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Empirical Project 6 Working in R

Don’t forget to also download the data into your working directory by following the steps in this project.

Getting started in R

For this project you will need the following packages:

• tidyverse, to help with data manipulation
• ggthemes, to change the look of charts easily.

We will also use the ggplot2 package to produce graphs, but that does come as part of the tidyverse package.

If you need to install either of these packages, run the following code:

install.packages(c("tidyverse","ggthemes"))

You can import the libraries now, or when they are used in the R walk-throughs below.

library(tidyverse)
library(ggthemes)

Part 6.1 Looking for patterns in the survey data

First download the data used in the paper to understand how this information was collected. The data is publicly available and free of charge, but you will need to create a user account in order to access it.

• Go to the World Management Survey main page.
• In the top right corner, click ‘Register’, and then click the ‘Register’ button on the next page.
• Fill in the form with the required details, then click ‘Register’.
• An account activation link will be sent to the email you provided. Click on it to activate your account.
• Unzip the files in the downloaded zip folder into your working directory (the folder you will be working from).
• You may also find it helpful to download the Bloom et al. paper ‘Management practices across firms and countries’.
1. To learn about how Bloom et al. (2012) conducted their survey, read the sections ‘How Can Management Practices Be Measured?’ and ‘Validating the Management Data’ (pages 5–9) of their paper.
• Briefly describe how the interviews with managers were conducted, and explain some measures the researchers took to improve the reliability and validity of their data. (There are a few technical terms that you may not understand, but these are not necessary for answering this question.)
• Three aspects of management practices were evaluated: monitoring, targets, and incentives. Do you think that these are the best criteria for assessing management practices? What (if any) important aspects of management are not included in this assessment? (You may also find it helpful to refer to the ‘Contingent Management’ section on pages 23–25 of the paper.)

Now we will create some charts to summarize the data and make comparisons across countries, industries (manufacturing, healthcare, retail, and education), and firm characteristics.

1. In ‘Manufacturing: 2004–2010 combined survey data (AMP)’, open the file ‘AMP_graph_manufacturing.csv’. Use this data on manufacturing firms to do the following:
• Create two tables, as shown in Figures 6.2a and 6.2b, and calculate the required values. The variables for individual criteria are ‘monitor’, ‘target’, and ‘people’. For each criterion, rank countries from highest to lowest. Do countries with a high overall rank also tend to rank highly on individual criteria?
Country Overall management (mean) Monitoring management (mean) Targets management (mean) Incentives management (mean)

Mean of management scores.

Figure 6.2a Mean of management scores.

Country Overall management (rank) Monitoring management (rank) Targets management (rank) Incentives management (rank)

Rank according to management scores.

Figure 6.2b Rank according to management scores.

• Create a bar chart showing the average overall management score (the variable management) for each country, ordered from highest to lowest. (Hint: You will need to sort your data from highest to lowest so it appears correctly in the chart.) Your chart should look similar to Figure 6.1.
• Compare your chart with Figure 1 in Bloom et al. (2012). Can you explain why your chart is slightly different? (Hint: See the note at the bottom of Figure 1.)

R walk-through 6.1 Importing data into R and creating tables and charts

Before uploading an Excel or csv file into R, first open the file in a spreadsheet software (like Excel) to understand how the file is structured. From looking at the file we learn that:

• the variable names are in the first row (no need use the skip option)
• missing values are represented by empty cells (hence we will use na.strings = "")
• the last variable is in column S, with short variable descriptions in column U: it is easier to import everything first and remove the unnecessary data afterwards.
man_data <- read.csv("AMP_graph_manufacturing.csv",na.strings = "")
str(man_data)

man_varinfo <- unlist(man_data$storage..display.....value[1:23]) # Keep the variable information man_data <- man_data[,!(names(man_data) %in% c("X","storage..display.....value"))] # Delete last two variables  Let’s look at the variables. man_varinfo  A few of the variables that have been imported as numbers are actually categorical (‘factor’) variables (mne_f, mne_d, and competition2004). We use the factor function to tell R how to treat these variables. man_data$mne_f <- factor(man_data$mne_f, labels=c("no MNE_f","MNE_f")) # Indicates what to call 0 and 1 entries man_data$mne_d <- factor(man_data$mne_d, labels=c("no MNE_d","MNE_d")) man_data$competition2004 <- factor(man_data$competition2004, labels=c("No competitors","A few competitors","Many competitors")) # Indicates what to call 1, 2, and 3 entries  When you create new labels, check the labels have been attached to the correct entries (the labels should be ordered from lowest to highest entry). To create the tables we use the tidyverse package, specifically piping operators ($>$). Refer to a short introduction on using piping operators.1 In addition to the mean values for the different categories and the overall score, we add a variable recording how many observations we have for each country (obs). Finally, we order the countries according to their overall score (highest to lowest). library(tidyverse) table_mean <- man_data %>% group_by(country) %>% summarize(obs = length(management), m_overall = mean(management), m_monitor = mean(monitor), m_target = mean(target), m_incentives = mean(people)) %>% arrange(desc(m_overall)) table_mean  ## # A tibble: 20 x 6 ## country obs m_overall m_monitor m_target m_incentives ## <fct> <int> <dbl> <dbl> <dbl> <dbl> ## 1 United States 1225 3.35 3.58 3.26 3.25 ## 2 Germany 646 3.23 3.57 3.22 2.98 ## 3 Japan 176 3.23 3.50 3.34 2.92 ## 4 Sweden 388 3.21 3.64 3.19 2.83 ## 5 Canada 385 3.17 3.55 3.07 2.94 ## 6 UK 1242 3.03 NA 2.98 2.86 ## 7 France 613 3.03 3.43 2.97 2.74 ## 8 Italy 289 3.03 3.26 3.10 2.76 ## 9 Australia 392 3.02 3.29 3.02 2.74 ## 10 New Zealand 106 2.93 3.18 2.96 2.63 ## 11 Mexico 189 2.92 3.29 2.88 2.71 ## 12 Poland 351 2.90 3.12 2.94 2.83 ## 13 Republic of Ireland 106 2.89 3.14 2.81 2.79 ## 14 Portugal 247 2.87 3.27 2.83 2.59 ## 15 Chile 317 2.83 3.14 2.72 2.67 ## 16 Argentina 249 2.76 3.08 2.68 2.56 ## 17 Greece 251 2.73 2.97 2.66 2.58 ## 18 China 746 2.71 2.90 2.63 2.69 ## 19 Brazil 569 2.71 3.06 2.69 2.55 ## 20 India 720 2.67 2.91 2.66 2.63  You will see that m_monitor for the UK is recorded as NA, because there is a NA entry for the monitor variable. The mean function, by default, will not produce a mean value if any observations are missing. Doing so allows you to investigate if there is a data issue. Here, this missing observation isn’t really an issue for our analysis. You use the option na.rm = TRUE in the mean function to calculate the mean, ignoring the missing observation(s). table_mean <- man_data %>% group_by(country) %>% summarize(obs = length(management), m_overall = mean(management), m_monitor = mean(monitor, na.rm = TRUE), m_target = mean(target), m_incentives = mean(people)) %>% arrange(desc(m_overall)) table_mean  ## # A tibble: 20 x 6 ## country obs m_overall m_monitor m_target m_incentives ## <fct> <int> <dbl> <dbl> <dbl> <dbl> ## 1 United States 1225 3.35 3.58 3.26 3.25 ## 2 Germany 646 3.23 3.57 3.22 2.98 ## 3 Japan 176 3.23 3.50 3.34 2.92 ## 4 Sweden 388 3.21 3.64 3.19 2.83 ## 5 Canada 385 3.17 3.55 3.07 2.94 ## 6 UK 1242 3.03 3.34 2.98 2.86 ## 7 France 613 3.03 3.43 2.97 2.74 ## 8 Italy 289 3.03 3.26 3.10 2.76 ## 9 Australia 392 3.02 3.29 3.02 2.74 ## 10 New Zealand 106 2.93 3.18 2.96 2.63 ## 11 Mexico 189 2.92 3.29 2.88 2.71 ## 12 Poland 351 2.90 3.12 2.94 2.83 ## 13 Republic of Ireland 106 2.89 3.14 2.81 2.79 ## 14 Portugal 247 2.87 3.27 2.83 2.59 ## 15 Chile 317 2.83 3.14 2.72 2.67 ## 16 Argentina 249 2.76 3.08 2.68 2.56 ## 17 Greece 251 2.73 2.97 2.66 2.58 ## 18 China 746 2.71 2.90 2.63 2.69 ## 19 Brazil 569 2.71 3.06 2.69 2.55 ## 20 India 720 2.67 2.91 2.66 2.63  Let’s make the table showing the ranks. We use the mutate function, which adds variables calculated from existing variables. table_rank <- table_mean %>% mutate(r_overall = rank(desc(m_overall)), r_monitor = rank(desc(m_monitor)), r_target = rank(desc(m_target)), r_incentives = rank(desc(m_incentives))) table_rank[c(1,7:10)] # Select the country variable (Column 1) and the columns with rank information (7 to 10)  ## # A tibble: 20 x 5 ## country r_overall r_monitor r_target r_incentives ## <fct> <dbl> <dbl> <dbl> <dbl> ## 1 United States 1 2 2 1 ## 2 Germany 2 3 3 2 ## 3 Japan 3 5 1 4 ## 4 Sweden 4 1 4 6 ## 5 Canada 5 4 6 3 ## 6 UK 6 7 8 5 ## 7 France 7 6 9 11 ## 8 Italy 8 11 5 9 ## 9 Australia 9 8 7 10 ## 10 New Zealand 10 12 10 16 ## 11 Mexico 11 9 12 12 ## 12 Poland 12 15 11 7 ## 13 Republic of Ireland 13 14 14 8 ## 14 Portugal 14 10 13 17 ## 15 Chile 15 13 15 14 ## 16 Argentina 16 16 17 19 ## 17 Greece 17 18 19 18 ## 18 China 18 20 20 13 ## 19 Brazil 19 17 16 20 ## 20 India 20 19 18 15  Now we use the ggplot set of functions (part of the tidyverse package uploaded earlier) to create a bar chart using the m_overall value in table_mean. ggplot(table_mean, aes(x=reorder(country,m_overall,mean),y=m_overall)) + geom_bar(stat = "identity",position="identity") + xlab("") + ylab("Average management practice score") + coord_flip() + theme_bw() # Note that x=reorder(country,m_overall,mean), presents the countries in order of the score. # Using x = country would have ordered the countries alphabetically which is the default option. # coord_flip() flips the x and y axis.  Management practices in manufacturing firms around the world. Figure 6.3 Management practices in manufacturing firms around the world. If you want to make the order the same as in Figure 6.1, use rev(table_mean$m_overall) and rev(table_mean$country) to reverse the order of the values. To look at how management quality varies within countries, instead of just looking at the mean we can use column charts to visualize the entire distribution of scores (as in Empirical Project 1). To compare distributions, we have to use the same horizontal axis, so we will first need to make a frequency table for each distribution to be used. Also, since each country has a different number of observations, we will use percentages instead of frequencies as the vertical axis variable. 1. For three countries of your choice and for the US, carry out the following: • Using the overall management score (variable ‘management’), create a frequency table similar to Figure 6.4 below for the US, and separately for each chosen country. The values in the first column should range from 1 to 5, in intervals of 0.2. Range of management score Frequency Percentage of firms (%) 1.00 1.20 4.80 5.00 Frequency table for overall management score. Figure 6.4 Frequency table for overall management score. • Plot a column chart for each country to show the distribution of management scores, with the percentage of firms on the vertical axis and the range of management scores on the horizontal axis. On each country’s chart, plot the distribution of the US on top of that country’s distribution, as shown in R walk-through 6.2. • Describe any visual similarities and differences between the distributions of your chosen countries and that of the US. (Hint: For example, look at where the distribution is centred, the percentage of observations on the left tail or the right tail of the distribution, and how spread out the scores are.) R walk-through 6.2 Obtaining frequency counts and plotting overlapping histograms To get frequency counts, use the cut function. temp_counts <- cut(man_data$management[man_data$country == "Chile"], breaks=seq(0,5,0.2)) table(temp_counts)  ## temp_counts ## (0,0.2] (0.2,0.4] (0.4,0.6] (0.6,0.8] (0.8,1] (1,1.2] (1.2,1.4] ## 0 0 0 0 0 0 1 ## (1.4,1.6] (1.6,1.8] (1.8,2] (2,2.2] (2.2,2.4] (2.4,2.6] (2.6,2.8] ## 3 6 24 15 30 25 49 ## (2.8,3] (3,3.2] (3.2,3.4] (3.4,3.6] (3.6,3.8] (3.8,4] (4,4.2] ## 52 28 27 21 20 9 6 ## (4.2,4.4] (4.4,4.6] (4.6,4.8] (4.8,5] ## 1 0 0 0  To create a complex chart like this we will call upon the help of geom_histogram member of the ggplot set of functions. Let’s first collect one pair of countries, Chile and the US. If you wanted to produce a histogram for the overall (management) rating, use the following code. g1 <- ggplot(subset(man_data, country == "Chile"),aes(management)) + geom_histogram(breaks=seq(0,5,0.2)) + xlab("Management score") + ylab("Frequency Count") print(g1)  Distribution of management scores, Chile. Figure 6.5 Distribution of management scores, Chile. Tip: Using ggplot, it is straightforward to add a second country to the chart. The way to learn is usually to search the Internet (here we searched for ‘r ggplot multiple histograms’). g1 <- ggplot(subset(man_data, country %in% c("Chile","United States")), aes(x=management, y = 0.2*..density.., fill = country)) + geom_histogram(breaks=seq(0,5,0.2), alpha=.5, position = "identity") + xlab("Management score") + ylab("Density") + ggtitle("Histogram for management score") + scale_fill_discrete(name ="Country") + theme_bw() # y = 0.2*..density.. - ..density.. is used to ensure the histograms are on the same scale, the result is a graph with an area of 1. As we want proportions we need to scale by multiplying with the bin width, here 0.2. # fill = country - ensures that R knows to print different histograms for different countries # breaks=seq(0,5,0.2) - sets the breakpoints # alpha=.5 - makes the bars semi-transparent # position = "identity" - ensures that the two histograms are overlaid and not stacked # scale_fill_discrete(name ="Country") gives a title to the legend print(g1)  Comparing the distribution of management scores for the US and Chile. Figure 6.6 Comparing the distribution of management scores for the US and Chile. box and whisker plot A graphic display of the range and quartiles of a distribution, where the first and third quartile form the ‘box’ and the maximum and minimum values form the ‘whiskers’. Another way to visualize distributions is a box and whisker plot, which shows some parts of a distribution rather than the whole distribution. We can use box and whisker plots to compare particular aspects of distributions more easily than when looking at the entire distribution. As shown in Figure 6.7, the ‘box’ consists of the first quartile (value corresponding to the bottom 25 per cent, or 25th percentile, of all values), the median, and the third quartile (75th percentile). The ‘whiskers’ are the minimum and maximum values. (In R, the ‘whiskers’ may not be the actual maximum or minimum, since any values larger than 1.5 times the width of the box are considered outliers and are shown as separate points.) Example of a box and whisker plot. Figure 6.7 Example of a box and whisker plot. 1. Using the same countries you chose in Question 3: • Make a box and whisker plot for each country and the US, showing the distribution of management scores. You can either make a separate chart for each country or show all countries in the same plot. To check that your plots make sense, compare your box and whisker plots to the distributions from Question 3. • Use your box and whisker plots to add to your comparisons from Question 3(c). R walk-through 6.3 Creating box and whisker plots We use exactly the same structure as for the overlapping histograms. A useful feature of ggplot is that using more or less the same structure, you can create a variety of graphs. In this example, we include a few more countries, as this can be done without overcrowding the figure. library(ggthemes) # Change the look of charts g2 <- ggplot(subset(man_data, country %in% c("Chile","United States","Brazil","Germany","UK")),aes(x = country,y = management)) + geom_boxplot() + ylab("Management score") + ggtitle("Box and whisker plots for management score") + theme_solarized() # x = Country – different countries go on the horizontal axis # y = management – management scores on the vertical axis # theme_solarized() – gives the plot a different look print(g2)  Box and whisker plots for a selection of countries. Figure 6.8 Box and whisker plots for a selection of countries. From the manufacturing data, firms in the US seem to be managed better (on average) than firms in other countries. To investigate whether this is the case in other sectors, we will use data gathered on hospitals and schools. 1. Using the data for hospitals and schools (AMP_graph_public.csv): • Create a table for hospitals and schools, showing the mean management score and criteria score (monitoring, targets, incentives) for each country, as in Figure 6.2a. • Make separate bar charts for hospitals and schools showing the mean overall management score for each country, sorted from highest to lowest, as in Figure 6.1. Are the country rankings similar to those in manufacturing? • Using your average criteria scores from Question 4(a), suggest some explanations for the observed rankings in either hospitals or schools. (You may find it helpful to research healthcare or educational policies and reforms in those countries to support your explanations.) Part 6.2 Are differences in management practices statistically significant? Using the management survey data collected by Bloom et al. (2012), we can compare average management scores across countries and industries. Rather than simply identifying differences between groups, we are also interested in whether these differences are statistically significant. confidence interval A range of values that is centred around the sample value, and is defined so that there is a specified probability (usually 95%) that it contains the ‘true value’ of interest. In Empirical Project 2, we assessed statistical significance using p-values. Now we will assess statistical significance using another method called confidence intervals. These two methods are equivalent, meaning that we would get the same conclusions about statistical significance whichever method we use. A 95% confidence interval is calculated from the data we observed and is designed so that the true value (for example, the mean of a population) will fall into the interval 95% of the time. Other common confidence intervals used in research studies are 90% and 99% confidence intervals, which are similarly defined. We will use 95% confidence intervals throughout this project. What do we mean by the ‘true value’? Remember that we usually work with data that is a small sample from the entire population of interest. For example, the World Management Survey collects information from a selection of all the firms in a particular country. Since we don’t have data on all the firms in every country, we cannot say with certainty that the average management score across all firms in Country A (the ‘true value’) is higher than that of Country B. However, based on the sample of firms we have from Country A and Country B, we can say whether any observed difference in means is likely to be due to chance, and assess how precisely we have estimated the ‘true value’ of the difference and individual means. To understand the principle behind confidence intervals, think about playing ring toss while blindfolded. You try to throw a ring so that it lands around a peg. The peg (the true value) is fixed in place, but depending on how wide your ring is (how spread out your sample is) and where you throw it (the mean of your sample), the ring may not land around the peg. Since you are blindfolded, you will never know if the ring actually landed around the peg. However, you know where to stand (at the sample mean) and how wide to make the ring so that it lands around the peg 95% of the time. As the name suggests, confidence intervals tell us how much confidence we can place in our estimates, in other words how precisely the sample mean is estimated. Wider confidence intervals suggest that our sample mean is estimated less precisely (the data is more spread out). Using the ring toss analogy, we are less sure that we are standing in the right place, so we need a wider ring to have the same chance (95%) of landing it around the peg. To sum up: A confidence interval is a range of values centred around the sample value and is defined so that there is a specified probability (usually 95%) that it contains the true value of interest. Rule of thumb for statistical significance When comparing two distributions, if neither mean is in the confidence interval for the other mean, the difference in means is statistically significant. This rule of thumb is handy when looking at charts. For a more definite conclusion, we can calculate the p-value (see Empirical Project 2) or construct a confidence interval for the difference in means. (This method involves more mathematics so we will discuss that in Empirical Project 8.) We will now build on the results from the Bloom et al. (2012) paper by using 95% confidence intervals to make comparisons between the mean overall management score for different countries and types of firms. The confidence interval for the population mean (mean management score for that country) is centred around the sample mean. To determine the width of the interval, we use the standard deviation and number of firms. 1. First look at manufacturing firms in different countries. Using the manufacturing data (AMP_graph_manufacturing.csv) for three countries of your choice and for the US: • Create a summary table for the overall management score as shown in Figure 6.9 below, with one row for each country. Country Mean Standard deviation Number of firms Summary table for manufacturing firms. Figure 6.9 Summary table for manufacturing firms. • Determine the width of the 95% confidence interval (this is the distance from the mean to one end of the interval). You should get a different number for each country. • Plot a bar chart showing the mean management score and add the confidence intervals to your chart. • Using your chart from Question 1(c) and the rule of thumb, determine which means (if any) are significantly different from the US mean. Use the width of the confidence intervals to describe how precisely each mean was estimated. Do your results change if you use a different specified probability (for example, 99%)? R walk-through 6.4 Calculating confidence intervals and adding them to a chart As in R walk-through 6.1, we use piping operators from the tidyverse package. table_stats <- man_data %>% filter(country %in% c("Chile","United States","Brazil","Germany","UK")) %>% group_by(country) %>% summarize(obs = length(management), mean_m = mean(management), sd_m = sd(management, na.rm = TRUE)) %>% arrange(rev(mean_m)) table_stats  ## # A tibble: 5 x 4 ## country obs mean_m sd_m ## <fct> <int> <dbl> <dbl> ## 1 United States 1225 3.35 0.643 ## 2 UK 1242 3.03 0.679 ## 3 Chile 317 2.83 0.599 ## 4 Germany 646 3.23 0.569 ## 5 Brazil 569 2.71 0.685  To get the confidence intervals, we use the t.test function. tUS <- t.test(subset(man_data,country == "United States",select = management)) tUS$conf.int[1:2] # tUS contains a lot of information; $conf.int[1:2] is the confidence interval.  ## [1] 3.312379 3.384448  We want to add these interval values to table_stats. The easiest way is to calculate the standard error for the sample mean and multiply this by 1.96 (m_err $= 1.96 \times \sqrt{s^2/(n − 1)}$), where 1.96 is the factor required to get a 95% confidence interval (assuming a normal distribution). The confidence interval is then [mean_m − m_err, mean_m + m_err]. table_stats <- man_data %>% filter(country %in% c("Chile","United States","Brazil","Germany","UK")) %>% group_by(country) %>% summarize(obs = length(management), mean_m = mean(management), sd_m = sd(management, na.rm = TRUE) , m_err = 1.96*sqrt(sd_m^2/(obs-1))) %>% arrange(rev(mean_m)) table_stats  ## # A tibble: 5 x 5 ## country obs mean_m sd_m m_err ## <fct> <int> <dbl> <dbl> <dbl> ## 1 United States 1225 3.35 0.643 0.0360 ## 2 UK 1242 3.03 0.679 0.0378 ## 3 Chile 317 2.83 0.599 0.0660 ## 4 Germany 646 3.23 0.569 0.0439 ## 5 Brazil 569 2.71 0.685 0.0563  Now we can use this information to make a bar chart: ggplot(table_stats, aes(y=mean_m, x=country)) + geom_bar(position=position_dodge(), stat="identity", colour="black", # Use black outlines size=.3) + # Add thinner lines for bars ylab("Mean management score") + xlab("") + geom_errorbar(aes(ymin=mean_m-m_err, ymax=mean_m+m_err), size=.6, # Add thinner lines for confidence intervals width=.5, position=position_dodge(.9)) + coord_cartesian(ylim=c(2,4)) + theme_bw() + theme(axis.text.x=element_text(size=rel(1.5)), axis.text.y=element_text(size=rel(1.3)))  Bar chart of mean management score in manufacturing firms for a selection of countries, with 95% confidence intervals. Figure 6.10 Bar chart of mean management score in manufacturing firms for a selection of countries, with 95% confidence intervals. 1. Using the data for hospitals or schools (AMP_graph_public.csv), using all available countries: • Create a summary table for the overall management score, with one row for each country. • Plot separate bar charts for hospitals and schools. Which (if any) countries have significantly better or significantly worse management practices (on average) than the US? 1. Look at the width of your confidence intervals and relate this to the standard deviation and number of observations. Are confidence intervals generally wider/narrower if the standard deviation is larger? How about if the number of observations is larger? With reference to the ring toss example, explain why we would expect there to be a relationship between the confidence interval width, standard deviation, and number of observations. Part 6.3 What factors affect the quality of management? Besides documenting and comparing management practices across industries and countries, another purpose of the World Management Survey was to investigate factors that affect management quality. One possible factor affecting differences in management is firm ownership. To look at the data for this factor in the healthcare and education sectors, we will focus on broad groups (public vs privately-owned firms), and for manufacturing firms we will focus on different kinds of private ownership. 1. Using the data for hospitals and schools (AMP_graph_public.csv): • Create a separate table for hospitals and schools, showing the average management score, standard deviation (StdDev), and number of observations, with country as the row variable, and ownership (public or private) and ind as the column variables. • Use your table to calculate the confidence interval width for each country and ownership type. • Plot a bar chart (one for hospitals and another for schools) showing the means and the confidence intervals for the management score. Describe the differences between public and private firms within countries and compare management scores for the same firm type across countries. (For example, is one type of firm generally better managed than the other? Are there similar patterns for hospitals and schools? If you have done Question 5 in Part 6.1, you may want to discuss whether the rankings change after conditioning on ownership type. Besides ownership type, management practices may vary depending on firm size, though it is difficult to predict what the relationship between these variables might be. Larger firms have more employees and could be more difficult to manage well, but may also attract more experienced managers. We will look at the conditional means for manufacturing firms, depending on whether they are above or below the median number of employees (calculated from the data), and see if there is a clear relationship. 1. Using the data for manufacturing firms (AMP_graph_manufacturing.csv): • Create a new variable that equals ‘Smaller’ if a firm has less than the median number of employees (330). In natural log terms, this corresponds to log employment of less than 5.80. • For two countries of your choice and the US, create a table showing the mean overall management score, standard deviation, and number of observations, with country and ownership as the row variables, and firm size as the column variable. (Note: When there is only one observation in a group, there is no standard deviation.) • Use your table to calculate the confidence interval width for each firm size and ownership type. • Plot a bar chart for each country, showing the means and the confidence intervals. Describe any patterns you observe across ownership types and firm size in each country. R walk-through 6.5 Calculating and adding conditional summary statistics and confidence intervals to a chart We will use many techniques encountered previously, but first we have to create a new variable that indicates whether a firm is large or small (size). A firm with lemp_firm > 5.8 is considered large. man_data$size <- factor(man_data\$lemp_firm > 5.8,labels=c("small","large"))

We choose Canada, Brazil, and the United States. Initially we add the new size variable and ownership as a grouping variable in the group_by command (as we did in R walk-through 6.1).

table_stats2 <- man_data %>% filter(country %in% c("Canada","United States","Brazil")) %>% group_by(country,ownership,size) %>% summarize(obs = length(management), mean_m = mean(management,na.rm = TRUE), sd_m = sd(management, na.rm = TRUE))
table_stats2

## # A tibble: 53 x 6
## # Groups:   country, ownership [?]
##    country ownership                  size    obs mean_m    sd_m
##    <fct>   <fct>                      <fct> <int>  <dbl>   <dbl>
##  1 Brazil  Dispersed Shareholders     small    28   3.06   0.667
##  2 Brazil  Dispersed Shareholders     large    45   3.48   0.731
##  3 Brazil  Family owned, external CEO small     8   2.82   0.725
##  4 Brazil  Family owned, external CEO large    10   2.99   0.688
##  5 Brazil  Family owned, family CEO   small    80   2.50   0.668
##  6 Brazil  Family owned, family CEO   large    41   2.70   0.645
##  7 Brazil  Founder                    small   124   2.35   0.524
##  8 Brazil  Founder                    large    72   2.66   0.591
##  9 Brazil  Government                 small     1   4    NaN
## 10 Brazil  Government                 large     2   2.44   1.18
## # ... with 43 more rows

Now we use the variable size as a column variable, so that we can see the summary statistics in two blocks of columns (separately for large and small firms). This is not a standard or straightforward procedure, but an Internet search (for ‘tidyverse spread multiple columns’) gives the following solution.

table_stats2_mc <- table_stats2 %>% gather(variable, value, -(country:size)) %>% unite(temp, size, variable) %>% spread(temp,value)
print(table_stats2_mc)

## # A tibble: 28 x 8
## # Groups:   country, ownership [28]
##    country ownership        large_mean_m large_obs large_sd_m small_mean_m
##    <fct>   <fct>                   <dbl>     <dbl>      <dbl>        <dbl>
##  1 Brazil  Dispersed Share~         3.48        45      0.731         3.06
##  2 Brazil  Family owned, e~         2.99        10      0.688         2.82
##  3 Brazil  Family owned, f~         2.70        41      0.645         2.50
##  4 Brazil  Founder                  2.66        72      0.591         2.35
##  5 Brazil  Government               2.44         2      1.18          4
##  6 Brazil  Managers                 2.51         7      0.631         2.64
##  7 Brazil  Other                    3.01        29      0.541         2.57
##  8 Brazil  Private Equity          NA           NA     NA             3.23
##  9 Brazil  Private Individ~         2.94        42      0.523         2.69
## 10 Canada  Dispersed Share~         3.52        53      0.582         3.43
## # ... with 18 more rows, and 2 more variables: small_obs <dbl>,
## #   small_sd_m <dbl>

To understand the logic of this command, go through it step by step: first apply gather and look at the result, then add the unite command and look at the result, and then add spread.

So far we have looked at correlations between firm characteristics and management practices, but have not made any causal statements. We will now discuss the difficulties with making causal statements using this data and examine how we might determine the direction of causation.

1. For each of the following variables, explain how it could affect management practices, and then explain how management practices could affect it:
• education level of managers (percentage with a college degree)
• number of competitors
• firm size (number of employees).
1. One way to establish the direction of causation is through a randomized field experiment. Read the discussion on pages 22–23 of the Bloom et al. paper (the section ‘Experimental Evidence on Management Quality and Firm Performance’) about one such experiment that was conducted in Indian textile factories.
• Briefly describe the idea behind a randomized field experiment, and explain, with reference to the results of the experiment in India, whether we can use it to determine the direction of causation between management practice and firm performance. The paper ‘Does Management Matter? Evidence from India’ provides more details about the experiment (pages 9–10 are particularly useful).
• Figure 12 in the paper shows productivity in treatment and control firms over time, with 95% confidence intervals. Use the information in the chart to describe the effect of the treatment on firm productivity.
1. University of Manchester’s Econometric Computing Learning Resource (ECLR). 2018. ‘R AnalysisTidy’. Updated 9 January 2018.


text:
Preprint Article Version 1 Preserved in Portico This version is not peer-reviewed

# On the Eternal Role of Planck Scale in Machian Cosmology

Version 1 : Received: 17 October 2018 / Approved: 18 October 2018 / Online: 18 October 2018 (08:42:57 CEST)
Version 2 : Received: 25 October 2018 / Approved: 25 October 2018 / Online: 25 October 2018 (09:51:27 CEST)
Version 3 : Received: 30 October 2018 / Approved: 2 November 2018 / Online: 2 November 2018 (02:30:57 CET)

How to cite: Seshavatharam, U.V.S.; Lakshminarayana, S. On the Eternal Role of Planck Scale in Machian Cosmology. Preprints 2018, 2018100407 (doi: 10.20944/preprints201810.0407.v1). Seshavatharam, U.V.S.; Lakshminarayana, S. On the Eternal Role of Planck Scale in Machian Cosmology. Preprints 2018, 2018100407 (doi: 10.20944/preprints201810.0407.v1).

## Abstract

Considering Planck mass as a baby universe and evolving universe as a growing Planck ball, a hypothetical spherical model of cosmology can be developed. In addition, by considering the famous relation GM = c2R, a Machian model of quantum cosmology can also be developed. Proceeding further, in all directions, cosmic expansion velocity can be referred to ‘from and about’ the baby universe. This hypothetical scenario can be made versatile by adding a heuristic idea: apart from normal expansion, ‘cosmic temperature seems to be redshifted by a factor $\left({\left({Z}_{T}\right)}_{t}+1\right)\cong \sqrt{1+\mathrm{ln}\left({H}_{pl}/{H}_{t}\right)}$; where (Hpl, Ht) represent Planck scale and time dependent Hubble parameters respectively.

## Subject Areas

Planck scale; Mach’s relation; quantum cosmology; critical density; ordinary matter; dark matter; thermal redshift; expansion velocity; Hubble’s law

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text:
# Trigonometry/Angles of a triangle sum to 180 Degrees

In any triangle the angles
always sum to $180^{\circ}\,$

## The Sum of AnglesEdit

In any triangle the angles always sum to $180^{\circ}\,$

This is a perhaps surprising fact.

Because $90^{\circ}\,$ is a right angle, it means that the sum of the angles of any triangle is the same as two right angles. If we 'tore the corners off' and placed them together at the same point, we could arrange them so that they exactly formed a straight line. There doesn't need to be anything special about the triangle. It works for any triangle.

Angles sum to 180o

Some examples that we had before of triangles are shown below

.
 Equilateral Triangle 45-45-90 Triangle 30-60-90 Triangle 50-60-70 Triangle 20-40-120 Triangle
• The first example shows an equilateral triangle. All of the sides are equal. All of the angles are equal. Each angle is 60 degrees. The sum of the angles is $60^{\circ}+60^{\circ}+60^{\circ}\,$ which is $180^{\circ}\,$.
• The second triangle shows a right angle triangle. One of the angles is a right angle. This right angle triangle has two sides the same length. It is symmetric. It fulfils our criteria for being an isosceles triangle. This is a particularly special isosceles triangle because it is isosceles and it is a right triangle. There is one angle of 90 degrees and each of the two remaining angles is 45 degrees. The sum of the angles is $45^{\circ}+45^{\circ}+90^{\circ}\,$ which is $180^{\circ}\,$.
• The third triangle is sometimes called the 30-60-90 triangle, because of its angles. It is actually half an equilateral triangle. The sum of the angles is $30^{\circ}+60^{\circ}+90^{\circ}\,$ which is $180^{\circ}\,$.

The pattern is pretty clear.

• Next we have a more arbitrary triangle. All the sides are different. The angles are 50, 60 and 70. The sum of the angles is $50^{\circ}+60^{\circ}+70^{\circ}\,$ which is $180^{\circ}\,$.
• Finally we have a triangle with an obtuse angle, that is one of the angles is larger than 90 degrees. The angles happen to be 20, 40 and 120, and the sum of the angles is $20^{\circ}+40^{\circ}+120^{\circ}\,$ which is $180^{\circ}\,$.

### The examples suggest it is true, but they don't prove it.Edit

We could keep on doing this for other triangles, and keep finding the same answer, unless we make a mistake. This might convince us that our statement that the angles sum to 180 is true for all triangles, but it does not prove that it is so. To prove it we need some kind of general argument that could convince a mathematician that it is true. How do we know it is always true?

How could it go wrong? Well, if we hadn't tried with a triangle with an obtuse angle, it might be the case that the formula only works for triangles which don't have obtuse angles. Even having tried the triangle with an obtuse angle we could have not been trying hard enough to find an example that doesn't work. For all we know the formula only works if the angles are multiples of 5 degrees.

### Proof will show it works for all trianglesEdit

The formula does in fact work for all triangles. We can for example make a triangle with angles of 33 degrees and 66 degrees and the third angle will have to be 81 degrees. Making more and more examples unfortunately doesn't get us anywhere closer to proving it is true of all triangles. We need a different approach. We'll show a proof later. The point of having a proof is to show that it is true for all triangles, not just the ones we've chosen to look at.

## ExercisesEdit

 A triangle has angles 30 degrees and 90 degrees, what is the third angle?
 A triangle has angles 15 degrees and 65 degrees, what is the third angle?
 A triangle has angles 100 degrees and 79.5 degrees, what is the third angle?   Do you think all the sides of this triangle will be about the same length?
 What is the measure of each angle of an equilateral triangle?
 The measure of one angle of an isosceles triangle is $\displaystyle 108^{\circ}$. What are the measures of the other two angles?


text:
# Black hole formation from collisions of cosmic fundamental strings

@article{Iengo2006BlackHF,
title={Black hole formation from collisions of cosmic fundamental strings},
author={Roberto Iengo and Jorge G Russo},
journal={Journal of High Energy Physics},
year={2006},
volume={2006},
pages={079-079}
}
• Published 13 June 2006
• Physics
• Journal of High Energy Physics
We develop the general formalism for joining, splitting and interconnection of closed and open strings. As an application, we study examples of fundamental cosmic string collisions leading to gravitational collapse. We find that the interconnection of two strings of equal and opposite maximal angular momentum and arbitrarily large mass generically leads to the formation of black holes, provided their relative velocity is small enough.
8 Citations

## Figures and Tables from this paper

We explain simple semi-classical rules to estimate the lifetime of any given highly excited quantum state of the string spectrum and apply them to identify new long-lived string states. Using
We make an attempt to dynamically study, in four space-time dimensions, the classical gravitational collapse of a macroscopic circular fundamental string, by a truncation of the Einstein equations
In this paper we present semiclassical computations of the splitting of folded spinning strings in ${\mathrm{AdS}}_{3}$, which may be of interest in the context of AdS/CFT duality. We start with a
Using integrability we reduce the problem of constructing general classical splitting string solutions on R ×S 3 to a series of Birkhoff factorization problems. Namely, given any incoming string
A bstractUsing integrability we reduce the problem of constructing general classical splitting string solutions on $$\mathbb{R}$$ × S3 to a series of Birkhoff factorization problems. Namely, given
A bstractWe study a three-point correlator of the three heavy vertex operators representing the circular winding string states which are point-like in AdS5 and rotating with two spins and two winding
• Physics
• 2008
We investigate the tunnelling of spin-1/2 particles through event horizons. We first apply the tunnelling method to Rindler spacetime and obtain the Unruh temperature. We then apply fermion

## References

SHOWING 1-10 OF 22 REFERENCES

• Physics
• 2003
Macroscopic fundamental and Dirichlet strings have several potential instabilities: breakage, tachyon decays, and confinement by axion domain walls. We investigate the conditions under which
• Physics
Physical review letters
• 2006
There are important kinematical constraints on the process of intercommuting of two straight strings in which they exchange partners but become joined by a third string and it is shown that the exchange cannot occur if the strings meet with very large relative velocity.
• Physics
• 1985
Preface 1. Introduction 2. Phase transitions in the early universe 3. Topological defects 4. String field theory 5. Superconducting strings 6. String dynamics 7. String gravity 8. String interactions
• Physics
• 2005
Cosmic strings are linear concentrations of energy that may be formed at phase transitions in the very early universe. At one time they were thought to provide a possible origin for the density
• Physics
• 2005
We present solutions of the equations of motion of macroscopic F and D strings extending along the non compact 4D sections of the conifold geometry and winding around the internal directions. The
• Physics
• 2003
We study the classical breaking of a highly excited (closed or open) string state on the leading Regge trajectory, represented by a rotating soliton solution, and we find the resulting solutions for
Brane inflation, where branes move towards each other in the brane world, has been shown to be quite natural in superstring theory. Inflation ends when branes collide and heat the universe,
• Physics
• 2006
We explain simple semi-classical rules to estimate the lifetime of any given highly-excited quantum state of the string spectrum in flat spacetime. We discuss both the decays by splitting into two
There are two main reasons for the recent renewal of interest in cosmic strings: Fundamental string-theory models suggest their existence; and there are at least two tentative observations of their


text:
1. ## matrices

X is a 2 by 2 matric

2 -3
4 -1

multiplies by unkonw matrix X

=

-12 -9
1 3

Find the matrix X

2. Make the unknow matrix and label each part, eg.

$\begin{bmatrix}
{2}\;\; {-3}\\
{4}\;\;{-1}

\end{bmatrix} \cdot\begin{bmatrix}
{a}\;\; {b}\\
{c}\;\;{d}

\end{bmatrix}=\begin{bmatrix}
{-12}\;\; {-9}\\
{1}\;\;\;{3}

\end{bmatrix}$

Perform the multiplication and solve for the variables, eg.
$2a-3c=-12$
$4a-c=1$

3. Hello, DINOCALC09!

$X$ is a 2x2 matrix such that: . $\begin{pmatrix}2 & \text{-}3 \\ 4 & \text{-}1\end{pmatrix}\,X \;=\;\begin{pmatrix}\text{-}12 & \text{-}9 \\ 1 & 3 \end{pmatrix}$

Find the matrix $X$
Let $X \:=\:\begin{pmatrix}w & x \\ y & z\end{pmatrix}$

Then we have: . $\begin{pmatrix}2 & \text{-}3 \\ 4 & \text{-}1\end{pmatrix}\,\begin{pmatrix}w & x \\ y & z \end{pmatrix} \;=\;\begin{pmatrix}\text{-}12 & \text{-}9 \\ 1 & 3\end{pmatrix}$

. . . . $\begin{pmatrix}2w-3y & 2x-3z \\ 4w - y & 4x-z\end{pmatrix} \;=\;\begin{pmatrix}-12 & -9 \\ 1 & 3\end{pmatrix}$

We have two systems of equations: . $\begin{Bmatrix}2w - 3y & = & \text{-}12 \\ 4w - y &=&1\end{Bmatrix}\quad\begin{Bmatrix}2x-3z &=&\text{-}9 \\ 4x-z &=&3\end{Bmatrix}$

Solve them . . .

You should get: . $X \;=\;\begin{pmatrix}\frac{3}{2} & \frac{9}{5} \\ 5 & \frac{21}{5}\end{pmatrix}$

4. Originally Posted by DINOCALC09
X is a 2 by 2 matric

2 -3
4 -1

multiplies by unkonw matrix X

=

-12 -9
1 3

Find the matrix X
Solve
$\left ( \begin{matrix} 2 & -3 \\ 4 & -1 \end{matrix} \right ) X = \left ( \begin{matrix} -12 & -9 \\ 1 & 3 \end{matrix} \right )$

Multiply both sides by the inverse of the coefficient matrix:
$\left ( \begin{matrix} 2 & -3 \\ 4 & -1 \end{matrix} \right ) ^{-1} = \frac{1}{2 \cdot -1 - (-3) \cdot 4} \left ( \begin{matrix} -1 & 3 \\ -4 & 2 \end{matrix} \right )$

So
$\frac{1}{10} \left ( \begin{matrix} -1 & 3 \\ -4 & 2 \end{matrix} \right ) \left ( \begin{matrix} 2 & -3 \\ 4 & -1 \end{matrix} \right ) X = \frac{1}{10} \left ( \begin{matrix} -1 & 3 \\ -4 & 2 \end{matrix} \right ) \left ( \begin{matrix} -12 & -9 \\ 1 & 3 \end{matrix} \right )$

$X = \frac{1}{10} \left ( \begin{matrix} 15 & 18 \\ 50 & 42 \end{matrix} \right )$

-Dan


text:
# History of LaTeX

 2005-01-25 11:44 UTC Revision 1 . . . . AlexSchroeder – where to put .sty files?


text:
# Max of conditional Negative Binomial

Suppose $$X|K = w$$ is a Negative Binomial with parameters $$r$$ and $$q$$. K follows a Binomial Distribution with parameters $$m$$ and $$p$$.

I want to calculate the expected value of $$Z = max(X_1, X_2,...,X_n)$$.

I know the conventional way of solving such a problem using CDF of X. However, due to this rare condition, the CDF of X comes out to be very complex. I tried using the PGF of X too, but that also doesn't have a decent expression. Is there any simple algebra trick that could help me in solving such expression?

• Perhaps you could use the law of total expectation: $E(Z)=E(E[Z|K])$ – gd1035 Dec 5 '18 at 19:47


text:
# Result of setFormula() not recognised (solved)

The following line of code inserts a formula using the function COUNTIF into cell F43.

`````` oSheet.getCellByPosition(5,42).setFormula("=COUNTIF(F33:F36,1)")
``````

The formula is inserted in the cell exactly as expected, but the result in the spreadsheet is Err:508.
(I tried “”=1"", but the result is the same.)
I can manually copy and paste the formula from the formula bar back in to the formula bar and it then works correctly.
A similar expression using a SUM function in place of COUNTIF works correctly.
Is there some limitation on which functions can be used in setFormula? Or am I missing something?

When the sheet is copied from elsewhere, Calc seems to be deciding that the column F is all DATES. This may have something to do with problem, but does not entirely explain it. The problem reproduces in another column. Now there is a second less important question: why is Col F becoming DATES?

Hi

You have to use same separator as `Tools``Options``Calc``Formula``Separators``Function`.

For example, with my french UI :

``````oSheet.getCellByPosition(5,42).setFormula("=COUNTIF(F33:F36;1)")
``````

So, is it `,` for you?

Regards

Thank you so much, Pierre-Yves, the semi-colon has fixed it, although, yes, the separator IS defined as comma, so there is a bit of mystery there!
Strange also that you can get away with a comma separator when entering the formula manually.
(I’ve fixed the date problem)


text:
# 11.1 Systems of linear equations: two variables

 Page 1 / 20
In this section, you will:
• Solve systems of equations by graphing.
• Solve systems of equations by substitution.
• Solve systems of equations by addition.
• Identify inconsistent systems of equations containing two variables.
• Express the solution of a system of dependent equations containing two variables.

A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions.

## Introduction to systems of equations

In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations    consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.

In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.

The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.

$\begin{array}{l}2\left(4\right)+\left(7\right)=15\text{ }\text{True}\hfill \\ 3\left(4\right)-\left(7\right)=5\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{True}\hfill \end{array}$

In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system    of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y -intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.

x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has


text:
# Re: [tlaplus] Trying to figure out how theorems work with this basic example

Thanks, that helped figure out quite a few things.

What I'm now trying to understand is the difference between ASSUME and THEOREM, or rather what to use THEOREM for. I've tried looking that up on Specifying Systems, but I'm still somewhat lost. Quoting from 17.7:

> More precisely, let A be the conjunction of all the assumptions in Ass. The module asserts that, for every theorem T in Thm, the formula A => T is valid.

So Thm defines what the assumptions in Ass imply. Is it just for the human reader? Or is there some way to use that in checking models or specs?

On Saturday, December 8, 2018 at 7:15:57 PM UTC+1, Stephan Merz wrote:
Hello,

first, TLC will not evaluate your definitions of Numbers because it involves a set comprehension whose base set (Nat) is infinite. I replaced it by

Numbers == 10 .. 99

which TLC can evaluate.

Concerning your question, you did not describe how exactly you ran the model checker. Your module does not contain a behavior specification, contrary to the most common use case for TLC. Perhaps you believe that TLC checks your THEOREM statement, this is not the case. If you just click on the green triangle in the model checking pane of the Toolbox, TLC doesn't find anything to do and simply terminates successfully.

You can however use TLC to check assumptions on constant module parameters: replacing THEOREM by ASSUME will either return in no output (e.g., when you choose the set {10, 14, 22} for input) or in the output "Assumption line ... is false." (when input is {10,16,22}, for example).

It is a little nicer to use TLC's facility of evaluating constant expressions. I replaced your THEOREM by the definition

ValidInput == input \subseteq NumbersThatDivideBy2Once

and then entered ValidInput in the field "Evaluate Constant _expression_" of the sub-pane "Model Checking Results". Running TLC will print TRUE or FALSE in the "Value" field.

You can use the same facility for evaluating the operator NumbersThatDivideBy2Once and make TLC output the entire set.

Below is the module that I used for my experiments.

Hope this helps,
Stephan

----------------------------- MODULE TestModule -----------------------------
EXTENDS Naturals

CONSTANT input

\* Numbers == { n \in Nat : n > 9 /\ n < 100 }

Numbers == 10 .. 99

DividesBy2(n) == (n % 2) = 0

DividesBy2Once(n) == DividesBy2(n) /\  ~DividesBy2(n \div 2)

NumbersThatDivideBy2Once == { n \in Numbers: DividesBy2Once(n) }

ValidInput == input \subseteq NumbersThatDivideBy2Once

=============================================================================

On 8 Dec 2018, at 18:53, dominy...@xxxxxxxxx wrote:

The below module declares a set of numbers that are in the range 10 to 99 that are divisible by 2 only once and call it NumbersThatDivideBy2Once. At the end it declares a theorem that the constant input is a subset of NumbersThatDivideBy2Once.

--------------------------- MODULE TestModule ---------------------------
EXTENDS Naturals

CONSTANT input

Numbers == { n \in Nat : n > 9 /\ n < 100 }

DividesBy2(n) == (n % 2) = 0

DividesBy2Once(n) == DividesBy2(n) /\  ~DividesBy2(n \div 2)

NumbersThatDivideBy2Once == { n \in Numbers: DividesBy2Once(n) }

THEOREM input \subseteq NumbersThatDivideBy2Once

=======================

How can I check if this theorem holds for a given input? If I run a model check with a provided set of numbers as input, even if some of those numbers are not part of NumbersThatDivideBy2Once I still get no errors.


text:
Properties

 Label 324.4.e.e Level $324$ Weight $4$ Character orbit 324.e Analytic conductor $19.117$ Analytic rank $0$ Dimension $2$ CM discriminant -3 Inner twists $4$

Related objects

Newspace parameters

 Level: $$N$$ $$=$$ $$324 = 2^{2} \cdot 3^{4}$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 324.e (of order $$3$$, degree $$2$$, not minimal)

Newform invariants

 Self dual: no Analytic conductor: $$19.1166188419$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ Defining polynomial: $$x^{2} - x + 1$$ x^2 - x + 1 Coefficient ring: $$\Z[a_1, \ldots, a_{13}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 108) Sato-Tate group: $\mathrm{U}(1)[D_{3}]$

$q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + ( - 37 \zeta_{6} + 37) q^{7}+O(q^{10})$$ q + (-37*z + 37) * q^7 $$q + ( - 37 \zeta_{6} + 37) q^{7} + 19 \zeta_{6} q^{13} - 163 q^{19} + ( - 125 \zeta_{6} + 125) q^{25} - 308 \zeta_{6} q^{31} + 323 q^{37} + ( - 520 \zeta_{6} + 520) q^{43} - 1026 \zeta_{6} q^{49} + (719 \zeta_{6} - 719) q^{61} + 127 \zeta_{6} q^{67} - 919 q^{73} + ( - 1387 \zeta_{6} + 1387) q^{79} + 703 q^{91} + ( - 523 \zeta_{6} + 523) q^{97} +O(q^{100})$$ q + (-37*z + 37) * q^7 + 19*z * q^13 - 163 * q^19 + (-125*z + 125) * q^25 - 308*z * q^31 + 323 * q^37 + (-520*z + 520) * q^43 - 1026*z * q^49 + (719*z - 719) * q^61 + 127*z * q^67 - 919 * q^73 + (-1387*z + 1387) * q^79 + 703 * q^91 + (-523*z + 523) * q^97 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q + 37 q^{7}+O(q^{10})$$ 2 * q + 37 * q^7 $$2 q + 37 q^{7} + 19 q^{13} - 326 q^{19} + 125 q^{25} - 308 q^{31} + 646 q^{37} + 520 q^{43} - 1026 q^{49} - 719 q^{61} + 127 q^{67} - 1838 q^{73} + 1387 q^{79} + 1406 q^{91} + 523 q^{97}+O(q^{100})$$ 2 * q + 37 * q^7 + 19 * q^13 - 326 * q^19 + 125 * q^25 - 308 * q^31 + 646 * q^37 + 520 * q^43 - 1026 * q^49 - 719 * q^61 + 127 * q^67 - 1838 * q^73 + 1387 * q^79 + 1406 * q^91 + 523 * q^97

Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/324\mathbb{Z}\right)^\times$$.

 $$n$$ $$163$$ $$245$$ $$\chi(n)$$ $$1$$ $$-\zeta_{6}$$

Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
109.1
 0.5 − 0.866025i 0.5 + 0.866025i
0 0 0 0 0 18.5000 + 32.0429i 0 0 0
217.1 0 0 0 0 0 18.5000 32.0429i 0 0 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 CM by $$\Q(\sqrt{-3})$$
9.c even 3 1 inner
9.d odd 6 1 inner

Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 324.4.e.e 2
3.b odd 2 1 CM 324.4.e.e 2
9.c even 3 1 108.4.a.b 1
9.c even 3 1 inner 324.4.e.e 2
9.d odd 6 1 108.4.a.b 1
9.d odd 6 1 inner 324.4.e.e 2
36.f odd 6 1 432.4.a.h 1
36.h even 6 1 432.4.a.h 1
72.j odd 6 1 1728.4.a.o 1
72.l even 6 1 1728.4.a.r 1
72.n even 6 1 1728.4.a.o 1
72.p odd 6 1 1728.4.a.r 1

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
108.4.a.b 1 9.c even 3 1
108.4.a.b 1 9.d odd 6 1
324.4.e.e 2 1.a even 1 1 trivial
324.4.e.e 2 3.b odd 2 1 CM
324.4.e.e 2 9.c even 3 1 inner
324.4.e.e 2 9.d odd 6 1 inner
432.4.a.h 1 36.f odd 6 1
432.4.a.h 1 36.h even 6 1
1728.4.a.o 1 72.j odd 6 1
1728.4.a.o 1 72.n even 6 1
1728.4.a.r 1 72.l even 6 1
1728.4.a.r 1 72.p odd 6 1

Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{4}^{\mathrm{new}}(324, [\chi])$$:

 $$T_{5}$$ T5 $$T_{7}^{2} - 37T_{7} + 1369$$ T7^2 - 37*T7 + 1369

Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2}$$
$5$ $$T^{2}$$
$7$ $$T^{2} - 37T + 1369$$
$11$ $$T^{2}$$
$13$ $$T^{2} - 19T + 361$$
$17$ $$T^{2}$$
$19$ $$(T + 163)^{2}$$
$23$ $$T^{2}$$
$29$ $$T^{2}$$
$31$ $$T^{2} + 308T + 94864$$
$37$ $$(T - 323)^{2}$$
$41$ $$T^{2}$$
$43$ $$T^{2} - 520T + 270400$$
$47$ $$T^{2}$$
$53$ $$T^{2}$$
$59$ $$T^{2}$$
$61$ $$T^{2} + 719T + 516961$$
$67$ $$T^{2} - 127T + 16129$$
$71$ $$T^{2}$$
$73$ $$(T + 919)^{2}$$
$79$ $$T^{2} - 1387 T + 1923769$$
$83$ $$T^{2}$$
$89$ $$T^{2}$$
$97$ $$T^{2} - 523T + 273529$$


text:
# Tag Info

The equation of motion of the particle is $$m \ddot{x}(t) = F(t)$$ where $x(t)$ is the position and $F(t)$ is the force. In the situation you describe, ("suddenly I hit the particle"), the force as a function of time can be written as $F(t) \propto \delta (t)$, with $\delta$ the Dirac distribution. Integrating once, you obtain that $$\dot{x}(t) \propto ... 0 What's wrong with this is the phrase 'suddenly I hit the particle'. What you are assuming by this is that you have some hard-edged and rigid object which you crash into the particle, which is also hard-edged. But you don't have either of those things: what you have is something which is both not rigid and whose surface is actually a little bit of EM field ... 0 This diagram shows part of a surface cut up into small elements (shaded pink) of area dA. The normal to the area is the vector \vec {dA} and is of magnitude dA. What is required is the projection of the magnetic field \vec B onto the vector \vec{dA} which is B_\bot = B \cos \phi in the diagram. To do this the dot product is taken \vec B ... 0$$ \int \mathbf{B} \cdot \,d\mathbf{A} $$just means$$ \int B\,dA\,\cos\theta. $$d\mathbf{A} points in the direction perpendicular to the surface, so if this is in the xy plane, it'll be in \mathbf{\hat{z}}. I.e. your usual integration over the area ("integrand") weighted by a factor of \cos\theta. \theta is the angle with the z axis. Keeping ... 0 The integration variable is the surface vector: flux is an oriented thing. It is the "normal" integration of the scalar product of the field and the surface vector:$$\Phi=\iint_S \vec{B}\cdot\vec{\mathrm dS}$$Or, if \vec{n} is the normal to the surface at the point of integration:$$\Phi=\iint_S \vec{B}\cdot\vec{n}\ \mathrm dS$$0 Your derivation is almost correct, and as I'll show below, one does sometimes make use of it. However, it relies on an unwritten assumption that makes it applicable only in a limited range of situations. That assumption is that the heat capacity c doesn't depend on the temperature. It is this assumption that allowed you to take the factor of mc outside ... 0 Your equation is correct only if:$$\mathrm dQ = mc\,\mathrm dT$$which is not generally true, indeed, common sense tells you that a change in temperature leads to conclusion that an object being heated up. But we do not encounter gases much in our life, which could be regarded as a general case. In reality your assumption is generally false, a good example ... -1 Here is a link that might be useful http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/temper2.html#c2 0 This is just complementary to Alex's answer. For the second integral the book provides an analysis in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral$$ \frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}} $$At the limit r\rightarrow \infty, the effect ... 0 A differential like dx in an equation like$$dz = dx + dy + dx^2 + dx\,dy is shorthand for a quantity that will eventually go to zero in a limit. If a differential is divided by another differential, then it has a chance of resulting in a finite, non-zero quantity. Let's assume $x$, $y$, and $z$ can be parameterized by a variable $t$. If we divide the ...


text:
#### You may install a binary package by installing ‘psycopg2-binary’ from PyPI [WINDOWS]

Hi guys I’m getting this error and I don’t know how to fix this

the command that I run in my console is the following:

pip install -r ..requirementslocal.txt

    It appears you are missing some prerequisite to build the package from source.

You may install a binary package by installing 'psycopg2-binary' from PyPI.
If you want to install psycopg2 from source, please install the packages
required for the build and try again.

For further information please check the 'doc/src/install.rst' file (also at
<http://initd.org/psycopg/docs/install.html>).

error: command 'C:Program Files (x86)Microsoft Visual Studio2019BuildToolsVCToolsMSVC14.29.30133binHostX86x64cl.exe' failed with exit code 2

I don’t know how to fix this I really need help

Source: Docker Questions


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8.38 What can be inferred from the magnetic moment values of the following complex species ?

Example                                    Magnetic Moment (BM)

$K_{4}[Mn(CN)_{6})$                                2.2

$[Fe(H_{2}O)_{6}]^{2+}$                                          5.3

$K_{2}[MnCl_{4}]$                                            5.9

Magnetic moment is given as -  Putting the value on n = 1, 2, 3, 4, 5  (number of unpaired electrons in d-orbital) we get the value of  are 1.732, 2.83, 3.87, 4.899, 5.92 respectively. By comparing with our calculation we get the values n nearest to 1. It means, in above compound d-orbital has one unpaired electron(), which means  is astrong field ligand that cause force pairing of the...

8.37 Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

Elements of the first transition series possess many properties different from those of heavier transition elements in the following ways- The atomic size of the 1st transition series is smaller than those of 2nd and 3rd series elements. But due to lanthanoid contraction, atomic size of the 2nd series elements are nearly the same as 3rd series element of the corresponding same vertical...

8.36 Write down the number of 3d electrons in each of the following ions: $Ti^{2+}, V^{2+}, Cr^{3+}, Mn^{2+}, Fe^{2+}, Fe^{3+}, Co^{2+}, Ni^{2+} and\; \; Cu^{2+}$. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

For  d-orvbital has two electron. So, filling of d-orbitals can be  In   d-orbital has three electron. So, the filling of d-orbital can be  Similarily (Ions) (No. of d electrons) (Filling of d-orbitals)

8.35 Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(iv) atomic sizes.

Generally, atomic sizes decrease from left to right across the period. In among the three transition series, the size of the second series element is bigger than that of the first transition element of the same vertical group. But the atomic size of the third transition element is nearly the same as the element of the second transition series element. This is because of Lanthanoid contraction.

8.35 Compare the general characteristics of the first series of the transition metalswith those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(iii) ionisation enthalpies

In all of the three transition series, the 1st ionisation energy increases from the left side to right side. But, there are some exceptions like the first ionisation enthalpies of the third transition series are more significant than those of the first and second transition series. This is happening due to the weak shielding effect of 4 electrons in the third series. Some elements in the...

8.35 Compare the general characteristics of the first series of the transition metalswith those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(ii) oxidation states

In each of the three transition series, the no. of oxidation state is minimum at the extremes and the highest at the middle of the row. In the first transition series, the +2 and +3 oxidation state are quite stable. Elements of first transition series metals form stable compounds of +2 and +3 oxidation state. But the stability of +2 and +3 oxidation state decreases in second and third...

8.35 Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(i) electronic configurations

Electronic configurations- In 1st, 2nd and 3rd transition metal series 3, 4 and 5 orbitals are used respectively. In first series copper and zinc show unusual electronic configuration.  In the second transition series different electron configuration shown by following metals,  (42) = 4d5 5s1, (43) = 4d6 5s1, (44) = 4d7 5s1, (45) = 4d8 5s1, (46) = 4d10 5s0, (47) = 4d10 5s1 In 3rd series there...

8.34 Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

Atomic number = 61, Promethium the electronic configuration is  atomic number = 91, protactinium the electronic configuration is  Atomic number = 101, Mendelevium the electronic configuration is  Atomic number = 109, Meitnerium the electronic configuration is

8.32 Name the members of the lanthanoid series which exhibit $+4$ oxidation states and those which exhibit $+2$ oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.

Members of the lanthanoids which exhibits +4 oxidation states are-  members who exhibit +2 oxidation states =  After losing 4 electrons attains stable configuration  and also the same thing happen to  In the case of  and ,  after losing two electrons they also get their stable electronic configuration.

8.31 Use Hund’s rule to derive the electronic configuration of $Ce^{3+}$ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.

Electronic configuration of Magnetic moment can be calculated as , where n= no. of unpaired electrons in Cerium n = 2 So, by putting the value of n we get

8.30 Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

The last element of the actinoid series is Lawrencium (). Its atomic number is 103. The electronic configuration of  is . The possible oxidation state of lawrencium is +3 because after losing 3 electrons it becomes a stable molecule.

8.29 The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.

Lanthanoid primarily shows three oxidation states +2, +3, and +4 and out of these +3 is most common in lanthanoids. they show limited no. of oxidation states due to the large difference in energies of 4, 5 and 6 orbitals. But, actinoids shows large no. of oxidation state because they have comparable energy difference in 5,6 and 7 orbitals. For example  and  exhibits +3, +4, +5 and +6 oxidation states.

8.28 What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104.

Inner transition metals are those in which the last electrons are filled in f-orbitals. The elements in which 4f and 5f are filled are called f block elements. 59, 95 and 102 are the inner transition elements.

8.27 What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

It is a solid solution of two or more elements in a metallic matrix. Alloys possess different physical properties than component materials. An important alloy of lanthanoid is mischmetal. uses- mischmetal is used in cigarettes and gas lighters Used in flame-throwing tanks It is used in tracer bullets and shells

8.26 Indicate the steps in the preparation of:

(ii) $KMnO_{4}$ from pyrolusite ore.

Potassium permanganate can be prepared from the fusion of pyrolusite ore() with an alkali metal hydroxide and an oxidising agent (like ). This gives dark green . It disproportionates in acidic or neutral medium to give permanganate.

8.26 Indicate the steps in the preparation of:

(i) $K_{2}Cr_{2}O_{7}$  from chromite ore.

(i) Potassium dichromate is obtained from the fusion of chromite ore  with sodium and potassium carbonate in the free supply of air. (ii) Sodium chromate is filtered and acidified with sulphuric acid () to form sodium dichromate,  can be crystallised (iii) Sodium dichromate is more soluble than potassium dichromate. So, treat the solution of dichromate with the potassium chloride() The...

8.25 Give examples and suggest reasons for the following feature of the transition metal chemistry:

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Oxygen is a strong oxidizing agent because of its small in size and high electronegativity. Thus oxo-anions of metals shows the highest oxidation state. For example- , here manganese shows +4 oxidation state.

8.25 Give examples and suggest reasons for the following feature of the transition metal chemistry:

(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.

Oxygen and fluorine are a strong oxidizing agent because of their small in size and high electronegativity. So, they help transition metals to exhibit the highest oxidation states. Examples of oxides and fluorides of transition metals are  and

8.25 Give examples and suggest reasons for the following feature of the transition metal chemistry:

(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.

The lowest oxidation states of transition metals are basic because some of their valence electrons are not participating in bonding. Thus they have free electrons, which they can donate and act as a base. In the higher oxide of transition metals, valence electron of their participate in bonding, so they are unavailable. But they can accept electrons and behave as an acid. For example (+2)-...

8.24 Which one of these is the most stable in aqueous solution?

$\\(i)Mn^{3+}\\(ii)Cr^{3+}\\(iii)V^{3+}\\(iv)Ti^{3+}$

is the most stable in the aqueous solution solution because it attains the configuration, which is stable configurtaion. Eelectronic configuration of  =
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text:
Many important files within Microsoft Windows have structures that are undocumented. One of the principals of computer forensics is that all analysis methodologies must be well documented and repeatable, and they must have an acceptable margin of error. Currently, there are a lack of open source methods and tools that forensic analysts can rely upon to examine the data found in proprietary Microsoft files.
Many computer crime investigations require the reconstruction of a subject’s internet activity. Since this analysis technique is executed regularly, we researched the structure of the data found in Internet Explorer activity files (index.dat files). The Pasco application (from the latin word meaning “browse”) was developed to examine the contents of Internet Explorer’s cache files.
Pasco will parse the information in an index.dat file and output the results in a field delimited manner so that it may be imported into your favorite spreadsheet program.

## Pasco Activation Key Latest

==========
Pasco Serial Key is a tool for forensic analysis of Internet Explorer’s cache files.
The cache files usually contain information that indicates a series of Internet Web sites visited during a single internet session.
The purpose of the Pasco Full Crack analysis is to extract and reconstruct the series of Internet pages visited by the user during a specific Internet session.
When Microsoft Internet Explorer is launched, a new cache file is created. This cache file is also named index.dat, and it contains three structures:
* SiteData: Contains an array of site names and a timestamp when the site was visited. The browser sends this information to the server on a regular basis.
* History: A list of all the site addresses visited. The browser sends the information to the server on demand.
* Cookies: An array of cookies the browser has stored on the client machine.
The following example shows an index.dat file that was generated on a system when a user was using IE8, and a few of the extracted data structures.
* SiteData
* History
* SiteData
* SiteData
* SiteData
“12/26/2013” 23:16:10
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”
“”

Pasco is a microapplication.
Pasco Input:
A list of index.dat files, comma-separated list of file names.
Pasco Output:
Field separated text with key-value formatted table. As a new user of Pasco, you should simply type in a few of the fields on your index.dat file. This will parse the entire file and output the results.
Pasco Performance:
Pasco takes a long time to parse a single index.dat file. When parsing multiple index.dat files, the Pasco application will output results on the order of seconds, as opposed to minutes or hours.
Pasco Filesize:
Pasco does not include parsing of any other index.dat files. Pasco’s features are predicated on the number of files it parses. As many index.dat files as possible is the best approach to parsing.
Pasco Installation:
To install Pasco, you must have Visual Studio Express installed.
cd desktop
unzip pasco.zip
Pasco must be run from the directory you just created.
Example: C:\Users$your-user-name]\Desktop\pasco.exe Source Code: Pasco is a free, open source application. The source code for Pasco is included within the Pasco zip file. Pasco parses index.dat files and outputs the results in a field-seperated, key-value format suitable for importing into your favorite spreadsheet program. If you like what you see, please let me know! The following URLs provide a list of additional information on Pasco: To install and use WiX tools, you need to know how to create and maintain the.wxs (Windows Installer XML) files. The.wxs files are the basis for creating Windows Installer packages. And without them, you are lost. This guide is for Windows Installer v2.0, but you can apply it to v1.1 and v3.0 too. In Windows XP/2003/vista/win7 and Windows 8 and 8.1, you can create a new Installer package using the Microsoft Windows Installer Package Designer. So first of all b7e8fdf5c8 ## Pasco Crack The index.dat file is created every time Internet Explorer’s cache folder is cleared by the user, or it can be created automatically by Internet Explorer itself when the cache needs to be cleared. In typical use, this file is created at approximately every 30 minutes. Pasco can examine the index.dat files, which store the history of browser activity, and may also report Internet Explorer’s activity to an analysis server. Reports may be created from the data stored in a single index.dat file or from multiple index.dat files. The reports will display the history of a subject’s web browsing, including site visits and form submissions. Pasco also contains the ability to save these reports to the local disk with the ability to transfer them to an analysis server if needed. Pasco is a command line tool, or terminal based, application. It is written in C# and has a command line interface for running various requests. Download Pasco for use on a Windows platform.Q: How to add raw LaTeX code in my question? I have a question in which a simple mathematical formula is to be put up. Is there any way that I can do this? I know I can put an image of it, but I just wanted to know if there is a better way to add it? A: You can put the code in the question directly by editing it. You won’t be able to do any formatting though. You can put some small code in the body text of the post, under the post text. This is usually a form like this: But note this is not valid TeX, it is just a way to show where the text goes in an image (usually an equation or a formula). If you have questions about how to insert a \begin{…}/\end{…} pair into the body of your question, you should post an actual \[…$ question. Only use this is you are posting a “real” question about LaTeX.

A:

You can wrap the code in \[. The following
This is a line of LaTeX.

\frac{1+\frac{1+\frac{1+\frac{1+\frac{1+\frac{1+\frac{1+\frac{1+

## What’s New In Pasco?

================
Pasco parses the contents of index.dat files. It is able to look through files with indexes between 0 and 9, as defined by the “magic” number between “.par” and “.end_par “.
Pasco can only parse the following index.dat file types:
1. Netbus files – Netbus files are created by Netscape, and normally contain complete browsing sessions.
2. WinCache files – This is the cache files created by MS Windows while running web browsers such as Internet Explorer.
Pasco input format, output format, help and sample program:
=============================================================
Pasco input:
————-
1) Index.dat file with a “.par” or “.end_par”
2) Normalized (ASCII) text file
Pasco output:
————
1) Parsed and formatted data file in ASCII format
2) EML file to contain formatted data
3) HTML file to contain the formatted data
Pasco help:
————–
1) pascohtml
2) pascoparse
Pasco example program:
———————-
// PASCOParse.CPP
#include “stdafx.h”
#include “Descriptor.h”
// ^^ this header files will contain all of the following function declarations
//.. of all object files
#include “Descriptor.h”
#include
#include
//.. and all of the following symbols
using namespace std;
// ^^ use the C++ namespace for object files
// use.CPP extension instead of.cpp for object files
// this file contains of a function named doWork()
void doWork(int argc, char *argv[]);
void doWork(int argc, char *argv[]) {
}
//.. end of function doWork()
// end of object file
//.. end of namespace
//.. end of C++ namespace
//.. end of.CPP extension
//.. end of current file
int _tmain(int argc, char* argv[])
{
// open the index.dat file to parse, or else use the default file which is
// created when you first run Pasco
Descriptor defaultDescriptor;
defaultDesc

## System Requirements:

OS: Vista and later.
Processor: Intel Core 2 Duo CPU or AMD equivalent, 2 GHz
Memory: 2 GB RAM
Graphics: DirectX 9.0c compatible video card with 128 MB Video RAM
DirectX: Version 9.0c compatible graphics card with 128 MB Video RAM
Hard Drive: 1 GB available space for installation
Sound Card: Compatible sound card


text:
English Információ A lap Pontverseny Cikkek Hírek Fórum

Rendelje meg a KöMaL-t!

VersenyVizsga portál

Kísérletek.hu

Matematika oktatási portál

C. 1253. Prove that if the measure of each side of a right-angled triangle is an integer, then the area of the triangle formed by the right-angled vertex and the two points dividing the hypotenuse into three equal parts is also an integer.

(5 points)

This problem is for grade 1 - 10 students only.

Deadline expired on 10 December 2014.

Statistics on problem C. 1253.
 154 students sent a solution. 5 points: 93 students. 4 points: 8 students. 3 points: 14 students. 2 points: 13 students. 1 point: 15 students. 0 point: 6 students. Unfair, not evaluated: 4 solutions. Unfair, not evaluated: 1 solution.

• Problems in Mathematics of KöMaL, November 2014

•  Támogatóink: Morgan Stanley


text:
# Simple digital filter design

06 Sep 2019 - tsp

## What are digital filters?

Digital filters employ the idea of simulating traditional analog filter responses with digital calculations. The basic idea is to sample a signal with an analog digital converter (ADC) and perform the calculations numerically. This allows a more flexible setup (for example on can change the filter response at runtime) as well as a more cheap setup since less analog components are used. These filters are normally limited only by Nygists sampling theorem that says that one cannot process signals with a frequency above half the sampling frequency after discretization as well as available processing power. Note that you should normally employ at least an analogous lowpass filter in front of your ADCs to remove sampling artifacts from high frequency components.

Digital filters are normally specified as a filter function that can be applied to the input signal. This specification is done in frequency space as a function $H(\omega)$ which is simply multiplied to the input signal $x(\omega)$. This leads to an output signal

$y(\omega) = x(\omega) H(\omega)$

Since calculation in frequency space would require one to transform the input signal into frequency space, multiply it and then perform a backtransformation which would not allow a continuous evaluation of the response one can use the convolution (since it’s known that the timedomain convolution of two signals equals the frequency domain multiplication):

$y(t) = x(t) * H(t) = \int_{-\infty}^{\infty} x(\tau) H(t-\tau) d\tau$

Later on we will do the convolution in discretized space - this can be done by moving from the integral (an infinite sum) to a discretized sum:

$y(n) = x(n) * H(n) = \sum_{\tau} x(\tau) H(n - \tau)$

## Recalling the (discrete) fourier transform

First we have to recall how fourier transformation worked. First we define our quantities:

• A function in time domain $f(t)$ where $t \in \mathfrak{R}$ is the continuous time and $f(t)$ can be any arbitrary amplitude.
• A function in frequency domain $f(\omega)$ with $\omega \in \mathfrak{R}$ being the continous normalized frequency.

Note that we have defined normalized frequency since the results can be used that way in a general way. For generalized normalization we have to know the sampling frequency $f_s$ at which our digital system works. Due to the Nyquist sampling theorem we know the highest frequency we can work with is $\frac{f_s}{2}$. Since the normal definition of the angular frequency would be $\omega = 2 \pi f$ we normalize using $\omega = 2 \pi \frac{f}{f_s}$. This yields a useable normalized frequency range from $-\pi$ to $\pi$.

The continous transformation between frequency and time domain is done via the fourier transformation. This is based on the assumption that each signal can be represented by an infinite series of sine and cosine functions:

$f(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(t) e^{-i 2 \pi \omega t} dt$

The factor $\frac{1}{\sqrt{2\pi}}$ compensated for our usage of angular frequency instead of frequency itself. To transform a signal back from frequency domain to time domain we use the inverse transformation:

$f(t) = \int_{-\infty}^{\infty} f(\omega) e^{-i 2 \pi \omega t} d\omega$

### Discrete transformations

Since we use a digital system and don’t have a continuous signal available we could use discrete transformations. This will be especially interesting if we don’t transform basic signals and want to calculate these numerically or if we want to do a frequency anlysis or digital synthesis of a signal. To move from continuous to discrete systems we simply introduce the sampling number $n$ (which is - using the sample count per unit time $N$ - is related to continuous time via $t = \frac{1}{N} n = \delta t n$). The same way we introduce the wave vector $k$ as counterpart to the continuous frequency $\omega$ - with definition $\omega_0 = \frac{2\pi}{N}$ and $\omega = \omega_0 k$.

Now we can simply substitute the integrals (which are just infinite sums anyways) by discrete sums. This will of course lead to known discretization error:

$f(k) = \sum_{n=0}^{N-1} f(n) e^{-i k \omega_0 n}$

$f(n) = \frac{1}{N} \sum_{k=0}^{N-1} f(k) e^{i k \omega_0 n}$

## Designing filters by ideal response functions

### Low pass filter

The ideal response for a lowpass filter is to filter all frequencies whose absolute value lies above a given threashold - the cutoff frequency $\omega_c$. We can use the impulse response of our filter:

[ H(\omega) = \left\{ \begin{align} 1, \mid\omega\mid \leq \omega_c \\ 0, else \end{align} \right. ]

This is the response of the filter that we expect when putting in an impulse $\delta(t)$ into out filter. Since we know that the fourier transform of $\delta(t)$ just leads to a constant:

$\int_{-\infty}^{\infty} \delta(t) e^{-i 2\pi \omega t} dt = e^{0} = 1$

Since we want to perform the convolution in time domain we can simply perform the backtransformation:

$H(t) = \int_{-\infty}^{\infty} H(\omega) e^{i2\pi \omega t} d\omega = \int_{-\omega_c}^{\omega_c} e^{i 2\pi t \omega} d\omega$ $= \frac{e^{i 2 \pi t \omega}}{2i \pi t} \mid_{-\omega_c}^{\omega_c} = \frac{1}{\pi t} \frac{e^{i 2\pi t \omega_c} - e^{-i 2\pi t \omega_c}}{2i}$

Using Eulers formula $e^{i\phi} = \cos(\phi) + i \sin(\phi)$ and it’s presentation $sin(\phi) = \frac{e^{i\phi} - e^{-i\phi}}{2i}$ we can conclude that

$H(t) = \frac{1}{\pi t} \sin(2 \pi t \omega_c)$

As we can see the function extends into both the positive and negative time domain. This means this filter is not causal since its output depends on future values. Such filters can be calculated exactly on previously calculated signals but not on samples filtered in real time.

Using $f_c = \frac{1}{2} f_s$ i.e. $\omega_c = 2 \pi \frac{f}{f_s} = \pi$ we could calculate the following response function:

Since - in applications that are not working with previously recorded data - we are using only old known values we have to make the filter causal. To do this we simply drop all values that depend on positive times. We also have to limit ourself to a fixed number of simples to we cut the coefficients after a given number of samples. The filter coefficients will simply be extracted from our transformed function at the given number of timesteps. This leads to a finite impulse response (FIR) filter. Note that such a construction will lead to errors in the filter function.

### Bandpass filter

Since we will construct our high pass filter (due to some mathematical constraints) as a bandpass filter we will first look at the filter coefficients for a bandpass filter. Again we start with our filter specification in frequency space:

[ H(\omega) = \left\{ \begin{align} 0, \mid\omega\mid < \omega_c \\ 1, \omega_c \leq \mid\omega\mid \leq \omega_N \\ 0, \mid\omega\mid > \omega_N \end{align} \right. ]

Again we simply perform the inverse fourier transformation:

$H(t) = \int_{-\infty}^{\infty} H(\omega) e^{i 2\pi \omega t} d\omega = \int_{-\omega_N}^{-\omega_C} e^{i2\pi\omega t} d\omega + \int_{\omega_C}^{\omega_N} e^{i2\pi\omega t} d\omega$

Again this leads to an expression that we can transform into a sum of sines:

$H(t) = \frac{1}{\pi t} \left[ -\frac{e^{i2\pi \omega_c t} - e^{-i2\pi \omega_c t}}{2i} + \frac{e^{i2\pi \omega_N t} - e^{-i2\pi \omega_N t}}{2i} \right]$

$H(t) = \frac{1}{\pi t} \left[ -\sin(2\pi \omega_c t) + \sin(2 \pi \omega_N t) \right]$

As before we see that the function is not causal (i.e. it contains terms depending on future events). This can be solves as above introducing the same artifacts as above. A simple FIR filter can be constructed by dropping all elements depending on future events and sampling the function on a finite interval.

### High pass filter

If we try to construct the highpass filter as above we get into a mathematical problem:

[ H(\omega) = \left\{ \begin{align} 1, \mid\omega\mid \geq \omega_c \\ 0, \omega < \omega_c \end{align} \right. ]

$H(t) = \frac{e^{-i2\pi t \omega_c t} - \overbrace{e^{-i 2 \pi t \infty}}^{\to 0}}{i 2 \pi t} + \frac{\overbrace{e^{i 2 \pi t \infty}}^{\to \infty} - e^{i 2 \pi \omega_c t}}{i 2 \pi t}$

As we see the second term diverges. Even when using the same trick as before we get into the problem of having to evaluate our sine at infinity. Since the limit of the sine doesn’t exist (it oscilates between $-1$ and $+1$) we cannot use this term to evaluate our filter expression. Of course we could argue that this term is not relevant because it’s representing the dependence of our filter on future events so the simple windowing method of simply cutting events from the future might resolve that problem. But there is another way that we can use to construct our filter in case we want to apply it to real-time sampled data: Make it a bandpass filter. Since we know that we are designing a time discrete digital filter we also know that there is an upper sampling frequency that we can depend on. That way we can set the upper frequency to the Nyquist frequency (i.e. half the sampling frequency) and use the above equation for bandpass filters to construct our digital highpass.