and 2pi
for r 2 and 0 and for z 4 - r^2
Yup.
Thus, your integral would be $\displaystyle \int_0^{2\pi}\int_0^2\int_0^{4-r^2}\,dz\,dr\,d\vartheta$
--Chris
• Oct 19th 2008, 01:44 PM
calculusgeek
thanks very much
i was wonder if you would know how to help me with changing to polar co ordinates in the double integral
arctan (y/x) dxdy
with the limit (x-1)^2+y^2<1
• Oct 19th 2008, 02:09 PM
Chris L T521
Quote:
Originally Posted by calculusgeek
thanks very much
i was wonder if you would know how to help me with changing to polar co ordinates in the double integral
arctan (y/x) dxdy
with the limit (x-1)^2+y^2<1
In converting from rectangular to polar,
$\displaystyle (x-1)^2+y^2<1\implies 0<r<2\cos\vartheta$ and $\displaystyle 0<\vartheta<\pi$
Thus, the double integral in polar coordinates will be $\displaystyle \int_0^{\pi}\int_0^{2\cos\vartheta}\vartheta r\,dr\,d\vartheta$, since $\displaystyle \tan^{-1}\left(\frac{y}{x}\right)=\tan^{-1}\left(\frac{r\sin\vartheta}{r\cos\vartheta}\righ t)=\tan^{-1}\left(\frac{\sin\vartheta}{\cos\vartheta}\right) =\tan^{-1}\left(\tan\vartheta\right)=\vartheta$
--Chris
p.s. From now on, ask new questions in a new thread :)
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https://allthingsstatistics.com/miscellaneous/can-coefficient-determination-negative/
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Can Coefficient of Determination be Negative?
-
The squared value of the correlation coefficient r is called the coefficient of determination, denoted as r2. It always has a value between 0 and 1. By squaring the correlation coefficient we retain information about the strength of the relationship but we lose information about the direction.
What does the coefficient of determination tell us?
The coefficient of determination tells us the proportion (or percentage) of the total variability of the dependent variable, y that is accounted for or explained by the independent variable x. If the value is 1, then the values of y are completely explained by x. There is a perfect association between x and y.
A value closer to 0 shows a low proportion of variation in y explained by x. On the other hand value closer to 1 shows that the variable x can predict the actual value of the variable y.
Can Coefficient of Determination be Negative? Why or Why Not?
The coefficient of determination is defined as the square of the correlation coefficient (r). Since the square of any real number can never be negative we conclude that the coefficient of determination can never be negative. Note that it can be equal to 0, but it cannot take any value lower than that.
On the other hand, the coefficient of correlation r can be either positive or negative depending on the relationship between the two variables. If an increase in the value of one variable leads to an increase in the other, then the correlation is positive. If an increase in the value of one variable leads to a decrease in the other, then the correlation is negative.
If the correlation between two variables is 0.496, what is the coefficient of determination?
As explained earlier we should square the correlation coefficient to obtain the coefficient of determination.
Coefficient of Determination = 0.4962 = 0.246
If the coefficient of determination is a positive value, then what can we say about the regression equation?
If the coefficient of determination is a positive value, then the regression equation can have either a positive slope or a negative slope. In short, we cannot make any assumptions about how the variables are correlated. If the variables are positively correlated, the regression equation will have a positive slope. If the variables are negatively correlated then the regression equation will have a negative slope.
The reason we cannot make any conclusions about the slope of the line, in this case, is that the coefficient of determination is always positive. So no extra information about the regression lines can be extracted from this fact.
What is the coefficient of determination in linear regression?
The coefficient of determination represents the ratio of SSR (Regression Sum of Squares) to SST (Total Sum of Squares). The SSE (Error Sum of Squares) can be calculated as the difference between the SSR and SST.
In a regression analysis if SSE = 200 and SSR = 300, then the coefficient of determination is equal to?
Since SSE (Error Sum of Squares) is equal to 200 and SSR (Regression Sum of Squares) is equal to 300 we can find the value of SST (Total Sum of Squares) by adding them.
SST = SSR + SSE = 200 + 300 = 500.
Coefficient of Determination = SSR/SST = 300/500 = 0.6
In a regression analysis if SST = 4500 and SSE = 1575, then the coefficient of determination is equal to?
Since SSE (Error Sum of Squares) is equal to 1575 and SST (Total Sum of Squares) is equal to 4500 we can find the value of SSR (Regression Sum of Squares) by subtracting them.
SSR = SST – SSE = 4500 – 1575 = 2925.
Coefficient of Determination = SSR/SST = 2925/4500 = 0.65
What does a coefficient of determination equal to zero indicate?
If the value of the coefficient of determination is equal to zero then it means that the coefficient of correlation of the two variables is equal to zero. This means that there is no linear relationship between the two variables.
It is important to note that even though there may be no linear relationship, the two variables can have a non-linear dependence on each other. For example, it is possible that one variable is equal to the square of the other variable.
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https://jm.edugain.com/articles/2/The-magic-of-Primes
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## The magic of Primes
We all know what a prime number is. It is a number that is divisible only by 1 and itself. But do you know how fascinating these numbers are?
You could call them the building block of all numbers.
Why do I say that?
We all know numbers that are not prime numbers are called composite numbers. But you may not know that all composite numbers can be built from prime numbers in a unique way by multiplication. In fact, this fact is called the “Fundamental Theorem of Mathematics” – that any number can be written as a factor of two or more primes in a unique way (by unique I mean there is only one way to write it as a factor of primes).
For example, take 96. If you factor it in primes, you will get 96 = 2 * 48 = 2 * 2 * 24 = 2 * 2 * 2 * 12 = 2 * 2 * 2 * 2 * 6 = 2 * 2 * 2 *2 * 2 * 3
So 96 can be represented as 2 * 2 * 2 * 2 * 2 * 3. There is no other way to factorize it using primes.
And this is true for all prime numbers. Isn’t it fascinating?
There are many fascinating things about prime numbers and their role as the building block for all numbers.
But let me tell you about one strange fact – there is no largest prime number. It is just not possible to have something called the “largest prime number”. A Greek Mathematician called Euclid proved it more than 2000 years ago.
There is an active search going on for finding larger and larger prime numbers. The largest one found so far is 243,112,609 ? 1. What is that? It is 2 multiplied by itself 43 million, one lakh, twelve thousand, six hundred and nine times, and one subtracted from that number. That number is 12 million digits long.
How big is that? Let me tell you – If you could write 80 numbers in a line, and your notebook page had 40 lines, and your notebook had 50 pages, it would take you 80 notebooks to write down this number. Of course, the search for still larger prime numbers continues.
1. Posted by Rohit Gautam| 2010-11-17 07:56:25
thanks for the gr8 and fascinating information abt the prime nos.
I HOpe we never get the largest prime no. as there is no largest no.. it is believed to have infinite nos.
2. Posted by Surbhit| 2011-01-26 14:54:32
I studied prime numbers in my class, but never realized the beauty and complexity of these numbers before reading this article.
Great work.
3. Posted by Malinda| 2011-04-19 07:17:49
THX that's a great aneswr!
4. Posted by Spike| 2011-04-19 07:28:57
Kudos to you! I hadn't thoguht of that!
5. Posted by SJS Inamdar| 2011-09-18 00:10:54
Mind blowing...!!!
6. Posted by SJS Inamdar| 2011-09-18 00:11:26
Marvellous...!
7. Posted by Alfarhan Zahedi| 2012-04-12 16:04:44
Really a good article.
thanks for the information.
8. Posted by Deependra| 2012-08-15 10:04:08
THIS IS GOOD ARTICLE.
9. Posted by Ayush| 2013-05-26 16:01:04
Thanks for the edugain for help in my educatin. it is a good article.
10. Posted by aditya| 2014-05-27 18:23:49
thanks. today i got a very interesting fact.
i like it .
11. Posted by CHANDRIMA| 2014-06-05 15:16:11
i like it very much
12. Posted by CHANDRIMA| 2014-06-05 15:21:44
Very fascinating wonderful story.I enjoyed reading it very much.
13. Posted by Ebasco| 2014-06-27 02:37:45
Properly, it is the Fundamental Theorem of ARITHMETIC (not mathematics) and it was proven by Gauss. I'm not sure any school textbook has tackled the concept properly since Dolciani, et al in Modern Introductory Analysis (1967). The extensions of the complexities of prime numbers include the Goldbach Conjecture and the Riemann Conjecture.
14. Posted by Tharkesh| 2014-07-08 21:11:23
WOW !! PRIME NO. FACTS ARE AMAZING.
15. Posted by ankush bhunia| 2014-12-17 09:34:38
that's cool
16. Posted by Bea Binu| 2016-03-06 21:53:30
Thanks! This was a lot of help!! Keep on writing!! : )
17. Posted by Shivaahnee| 2016-12-26 01:11:16
Cool info! Never thought that way
18. Posted by Sejal | 2017-12-19 10:55:34
That's great
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https://discuss.leetcode.com/topic/102918/merge-k-sorted-list
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# Merge k Sorted List
• why does priority queue method need extra space?? the space should be o(1)
• We often implement priority queue by heap, and the heap costs O(k) space. It's such a small space cost generally, but it might be large if we have lots of linked-list to merge.
• What would be the implementation (preferably Python) of the compare one by one?
• @NickPuljic The Priority Queue method actually do the same thing as comparing one by one: select the smallest node among k heads of list. If you like, you can just compare the k nodes one by one which costs O(k).
• class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if(lists.length == 0)
return null;
ListNode result = null;
ListNode x = null;
int min = Integer.MAX_VALUE;
int minIndex = 0;
while(true) {
boolean zerofound = true;
for(int i = 0; i < lists.length; i++) {
if(lists[i] == null) continue;
zerofound = false;
if(lists[i].val < min) {
min = lists[i].val;
minIndex = i;
}
}
`````` if(zerofound == true)
break;
if(result == null) {
result = lists[minIndex];
x = result;
} else {
x.next = lists[minIndex];
x = x.next;
}
lists[minIndex] = lists[minIndex].next;
min = Integer.MAX_VALUE;
}
return result;
}
``````
}
Can anybody explain what is wrong with this solution this is compare one by one. This gives runtime error for the 130th test case which is the last test case
• The expected answer got wrong:
When `[[1,2,3],[22,1]]` is input, `[1,1,2,3,22]` should be returned, not `[1,2,3,22,1]`.
• Note [22,1] is not sorted ascending in your input [[1,2,3],[22,1]]
• @priyekant I had the same issue with my one by one solution. I think one by one solution is just too slow for the test time cutoff.
• In the merge with divide and conquer, when merging the two lists, why if l2 is bigger do we swap l1 and l2 pointers? My code was accepted without that.
`````` while l1 and l2:
if l1.val <= l2.val:
point.next = l1
l1 = l1.next
else:
point.next = l2
l2 = l2.next
point = point.next``````
• @Hermann0 It does not do the same thing. Priority queue gets the smallest with O(lgK), while compare one by one costs O(K)
• why is the space is O(1)? Don't we add up the tmp space for every recursive call in the process of merge???
• @danielwx I have the same question. I think it should be O(log k) for recursive stack, just like merge sort. Read more here: https://stackoverflow.com/questions/24171242/why-is-mergesort-space-complexity-ologn-with-linked-lists
• @dragonfly1912 Yep, I think it is O(logk). Thank you for your reference!
• public ListNode mergeKLists(ListNode[] lists) {
//this is Approach #2 ,it is ok in my computer,
//but it is Time Limit Exceeded,why?
int k = lists.length;
ListNode newNode =new ListNode(0);
ListNode first =newNode;//当前指针
int index =0;
boolean flag =true;
if(k==0) return null;
while(flag) {//Traverse all nodes and stop
int nullNum =0;//the number of Null node
int temp =Integer.MAX_VALUE;
for(int i = 0;i<k;i++) {//遍历k格链表，获取当前节点
if(lists[i]==null) {
nullNum++;
}else if(lists[i].val<=temp) {
temp =lists[i].val;
index =i;
}
}
first.next =lists[index];
first = first.next;
if(lists[index]!=null) {
lists[index] =lists[index].next;
}
flag =nullNum==k?false:true;//Traverse all nodes
}
return newNode.next;
}
• //The answer is accepted！But,I do not know how to calculate the time complexity.It may be O(Nlogk) or others.The same as approach#3?
int k =lists.length;
ListNode dummy =new ListNode(0);
ListNode first =dummy;
Map<Integer,Integer> map =new TreeMap<>(
new Comparator<Integer>() {
`````` @Override
public int compare(Integer o1, Integer o2) {
return o1-o2;
}
});
for(int i =0;i<k;i++) {
ListNode temp =lists[i];
while(temp!=null) {
if(!map.containsKey(temp.val)) {
map.put(temp.val, 1);
}else {
map.put(temp.val, map.get(temp.val)+1);
}
temp = temp.next;
}
}
for(Integer val:map.keySet()) {
for(int i=0;i<map.get(val);i++) {
first.next =new ListNode(val);
first =first.next;
}
}
return dummy.next;``````
• `````` while l1 and l2:
if l1.val <= l2.val:
point.next = l1
l1 = l1.next
else:
point.next = l2
l2 = l1
l1 = point.next.next
point = point.next
``````
For the else section why is the l2 and l1 switched? Isn't this unnecessary? Not to mention... Doesn't this mess up "point.next = l2" and make it essentially "point.next = l1" since "l2 = l1" in the following line?
• Simple Java implementation,time complexityO(NlogN), Space complexityO(N):
public ListNode mergeKLists(ListNode[] lists) {
`````` ArrayList<Integer> aList = new ArrayList<Integer>();
for(int i=0;i<lists.length;i++){
try{
while(lists[i].next!=null){
lists[i]=lists[i].next;
}
}catch (NullPointerException e){}
}
Collections.sort(aList);
ListNode resultList=new ListNode(0);
for(int j=0;j<aList.size();j++){
resultList.next = new ListNode(aList.get(j));
resultList = resultList.next;
}
| 1,366 | 4,822 |
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| 3.5625 | 4 |
CC-MAIN-2018-05
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longest
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https://www.teachoo.com/7910/2082/Example-17/category/Examples/
| 1,723,554,159,000,000,000 |
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| 790,537,019 | 21,559 |
Examples
Chapter 2 Class 8 Linear Equations in One Variable
Serial order wise
### Transcript
Example 4 Solve 5x – 2 (2x – 7) = 2 (3x – 1) + 7/( 2)5x − 2 (2x − 7) = 2 (3x − 1) + 7/2 5x − 4x + 14 = 2 (3x − 1) + 7/2 5x − 4x + 14 = 6x − 2 + 7/2 x + 14 = 6x − 2 + 𝟕/𝟐 x + 14 = (12𝑥 − 4 + 7)/2 x + 14 = (𝟏𝟐𝒙 + 𝟑)/𝟐 2 (x + 14) = 12x + 3 2x + 28 = 12x + 3 12x + 3 = 2x + 28 12x − 2x + 3 = 28 10x + 3 = 28 10x = 28 − 3 10x = 25 x = 25/10 x = 𝟓/𝟐 L.H.S 5𝑥−2(2𝑥−7) = 5× 5/2−2(2×5/2−7) = 25/2−2(5−7) = 25/2−2 (−2) =25/2 + 4 =(25 + 8)/2 =𝟑𝟑/𝟐 R.H.S 2(3𝑥−1)+7/2 = 2(3× 5/2−1)+7/2 = 2(15/2−1)+7/2 =2((15 − 2)/2)+7/2 =2(13/2)+7/2 = 26/2+7/2 =𝟑𝟑/𝟐 ∴ LHS = RHS Hence Verified.
| 469 | 661 |
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| 4.125 | 4 |
CC-MAIN-2024-33
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latest
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| 0.521321 |
https://math.stackexchange.com/questions/4471676/why-d-4-is-the-biggest-group-generated-by-relations-langle-f-r-f2-1-r4
| 1,702,001,965,000,000,000 |
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| 440,485,766 | 36,713 |
# Why $D_4$ is the biggest group generated by relations $\langle f,r | f^2 =1 ,r^4 =1 , fr=r^3f \rangle$? [duplicate]
I want to find the presentation of group $$D_4= \{1, f, r,r^2,r^3, rf, r^2f, r^3f \}$$. $$r$$ is the rotation of a square counterclockwise by 90 degree and $$f$$ is the action that flips the square. Here $$f$$ has order $$2$$ and $$r$$ has order $$4$$.
I have found that the relations $$f^2 =1 \\ r^4 =1 \\ fr=r^3f$$ can deduce all the operation results of elements in $$D_4$$. However, I am not sure $$D_4 =\langle f,r | f^2 =1 ,r^4 =1 , fr=r^3f \rangle$$ is the presentation of $$D_4$$. In other words, is $$D_4$$ the biggest group having these relations?
I feel that equality is key to this question. One must show there are no other equalities of two elements except those deduced from the abovementioned generating relations.
• See Theorem 1.1 in kconrad.math.uconn.edu/blurbs/grouptheory/dihedral2.pdf for this result for all dihedral groups $D_n$ when $n \geq 3$.
– KCd
Jun 13, 2022 at 13:49
• I vote to reopen the question. This question is not a duplicate of the question Why must a group with presentation The question at the link and the first answer to it has to do with defining the concept of generators and relations on these generators. The question under discussion is about computing the presentation of a particular group. Jun 14, 2022 at 5:46
The usual reasoning is this.
Hints.
Let $$F=\operatorname{gr}(x,y)$$ be a free group with free generators $$x$$ and $$y$$, and $$H$$ be a normal subgroup generated by $$x^2$$, $$y^4$$, and $$(xy)^2$$.
1. Consider the homomorphism $$\phi:F\to D_4$$ defined by the formulas $$\phi(x)=f$$, $$f(y)=r$$. Clearly, $$H\leq\operatorname{Ker}(\phi)$$ and $$\operatorname{Im}(\phi)=D_4$$. Therefore the factor-group $$|F/H|\geq |D_4|$$
2. Let us prove that $$|F/H|\leq8$$. To do this, work with words in the group $$F$$. Using the fact that $$x^2,y^4,(xy)^2\in H$$ we prove that $$x^{k_1}y^{l_1}\ldots x^{k_s}y^{l_s}H=x^ky^lH,$$ where $$k_i,l_i,k,l\in\mathbb{Z}$$ and $$0\leq k<2$$, $$0\leq l<4$$.
3. It follows from 1 and 2 that $$F/H\cong D_4$$.
• Thanks for your answer! The concept of free group and quotient group is new to me. But this is very helpful. Is it also true that $F/H$ is isomorphic to any group of order 8 with the relation $x^2=x^4=(xy)^2=1$, so that the biggest group with these relations is unique? Jun 14, 2022 at 5:05
• btw should the notations in (2.) be $k<2$ and $l<4$ instead of $\leq$? Jun 14, 2022 at 5:08
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| 3.578125 | 4 |
CC-MAIN-2023-50
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latest
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| 0.862051 |
https://www.physicsforums.com/threads/buoyancy-homework-help.541644/
| 1,511,476,638,000,000,000 |
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| 851,736,570 | 15,348 |
# Buoyancy homework help
1. Oct 18, 2011
### physics kiddy
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Hello everyone,
Suppose, we have a stone tied to one end of a rubber string. We suspend the stone by holding the balance or the string.Then, note the reading on the spring balance. Now, slowly dip the stone in water in a container. Our teacher told that the stone faces buoyancy as soon as it is dipped. When the stone is fully immersed in water, no further decrease in elongation is observed in the string. Please explain why does that happen.
2. Oct 18, 2011
### lurflurf
Re: Buoyancy
What have you tried? What determines the amount of buoyancy on an object?
3. Oct 18, 2011
### physics kiddy
Re: Buoyancy
Obviously, weight of an object = weight of water displaced by it = buoyant force.
4. Oct 19, 2011
### lurflurf
Re: Buoyancy
Good. So how does the buoyant force change as the object is
000% submerged
001% submerged
037% submerged
084% submerged
099% submerged
100% submerged 1 mm under the surface
100% submerged 10 km under the surface
in particular when is the buoyant force maximal?
5. Oct 19, 2011
### physics kiddy
Re: Buoyancy
That's the problem. I don't know when it is maximum and minimum.
6. Oct 19, 2011
### lurflurf
Re: Buoyancy
So you said "Obviously, ***weight of an object ***= weight of water displaced by it = buoyant force." Which was mostly right, as in general we do not have weight of an object = weight of water displaced by it. So we start lowering the object into water. For simplicity assume that we only lower it, and do not for example alternate raising with lowering. At each instant the buoyancy depends only upon the fraction of the object submerged. Try to write an equation for this. In particular that buoyancy does not change when the fraction submerged does not, including when the object is raised or lowed without changing the fraction submerged.
| 496 | 1,988 |
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| 4.0625 | 4 |
CC-MAIN-2017-47
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longest
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en
| 0.932185 |
https://metanumbers.com/6882
| 1,611,147,267,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703520883.15/warc/CC-MAIN-20210120120242-20210120150242-00643.warc.gz
| 461,913,258 | 7,455 |
## 6882
6,882 (six thousand eight hundred eighty-two) is an even four-digits composite number following 6881 and preceding 6883. In scientific notation, it is written as 6.882 × 103. The sum of its digits is 24. It has a total of 4 prime factors and 16 positive divisors. There are 2,160 positive integers (up to 6882) that are relatively prime to 6882.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 24
• Digital Root 6
## Name
Short name 6 thousand 882 six thousand eight hundred eighty-two
## Notation
Scientific notation 6.882 × 103 6.882 × 103
## Prime Factorization of 6882
Prime Factorization 2 × 3 × 31 × 37
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 6882 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 6,882 is 2 × 3 × 31 × 37. Since it has a total of 4 prime factors, 6,882 is a composite number.
## Divisors of 6882
1, 2, 3, 6, 31, 37, 62, 74, 93, 111, 186, 222, 1147, 2294, 3441, 6882
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 14592 Sum of all the positive divisors of n s(n) 7710 Sum of the proper positive divisors of n A(n) 912 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 82.9578 Returns the nth root of the product of n divisors H(n) 7.54605 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 6,882 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 6,882) is 14,592, the average is 912.
## Other Arithmetic Functions (n = 6882)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 2160 Total number of positive integers not greater than n that are coprime to n λ(n) 180 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 888 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 2,160 positive integers (less than 6,882) that are coprime with 6,882. And there are approximately 888 prime numbers less than or equal to 6,882.
## Divisibility of 6882
m n mod m 2 3 4 5 6 7 8 9 0 0 2 2 0 1 2 6
The number 6,882 is divisible by 2, 3 and 6.
• Arithmetic
• Abundant
• Polite
• Square Free
## Base conversion (6882)
Base System Value
2 Binary 1101011100010
3 Ternary 100102220
4 Quaternary 1223202
5 Quinary 210012
6 Senary 51510
8 Octal 15342
10 Decimal 6882
12 Duodecimal 3b96
16 Hexadecimal 1ae2
20 Vigesimal h42
36 Base36 5b6
## Basic calculations (n = 6882)
### Multiplication
n×i
n×2 13764 20646 27528 34410
### Division
ni
n⁄2 3441 2294 1720.5 1376.4
### Exponentiation
ni
n2 47361924 325944760968 2243151844981776 15437370997164582432
### Nth Root
i√n
2√n 82.9578 19.0212 9.10812 5.85522
## 6882 as geometric shapes
### Circle
Radius = n
Diameter 13764 43240.9 1.48792e+08
### Sphere
Radius = n
Volume 1.36531e+12 5.95167e+08 43240.9
### Square
Length = n
Perimeter 27528 4.73619e+07 9732.62
### Cube
Length = n
Surface area 2.84172e+08 3.25945e+11 11920
### Equilateral Triangle
Length = n
Perimeter 20646 2.05083e+07 5959.99
### Triangular Pyramid
Length = n
Surface area 8.20333e+07 3.8413e+10 5619.13
## Cryptographic Hash Functions
md5 5317b6799188715d5e00a638a4278901 65c013698dcd79eb85943972444a7b629e4476f0 f615d968944f7f902ee90d76ce930ee3be500b4dd03aba9f7bad7c907822cde0 e0d4aa10a119432ea5fffd39d012f226ba88cc0af5712a09dd69b4eb6093838c5485656f9a7e828ac300f69c952b874e29d090d2aecd5f7549b78ce2d37ad6e0 2fd02f55e9e02d9d67412e79c8aec2fdd02a137b
| 1,482 | 4,092 |
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| 3.78125 | 4 |
CC-MAIN-2021-04
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| 0.807333 |
https://www.had2know.org/academics/cone-construction-calculator.html
| 1,709,601,475,000,000,000 |
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| 769,697,615 | 12,140 |
# Cone Construction Calculator
R = H = L = θ = °
A cone is constructed by taking a circle sector of radius L and angle θ and joining the two straight edges together. And any cone without a closed base can be formed back into a circle sector by cutting a straight line from the nape to the base and flattening the resulting figure.
If you are given L and θ, you can plug these values into equations to find the height (H) and base radius (R) of the corresponding cone. Likewise, if you are given H and R, you can plug these values into equations to find L and θ. These formulas are explained below the diagrams.
You can also use the convenient calculator on the left to find H and R or L and θ. Just enter the measurements in the appropriate column and click the corresponding button to compute.
### Figuring L and θ Given H and R
The relation among L, H, and R comes from the Pythagorean Theorem: H² + R² = L², or equivalently, L = sqrt(H² + R²).
To find θ given H and R, note first that the circumference of the cone's base is 2πR, which is also the arc length of the flattened sector. The circumference of the larger circle from which the sector is cut is 2πL. A sector's angle is equal to 360° times the ratio of the arc length to the full circumference. Thus, θ = 360R/L, or equivalently, θ = 360R/sqrt(H² + R²).
### Figuring H and R Given L and θ
The equations for L and θ above can be manipulated to solve for H and R. First, notice that
Lθ = sqrt(H² + R²)*360R/sqrt(H² + R²) = 360R,
Thus, R = Lθ/360. If you substitute this expression for R into the equation L = sqrt(H² + R²), you can solve for H in terms of L and θ. This gives you
H = L*sqrt[1 - (θ/360)²]
### Example
A circular sector has a radius of 10 cm and an angle of 210°. If the figure is formed into a cone, what will its height and base radius be? First we have L = 10 and θ = 210. Using the equations for H and R gives us
R = 10*210/360 = 5.8333 cm
H = 10*sqrt[1 - (210/360)²] = 8.1223 cm
| 543 | 1,976 |
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| 4.71875 | 5 |
CC-MAIN-2024-10
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| 0.910106 |
https://www.shaalaa.com/question-bank-solutions/the-maximum-value-x1-x-x-0-graph-maxima-minima_46634
| 1,597,448,018,000,000,000 |
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| 827,041,669 | 9,813 |
Share
# The Maximum Value of X1/X, X > 0 is - Mathematics
#### Question
The maximum value of x1/x, x > 0 is __________ .
##### Options
• e^(1/e)
• (1/e)^e
• 1
• none of these
#### Solution
$e^\frac{1}{e}$
$\text { Given }: f\left( x \right) = x^\frac{1}{x}$
$\text { Taking log on both sides, we get }$
$\log f\left( x \right) = \frac{1}{x}\log x$
$\text { Differentiating w . r . t . x, we get }$
$\frac{1}{f\left( x \right)}f'\left( x \right) = \frac{- 1}{x^2}\log x + \frac{1}{x^2}$
$\Rightarrow f'\left( x \right) = f\left( x \right)\frac{1}{x^2}\left( 1 - \log x \right)$
$\Rightarrow f'\left( x \right) = x^\frac{1}{x} \left( \frac{1}{x^2} - \frac{1}{x^2}\log x \right) . . . \left( 1 \right)$
$\Rightarrow f'\left( x \right) = x^\frac{1}{x} - 2 \left( 1 - \log x \right)$
$\text { For a local maxima or a local minima, we must have }$
$f'\left( x \right) = 0$
$\Rightarrow x^\frac{1}{x} - 2 \left( 1 - \log x \right) = 0$
$\Rightarrow \log x = 1$
$\Rightarrow x = e$
$\text { Now,}$
$f''\left( x \right) = x^\frac{1}{x} \left( \frac{1}{x^2} - \frac{1}{x^2}\log x \right)^2 + x^\frac{1}{x} \left( \frac{- 2}{x^3} + \frac{2}{x^3}\log x - \frac{1}{x^3} \right) = x^\frac{1}{x} \left( \frac{1}{x^2} - \frac{1}{x^2}\log x \right)^2 + x^\frac{1}{x} \left( - \frac{3}{x^3} + \frac{2}{x^3}\log x \right)$
$\text { At }x = e:$
$f''\left( e \right) = e^\frac{1}{e} \left( \frac{1}{e^2} - \frac{1}{e^2}\log e \right)^2 + e^\frac{1}{e} \left( - \frac{3}{e^3} + \frac{2}{e^3}\log e \right) = - e^\frac{1}{e} \left( \frac{1}{e^3} \right) < 0$
$\text { So, x = e is a point of local maxima }.$
$\therefore \text { Maximum value } = f\left( e \right) = e^\frac{1}{e}$
Is there an error in this question or solution?
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## AVERAGE: Excel Formulae Explained
Microsoft Excel, a powerful tool in the world of data analysis and computation, offers a plethora of functions to simplify complex calculations and data manipulation. One such function is the AVERAGE function, a statistical formula that calculates the arithmetic mean of a range of cells. This article aims to provide a comprehensive understanding of the AVERAGE function, its syntax, usage, variations, and potential errors.
## Understanding the AVERAGE Function
The AVERAGE function is a built-in formula in Excel that calculates the arithmetic mean of a given set of numbers. It adds all the numbers in the range and then divides the sum by the count of numbers. The function ignores text, logical values, and empty cells in the range.
Its syntax is straightforward: =AVERAGE(number1, [number2], ...). The function can accept up to 255 arguments, where 'number1' is required, and the rest are optional. The arguments can be numbers, cell references, ranges, arrays, or a combination of the above.
## Using the AVERAGE Function
### Basic Usage
Let's start with a simple example. Suppose we have a range of cells (A1:A5) containing the numbers 1, 2, 3, 4, and 5. To calculate the average of these numbers, we would use the formula =AVERAGE(A1:A5). Excel would add all the numbers and divide the sum by 5 (the count of numbers), returning the result 3.
The AVERAGE function can also handle multiple ranges and individual numbers. For example, =AVERAGE(A1:A5, 10, B1:B3) would calculate the average of all the numbers in cells A1 through A5, the number 10, and the numbers in cells B1 through B3.
### Handling Non-Numeric Values
As mentioned earlier, the AVERAGE function ignores text, logical values, and empty cells. For instance, if the range A1:A5 contains the numbers 1, 2, 3, and the text values 'four' and 'five', the formula =AVERAGE(A1:A5) would return 2, as it only considers the numeric values.
However, if you want to include logical values or text representations of numbers in your calculations, you can use the AVERAGEA function. The syntax is the same as the AVERAGE function, but it treats logical values and text representations of numbers as numbers. TRUE is treated as 1, FALSE as 0, and text numbers as the numbers they represent.
## Variations of the AVERAGE Function
### AVERAGEIF and AVERAGEIFS
Excel also provides variations of the AVERAGE function to handle more complex scenarios. The AVERAGEIF function calculates the average of numbers in a range that meet a specified criterion. Its syntax is =AVERAGEIF(range, criteria, [average_range]).
For instance, =AVERAGEIF(A1:A5, ">2") would calculate the average of the numbers in cells A1 through A5 that are greater than 2. If an 'average_range' is specified, the function averages the numbers in this range corresponding to the cells in 'range' that meet the 'criteria'.
The AVERAGEIFS function works similarly, but it can handle multiple criteria. Its syntax is =AVERAGEIFS(average_range, criteria_range1, criteria1, [criteria_range2, criteria2], ...). It calculates the average of numbers in 'average_range' that meet all the specified criteria.
### AVERAGEWEEK, AVERAGEMONTH, and AVERAGEYEAR
These are not built-in functions in Excel, but common tasks that can be achieved using a combination of other functions. For example, to calculate the average of numbers in a particular week, month, or year, you can use the AVERAGEIFS function with date criteria.
## Potential Errors with the AVERAGE Function
While the AVERAGE function is straightforward to use, there are a few potential pitfalls to be aware of. One common error is the #DIV/0! error, which occurs when the function tries to divide by zero. This happens when there are no numeric values in the specified range.
Another potential issue is the inclusion of hidden or filtered cells in the calculation. By default, the AVERAGE function includes hidden or filtered cells in the calculation. If you want to exclude these cells, you can use the SUBTOTAL or AGGREGATE function instead.
Lastly, remember that the AVERAGE function ignores text values and logical values. If you want to include these in your calculation, consider using the AVERAGEA function or converting the values to numbers.
## Conclusion
The AVERAGE function in Excel is a powerful tool for calculating the arithmetic mean of a set of numbers. With its variations like AVERAGEIF and AVERAGEIFS, it can handle a wide range of scenarios and data types. Understanding how to use this function effectively can greatly enhance your data analysis and computation tasks in Excel.
## Take Your Data Analysis Further with Causal
While mastering the AVERAGE function in Excel is a significant step towards enhancing your data analysis skills, why not elevate your capabilities even further? Causal is designed specifically for number crunching and data manipulation, offering intuitive tools for modelling, forecasting, and scenario planning. With Causal, you can also create compelling visualizations and interactive dashboards to present your findings. It's quick to learn and free to start. Ready to transform your data analysis experience? Sign up today and discover the power of Causal.
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ratio calculator pulley
# ratio calculator pulley
Mechanical advantage of a pulley. As visible in the figure above, a pulley is a system made of wheels and ropes that connect them. Two wheels make one pulley. To calculate the mechanical advantage of such a system, simply use the following equation: MA = 2 * n. where n is the number of pulleys
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### Pulley Calculator. RPM, Belt Length, Speed, Animated Diagrams
If you know any 3 values (Pulley sizes or RPM) and need to calculate the 4th, enter the 3 known values and hit Calculate to find the missing value. For example, if your small pulley is 6" diameter, and spins at 1000 RPM, and you need to find the second pulley size to spin it at 500 RPM, Enter Pulley1 Size 6, Pulley 1 RPM 1000, Pulley 2 RPM 500, and hit Calculate to find the second pulley diameter.
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### Pulleys Diameters and Speed Engineering ToolBox
Single Belt Transmission one driving pulley and one driven pulley. For a system with two shafts and two pulleys as indied with pulley 1 and 2 in the figure above: d 1 n 1 = d 2 n 2 (1) where. d 1 = driving pulley diameter (inch, mm) n 1 = revolutions of driving pulley (rpm rounds per minute)
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### Pulley Size Calculations For Rpms The HobbyMachinist
Jun 29, 2016 · Take the stated motor rpm (or measured) and simply apply the appropriate ratio to the diameter of the motor pulley to get the speed you want. To make this easier: Motor rpm = 2000 rpm Motor drive pulley dia. = 2" (just for sake of argument) 2000 rpm, use a 2" driven pulley (1/1) 1000 rpm, use a 4" driven pulley (1/2)
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### Power Pulleys Powermaster
It is important that any dress up pulley sets do not deviate from this ratio. Typically, a street driven car should have a pulley ratio of at least 3:1. If the vehicle has an automatic transmission with a low idle and the vehicle spends a lot of time cruising, then a higher pulley ratio perhaps 3.5:1 should be used.
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### Supercharger Boost Calculator Performance Trends
A FREE calculator that determines the Supercharger boost based on Displacements, Belt Ratio and Efficiencies. Performance Trends Inc Producing Quality Computer Tools
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### Sheave Ratio Calculation Dorris
Sheave Ratio Calculation. To calculate the sheave ratio required, first divide the motor speed by the gear drive output speed to find the total ratio required. Select the nominal gear drive ratio that is closest to the total ratio required. Then divide the total ratio required by the actual gear drive ratio for the model selected to find the
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### Gear Ratios, RPM and Tyre Diameter to KPH with Shift
Enter Diff + Gear Ratios and tyre diameter, then drag RPM slider to calculate speed from RPM. Drag Max RPM slider to adjust the range of the RPM and KPH Gauges. Enter Shift at RPM, select gear to shift to and hit Auto Shift to run up through the gears recording shift point RPM and KPH. For motorcycles, check Sprockets and enter Primary Drive Ratio and front and rear sprocket tooth count to
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### How to measure pulley size! YouTube
Feb 05, 2018 ·� Cobra belt chart https://lmr.com/products/20032004Fo This is a quick video to show you how to calculate how big your pulley is. Pullies are measured by
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### Calculators ProCharger
What is your ProCharger SuperCharger pulley size (in inches) What is your Crank Damper pulley size diameter (in inches) What is your maximum Engine Rpm . Gear Drive ratios CrankDrive (Example – 1.76) RaceDrive (1.20, 1.25, 1.30, 1.35, 1.40, 1.45, 1.53, 1.56) What is your maximum Engine Rpm .
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### Pulley Calculator. RPM, Belt Length, BeltSurface Speed
If you know any 3 values (Pulley sizes or RPM) and need to calculate the 4th, enter the 3 known values and hit Calculate to find the missing value. For example, if your small pulley is 80mm diameter, and spins at 1000 RPM, and you need to find the second pulley size to spin it at 400 RPM, Enter Pulley1 80, Pulley 1 RPM 1000, Pulley 2 RPM 400, and hit Calculate to find the second pulley diameter.
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### How to Calculate the Pulley & Belt Ratio Our Pastimes
The result will be the pulley and belt ratio. For example: a 2inch drive pulley and a 4inch load pulley yields a pulley and belt ratio of 1/2. This means that for every 1 turn of the drive pulley, the load pulley will make 1/2 of a full turn. The speed of the pulley will also change in the same way.
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### Timing Belt Pulley Design Calculator Daycounter
Timing Belt Pulley Design Calculator See our other Engineering Calculators. The dimensions on timing belt pulleys are not inherently intuitive. Normally the belt is wrapped around two pulleys with 180° of the belt touching each of the two pulleys.
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### Calculators for Contractors, Builders, Remodelers
Calculator Directory Pitch To Angle Rise Run Angle Pitch Measure Pitch Angle Square Layout Scale Calculator Triangles Circles Visual Trigonometry Pulley Belt Length RPM Sprockets Chain Length RPM Gear Ratio and RPM to Speed Wood Linear to Cubic Wood Linear to Square Large Text Templates Golden Ratio Find Scale From Image Deck Boards
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### Pulley Diameter and Belt Size Calculator Telus
calculated from pulley center to pulley center distance. Something to Consider With a two pulley system, there is a small lose of power. When the pulley system is a gear up arrangement, there is a greater power lose. Links. Sprocket Calculator Worm Gear RPM Calculator Tool Tips for Handy People Amazon Pulley Wheels Amazon Pulley Belts Amazon
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### Pulley Diameter and Belt Size Calculator Telus
calculated from pulley center to pulley center distance. Something to Consider With a two pulley system, there is a small lose of power. When the pulley system is a gear up arrangement, there is a greater power lose. Links. Sprocket Calculator Worm Gear RPM Calculator Tool Tips for Handy People Amazon Pulley Wheels Amazon Pulley Belts Amazon
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### How to Calculate First, Second and Third Pulley Systems
The article discusses three popular systems of pulleys and discretely explains the involved operational mechanisms. The systems basically include different groups of pulley/string mechanisms, each involving a specific pattern of movement for lifting the attached weight in response to the applied effort. All the systems basically are designed with one common focus that is to make the involved
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### Pulley Size Calculations For Rpms The HobbyMachinist
Jun 29, 2016 · Take the stated motor rpm (or measured) and simply apply the appropriate ratio to the diameter of the motor pulley to get the speed you want. To make this easier: Motor rpm = 2000 rpm Motor drive pulley dia. = 2" (just for sake of argument) 2000 rpm, use a 2" driven pulley (1/1) 1000 rpm, use a 4" driven pulley (1/2)
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### Pulley Torque calculations Physics Forums
Apr 29, 2013 · In my initial question I was trying to calculate a different torque at the motor pulley, which cannot happen, but I can have a different resultant force at the larger radius. So continuing on through the drive train, pulley #2 (34in one from above) and pulley
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### Supercharger Boost Calculator Performance Trends
A FREE calculator that determines the Supercharger boost based on Displacements, Belt Ratio and Efficiencies. Performance Trends Inc Producing Quality Computer Tools
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### Gear Ratios, RPM and Tyre Diameter to KPH with Shift
Enter Diff + Gear Ratios and tyre diameter, then drag RPM slider to calculate speed from RPM. Drag Max RPM slider to adjust the range of the RPM and KPH Gauges. Enter Shift at RPM, select gear to shift to and hit Auto Shift to run up through the gears recording shift point RPM and KPH. For motorcycles, check Sprockets and enter Primary Drive Ratio and front and rear sprocket tooth count to
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### Timing Belt Calculator Belt Length Calculator B&B
Calculate the length of your timing belt with B&B Manufacturing''s belt length calculator. Stock and custom belts available. Speed Ratio : Pitch Diameter Large Pulley (A) (in) Pitch Diameter Small Pulley (B) (in) Desired Belt Pitch Length (in) Next Available Belt Pitch Length (in)
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### FingerTech Robotics HowTo Pulley & Belt Calculator
Pulley Ratio: (Pulley 1 Pitch Diameter / Pulley 2 Pitch Diameter) OR (Pulley 1 RPM / Pulley 2 RPM) Center Distance: Distance between the center of rotation of Pulley 1 and Pulley 2. Belt Length: If you were to cut a belt and lay it flat, it would be this long. A 3mm pitch belt has 1 tooth every 3mm, so a 100T belt will be 300mm long.
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### Calculators ProCharger
What is your ProCharger SuperCharger pulley size (in inches) What is your Crank Damper pulley size diameter (in inches) What is your maximum Engine Rpm . Gear Drive ratios CrankDrive (Example – 1.76) RaceDrive (1.20, 1.25, 1.30, 1.35, 1.40, 1.45, 1.53, 1.56) What is your maximum Engine Rpm .
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### Belt and Pulley Ratio Calculator CalcuNation.com
A ratio of 2 to 1 would be signified by 2:1. A ratio of 1/4 to one would be signified by .25:1. To find the total ratio, use the pulley ratio formula: Ratio = (Radius of Driven Pulley) / (Radius of Drive Pulley) Example: A handcrank is attached to a drive pulley of 2 inches in radius. The belt runs from the drive pulley to a driven pulley.
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### Timing Belts and Pulleys: Speed Ratio Formula MISUMI Blog
Jun 02, 2016 · Despite its sizing importance, the concept of the speed ratio is very simple to grasp, yet is extremely useful for the sizing of pulley drive systems. The speed ratio is defined as the ratio of the large to small pulley size and can be calculated simply by dividing the number of teeth in the large pulley by the number of teeth in the small pulley.
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### supercharger overdrive pulley ratio calculator racingcalcs
This calculates the supercharger overdrive based on the number of teeth on the bottom and top pulleys. variable description bottom pulley number of teeth the number of teeth on the bottom pulley top pulley number of teeth the number of teeth on the top pulley percent overdrive the total percentage of overdrive (the percentage that supercharger overdrive pulley ratio calculator Read More »
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### Pulley size and RPM Grootfontein
This is an online calculator for pulley size and revolutions per minute (RPM).
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### RPM and Pulley Calculator mgfic.com
A rpm pulley calculator to determine the motor or drive speeds or size of a pulleys and gears. RPM Pulley Calculator 2.0 This page has been so popular that I decided to package the form into a desktop appliion. It has the same funcitonality as the form you see below except that it runs on your desktop with no internet connection needed.
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### Calculating Output Speed Using Pulley VibrAlign
Jun 27, 2019 · Your current ratio is 3.5/9 = .388 so any combination of pulleys that provides that ratio will maintain the speed. If you use an 8″ pulley on the fan than a 3″ pulley on the motor will be close. 3/8 = .375 ratio. If the motor is 1800 rpm your current fan speed is 1800(.388) = 698 rpm. A 3″ to 8″ pulley would give you 1800(.375) = 675
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### The Mechanical Advantage of Block & Tackle Sciencing
Apr 24, 2017 · A block and tackle pulley is a machine that greatly reduces the amount of force necessary to move or lift an object such as a heavy crate. A standard pulley is composed of a single wheel on an axle with a rope running over it. On its own, a pulley can only change the direction of a force applied to an object. A system
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### Ratio Calculator
Calculator Use. The ratio calculator performs two types of operations: Solve ratios for the missing value when comparing ratios or proportions. Compare ratios and evaluate as true or false to answer whether ratios or fractions are equivalent.
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### VBelt Calculation Basics Search Autodesk Knowledge
The first pulley is considered a driver pulley. The rest of the pulleys are driven pulleys or idlers. Input power can be split among several driven pulleys by using a power ratio factor for each pulley. The forces and torques are calculated accordingly. Arc of contact correction factor c 1 The arc of contact correction factor corrects the power rating of the Vbelt for pulleys where the arc of
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### Pulley Belt Length, Speed and RPM Calculator YouTube
Dec 11, 2011 · Calculate Pulley RPM, Belt Length and Speed. Free to use online Visual Animated Calculator
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### Calculators ProCharger
What is your ProCharger SuperCharger pulley size (in inches) What is your Crank Damper pulley size diameter (in inches) What is your maximum Engine Rpm . Gear Drive ratios CrankDrive (Example – 1.76) RaceDrive (1.20, 1.25, 1.30, 1.35, 1.40, 1.45, 1.53, 1.56) What is your maximum Engine Rpm .
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### Pulley Calculator Omni
This pulley calculator analyzes a system of two pulleys joined by a conveyor belt (also called a belt drive). You can use it to calculate the pulley RPM (revolutions per minute), but also its diameter, and some properties of the whole system (such as the pulley speed, belt tension or torque). You can use this tool right away or continue reading
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Mechanical advantage of a pulley. As visible in the figure above, a pulley is a system made of wheels and ropes that connect them. Two wheels make one pulley. To calculate the mechanical advantage of such a system, simply use the following equation: MA = 2 * n. where n is the number of pulleys
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### Two Pulley Calculator Home Metal Shop Club
Pulley and Belt Calculator Check 1 or 2 Boxes Enter Values. Left Diameter Right Diameter Ratio (Left / Right) Check 1 Box Enter Value. Center Separation Belt Length Calculated Values are Red color Developed by Dick Kostelnicek 04/2012 Belt Slope (degrees) Change to Cross Belt
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### Pulley and RPM Calculator Engineering Information
Pulley and Speed Calculator: Below is a small calculator that will solve the ratio for you. Check the circle next to the item you are solving for and enter the remain three items in the spaces provided. If you have any problems with the pulley calculator please email us.
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1. ## Taylor Series Problem
Find the first five terms of the Taylor series for f(x) = lnx about x = 1.
Okay, so I know the general formula for a Taylor series, and roughly how to do this, but I've seen conflicting stuff in my notes and whenever I've looked at examples (it's not really "conflicting", I just don't understand, I'm sure).
Could someone walk me through this one so I can make sure I understand the proper steps and everything? I really appreciate it.
2. ## Re: Taylor Series Problem
$f(x)=ln(x)$
$f'(x)=\frac{1}{x}$
$f''(x)=\frac{-1}{x^2}$
$f^3(x)=\frac{2}{x^3}$
$f^4(x)=\frac{-6}{x^4}$
$f^5(x)=\frac{24}{x^5}$
Substitute $x=1$ into all of these to get (respectively) $0, 1, -1, 2, -6, -24$
Then, just substitute into the formula:
$f(x)\approx f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2!}+\frac{f^3(a)(x-a)^3}{3!}+\frac{f^4(a)(x-a)^4}{4!}+\frac{f^5(a)(x-a)^5}{5!}$
So, for example, for this term:
$\frac{f^4(a)(x-a)^4}{4!}$
We'd substitute in as:
$\frac{-6(x-1)^4}{4!}$
$=\frac{-1}{4}(x-1)^4$
Hope this helps.
3. ## Re: Taylor Series Problem
Thank you! That clears up a lot of my confusion.
4. ## Re: Taylor Series Problem
Originally Posted by bobsanchez
Find the first five terms of the Taylor series for f(x) = lnx about x = 1.
Okay, so I know the general formula for a Taylor series, and roughly how to do this, but I've seen conflicting stuff in my notes and whenever I've looked at examples (it's not really "conflicting", I just don't understand, I'm sure).
Could someone walk me through this one so I can make sure I understand the proper steps and everything? I really appreciate it.
A fast solution starts from the well known series expansion...
$\ln (1+\xi)= \xi - \frac{\xi^{2}}{2} + \frac{\xi^{3}}{3} - \frac{\xi^{4}}{4} + ...$ (1)
... and now You set in (1) $x=1 + \xi \implies \xi= x-1$ ...
Kind regards
$\chi$ $\sigma$
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# Economics formula [closed]
This is an economics question, but if I can get the correct answer to this formula I can answer the question. Any help is appreciated, thanks!
If the demand $Q_x^d$ for a product given the price $P_x$ is
$$\ln Q_x^d = 10 – 5 \ln P_x$$
then product $x$ is:
1. Elastic
2. Inelastic.
3. Unitary elastic.
-
## closed as off-topic by Jonas Meyer, Micah, hardmath, Daniel Rust, voldemortDec 13 at 18:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Micah, hardmath, Daniel Rust, voldemort
If this question can be reworded to fit the rules in the help center, please edit the question.
Since economics uses the term "elastic demand" in a technical sense, you should provide the definition here. Asking at Math.SE about how economists define the term is more than likely off-topic, which is why you need to assume responsibility for it. – hardmath Dec 13 at 18:02
The price elasticity $E$ of a product which has price $P$ given a demand $Q$ is defined as the percentage change in demand for each percentage change in price, i.e.
$$E = \frac{d \ln Q}{d \ln P}\qquad (1)$$
or alternatively
$$E = \frac{P}{Q} \frac{dQ}{dP}\qquad (2)$$
Since in your case you have
$$\ln Q = 10 - 5\ln P$$
You should be able to differentiate $\ln Q$ with respect to $\ln P$ (equation (1)) to get the result you require.
Alternatively, you could express $Q$ in terms of $P$ by exponentiating both sides of the formula, giving
$$Q = e^{10 - 5\ln P} = \frac{e^{10}}{P^5}$$
and apply equation (2) to find $E$, which may be easier if you're not that experienced with differentiation.
Once you have calculated $E$, you can decide whether the good in question is elastic or inelastic by noting that elastic goods have $|E|>1$ and inelastic goods have $|E|<1$.
-
... and presumably unitary elastic goods have $|E|=1$. – Henry Sep 27 '11 at 23:43
(+1)... even though this is quite generous (considering how terrible the OP was, and that the other Chris has done zero to contribute to the solution of his own problem in the past two hours!) – The Chaz 2.0 Sep 28 '11 at 0:05
I feel like it's better to show first-timers the benefit of the doubt (fix their questions, give them good answers etc) since they don't know the ropes yet, and there are many reasons that someone might not look at a question they've posted for hours or even days after the original posting. – Chris Taylor Sep 28 '11 at 7:02
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Question
An elevator descends into a mine shaft at the of 6 Min/m . If the descent starts from 10m above the ground level, how long will it take to reach- 350m
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Solution
Distance descended is denoted by a negative integer. Initial height = +10 m Final depth = −350 m Total distance to be descended by the elevator = (−350) − (+10) = −360 m Time taken by the elevator to descend −6 m = 1 min Thus, time taken by the elevator to descend −360 m = (−360) ÷ (−6) = 60 minutes = 1 hour
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# Sudokus #193-#196 for April
### By: Dan LeKander
Volume 19, Issue 4, April 2024
Here is some Sudoku fun when the temperatures start to rise!
Note: PUZZLE PREPARATION
Prior to utilizing techniques 1-8, complete the 5 Steps of Puzzle Preparation …
1. FILL IN DATA FROM OBSERVATIONS
4. MARK UNSOLVED CELLS WITH OPTIONS THAT CANNOT EXIST IN THOSE CELLS
5. FILL IN THE OPTIONS FOR THE UNSOLVED CELLS
### Clueless?
We start with a Sudoku puzzle in progress, where it appears that there are no more obvious or not-so-obvious clues. Can you find the hidden clue in Puzzle #193
(The answer follows the conclusion of Puzzle #196, the feature puzzle for this month)
### Simply a Logic Puzzle
Difficult rating … 2.5/10
(Rating based on puzzles not requiring advanced techniques)
### Impossible to solve?
Impossible? No. Difficult? Extremely. Puzzle #195 may be a challenge for you. Advanced Techniques (Steps 6 and/or 7) are required. Difficult rating 9/10.
### Feature Puzzle
Difficult rating … 6/10
(Rating based on puzzles requiring advanced techniques)
(hint … If you need a refresher on Step 6, please click the April 2023 issue)
(hint … if you need a refresher on Step 7, please click the January 2023 issue)
May the gentle winds of Sudoku be at your back,
Dan LeKander
Clue for Puzzle #193 What is your next move? The short answer is C6R2=5. If you are having difficulty determining this, follow along. In box 1 C1R1 are limited to options 5 & 6. Now we know that C3R1, C5R1 & C6R1 cannot be a 5. In box 5 C5R4 & C5R6 are the only two unsolved cells that can be a 5. Therefore, we can see that C5R2 cannot be a 5, leaving C6R2 as the only cell that can be a 5 in box 2.
Editor’s Note: Dan would like to hear more from his readers regarding questions, comments, and suggestions. To confidentially reach Dan, please e-mail him at this address …
[email protected]
Recently someone wrote to ask how to fill them out online. Unfortunately, you need to print them yourself, but that is easy to do and I know you will have just as much fun. Also you can use a grid sheet and copy numbers onto the grid.
And, if you have not already done so, we suggest you purchase Dan’s book: “3 Advanced Sudoku Techniques, That Will Change Your Game Forever!” Purchase of a book includes a 50-page blank grid pad, 33 black and two green tokens. The book is available by contacting [email protected].
### Past Issues
Dan recommends the following Steps to complete the puzzles.
Step 1: Sudoku Pairs, Triplets and Quads – See September 2015
Step 2: Turbos & Interaction – See October 2015
Step 3: Sudoku Gordonian Rectangles and Polygons – See November 2015
Step 4: XY-Wings & XYZ Wings – See December 2015
Step 5: X-Wings – See January 2016
Posted in: Volume 19, Issue 4, April 2024, Sports
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• Nov 2nd 2008, 06:38 AM
euclid2
1.Check that the following expressions are equivalent using the values x=-2,-1,0,1,2
12x^2-27/3 and -(-2x+3)(2x+3)
2.Prove the equivalence of the equations using algebraic manipulation
• Nov 2nd 2008, 06:48 AM
skeeter
I assume you did part 1 ... here is part 2.
$\displaystyle \frac{12x^2-27}{3} = 4x^2-9 = (2x-3)(2x+3) = -(-2x+3)(2x+3)$
btw, next time use parentheses like so ... (12x^2 - 27)/3
what you posted is ... 12x^2 - 27/3 = 12x^2 - 9
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# Expected time spent in a node between visits to another node in an asymmetric random walk on integers
Recall that the first passage time to state $$i$$ is the random variable $$T_i$$ defined by $$\begin{equation*} T_i(\omega)=\mathrm{inf}\{n\geq1:X_n(\omega)=i\} \end{equation*}$$ where inf$$\emptyset=\infty$$. Let $$(X_n)_{n\geq0}$$ be a simple random walk on $$\mathbb{Z}$$ with $$p_{i,i-1}=q. Find $$\begin{equation*} \gamma^0_1=\mathbb{E}_0(\sum_{n=0}^{T_0-1}1_{X_n=1}). \end{equation*}$$
My efforts:
$$\begin{equation*} \gamma^0_1=\sum^\infty_{n=0}\mathrm{Pr}(X_n=1,n+1\leq T_0). \end{equation*}$$ $$\begin{equation*} \mathrm{Pr}(X_{2k}=1,2k+1\leq T_0)=0. \end{equation*}$$ $$\begin{equation*} \mathrm{Pr}(X_{2k+1}=1,2k+2\leq T_0)=c_kp^{k+1}q^k. \end{equation*}$$
We only need to determine $$c_k$$, which is a complicated counting problem. I know $$c_0=1,c_1=1,c_2=2,c_3=4$$. We only need to consider the walk to the right hand side of 0. In the case of $$X_{2k+1}=1,2k+2\leq T_0$$, the farthest place that the walker can reach is $$k+1$$. Of course the walker does not necessarily reach $$k+1$$. It can just repeat between some nodes, e.g., $$k-1$$ and $$k$$.
This is a transient random walk. So we cannot use the stationary distribution to calculate this.
The generating function method of calculating number of steps returning to 0 starting from 1 doesn't help either. Number of steps taken to return to 0 and number of visits to 1 are two very different things, although you must pass 1 so as to return to 0.
There seems to be no way to avoid counting. So my problem is: what is a smart way to count $$c_k$$?
$$\begin{equation*} \gamma^0_0=1. \end{equation*}$$ $$\begin{equation*} \gamma^0_i=q\gamma^0_{i+1}+p\gamma^0_{i-1},i\neq0. \end{equation*}$$ Thus $$\begin{equation*} \gamma^0_i=A+(1-A)(\frac{p}{q})^i. \end{equation*}$$ Consider the case of $$i>0$$. To visit $$i+1$$ and return to 0, the walker must visit $$i$$ at least twice. $$\gamma^0_i$$ is a non-increasing function of $$i$$. But $$(p/q)^i$$ is an increasing function. The best we can do is setting $$A=1$$ so that $$\gamma^0_i=1$$ for all $$i>0$$.
Consider the case to the left of 0. Still let $$i>0$$. We have $$\begin{equation*} \gamma^0_{-i}=A+(1-A)(\frac{q}{p})^i. \end{equation*}$$ lim$$_{i\rightarrow\infty}\gamma^0_{-i}=0$$, so $$A=0$$ and $$\gamma^0_{-i}=(q/p)^i$$ for all $$i>0$$.
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Circle, Ellipse, Parabola, Hyperbola. These are all names familiar to anyone who has had high school analytic geometry. They are the four conic sections, known to the ancient Greeks. The word section means to cut or divide into sections, so conic sections are cuts, or cross sections of a cone. Though they can arise in any cone, traditionally they are considered as coming from a right circular cone (a cone whose axis is perpendicular to its base). The model pictured in figure 1 is of a wooden right circular cone and shows the conics sections as they arise when the cone is cut by a plane.
Fig 1. Conic Section model showing, from top to bottom, and ellipse, a parabola, hyperbola and circle (base of cone). SI image DOR2013-17883
A circle is formed when a plane cuts the cone parallel to the base or perpendicular to the axis of the cone. The other three conics are formed when the cutting plane is no longer parallel to the base (or perpendicular to the axis). After the circle, the next conic section encountered in school is the ellipse. The word ellipse comes from the Greek elleipsis and means “to fall short”. An ellipse is formed when the cutting plane meets the base at an angle less than, or falls short of, the angle formed by the base and the side of the cone. In figure 1 above, notice how the angle the section makes with the base, if it were extended, is more shallow than the angle the side of the cone makes with the base. Thus it “falls short”. The cutting plane cuts a complete closed curve which is depicted in the top slice of the model above and in figure 2 below.
Fig 2. One piece of the conic section model showing an ellipse. SI image DOR2013-17886
Parabola comes from the Greek parabole and means “comparison”. A parabola is formed when the cutting plane is inclined to the base at the same angle as the side of the cone (middle cut in figure 1). A parabola does not form a closed curve but a U shape. Finally, the word hyperbola comes from the Greek hyperbole which means to “go beyond” and is formed when the plane cuts the cone at an angle greater than the angle at which the side of the cone meets the base (bottom cut in figure 1). Hyperbolas form very open U or corner shapes. Both a parabola and hyperbola are show in the figure 3 below which shows the middle piece of the conic section model of figure 1.
Fig 3. Section of cone showing a parabola on the back side and a hyperbola on the front side. SI image DOR2013-17887
The conics were known to the ancient Greeks. Arches and bridges have been constructed using ellipses and parabolas since the Romans. But scientific applications of the conics (other than circles) were not discovered until comparatively recently. For example, Galileo (1683) realized that any projectile follows a parabolic path, while Kepler (1609) discovered that all planets follow elliptical paths around the sun, as opposed to perfect circles as believed since antiquity. In fact, all orbital motion under the influence of gravity can be described using one of the four conic sections based on the mass and speed of the body in orbit. Today, the conics are used to describe the motion of myriads of objects from sub-atomic particles to satellites and whole galaxies.
Ellipsographs (also known as elliptographs) are devices used to draw ellipses. Why would you need to draw an ellipse? These curves arise most often in the areas of architectural and engineering drawing as well as in art and graphic design when drawing in perspective. In fact, German artist Albrecht Dürer, known for is precise perspective drawings, invented a compass to draw ellipses in 1540. When drawing plans or blueprints in perspective, a circle when viewed from an angle appears as an ellipse. Many windows, vaulted ceilings, stairs, bridges and arches are elliptical in design and need to be rendered accurately in technical drawings.
An ellipse is an elongated circle with a center C as well as two foci designated by F. The standard equation for an ellipse in Cartesian coordinates is:
where a is the length of the semi-major axis and b is the length of the semi-minor.
Fig 4. Diagram of an ellipse.
The standard description of an ellipse is the set of all points whose distance to the two foci is a constant. In the diagram above, the sum of the distances r1 + r2 is the same for any point on the ellipse. The eccentricity (e) of an ellipse is a number between 0 and 1 that indicates how elongated the curve is. The eccentricity is the ratio of the distance from one focus to the center compared to the length of the semi major axis (e=f/a). For a circle, the foci are at the center and so the ratio is zero (e=0/a=0). Thus a circle is a special case of an ellipse where the two foci have come together. The larger e, the more elongated the ellipse is. If you let e equal 1, your ellipse would end up being a line segment, it would have zero width. An infinite number of different ellipses can be formed by changing the separation between the two foci (which changes how long it is) or by changing the distance r1 + r2 (which changes how wide the ellipse is).
The easiest way to draw an ellipse it to take a length of string and tie it to make a loop and place it around two tacks pushed into a piece of paper. With a pencil pull the string taught and start drawing, keeping the sting at the same tension the whole while (figure 5). Since the length of string does not change, the total distance to the two foci remains constant. The curve that results is an ellipse. You can draw a perfect circle by simply using only one pin.
The easiest way to draw an ellipse it to take a length of string and tie it to make a loop and place it around two tacks pushed into a piece of paper. With a pencil pull the string taught and start drawing, keeping the sting at the same tension the whole while (figure 5). Since the length of string does not change, the total distance to the two foci remains constant. The curve that results is an ellipse. You can draw a perfect circle by simply using only one pin.
Fig 5. Image of drawing an ellipse using string and pins by the author.
There are several mechanical devices that draw an ellipse, but most of them use a method equivalent to this simple string procedure. The simplest ellipsograph used in technical drawing is a template or elliptic curve made of wood or plastic created using a variation on the string method. This type of ellipsograph is static and can only draw one size ellipse but is easy and inexpensive to make. A draftsman may have a set of several sizes of these templates. Item 82.0795.38 is one such elliptic curve and was most likely used in the classroom to draw an ellipse on the chalkboard.
The most common device for drawing ellipses is by using an elliptical compass or elliptical trammel, often referred to as the Trammel of Archimedes. A simple version can be found in handmade toy shops and is often referred to as the “do-nothing machines” or a grinder (fig 5). Animations and directions for building one are readily available on-line.
Fig 6. Trammel of Archimedes. (US Public domain)
As the handle is turned, the two small wood blocks slide back and forth in their tracks. As one moves away from the center, the other moves toward the center. The sum of the distance of each block to the handle remains constant and is twice the semi-major axis, a. A pencil attached to the end of the handle would trace an ellipse. The vast majority of ellipsographs produced are based on the same theory. However, other than the simple trammel, there are more elaborate ellipsographs that produce more precise drawings. The Smithsonian National Museum of American History has eight ellipsographs in its collections. Apart from the single elliptic curve mentioned above, there are three simple trammels (objects 1985.0112.227, 304722.14, 1987.0379.02) and four high precision items (objects 315255, 308981, 308910, 314861). Three of the items are patent models. All date from the mid-nineteenth century to the mid-twentieth century.
Resources
Gunther, R.T., Handbook of the Museum of the History of Science in the Old Ashmolean Building Oxford, Oxford University Press, Oxford, 1935, p. 66.
Hambly, M., Drawing Instruments 1590-1980, Sotheby’s Publications, 1988, pp. 89-95.
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# Negative Velocity
Velocity is a vector, meaning it has both magnitude and direction. The "direction" of velocity is either positive or negative. Positive and negative velocities describe motion in opposite directions. In Dance Dance Revolution, negative would either be juking left or back and positive would be moving towards the forward and right arrows. However, if the player would rotate 90 degrees, the relative negative and positive directions might change. DDR would enter bazaar-o world, but it could happen. The context of a problem will tell you how to interpret positive and negative velocities for that problem.
### Sample Problems
1) Let v(t) be the velocity of a bug crawling back and forth on the x-axis. When v(t) is positive, the bug is moving to the right:
When v(t) is negative, the bug is moving to the left.
2) Calvin is biking away from his house with velocity v(t).
When v(t) is negative it means Calvin is biking back towards his house.
3) A bird is flying North with velocity v(t).
When v(t) is negative it means the bird is flying South.
Speed is the magnitude, or absolute value, of velocity. This means speed can't be negative!
### Sample Problem
Let v(t), in units per second, be the velocity of a bug crawling back and forth on the x-axis. When v(t) = -5 it means the bug is crawling left at a speed of 5 units per second.
When a question asks "how fast" something is going, you're being asked for the speed.
### Sample Problem
When Calvin's velocity is -7 miles per hour, how fast is he going?
The question is asking for Calvin's speed, which is 7 miles per hour. The negative sign tells which direction Calvin is traveling, but doesn't have anything to do with his speed.
Now that we know what negative velocities mean, it's time to bring the integrals into it again and figure out what integrals of negative velocities mean.
When velocity is non-negative, we know that
When velocity is negative, the integral of velocity is also negative. We can think of this negative value as a weighted distance.
The number part tells the distance, and the negative sign tells the direction, which is the opposite of the direction travelled when v(t) is positive.
Take the area between v(t) and the t-axis that's above the axis. This is the total distance the bug travels to the right.
Take the area that's between v(t) and the t-axis, below the t-axis. This is the total distance the bug travels to the left.
Subtract:
(distance bug travels right) – (distance bug travels left) = 9 – 19 = -10.
This is the net change in the bug's position from time t = 0 to time t = 10.
We found this net change by taking a weighted sum of the areas between v(t) and the t-axis, which means we also just found the integral of v(t) from 0 to 10:
Whatever the units, when velocity is 0, speed is also 0. If something isn't moving at all, it's not moving in any direction. If v(t) is Calvin's velocity away from his house, measured in mph, then when v(t) = 0 Calvin isn't moving away from his house and he's not moving towards his house either.
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Question
# If you have two lawns - one in front of your house and the other at the back of the shape of a rectangle, find the perimeter if its lengths and breadth s are given as follows: i.l = 10 cm, b = 8 cm ii. l = 18 cm, b = 12 cm i. 30 cm, ii. 40 cm i. 36 cm, ii. 60 cm i. 40 cm, ii. 56 cm i. 34 cm, ii. 64 cm
Solution
## The correct option is B i. 36 cm, ii. 60 cm The general algebraic formula for perimeter of rectangle = 2(l + b) i.l = 10 cm, b = 8 cm Perimeter = 2(l + b) = 2(10 + 8) = 2 × 18 = 36 cm. ii. l =18 cm, b = 12 cm Perimeter = 2(l + b) = 2(18 +12) = 2 × 30 = 60 cm.
Suggest corrections
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# proof of infinite product of sums $1\tmspace-{.1667em}+\tmspace-{.1667em}a_{i}$ result without exponentials
In this entry, we show how the proof presented in the parent entry may be modified so as to avoid use of the exponential function. This modification makes it more elementary by not requiring that one first develop the theory of the exponential function before proving this basic result about infinite products. Note that it is only necessary to redo the part of the result which states that, if the series converges, then the product also converges because the proof of the opposite implication did not involve the exponential function.
We begin with a simple inequality. Suppose that $a$ and $b$ are real numbers such that $0\leq a$ and $0\leq b<1/2$. Then we have $2ab\leq a$, hence
$\displaystyle(1+a)(1+2b)$ $\displaystyle=1+2b+a+2ab$ $\displaystyle\leq 1+2b+a+a$ $\displaystyle=1+2(a+b).$
Now suppose that the series $a_{1}+a_{2}+a_{3}+\cdots$ converges to a value $S$. Since the convergence of an infinite series or product is not affected by removing a finite number of terms we may, without loss of generality, assume that $S<1/2$. Then, since the terms $a_{n}$ are nonnegative for all $n$, for each partial sum $s_{n}$ we will have $0\leq s_{n}<1/2$.
Clearly, $t_{1}\leq 1+2s_{1}$. Suppose that, for some $n$, we have $t_{n}\leq 1+2s_{n}$. Then, using the definitions of $t_{n}$ and $s_{n}$ along with the inequality demonstrated above, we conclude that
$\displaystyle t_{n+1}$ $\displaystyle=t_{n}(1+a_{n+1})$ $\displaystyle\leq(1+2s_{n})(1+a_{n+1})$ $\displaystyle\leq 1+2(s_{n}+a_{n+1})$ $\displaystyle=1+2s_{n+1}$
Hence, if $t_{n}\leq 1+2s_{n}$, then $t_{n+1}\leq 1+2s_{n+1}$ as well. By induction, we conclude that $t_{n}\leq 1+2s_{n}$ for all $n$.
Thus, for all $n$, we have $t_{n}\leq 1+2s_{n}\leq 1+2S$. Substituting this inequality for the inequality $t_{n}\leq e^{s_{n}}\leq e^{S}$ in the parent entry, the rest of the proof proceeds in exactly the same manner.
Title proof of infinite product of sums $1\tmspace-{.1667em}+\tmspace-{.1667em}a_{i}$ result without exponentials ProofOfInfiniteProductOfSums1aiResultWithoutExponentials 2013-03-22 18:40:38 2013-03-22 18:40:38 rspuzio (6075) rspuzio (6075) 11 rspuzio (6075) Result msc 40A20 msc 26E99
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How This Developer Solves a Puzzle
## How This Developer Solves a Puzzle
My wife purchased a puzzle for our family, and her Aunt and Uncle that we’re staying with on vacation to do. The puzzle consists of nine square pieces each of which contain half of an image of a cat on each side. The completed puzzle is a 3×3 grid of the pieces where all of the cats match up. After an hour or so of trying to randomly try pieces and combinations, I decided to come up with a plan that would definitively solve the puzzle. I decided to break the problem down into a set of comparisons the computer could do – and write a program to solve the puzzle.
Before I get to far, yes, I was informed by my wife that this was cheating. However, I’m not sure I agree (feel free to provide your comments if you would like). I don’t agree because puzzle solving should be about using all of your skills to be able to create the solution – not just simple trial and error. If you believe I cheated in finding the answer, that’s OK. If you prefer to think, as I do, that what I did was find an ingenious way to solve the puzzle read on.
Since I’m not an expert in image recognition, I decided the first step was to inventory the pieces and convert the images into something easier to handle. I decided that each cat had a left and a right side of the image. Since there were four different cats I’d number them 1 (Black), 2 (Orange), 3 (Gray), and 4 (Tan). I would then label the left side of the cat as the A (or true) side and the right side of the cat as the B (or false) side. Then I could create a inventory of each piece with it’s numberic identifier for the piece (which I randomly assigned) and the number for the cat image on each side: 3b, 2a, 4b, 1a… With this conversion from physical objects into numbers in hand I was ready to create some classes and structures.
I decided at the most basic I was matching sides so I created a class with the Cat Number and Cat half variables. The cat number had to match and the cat half had to be opposite. Because of this I used a boolean for the cat half. I also created a method for the class called match, which took in another puzle side and returned a simple true-false to indicate whether the sides could be matched together. One key here is that I returned true of there wasn’t a piece on the other edge. I knew there wasn’t a piece because the cat number was zero.
I also needed a puzzle piece to contain the piece’s identifier and the sizes of the piece. I also created a puzzle placement class that considered the orientation of the piece – since pieces could be rotated you didn’t know which side should be up. Finally I created a puzzle solution class to hold attempted solutions. The puzzle solution class had a method to return whether two solutions matched. This was used to prevent solutions from being tried multiple times by the program.
The main code of the program consisted of some lines to initialize variables, followed by a set of nested loops. The outer loop was a solutions loop where I recorded each tested solution. Inside of that I had loops for x and y positioning and inside of that pieces and inside of that orientation. Through these loops I’d test a piece at a time in each orientation to see if I could get it to match on all four sides. One key here was my array of positions was actually five by five and not three by three. Why? I kept a set of empty positions all around the pieces so that I didn’t have to worry about running into array bounds issues. It was wasteful of memory to be sure – but it’s efficient from a coding perspective and given this was quite literally throw away code, it seemd like a reasonable thing.
After working out some of the logic and a few minor logic bugs I ended up with a program that tried 283 solutions before finally settling on and displaying the answer on how to solve the puzzle to me. I should say that the answer was displayed instantaneously – at least to me it was instantaneously. I’m sure there were some number of milliseconds involved but to me they weren’t even measurable. I tested the solution on the real puzzle and it worked.
I ended up having about an hour in the code – and I’m certain that solving the puzzle by hand would have taken much longer than this given the complexity of the problem so I was happy with my puzzle solution.
Clearly you can’t use this technique to solving every puzzle. The typical jigsaw puzzle doesn’t lend itself quite as nicely to this sort of problem – but it was a fun exercise to convert a real world puzzle into an algorithmic one that the computer could solve.
The completed puzzle looks like this:
If you want to look at the code it’s available here.
## 1 Comment
1. Not cheating. There’s a long history of using computers to solve puzzles (Bill Cutler’s complete analysis of the six-piece burr is perhaps the most famous), though I will say that computer based solving takes come of the joy of puzzling.
There is a very powerful puzzle solving program called Burr Tools that is available for free. An amazing resource for solving many types of puzzles.
Something to consider – your program is brute force, mindlessly plugging through all options until success is found. Here’s a challenge: can you write a program that solves the puzzle without making any guesses? This would require creating logic rules to place the pieces and would be much more interesting.
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