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Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01c247dv41r
 Title: Rock-paper-scissors games in expanding populations: the effects of cyclic dominance and domain growth on diversity Authors: Maslan, Anne Margaret Advisors: Wingreen, Ned S. Department: Chemical and Biological Engineering Class Year: 2015 Abstract: Discovering mechanisms for the maintenance of species diversity is central in ecology. Conditions of cyclic dominance arise frequently in nature and are thought to be a major promoter of diversity. Rock-paper-scissors (RPS) games can be used to characterize conditions of cyclic dominance involving three competing strategies: rock crushes scissors, scissors cuts paper, and paper wraps rock. The role of RPS in maintaining diversity among competing strategies has been detailed in spatially explicit models of fixed-size microbial populations. However, we find that RPS does not necessarily contribute to diversity in growing populations. To explore the effects of population growth, we use agent-based modeling (ABM). We consider two regimes: surface-only growth and bulk growth. In the case of surface-only growth, RPS increases the rate of interface diffusion, thereby accelerating the formation of sectors at the expanding front and decreasing diversity. In the case of bulk growth, we observe both regions characterized by RPS attacks that keep domain sizes in check and regions dominated by domain growth. When domains of sufficient size become established early on and attack rates are sufficiently low, growth in the bulk of these domains outweighs attacks at the interface; coexistence of these established domains can result. Therefore, while RPS games can promote diversity in expanding populations, the growth of domains can lead to an entirely different mechanism for the maintenance of diversity. We conclude that diversity in growing microbial populations can stem not only from the dynamics associated with RPS games but also from the formation of domains that become too big to fail. Extent: 54 URI: http://arks.princeton.edu/ark:/88435/dsp01c247dv41r Type of Material: Princeton University Senior Theses Language: en_US Appears in Collections: Chemical and Biological Engineering, 1931-2016

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# Regression Modeling With Proportion Data (Part 1)

## Predicting Attendance in the German Handball-Bundesliga

As a data scientist, one often encounters dependent variables that are proportions: for example, the number of successes divided by the number of attempts, party vote, proportion of money spent for something, or the attendance rate of public events. Modeling and predicting such variables in a regression framework is possible, but one has to go beyond the standard linear model, because the data are restricted to the range between 0 and 1. Popular logistic regression is not suitable either, because it permits only 0s and 1s, but not an attendance rate of .80 or 80 %. In this blog post, I will compare different models that are available for proportions and illustrate them to predict the attendance rate of matches of the German Handball-Bundesliga.

## Modeling Proportion Data

As a starting point, a linear regression model without a link function may be considered to get one started. The model is obviously wrong, because it will easily make predictions smaller than 0 or larger than 1. Nevertheless, it may work okay especially for intermediate proportions.

A second option is a binomial or quasi-binomial model. Therein, the proportion is conceived of as the outcome of multiple binomial trials. A good example are the shots of a basketball player, where one may either model each individual shot using a logistic model for outcomes of 0 and 1. Or one may aggregate all attempts in a match and model the proportion of successful shots, which is a value in the interval of [0, 1], using a (quasi-)binomial model.

A related option is a Poisson model for count data that, for example, may be used to model the number of occurrences of a specific symptom per week or month. However, the model is not quite the right choice if the count variable may take on values of several thousand spectators of a sport event. More importantly, it can fail to capture, for example, the difference between 1,000 spectators in a stadium with a capacity of 1,000 (i.e., 100 %) and 2,000 spectators in a stadium of 5,000 (i.e., 40 %).

The third option considered is beta regression which assumes that the dependent variable is beta-distributed. This model is very flexible and ideally suited for original proportions or rates. However, it should be noted that it assumes values in the interval (0, 1), that is, 0 and 1 are excluded. There are two variants of beta regression: One predicts only the mean of the dependent variable whereas the second variant models also the dispersion parameter phi. This sounds attractive for the current problem, because handball teams like Flensburg almost always have attendance rates of over 90 % (small variance) while the attendance rates in other stadiums may show more variability.

## Application: Handball-Bundesliga

Data from the German Handball-Bundesliga were obtained for the current and the last two seasons from https://www.dkb-handball-bundesliga.de. The data needed some wrangling and cleansing and are not perfectly valid (e.g., some entries exhibit attendance rate larger than 100 %), but will suffice for the present purpose. You can find the respective R code on my Github.

### Setup

library("conflicted")
suppressPackageStartupMessages(library("tidyverse"))
conflict_prefer("filter", "dplyr")
#> [conflicted] Will prefer dplyr::filter over any other package
library("betareg")
library("broom")
library("kableExtra")
theme_set(theme_light(base_size = 12))
theme_update(panel.grid.minor = element_blank())

# Function to transform attendance rates for beta regression because values of 0
# and/or 1 may occur
transform_perc <- function(percentage_vec) {
# See Cribari-Neto & Zeileis (2010)
(percentage_vec * (length(percentage_vec) - 1) + 0.5)/length(percentage_vec)
}

### Selected Variables

Dependent variable:

• Beta regression: Attendance rate; values were transformed to the interval (0, 1) using transform_perc()
• Quasi-binomial regression: Attendance rate in the interval [0, 1]
• Linear regression: Attendance (i.e., count)
• In all cases, entries where the attendance was larger than the capacity were replaced with the maximum capacity.

Independent variables:

• Home team: Different teams have different average attendance rates and/or stadium capacities. For the beta model, home team was also used to predict the dispersion parameter phi.
• Away team: Generally, more successful away teams like Kiel attract more spectators.
• Year/season: Used to capture potential differences across the three seasons.
• Weekday: Matches on weekends probably attract more spectators.
• All variables are categorical variables and appropriate coding schemes (i.e., dummy or effect coding where chosen).

## Initial Results for Beta Regression

# Use data from 2016/17 and 2017/18
df21 <- filter(df1, season_id != 55149) %>%
mutate(year = droplevels(year),
home_team = droplevels(home_team),
away_team = droplevels(away_team)) %>%
mutate(y = transform_perc(perc))

# Use effect coding, contr.sum(), for teams so that intercept corresponds to
# average across all teams
contrasts(df21$home_team) <- contr.sum(levels(df21$home_team))
colnames(contrasts(df21$home_team)) <- head(levels(df21$home_team), -1)
contrasts(df21$away_team) <- contr.sum(levels(df21$away_team))
colnames(contrasts(df21$away_team)) <- head(levels(df21$away_team), -1)

# Dummy coding
contrasts(df21weekday) <- contr.treatment(7, base = 6) The first model I want to show is a beta regression model. As you can see below, the intercept is equal to 2.08 on the logit scale. Thus, the model predicts an average attendance rate of 89 % (plogis(2.08)). Average here means averaged across all teams (since effect coding was used) and predicted for a match on Saturday in the season 2016/17 (dummy coding used for weekday and year). Furthermore, there are huge differences across teams. For example, the predicted value for a home match of Flensburg is 2.08 + 1.42 corresponding to an attendance rate of 97 % (plogis(3.5)) but only 84 % (plogis(2.08 - 0.42)) for Lemgo or 63 % for Lübbecke. Likewise, the attendance rate differs by away team, where, for example, matches against Kiel are more attractive than matches against Erlangen. Apart from that, the precision or variability of the predicted mean also varies as a function of home team. For example, the precision is larger (and variability lower) for Flensburg than for Lemgo. m1 <- betareg(y ~ home_team + away_team + weekday + year | home_team, data = df21) m1_est <- tidy(m1) # Print some estimates slice(m1_est, 1, 49, seq(2, 69, 8)) #> # A tibble: 11 x 6 #> component term estimate std.error statistic p.value #> <chr> <chr> <dbl> <dbl> <dbl> <dbl> #> 1 mean (Intercept) 2.08 0.0815 25.5 2.54e-143 #> 2 precision (Intercept) 2.40 0.0643 37.4 2.44e-305 #> 3 mean home_teamSG Flensburg-~ 1.42 0.193 7.39 1.48e- 13 #> 4 mean home_teamTBV Lemgo Lip~ -0.422 0.125 -3.38 7.23e- 4 #> 5 mean home_teamTuS N-Lübbecke -1.53 0.168 -9.08 1.07e- 19 #> 6 mean away_teamTHW Kiel 0.728 0.123 5.94 2.78e- 9 #> 7 mean away_teamHC Erlangen -0.161 0.119 -1.35 1.77e- 1 #> 8 mean weekday1 0.291 0.257 1.13 2.57e- 1 #> 9 precision home_teamSG Flensburg-~ 0.491 0.270 1.82 6.85e- 2 #> 10 precision home_teamTBV Lemgo Lip~ 0.0346 0.235 0.147 8.83e- 1 #> 11 precision home_teamTuS N-Lübbecke -0.374 0.313 -1.19 2.32e- 1 ### Illustrative Plot of Estimates The model estimates can also be illustrated graphically. The parameters mean (mu) and precision (phi) from above correspond to a beta distribution, which is specific to each home team in the fitted model. To calculate the densities of the beta distribution using dbeta(), I first have to transform the estimates from the model. # Function to calculate a and b parameter of beta distribution from mu and phi # used in betareg with link function logit for mu and log for phi calc_ab <- function(mu, phi) { mu <- plogis(mu) phi <- exp(phi) b <- phi - mu*phi a <- phi - b return(cbind(a = a, b = b)[, 1:2]) } # Calculate predicted beta distributions for Flensburg and Lemgo preds1 <- # Estimates of mu and phi from above calc_ab(2.08 + c(1.42, -.42), 2.4 + c(.49, .03)) %>% asplit(1) %>% setNames(c("SG Flensburg-Handewitt", "TBV Lemgo Lippe")) %>% # Calculate densities via dbeta() lapply(function(y) data.frame(x = seq(0, 1, .001), y = dbeta(seq(0, 1, .001), y[1], y[2]))) %>% enframe("home_team") %>% unnest(value) # Print some values slice(preds1, seq(701, by = 100, length.out = 4)) #> # A tibble: 4 x 3 #> home_team x y #> <chr> <dbl> <dbl> #> 1 SG Flensburg-Handewitt 0.7 0.0133 #> 2 SG Flensburg-Handewitt 0.8 0.145 #> 3 SG Flensburg-Handewitt 0.9 1.40 #> 4 SG Flensburg-Handewitt 1 Inf Below, I plotted the model-implied beta distributions for the two teams, which are illustrated using the green lines. Furthermore, the plot shows the predicted means (green point on the x-axis), the raw densities calculated by geom_density() across the 34 matches per team, as well as the observed attendance rate for each match (placed below the densities by geom_rug()). In summary, the plot nicely illustrates the estimates from above, namely, very high and homogeneous attendance rates for Flensburg and lower and more varied attendance rates for Lemgo. This aligns quite well with the observed data. Note also that I picked these two teams more or less at random, but estimates are of course available for all teams in the data set. # Plot observed data using densities and rugs p1 <- df21 %>% filter(home_team %in% c("SG Flensburg-Handewitt", "TBV Lemgo Lippe")) %>% ggplot(aes(x = perc, fill = home_team)) + geom_density(trim = TRUE) + geom_rug() + facet_wrap(~ home_team) + coord_cartesian(xlim = c(.5, 1), ylim = c(0, 12)) + scale_fill_viridis_d() + labs(y = "Density", x = "Attendance Rate", fill = NULL, title = "Beta Regression: Observed and Fitted Attendance Rates") + theme(legend.position = "none") # Add lines corresponding to estimated distributions p1 <- p1 + geom_line(data = preds1, aes(x = x, y = y), inherit.aes = FALSE, size = 1, color = rgb(0, 155, 149, maxColorValue = 255)) # Add predicted means p1 <- p1 + geom_point(data = data.frame(y = 0, x = plogis(2.01 + c(1.49, -.35)), home_team = c("SG Flensburg-Handewitt", "TBV Lemgo Lippe")), aes(x = x, y = y), show.legend = FALSE, size = 3, color = rgb(0, 155, 149, maxColorValue = 255)) p1 ### Residuals Finally, I want to have a quick look at the residuals (which are on the logit scale). I decided to plot them chronologically, that is, against match day. As you see, residuals tend to get higher across the course of a season indicating that attendance rates near the end of the season are slightly higher than predicted. Thus, it seems useful to add variables to the model that can capture this effect (see below). m1_res <- augment(m1, data = df21) ggplot(m1_res, aes(x = round_number, y = .resid)) + geom_point() + stat_smooth(method = "loess", formula = "y ~ x") + facet_wrap(~ year) + labs(y = "Residuals", x = "Match Day", title = "Residuals as a Function of Match Day") ## Model Comparisons ### Models Considered In the following, I will compare different models: • The models will either model the variable matchday using a linear trend. Or the trend over time will be captured by the categorical variable month (thus allowing different attendance rates in each month). More information about the coding for variable month is provided in part 2. • Furthermore, I will compare different different link functions or forms: • Linear model (lm) • Quasi-binomial model (binomial) • Beta regression (beta) • Beta regression with dispersion parameter and logit link (beta+ logit) • Beta regression with dispersion parameter and loglog link (beta+ loglog) The performance of the models will be evaluated relative to the training data set from above (season 2016/17 and 2017/18) and to a holdout or cross-validation data set (season 2018/19). Furthermore, I will compare the models relative to simply predicting the average attendance rate of the home team. res11 <- list() ### Average ### # This "model" simply predicts for each match the yearly average attendance rate # of the home team res11average_no <-
glm(perc ~ home_team*year, data = df21,
family = quasibinomial(logit), weights = capacity)

### beta+dispersion (logit) ###

res11$betaDsplogit_lin <- betareg(y ~ home_team + away_team + weekday + year + half2 + matchday | home_team, data = df21) res11$betaDsplogit_mon <-
betareg(y ~ home_team + away_team + weekday + year + half2 +
Oct + Nov + Dec + Mar + Apr + May | home_team, data = df21)

### beta+dispersion (loglog) ###

res11$betaDsploglog_lin <- update(res11$betaDsplogit_lin, link = "loglog")
res11$betaDsploglog_mon <- update(res11$betaDsplogit_mon, link = "loglog")

### beta (logit) ###

res11$beta_lin <- betareg(y ~ home_team + away_team + weekday + year + half2 + matchday, data = df21) res11$beta_mon <-
betareg(y ~ home_team + away_team + weekday + year + half2 +
Oct + Nov + Dec + Mar + Apr + May, data = df21)

### lm - linear model ###

res11$lm_lin <- lm(attendance ~ home_team + away_team + weekday + year + half2 + matchday, data = df21) res11$lm_mon <-
lm(attendance ~ home_team + away_team + weekday + year + half2 +
Oct + Nov + Dec + Mar + Apr + May, data = df21)

### quasi-binomial ###

res11$quasibinomial_lin <- glm(perc ~ home_team + away_team + weekday + year + half2 + matchday, data = df21, family = quasibinomial(logit), weights = capacity) res11$quasibinomial_mon <-
glm(perc ~ home_team + away_team + weekday + year + half2 +
Oct + Nov + Dec + Mar + Apr + May, data = df21,
family = quasibinomial(logit), weights = capacity)

res12 <- enframe(res11)

res12
#> # A tibble: 11 x 2
#>    name              value
#>    <chr>             <list>
#>  1 average_no        <glm>
#>  6 beta_lin          <betareg>
#>  7 beta_mon          <betareg>
#>  8 lm_lin            <lm>
#>  9 lm_mon            <lm>
#> 10 quasibinomial_lin <glm>
#> 11 quasibinomial_mon <glm>

### Model Performance

For all models stored in res12, I will make predictions for the training data df21 and for the holdout data df22. I will then compare these predictions with the actual observations using the following three metrics:

• Mean absolute error (MAE)
• Root mean squared error (RMSE)
• R squared (R2)

R2 should be high, and MAE and RMSE should be relatively low. These latter two measures are provided in the metric of the original attendance values, that is an MAE of 500 means that the predictions are on average off by a value of 500 spectators.

# Holdout data
df22 <- filter(df1, season_id == 55149) %>%
filter(!home_team %in% c("SG BBM Bietigheim")) %>%
filter(!away_team %in% c("SG BBM Bietigheim")) %>%
mutate(year = "17/18") %>%
filter(date < as.Date("2019-02-01"))

### Make predictions ###

res13 <- res12 %>%
separate(name, into = c("mod", "matchday")) %>%
mutate(matchday = factor(matchday)) %>%
# Make "predictions" for data set used
mutate(olddata = map(value, ~augment(.x, data = df21, type.predict = "resp"))) %>%
# Make predictions for data from 2018/19 (cross validation)
mutate(newdata = map(value, ~augment(.x, newdata = df22, type.predict = "resp"))) %>%
# Calcuate predictions and residuals for actual attendance number rather
# than attendance rate
mutate(olddata = map(olddata, function(x = .x){
if (x$.fitted[1] > 1) { x$.pred <- round(x$.fitted) } else { x$.pred <- round(x$.fitted*x$capacity)
}
x$.res <- x$attendance - x$.pred return(x) })) %>% mutate(newdata = map(newdata, function(x = .x){ if (x$.fitted[1] > 1) {
x$.pred <- round(x$.fitted)
} else {
x$.pred <- round(x$.fitted*x$capacity) } x$.res <- x$attendance - x$.pred
return(x)
}))

### Performance measures (yardstick package) ###

res14 <- res13 %>%
mutate(old = map(olddata, ~yardstick::metrics(.x, attendance, .pred))) %>%
mutate(new = map(newdata, ~yardstick::metrics(.x, attendance, .pred))) %>%
# Wrangle into long format
select(-value, -olddata, -newdata) %>%
unnest(c(old, new), names_sep = "_") %>%
select(-old_.estimator, -new_.estimator) %>%
pivot_longer(c(old_.metric, old_.estimate, new_.metric, new_.estimate),
names_sep = "_", names_to = c("data", ".value"))

tail(res14)
#> # A tibble: 6 x 5
#>   mod           matchday data  .metric .estimate
#>   <chr>         <fct>    <chr> <chr>       <dbl>
#> 1 quasibinomial mon      old   rmse      652.
#> 2 quasibinomial mon      new   rmse      995.
#> 3 quasibinomial mon      old   rsq         0.926
#> 4 quasibinomial mon      new   rsq         0.842
#> 5 quasibinomial mon      old   mae       419.
#> 6 quasibinomial mon      new   mae       538.

### Polish labels etc ###

res15 <- res14 %>%
mutate(matchday = fct_recode(matchday, Linear = "lin", Month = "mon"),
data = fct_rev(fct_recode(data, "Holdout Data" = "new", "Training Data" = "old")),
.metric = fct_recode(.metric,
RMSE = "rmse", R2 = "rsq", MAE = "mae"),
mod = sub("Dsp(.*)", "+ (\\1)", mod),
mod = factor(mod,
levels = c("average", "lm", "quasibinomial", "beta", "beta+ (logit)",
"beta+ (loglog)")))

As shown in the plot below, the quasi-binomial model shows the best performance in the training data set. The beta models perform slightly worse but outperforms the quasi-binomial with respect to R2 in the holdout data set. The linear model performs worst (with the exception for RMSE in the holdout data), even worse than predicting just averages without a real model. When comparing the different variants of the beta model, the enhanced models (beta+), which also model the precision in addition to the mean, perform slightly better, especially in the holdout data set. The difference between the logit and the loglog link is tiny and I would prefer the logit model here because it is more common and thus easier to communicate.

The time effect is slightly better captured by the categorical predictor months compared to a linear trend, more on that in part 2.

In summary, both the quasi-binomial and the beta model are able to capture the relevant aspects of the attendance rates of matches of the German Handball-Bundesliga. The beta model, however, has the advantage that it can provided prediction intervals if desired whereas the intervals of the quasi-binomial model are way too narrow with data of several thousand spectators (not shown herein).

ggplot(res15, aes(y = .estimate, x = mod, color = matchday)) +
geom_point(position = position_dodge(width = .5), size = 3) +
facet_wrap(data ~.metric, scales = "free_y") +
scale_color_viridis_d(begin = .2) +
labs(x = "Model", y = "", color = "Time\nTrend",
title = "Comparison of Different Models for Proportion Data") +
theme(axis.text.x = element_text(angle = 45, hjust = 1))

## Prediction of Future Matches

Last, I want to make some real predictions, that is, predict the attendance of future matches.

### Model: Beta and Quasibinomial ###

df31 <- filter(df1, date < "2019-05-01") %>%
mutate(y = transform_perc(perc))

contrasts(df31$home_team) <- contr.sum(levels(df31$home_team))
contrasts(df31$away_team) <- contr.sum(levels(df31$away_team))

res31 <- betareg(y ~ home_team + away_team + weekday + half2*year +
Oct + Nov + Dec + Mar + Apr + May | home_team, data = df31)
res32 <- glm(perc ~ home_team + away_team + weekday + half2*year +
Oct + Nov + Dec + Mar + Apr + May, data = df31,
family = quasibinomial(logit), weights = capacity)

# newdata
df32 <- df1 %>%
filter(season_id == 55149) %>%
group_by(home_team) %>%
mutate(avg = round(mean(ifelse(attendance == 0, NA, attendance), na.rm = TRUE), -1)) %>%
filter(round_number %in% 30:31 & date > as.Date("2019-05-10"))

### Make predictions ###

tab3 <- enframe(list(beta = res31, binomial = res32), "Model") %>%
mutate(pred = map(value, ~augment(.x, newdata = df32,
type.predict = "resp"))) %>%
select(-value) %>%
unnest(pred) %>%
mutate(prediction = round(.fitted*capacity, -1)) %>%
select(Model, home_team, away_team, date, capacity, .fitted, prediction, avg) %>%
pivot_wider(names_from = Model, values_from = .fitted:prediction)

### Prediction interval for beta regression ###

# predict(res31,
#         newdata = filter(df1, season_id == 55149 & round_number == 31),
#         type = "quantile", at = c(.025, .975))

### Make table ###

tab3 %>%
mutate_at(vars(home_team, away_team),
~sub("(-Handewitt)|(Die Eulen )|(FRISCH AUF! )", "", .)) %>%
mutate_at(vars(home_team, away_team), ~sub("^(\\w{2,4}\\s)+[0-9]*\\s*", "", .)) %>%
mutate_at(vars(home_team, away_team), ~sub("Burgdorf", "B.", .)) %>%
mutate_at(vars(home_team, away_team), ~sub("Rhein-Neckar", "R.-N.", .)) %>%
arrange(date, avg) %>%
mutate(home_team = sub("(Die Eulen )|(([A-Z]{2,3}\\s)+[0-9]*\\s*)", "", home_team)) %>%
mutate(away_team = sub("(Die Eulen )|(([A-Z]{2,3}\\s)+[0-9]*\\s*)", "", away_team)) %>%
select(-.fitted_beta, -.fitted_binomial) %>%
knitr::kable(digits = 2,
col.names = c("Home Team", "Away Team", "Date", "Capacity",
"Average", "Beta", "Quasi-B.")) %>%
kable_styling(bootstrap_options = c("striped", "hover", "condensed", "responsive")) %>%
add_header_above(c(" " = 5, "Predicted Attendance" = 2))
Predicted Attendance
Home Team Away Team Date Capacity Average Beta Quasi-B.
Gummersbach Lemgo Lippe 2019-05-11 4050 3150 3770 3760
Leipzig Bergischer HC 2019-05-11 6500 4250 5100 4940
Melsungen R.-N. Löwen 2019-05-12 4300 3910 4190 4200
Magdeburg Wetzlar 2019-05-12 7221 6410 6550 6600
Füchse Berlin Ludwigshafen 2019-05-12 9000 7470 7900 7920
Kiel Flensburg 2019-05-12 10285 10280 10280 10280
Hannover-B. Gummersbach 2019-05-14 10000 5400 8570 8140
Ludwigshafen Stuttgart 2019-05-15 2268 2160 2160 2160
Minden Leipzig 2019-05-15 4059 2730 3070 3110
Lemgo Lippe Magdeburg 2019-05-15 4861 3640 4260 4180
Erlangen Bietigheim 2019-05-16 7850 4330 4500 3630
Flensburg Melsungen 2019-05-16 6300 6030 6090 6110
R.-N. Löwen Göppingen 2019-05-23 13200 7940 8680 7980

format: plain


text:
# Every $\lambda-$pure morphism in a locally $\lambda-$presentable category is a regular monomorphism

Consider the page from the book by Adamek & Rosicky: Locally presentable and accessible categories. given below. I need to derive this: Here is the statement

(2) to prove the universal property of $$f$$, it is sufficient to show that for each $$\lambda-$$presentable object $$H$$, each morphism $$h:H\to B$$ merging $$p$$ and $$q$$ factorizes uniquely through $$f$$.We know that $$f$$ factorizes through some $$b_i$$, say $$h=b_i\cdot h'$$. Then $$d_i^*\cdot (p_i\cdot h')=p\cdot b_i\cdot h'.$$

My question is why this equation holds? All the relevant letters are given here:

And here is the ERRATA to that page from the book.

The equation follows from

$$d^*_i \cdot p_i = p \cdot b_i$$

and this equation holds because of the definition of $$p\colon B\to B^*$$.

Recall that two diagrams of shape $$I$$ are introduced on page 88:

• $$D\colon I\to {\cal K}$$ with $$D_i=B_i$$ and $$D(i,j)=b_{i,j}$$

• $$D^*\colon I\to {\cal K}$$

These have colimits $$B=\mathrm{colim} B_i$$ and $$B^*=\mathrm{colim} D^*_i$$, together with maps $$b_i\colon B_i\to B$$ and $$d^*_i\colon D^*_i \to B^*$$.

The family of maps $$p_i\colon B_i\to D^*_i$$ forms a natural transformation from $$D$$ to $$D^*$$. Therefore the maps $$d^*_i \cdot p_i\colon B_i \to B^*$$ form a cocone which factors through the colimit cocone via a unique map $$p: B\to B^*$$. Similar for the $$q_i\colon B_i\to D^*_i$$ and $$q$$.

• Thank you for your detailed and clear answer. I just do not follow where (and if) we have used the uniqueness of such a $p$? – user122424 Feb 14 '19 at 17:09
• As far as I can see, uniqueness of such a $p$ was not used so far. – Marc Olschok Feb 14 '19 at 18:35


text:
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A while ago I developed and shared an emoji decoder because I was facing problems when retrieving data from Twitter and Instragram. In a nutshell, the issue is that R encodes emojis in a way that makes it a hassle identifying them. This is where the decoder/dictionary comes into play.

After I put together my decoder, new emojis have been released. For example, certain emojis came with different skin colours, which adds a little bit of complexity to the analysis. In fact, emojis with skin tone information consist of two unicode codepoints: the codepoint for the emoji (e.g. “U+1F466” for boy) and the codepoint for the respective skin tone (e.g. “U+1F3FB” for light skin tone). The emoji “boy: light kin tone” thus appears as “U+1F466 U+1F3FB”. The descriptions between the plain emoji and the emoji with skin tone information differ slightly (e.g. “princess” vs “princess: light skin tone”), in which case simple string matching would fail to identify these as being the same emoji. It’s gonna require some special attention when cleaning the data. More on that in the code. This list is a more complete version than the one I used for my decoder. Felipe released a new decoder based on the new list which I will use in the post.

Alright, so with the decoder at hand, we’re able to identify the emojis in, say, a tweet retrieved with the twitteR package. What now? Quite some people contacted me since I released the article asking for advice concerning emojis analysis and I’d like to cover some questions in this post. The whole code I used for the analysis in this article is available here.

Most used emoji

One such question was how to determine a users most used emoji.

We’ll start collecting some sample data to perform our analysis on. In case you didn’t already know, Paris Hilton is my favorite victim when it comes to emojis analysis or social media analysis, for the simple reason that she uses as lot ot them and shares a lot of content. The emojis_matching function is the heart of all the analysis performed in this post.

Paris Hilton’s favorite emoji is: SPARKLES! Who would have thought 😉 Her alltime favorite MUSICAL NOTES landed on place 7 (status quo 2017-03-24).

Tweet with most emojis

Another possible use case is to determine which tweet contains the most emojis. We can use the emojis_matchingfunction for this question again and than arrange by descending count. Done. Easy peasy lemon squeezy. This is how the output looks like:

From here, it’s easy to calculate the average number of emojis per tweet:

Sentiment analysis with emojis

Doing some research for this article, I came accross this paper, which extensively analyzes the valence (positive, negative, neutral) of emojis in a scientific manner. The authors made a csv file available containing all the emojis with their respective valences. I will base the sentiment analysis on this file.

For a reason I ignore, the csv file available doesn’t contain the sentiment score. The article gives guidance as how to compute the senteiment score of each emoji based on the data in the csv file, but at this point I prefer to scrape the list with the sentiment scores. It’s available here. After merging it with our firt emoji list to get the R encoding info, it looks like this:

Ok, so we have a list of emojis with their respective unicode codepoints and sentiment scores. The next step consists of matching sentiments to the tweets.

What we get is a list of the tweets with there respective, aggregated sentiment scores:

Note that the score od a single tweet is the sum of the sentiment score of all emojis in the tweet. The higher the score, the more positive the tweet. One could put this number in relation to the number of emojis in the tweet or choose a more binary format like “positve” “negative”. Most of Paris Hilton’s tweets are positive, which was to be expected, with only two tweets identified as having a negative valence. Some tweets don’t have any sentiment score, this is due to the fact that they didn’t contain any (identifiable) emoji.

Emojis associated with words in tweets

One question that came to my mind was: what words appear in the same tweets as emojis? What are the top n words associated with each emoji? Before we can perform this kind of text analysis, we need to do the usual house keeping: clean the texts from links, strange characters, punctuation etc. I recommand having a look at the code to see what I exactly did, especially at the cleaning pipe. Furthermore, I wrote the function wordFreqEmojis that outputs a data frame of emojis with the top 5 words (default value, can be changed) they are used most frequently with.

Browsing through words_emojis, it’s obvious that the data makes a lot of sense. The emojis associated with the word “adios” are for instance “sun”, “water wave”, “bikini” and “airplane”. These words all indicate the twitter user is travelling.

In natural language programming, there are literally endless possibilities. One could fine tune the results by using other stopwords, working with word stems or considering ngrams instead of single words just to name a few. Besides words, one can also find cooccuring emojis.

List of further emojis analysis ideas

Summing up, here are some ideas for further analysis I didn’t implement in this article but can be done with emojis:

• combine traditional text based sentiment analysis with emojis basd sentiment analysis
• analyse coocccurence of emojis
• identify topics based on emojis
• track trends in emojis use (weekdays, time of the day, “happy” emojis vs “sad” emojis, etc.)

Final thoughts

Emoji analysis is unlikely to make a good job at replacing natural language processing in a sentiment analysis context. Emojis can help easily identify positive content, but they’re not so good at identifying negative or serious, business related content as far as I can tell. It makes sense since most of the emojis have a positive meaning. Also, not everyone makes the same use of emoji and not every positive tweet contains an emoji, so again, I don’t think it’s a good idea to base your sentiment analysis exclusively on emojis. I’d rather suggest to perform traditional sentiment analysis and enrich it with emoji data. Also, emoji analysis can become quite difficult due to the constantly growing number of emojis. Some of them have more than one unicode codepoint, this can be a challenge in the analysis.

In this article, I only showed how to perform simple positive/negative sentiment analysis and very basic association with words, but one could come up with much more detailed approaches. One could harvest much more information from emojis than just their level of positiveness. I’m thinking activities, different sentiments, patriotism, fondness for children, frequency of travelling, etc.

Emojis Analysis in R was originally published by Kirill Pomogajko at Opiate for the masses on March 24, 2017.


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Paley construction

The Paley construction uses quadratic residues in a finite field GF(q) where q is a power of an odd prime number. There are two versions of the construction depending on whether q is congruent to 1 or 3 (mod 4).

Contents 1 Quadratic character and Jacobsthal matrix 2 Paley construction I 3 Paley construction II 4 Examples 5 The Hadamard conjecture 6 See also 7 References Quadratic character and Jacobsthal matrix Let q be a power of an odd prime. In the finite field GF(q) the quadratic character χ(a) indicates whether the element a is zero, a non-zero perfect square, or a non-square: {displaystyle chi (a)={begin{cases}0&{text{if }}a=0\1&{text{if }}a=b^{2}{text{ for some non-zero finite field element }}b\-1&{text{if }}a{text{ is not the square of any finite field element.}}end{cases}}} For example, in GF(7) the non-zero squares are 1 = 12 = 62, 4 = 22 = 52, and 2 = 32 = 42. Hence χ(0) = 0, χ(1) = χ(2) = χ(4) = 1, and χ(3) = χ(5) = χ(6) = −1.

The Jacobsthal matrix Q for GF(q) is the q×q matrix with rows and columns indexed by finite field elements such that the entry in row a and column b is χ(a − b). For example, in GF(7), if the rows and columns of the Jacobsthal matrix are indexed by the field elements 0, 1, 2, 3, 4, 5, 6, then {displaystyle Q={begin{bmatrix}0&-1&-1&1&-1&1&1\1&0&-1&-1&1&-1&1\1&1&0&-1&-1&1&-1\-1&1&1&0&-1&-1&1\1&-1&1&1&0&-1&-1\-1&1&-1&1&1&0&-1\-1&-1&1&-1&1&1&0end{bmatrix}}.} The Jacobsthal matrix has the properties Q QT = q I − J and Q J = J Q = 0 where I is the q×q identity matrix and J is the q×q all 1 matrix. If q is congruent to 1 (mod 4) then −1 is a square in GF(q) which implies that Q is a symmetric matrix. If q is congruent to 3 (mod 4) then −1 is not a square, and Q is a skew-symmetric matrix. When q is a prime number and rows and columns are indexed by field elements in the usual 0, 1, 2, … order, Q is a circulant matrix. That is, each row is obtained from the row above by cyclic permutation.

Paley construction I If q is congruent to 3 (mod 4) then {displaystyle H=I+{begin{bmatrix}0&j^{T}\-j&Qend{bmatrix}}} is a Hadamard matrix of size q + 1. Here j is the all-1 column vector of length q and I is the (q+1)×(q+1) identity matrix. The matrix H is a skew Hadamard matrix, which means it satisfies H+HT = 2I.

Paley construction II If q is congruent to 1 (mod 4) then the matrix obtained by replacing all 0 entries in {displaystyle {begin{bmatrix}0&j^{T}\j&Qend{bmatrix}}} with the matrix {displaystyle {begin{bmatrix}1&-1\-1&-1end{bmatrix}}} and all entries ±1 with the matrix {displaystyle pm {begin{bmatrix}1&1\1&-1end{bmatrix}}} is a Hadamard matrix of size 2(q + 1). It is a symmetric Hadamard matrix.

Examples Applying Paley Construction I to the Jacobsthal matrix for GF(7), one produces the 8×8 Hadamard matrix, 11111111 -1--1-11 -11--1-1 -111--1- --111--1 -1-111-- --1-111- ---1-111.

For an example of the Paley II construction when q is a prime power rather than a prime number, consider GF(9). This is an extension field of GF(3) obtained by adjoining a root of an irreducible quadratic. Different irreducible quadratics produce equivalent fields. Choosing x2+x−1 and letting a be a root of this polynomial, the nine elements of GF(9) may be written 0, 1, −1, a, a+1, a−1, −a, −a+1, −a−1. The non-zero squares are 1 = (±1)2, −a+1 = (±a)2, a−1 = (±(a+1))2, and −1 = (±(a−1))2. The Jacobsthal matrix is {displaystyle Q={begin{bmatrix}0&1&1&-1&-1&1&-1&1&-1\1&0&1&1&-1&-1&-1&-1&1\1&1&0&-1&1&-1&1&-1&-1\-1&1&-1&0&1&1&-1&-1&1\-1&-1&1&1&0&1&1&-1&-1\1&-1&-1&1&1&0&-1&1&-1\-1&-1&1&-1&1&-1&0&1&1\1&-1&-1&-1&-1&1&1&0&1\-1&1&-1&1&-1&-1&1&1&0end{bmatrix}}.} It is a symmetric matrix consisting of nine 3×3 circulant blocks. Paley Construction II produces the symmetric 20×20 Hadamard matrix, 1- 111111 111111 111111 -- 1-1-1- 1-1-1- 1-1-1- 11 1-1111 ----11 --11-- 1- --1-1- -1-11- -11--1 11 111-11 11---- ----11 1- 1---1- 1--1-1 -1-11- 11 11111- --11-- 11---- 1- 1-1--- -11--1 1--1-1 11 --11-- 1-1111 ----11 1- -11--1 --1-1- -1-11- 11 ----11 111-11 11---- 1- -1-11- 1---1- 1--1-1 11 11---- 11111- --11-- 1- 1--1-1 1-1--- -11--1 11 ----11 --11-- 1-1111 1- -1-11- -11--1 --1-1- 11 11---- ----11 111-11 1- 1--1-1 -1-11- 1---1- 11 --11-- 11---- 11111- 1- -11--1 1--1-1 1-1---.

The Hadamard conjecture The size of a Hadamard matrix must be 1, 2, or a multiple of 4. The Kronecker product of two Hadamard matrices of sizes m and n is an Hadamard matrix of size mn. By forming Kronecker products of matrices from the Paley construction and the 2×2 matrix, {displaystyle H_{2}={begin{bmatrix}1&1\1&-1end{bmatrix}},} Hadamard matrices of every allowed size up to 100 except for 92 are produced. In his 1933 paper, Paley says “It seems probable that, whenever m is divisible by 4, it is possible to construct an orthogonal matrix of order m composed of ±1, but the general theorem has every appearance of difficulty.” This appears to be the first published statement of the Hadamard conjecture. A matrix of size 92 was eventually constructed by Baumert, Golomb, and Hall, using a construction due to Williamson combined with a computer search. Currently, Hadamard matrices have been shown to exist for all {displaystyle scriptstyle m,equiv ,0mod 4} for m < 668. See also Paley biplane Paley graph References Paley, R.E.A.C. (1933). "On orthogonal matrices". Journal of Mathematics and Physics. 12: 311–320. doi:10.1002/sapm1933121311. Zbl 0007.10004. L. D. Baumert; S. W. Golomb; M. Hall Jr (1962). "Discovery of an Hadamard matrix of order 92". Bull. Amer. Math. Soc. 68 (3): 237–238. doi:10.1090/S0002-9904-1962-10761-7. F.J. MacWilliams; N.J.A. Sloane (1977). The Theory of Error-Correcting Codes. North-Holland. pp. 47, 56. ISBN 0-444-85193-3. Categories: Matrices

Si quieres conocer otros artículos parecidos a Paley construction puedes visitar la categoría Matrices.

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# Professional Development Seminar MAT/STA 587

Policy documents:
Department policies
Instructor:
Nándor Sieben
schedule
Syllabus
Agenda:
1. (8/29)
- TOPIC: Intro to LyX
- HW: produce the following quiz . Replace my name with yours. Keep all the features, including typefaces and placements. Turn in a hard copy on which I can make comments. Sample lyx file (right click) to give you a starting point. Detexify might be useful to find latex macros.
2. (9/5)
- DUE: quiz
- TOPIC: Professional article writing
- HW: write an article about the quadratic formula. You need to have all the parts we discussed in class including sections, theorems, proofs, examples, figures, cross references, citations with bibtex, etc. Make sure you include everything listed here. Solving an equation usually takes two steps. First we show that the equation implies that a solution must be one of a few candidates. Then we check which of the solution candidates actually work. Please read the equations article about this. Please also read the writing style article and try to follow the recommendations.
3. (9/12)
- DUE: nothing
- TOPIC: Figures with Tikz
- HW: Recreate the tikz figure about the derivative of the inverse. Use tikz to add your name to the figure at the top right corner.
Continue working on the quadratic article.

4. (9/19)
- DUE: tikz figure
- TOPIC: GeoGebra Overleaf
- HW: Recreate the Geogebra figure. Use Geogebra to add your name to the figure at the top right corner. Can you spot a theorem in the figure? (not the Pythagorean Theorem)
5. (9/26)
- TOPIC: Basic Web page design, HTML reference NAU allows you to create web pages on the dana server. Follow the instructions to map \\dana.ucc.nau.edu\webhome . Then create a file called index.html on this drive. You can access this file using the address http://dana.ucc.nau.edu/UserID
- HW: Create a professional looking personal web page. Include at least one picture, some links, and some math. Send me the link to the site by email.

6. (10/3)
- DUE: geogebra figure
- TOPIC: Beamer
- HW: Create a presentation about the Law of Sines. State and prove the full version of the theorem as discussed in class. State the theorem required for the proof. Include an example. Use GeoGebra to create the figures. Use the beamer lyx template, or the lyx template (also in pdf format) .
7. (10/10)
- DUE: Nothing.
- TOPIC: Guest speaker: Jeannie Backer (U.S. Geological Survey)
- HW: Create a Resume or Curriculum Vitae.
8. (10/17)
- DUE: Beamer presentation. Send me the pdf by email.
- TOPIC: Guest speaker: Terry Seckel (Nationwide E&S/Specialty)
- HW:
9. (10/24)
- DUE: CV or Resume. Please turn in a hardcopy
- TOPIC: Guest speaker: Robert Buscaglia: Introduction to R: class notes 1
You may find the following useful: A Sufficient Introduction to R.
- HW: R homework 1 from the class notes.
10. (10/31)
- DUE: R homework 1
- TOPIC: Guest speaker: Robert Buscaglia: Introduction to R: class notes 2
- HW: R homework 2 from the class notes
11. (11/7)
- DUE: R homework 2
Sage tutorial
another Sage tutorial
- HW: Sage worksheet
12. (11/14)
- DUE: Sage worksheet
- TOPIC: Guest speaker: Michael Klinefelter, W.L. Gore & Associates, Inc.
- HW: Use LyX/LaTeX to write a professional sample solution to the following problem: "Solve the inequality $$| x + 1 | + | 2x - 3 | > 6$$ on the set of real numbers." Use Tikz/Geogebra when appropriate. Use 1-inch margins and 12 point font. Everything should read as a sentence. Equations written row-after-row are not appropriate without an explanation of their logical connection. Include the statement of the problem. This is an elementary problem. Avoid relying on analysis terms like increasing, continuous, Intermediate Value Theorem. This means you cannot use test points to justify that a function has a certain sign on a whole interval. You need to do case analysis and the definition of absolute value as a piecewised defined function. Use interval notation whenever it is appropriate.
13. (11/21)
- DUE: Professional sample solution. Please turn in a hardcopy.
- TOPIC: Mathematical writing. Job application material.
- HW: Write a formal letter of application for the job description. Keep in mind the following: An application is automatically rejected if the minimum qualifications are not met. The screening committee assigns a numerical score for every preferred qualification using an evaluation grid . Addressing these points is crucially important.
14. (11/28)
- DUE: Formal letter of application. We use this application letter on this day.
- TOPIC: Work as a Search Committee. Bring a hard copy of your letter of application and your CV.
- HW: Finish all assignments.
15. (12/5)
- DUE:
- TOPIC: Problem solving.
- HW: Finish all assignments.


text:
Opuscula Math. 33, no. 2 (), 237-354
http://dx.doi.org/10.7494/OpMath.2013.33.2.237
Opuscula Mathematica

# Existence of solution of sub-elliptic equations on the Heisenberg group with critical growth and double singularities

Abstract. For a class of sub-elliptic equations on Heisenberg group $$\mathbb{H}^N$$ with Hardy type singularity and critical nonlinear growth, we prove the existence of least energy solutions by developing new techniques based on the Nehari constraint. This result extends previous works, e.g., by Han et al. [Hardy-Sobolev type inequalities on the H-type group, Manuscripta Math. 118 (2005), 235–252].
Keywords: sub-elliptic equations, Heisenberg group, Least energy solutions.
Mathematics Subject Classification: 35H20, 35J60.


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# A 500 V shunt motor runs at its normal speed of 250 RPM when the armature current is 200 A. The resistance of armature is 0.12 Ω. Calculate the speed when resistance is inserted in the field reducing the shunt field to 80% of normal value and the armature current is 100A.

A 500 V shunt motor runs at its normal speed of 250 RPM when the armature current is 200 A. The resistance of armature is 0.12 Ω. Calculate the speed when resistance is inserted in the field reducing the shunt field to 80% of normal value and the armature current is 100A.

### Right Answer is: 320 RPM

#### SOLUTION

Given that

Supply voltage Vs = 500 V
Speed N1 = 250 RPM
Armature current Ia1= 200 A
Armature Resistance R1 = 0.12Ω

Back EMF of DC Motor is given as

Eb = V − IaRa

Eb = 500 − 200 × 0.12 = 478V

Now when the resistance is inserted in the field, Flux reduce to 80%

Flux Φ2 = 0.8Φ1

Armature current Ia2 = 100A

Back EMF

Eb = V − IaRa

Eb = 500 − 100 × 0.12 = 488V

As we Know that In DC motor the back EMF is directly proportional to the speed and flux.

E ∝ Nφ

$\begin{array}{l}\dfrac{{{E_{b2}}}}{{{E_{b1}}}} = \dfrac{{{N_2}}}{{{N_1}}} \times \dfrac{{{\Phi _2}}}{{{\Phi _1}}}\\\\{N_2} = {N_1} \times \dfrac{{{E_{b2}}}}{{{E_{b1}}}} \times \dfrac{{{\Phi _2}}}{{{\Phi _1}}}\\\\{N_2} = 250 \times \dfrac{{488}}{{476}} \times \left( {\dfrac{{{\Phi _1}}}{{0.8{\Phi _1}}}} \right)\\\\{N_2} = 320{\rm{ R}}{\text{.P}}{\rm{.M}}\end{array}$

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# Texas Go Math Grade 8 Lesson 2.2 Answer Key Scientific Notation with Negative Powers of 10

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Lesson 2.2 Answer Key Scientific Notation with Negative Powers of 10.

## Texas Go Math Grade 8 Lesson 2.2 Answer Key Scientific Notation with Negative Powers of 10

Essential Question
How can you use scientific notation to express very small quantities?

You can use what you know about writing very large numbers in scientific notation to write very small numbers in scientific notation.

A typical human hair has a diameter of 0.000025 meter. Write this number in scientific notation.

A. Notice how the decimal point moves in the list below. Complete the list.

B. Move the decimal point in 0.000025 to the right as many places as necessary to find a number that is greater than or equal to 1 and less than 10. What number did you find? ____
D. Combine your answers to B and c to represent 0.000025 in scientific notation. ____

Reflect

Question 1.
When you move the decimal point, how can you know whether you are increasing or decreasing the number?
When the decimal point moves to the right, the number is increasing and if it moves to the left, the is decreasing.

Question 2.
Explain how the two steps of moving the decimal and multiplying by a power of 10 leave the value of the original number unchanged.
The power of 10 represents the number of times the decimal is moved, hence multiplying will leave the value of the original number unchanged.

Reflect

Question 3.
Critical Thinking When you write a number that is less than 1 in scientific notation, how does the power of 10 differ from when you write a number greater than 1 in scientific notation?
When we write a number that is less than 1 in scientific notation, the power of 10 is negative, whereas when we write a number that is greater than 1 in scientific notation, the power of 10 is positive

Write each number in scientific notation.

Question 4.
0.0000829
0.0000829 Given
8.29 Place the decimal point
5 Count the number of places the decimal point is moved.
8.29 × 10-5 Multiply 8.29 times a power of 10.
Since 0.0000829 is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
8.29 × 10-5

Question 5.
0.000000302
0.000000302 Given
3.02 Place the decimal point
7 Count the number of places the decimal point is moved.
3.02 × 10-7 Multiply 3.02 times a power of 10. Since the original number is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
3.02 × 10-7

Question 6.
A typical red blood cell in human blood has a diameter of approximately 0.000007 meter. Write this diameter in scientific notation. ________________________
0.000007 Given
7.0 Place the decimal point
6 Count the number of places the decimal point is moved.
7.0 × 10-6 Multiply 7.0 times a power of 10. Since the original number is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
7.0 × 10-6

Reflect

Question 7.
Justify Reasoning Explain whether 0.9 × 10-5 is written in scientific notation. If not, write the number correctly in scientific notation.
In order for a number to be written in scientific notation they must be of the format:
c × 10n (1)
where c is a decimal number greater or equal to 1 and less than 10 and n is an integer
Since in the given number, 0.9 < 1 this number is not written in scientific notation.
A correct way to write that number in scientific notation would be to move the decimal point one place to the
right because of that the exponent would also decrease by one:
9× 10-6 (2)

Question 8.
Which number is larger, 2 × 10-3 or 3 × 10-2? Explain.
2 × 10-3 Given
and
3 × 10-2
3 × 10-2 is larger when comparing exponents, -2 > -3

Write each number in standard notation.

Question 9.
1.045 × 10-6
1.045 × 10-6 Given
6 Use the exponent of the power of 10 to see how many places to move the decimal point.
places
0.000001045 Place the decimal point Since you are going to write a number less than 1.045, move the decimal to the left. Add placeholder zeros if necessary.
The number
1.045 × 10-6 in standard notation is 0.000001045

Question 10.
9.9 × 10-5
9.9 × 10-5 Given
5 Use the exponent of the power of 10 to see how many places to move the decimal point.
places
0.000099 Place the decimal point Since you are going to write a number less than 9.9, move the decimal to the left. Add placeholder zeros if necessary.
The number
9.9 × 10-5 in standard notation is 0.000099

Question 11.
Jeremy measured the length of an ant as 1 × 10-2 meter. Write this length in standard notation.
1 × 10-2 Given
2 Use the exponent of the power of 10 to see how many places to move the decimal point.
places
0.01 Place the decimal point Since you are going to write a number less than 1, move the decimal to the left. Add placeholder zeros if necessary.
The number
1 × 10-2 in standard notation is 0.01 meter
0.01

Write each number in scientific notation. (Explore Activity and Example 1)

Question 1.
0.000487
Hint: Move the decimal right 4 places.
0.000487 Given
4.87 Place the decimal point
4 Count the number of places the decimal point is moved.
4.87 × 10-4 Multiply 4.87 times a power of 10.
Since 0.000487 is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
4.87 × 10-4

Question 2.
0.000028
Hint: Move the decimal right 5 places.
Move the decimal point as many places as necessary to find a number that is greater that or equal to 1 and less
than 10:
2.8 Place the decimal point (1)
5 places Count the number of places you moved the decimal point (2)
2.8 × 10-5 Multiply the number from step (1) with 10-5 (3)
Note: in step (3) you multiplied 2.8 with 10-5 because you moved the decimal point 5 places to the right
2.8 × 10-5

Question 3.
0.000059
0.000059 Given
5.9 Place the decimal point
5 Count the number of places the decimal point is moved.
5.9 × 10-5 Multiply 5.9 times a power of 10. Since 0.000059 is less than 1, the decimal point moves to the right and the exponent on 10 is negative
5.9 × 10-5

Question 4.
0.0417
0.0417 Given
4.17 Place the decimal point
2 Count the number of places the decimal point is moved.
4.17 × 10-2 Multiply 4.17 times a power of 10. Since 0.0417 is less than 1, the decimal point moves to the right and the exponent on 10 is negative
4.17 × 10-2

Question 5.
Picoplankton can be as small as 0.00002 centimeter.
0.00002 Given
2.0 Place the decimal point
5 Count the number of places the decimal point is moved.
2 × 10-5 Multiply 2 times a power of 10. Since 0.00002 is less than 1, the decimal point moves to the right and the exponent on 10 is negative
2 × 10-5 centimeter

Question 6.
The average mass of a grain of sand on a beach is about 0.000015 gram.
0.000015 gram Given
1.5 Place the decimal point
5 Count the number of places the decimal point is moved.
1.5 × 10-5 Multiply 1.5 times a power of 10. Since 0.000015 is less than 1, the decimal point moves to the right and the exponent on 10 is negative
1.5 × 10-5 gram

Write each number in standard notation. (Example 2)

Question 7.
2 × 10-5
Hint: Move the decimal left 5 places.
2 × 10-5 Given
5 Use the exponent of the power of 10 to see how many places to move the decimal point.
places
0.00002 Place the decimal point. Since you are going to write a number less than 2, move the decimal to the left. Add placeholder zeros if necessary.
The number
2 × 10-5 in standard notation is 0.00002

Question 8.
3.582 × 10-6
Hint: Move the decimal left 6 places.
3.582 × 10-6 Given
6 Use the exponent of the power of 10 to see how many places to move the decimal point.
places
0.000003582 Place the decimal point. Since you are going to write a number less than 3.582, move the decimal to the left. Add placeholder zeros if necessary.
The number
3.582 × 10-6 in standard notation is 0.000003582

Question 9.
8.3 × 10-4
8.3 × 10-4 Given
4 Use the exponent of the power of 10 to see how many places to move the decimal point.
places
0.00083 Place the decimal point. Since you are going to write a number less than 8.3, move the decimal to the left. Add placeholder zeros if necessary.
The number
8.3 × 10-4 in standard notation is 0.00083

Question 10.
2.97 × 10-2
2.97 × 10-2 Given
2 Use the exponent of the power of 10 to see how many places to move the decimal point.
places
0.0279 Place the decimal point. Since you are going to write a number less than 2.97, move the decimal to the left. Add placeholder zeros if necessary.
The number
2.7 × 10-2 in standard notation is 0.0297

Question 11.
9.06 × 10-5
9.06 × 10-5 Given
5 Use the exponent of the power of 10 to see how many places to move the decimal point.
places
0.0000906 Place the decimal point. Since you are going to write a number less than 9.06, move the decimal to the left. Add placeholder zeros if necessary.
The number
9.06 × 10-5 in standard notation is 0.0000906

Question 12.
4 × 10-5
(1) First, use the exponent of the power of 10 to see how many places to move the decimal point. Since we have 10-5 we have to move the decimal point 5 places.
(2) Place the decimal point, since you are going to write a number less than 4, move the decimal point to the left Add placeholder zeros if necessary:
4 × 10-5 = 0.00004.
0.00004

Question 13.
The average length of a dust mite is approximately 0.0001 meter. Write this number in scientific notation. (Example 1)
Move the decimal point as many places as necessary to find a number that is greater than or equal to 1 and less
than 10:
1 Place the decimal point (1)
4 places Count the number of places you moved the decimal point (2)
1 × 10-4 Multiply the number from step (1) with 10-4 (3)
Note: in step (3) you multiplied 1 with 10-4 because you moved the decimal point 5 places to the right because 0.0001 is smaller than 1.
1 × 10-4 meter.

Question 14.
The mass of a proton is about 1.7 × 10-24 gram. Write this number in standard notation. (Example 2)
1.7 × 10-24 gram Given
24 Use the exponent of the power of 10 to see how many places to move the decimal point
places

0.0000000000000000000000017 Place the decimal point Since you are going to write a number less than 1.7, move the decimal to the left Add placeholder zeros if necessary.
The number
1.7 × 10-24 in
standard notation is
0.0000000000000000000000017 gram

Essential Question Check-In

Question 15.
Describe how to write 0.0000672 in scientific notation.
(1) Place the decimal point such that the new number is larger or equal to 1 but less than 10.
0.0000672 ⇒ 6.72
(2) Count the number of places you moved the decimal. point: 5 places.
(3) Multiply 6.72 by 10-5 (because you moved the decimal 5 places to the right the exponent is negative)
6.72 × 10-5

Use the table for problems 16-21. Write the diameter of the fibers in scientific notation.

Question 16.
Alpaca
__________
0.00277 Fiber diameter of Alpaca
2.77 Place the decimal point
3 Count me number 0f places the decimal point is moved.
2.77 × 10-3 Multiply 2.77 times a power of 10. Since 0.00277 is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
2.77 × 10-3

Question 17.
Angora rabbit
_________
0.0013 Fiber diameter of Angora rabbit
1.3 Place the decimal point
3 Count me number 0f places the decimal point is moved.
1.3 × 10-3 Multiply 1.3 times a power of 10. Since 0.0013 is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
1.3 × 10-3

Question 18.
Llama
_____
0.0035 Fiber diameter of Llama
3.5 Place the decimal point
3 Count me number 0f places the decimal point is moved.
3.5 × 10-3 Multiply 3.5 times a power of 10. Since 0.0035 is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
3.5 × 10-3

Question 19.
Angora goat
______
0.0045 Fiber diameter of Angora goat
4.5 Place the decimal point
3 Count me number 0f places the decimal point is moved.
4.5 × 10-3 Multiply 4.5 times a power of 10. Since 0.0045 is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
4.5 × 10-3

Question 20.
Orb web spider
0.015 Fiber diameter of Orb web spider
1.5 Place the decimal point
2 Count me number 0f places the decimal point is moved.
1.5 × 10-2 Multiply 1.5 times a power of 10. Since 0.0045 is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
1.5 × 10-2

Question 21.
Vicuna
0.0008 Fiber diameter of Vicuna
8.0 Place the decimal point
4 Count me number 0f places the decimal point is moved.
8.0 × 10-4 Multiply 8.0 times a power of 10. Since 0.0008 is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
8.0 × 10-4

Question 22.
Make a Conjecture Which measurement would be least likely to be written in scientific notation: the thickness of a dog hair, the radius of a period on this page, the ounces in a cup of milk? Explain your reasoning.
Both the thickness of a dog hair and the radius of a period on a page are very small, numbers (Lengths), something that we tend to write in scientific notation (because it is easier to work with when written in scientific notation). On the other hand, it is likely that the number of ounces in a cup of milk is a number that we can easily write and work with, so we don’t need to write it in scientific notation, so it is reasonable to suspect that it is the least likely to be written in scientific notation.
The number of ounces in a cup of milk

Question 23.
Multiple Representations Convert the length 7 centimeters to meters. Compare the numerical values when both numbers are written in scientific notation.
There are 100 centimeters in 1 meter, or 1 centimeter = 0.01 meters
Because of this 7 centimeters = 0.07 meters
Write 0.07 in scientific notation by moving the decimal point two places to the right:
0.07 = 7 × 10-2 meter
Since 7 centimeters is a number already greater or equal to 1 and less than 10 to get a scientific notation we just
multiply it with 100 = 1
7 = 7 × 100 centimeters
Finally we can see that when in scientific notation the first factor is the same but the exponents of the second factor differ by 2.

In scientific notation the first factor is the same but the exponents of the second factor differ by 2.

Question 24.
Draw Conclusions A graphing calculator displays 1.89 × 10-12 as 1.89E12. How do you think it would display 1.89 × 10-12? What does the E stand for?
We are told that 1.89. 1012 will be displayed as 1.89e12 in a scientific calculator.
We can conclude that here e stands for exponent of 10 and 1.89 . 10-12 will be displayed as 1.89e -12

Question 25.
Communicate Mathematical Ideas When a number is written in scientific notation, how can you tell right away whether or not it is greater than or equal to 1 ?
In a scientific notation, if the exponent of 10 is negative, the number is smaller than 1. Otherwise, it is greater or equal to 1.

Question 26.
The volume of a drop of a certain liquid is 0.000047 liter. Write the volume of the drop of liquid in scientific notation.
0.000047 liter Given
4.7 Place the decimal point
5 Count the number of places the decimal point is moved.
4.7 × 10-5 Multiply 4.7 times a power of 10. Since 0.000047 is less than 1, the decimal point moves to the right and the exponent on 10 is negative.
4.7 × 10-5

Question 27.
Justify Reasoning If you were asked to express the weight in ounces of a ladybug in scientific notation, would the exponent of the 10 be positive or negative? Justify your response.
Since the weight of a ladybug in ounces ¡s a number smaller than 1 we would need to move the decimal point to the right. Therefore the exponent of the 10 would be a negative number.

Negative, since the weight of a ladybug is smaller than 1.

Physical Science The table shows the length of the radii of several very small or very large items. Complete the table.

Question 28.

(1) Move the decimal point to the left and remove extra zeros:
1,740,000 ⇒ 1.74
(2) Count the number of places you moved the decimal. point: 6 places
(3) Multiply 1.74 by 106 (because we moved the decimal point 6 places to the left 1.74 × 106
1.74 × 106 meters

Question 29.

Question 30.

Question 31.

Question 32.

Question 33.

Question 34.
List the items in the table in order from the smallest to largest.

Compare the powers of 10 and when they are same, compare the first factor. Based on this rule, the items are arranged from smallest to largest.

Texas Go Math Grade 8 Lesson 2.2 H.O.T. Focus On Higher Order Thinking Answer Key

Question 35.
Analyze Relationships Write the following diameters from least to greatest. 1.5 × 10-2m 1.2 × 102m 5.85 × 10-3 m 2.3 × 10-2 m 9.6 × 10-1 m
Compare the powers of 10 (smaller the exponent the smaller the number), if some are the same then compare the
first factors:
5.85 × 10-3 < 1.5 × 10-2 < 2.3 × 10-2 < 9.6 × 10-1 < 1.2 × 102

5.85 × 10-3, 1.5 × 10-2, 2.3 × 10-2, 9.6 × 10-1, 1.2 × 102

Question 36.
Critique Reasoning Jerod’s friend Al had the following homework problem:
Express 5.6 × 10-7 in standard form.
Al wrote 56,000,000. How can Jerod explain Al’s error and how to correct it?
5.6 × 10-7 Given
Error is that the decimal is moved in the moved in right direction and not left direction.
Since the exponent of 10 is -7, the decimal should be moved to left.

7 Use the exponent of the power of 10 to see how many places to move the decimal point
places
0.00000056 Place the decimal point Since you are going to write a number less than 5.6, move the decimal to the left. Add placeholder zeros if necessary.

Question 37.
Make a Conjecture Two numbers are written in scientific notation. The number with a positive exponent is divided by the number with a negative exponent. Describe the result. Explain your answer.
$$\frac{a^{m}}{a^{n}}$$ = am-n
$$\frac{a \times 10^{m}}{b \times 10^{-n}}$$ = $$\frac{a}{b}$$ × 10m+n


text:
Weather

# A little warmer, hazy

Temperatures climb a bit and air quality gets worse
Posted at 7:21 PM, Nov 15, 2022
and last updated 2022-11-15 22:21:31-05

Temperatures have been a little on the cool side lately.

In fact, in Bakersfield our high temperatures have been below average for 9 days in a row, with highs as cool as 58° during that stretch.

We could break that pretty soon, though.

Highs are expected to be near average on Wednesday, with a projected high of 66° in Bakersfield, just one degree shy of the average of 67°.

The warmer temperatures will be a result of an area of high pressure building in.

That will also help trap more pollution in the Valley, so our skies are expected to remain hazy.

Air Quality is expected to be unhealthy for sensitive groups on Wednesday.


text:
11,052 Pages

Bremner-Macleod numbers are the smallest positive integer solutions $$(a,b,c)$$ of the following Diophantine equation:

$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=N$$ where $$N$$ is a positive integer. The equation and the sizes of its solutions were discussed by Andrew Bremner and Allan Macleod.[1]

## Size

For all odd $$N$$ and some even $$N$$, the equation has no positive integer solutions.

The Bremner-Macleod number for $$N=4$$ is (a,b,c) =

(154476802108746166441951315019919837485664325669565431700026634898253202035277999,

36875131794129999827197811565225474825492979968971970996283137471637224634055579,

4373612677928697257861252602371390152816537558161613618621437993378423467772036)

For $$N=6$$ it is

(20260869859883222379931520298326390700152988332214525711323500132179943287700005601210288797153868533207131302477269470450828233936557,

2250324022012683866886426461942494811141200084921223218461967377588564477616220767789632257358521952443049813799712386367623925971447,

1218343242702905855792264237868803223073090298310121297526752830558323845503910071851999217959704024280699759290559009162035102974023)

For $$N=10$$ it is

(4862378745380642626737318101484977637219057323564658907686653339599714454790559130946320953938197181210525554039710122136086190642013402927952831079021210585653078786813279351784906397934209,

269103113846520710198086599018316928810831097261381335767926880507079911347095440987749703663156874995907158014866846058485318408629957749519665987782327830143454337518378955846463785600977,

221855981602380704196804518854316541759883857932028285581812549404634844243737502744011549757448453135493556098964216532950604590733853450272184987603430882682754171300742698179931849310347)

For $$N\le200$$, the amounts of digits in $$\max\{a,b,c\}$$ are shown below:

N # digits
4 81
6 134
10 190
12 2707
14 1876
16 414
18 10323
24 33644
28 81853
32 14836
38 1584369
42 886344
46 198771
48 418086
58 244860
60 9188
66 215532
76 23662
82 85465
92 252817
102 625533
112 935970
116 112519
126 196670
130 8572242
132 3607937
136 26942
146 259164
156 12046628
158 15097279
162 1265063
178 398605460
182 2828781
184 20770896
186 5442988
196 11323026
198 726373
200 71225279

For $$N=896$$, $$a$$ has more than 2.187 trillion digits.

## Sources

1. Andrew Bremner and Allan Macleod, An unusual cubic representation problem
Community content is available under CC-BY-SA unless otherwise noted.


text:
# Search for New Physics Phenomena in Fermion-Pair Production at LEP

3 CMS
IPNL - Institut de Physique Nucléaire de Lyon
Abstract : The measurements of hadron and lepton--pair production cross sections and lepton--pair forward--backward asymmetries performed with the L3 detector at centre--of--mass energies between 130 GeV and 172 GeV are used to search for new physics phenomena. New physics effects involving four fermion vertices -- contact interactions -- are looked for in all channels. For hadron production the exchange of virtual leptoquarks and scalar quarks is studied. No evidence for deviations from the Standard Model expectations is found. Lower limits on the scale $\Lambda$ of contact interactions in the range 1.2 -- 7.1 TeV are obtained at the 95\% confidence level for various models. Upper limits on the coupling strengths of leptoquarks and scalar quarks are derived.
Document type :
Journal articles
Complete list of metadatas

Cited literature [30 references]

http://hal.in2p3.fr/in2p3-00000036
Contributor : Sylvie Flores <>
Submitted on : Friday, November 6, 1998 - 12:03:02 PM
Last modification on : Wednesday, October 30, 2019 - 1:50:01 AM
Long-term archiving on : Friday, May 29, 2015 - 2:51:17 PM

### Identifiers

• HAL Id : in2p3-00000036, version 1

### Citation

M. Acciarri, O. Adriani, M. Aguilar-Benitez, S. Ahlen, J. Alcaraz, et al.. Search for New Physics Phenomena in Fermion-Pair Production at LEP. Physics Letters B, Elsevier, 1998, 433, pp.163-175. ⟨in2p3-00000036⟩

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text:
# Coordinates of point inside of circle

I'm sorry if what I'm asking has already been answered or is really easy but I struggled a little and haven't been able to come up with an idea.

TThe context is as follows: a robot is being placed inside a circle. I need to find the position the robot is being placed given that I know the center and the radius of the circle and that I have a distance sensor which gives me the distance to a point on the circle circumference.

I believe there isn't enough information to calculate this point. If that is indeed the case could you please point me to the kind of information i'm missing?.

-
If the circle has radius $r$ and the robot measures distance $d<r$ to the circumference, then all you can say is that the robot is on the circle of radius $r-d$ around the center. –  Hagen von Eitzen Oct 21 '12 at 14:42
@Hagen: It says "distance to a point on the circumference", not "distance to the circumference". It's not clear in the question whether this is the distance to a fixed point, or to the nearest point, or to an arbitrary point, and if arbitrary, whether the sensor can be used more than once. –  joriki Oct 21 '12 at 17:01
the distance is to an arbitrary point that gets picked by the senzor –  Jack Blaze Oct 22 '12 at 16:29

In this question, you are basically trying to set up a co-ordinate system. For this, you need two things: 1. Origin 2.A fixed line (to function as your X Axis).

While you do have the origin, you need to set a line as your fixed axis.

Once you have set up the fixed line, you could use trigonometry (polar co ordinates) to find the position of the required point.

-
your ideea seems interesting to pursue can you elaborate a bit on that? thanks. –  Jack Blaze Oct 22 '12 at 16:32
If we had a compass with no N/S/E/W directions marked on it, we would be confused as to which direction was which. We would need one mark on the compass pointing North to guide us. This would tell us that the point opposite the North was south, 45* left of it was North East and so on. Similarly, in your question, we would need the X axis to help us 'fix' the co-ordinate system –  RahulHP Oct 23 '12 at 4:52
Once we have fixed the X axis, we can then use polar co ordinates (r cos(theta), r sin(theta)) to obtain the information about the point –  RahulHP Oct 23 '12 at 4:56

Well it depends what the problem means, it is a little vague as it is stated.

Does you distance sensor gives you the distance to a fixed point on the circle? Or can you use it multiple times to get the distance to more points on the circle?

Note that if you can use the sensor twice, for two fixed points on the circle, then you can most of the time determine the point (what I mean by most of the time is that if the two points are diametrally opposed you can only reduce the problem to two possibilities, while if the two points are closed you can find the robot)... Three points always determine the possition, no matter which three points are....

-
thanks for the hint search . after the first 5 seconds the robot is allowed to move so more samples could be gathered however there is no guarantee that the would come from different points of the circumference –  Jack Blaze Oct 22 '12 at 16:32


text:
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# For each customer, a bakery charges p dollars for the first loaf of

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For each customer, a bakery charges p dollars for the first loaf of  [#permalink]

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Updated on: 06 Jun 2019, 04:53
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For each customer, a bakery charges p dollars for the first loaf of bread bought by the customer and charges q dollars for each additional loaf bought by the customer. What is the value of p ?

(1) A customer who buys 2 loaves is charged 10 percent less per loaf than a customer who buys a single loaf.

(2) A customer who buys 6 loaves of bread is charged 10 dollars.

Originally posted by dwag on 19 Sep 2006, 15:19.
Last edited by Bunuel on 06 Jun 2019, 04:53, edited 3 times in total.
Edited the question and added the OA.
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Re: For each customer, a bakery charges p dollars for the first loaf of  [#permalink]

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13 Aug 2013, 01:27
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fozzzy wrote:
What would be the equations for this question?

For each customer, a bakery charges p dollars for the first loaf of bread bought by the customer and charges q dollars for each additional loaf bought by the customer. What is the value of p ?

(1) A customer who buys 2 loaves is charged 10 percent less per loaf than a customer who buys a single loaf:

Price of 2 loaves = $(p+q). Price per loaf =$(p+q)/2

Price of a single loaf = $p. Given that (p+q)/2=0.9p. Two unknowns. Not sufficient. (2) A customer who buys 6 loaves of bread is charged 10 dollars --> p+5q=10. Not sufficient. (1)+(2) We have two distinct linear equations with two unknowns: (p+q)/2=0.9p and p+5q=10, thus we can solve for both p and q. Sufficient.. Answer: C. Hope it's clear. _________________ ##### General Discussion Director Joined: 02 Jun 2006 Posts: 850 Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 19 Sep 2006, 16:34 2 Answer: C Cost for first loaf = p Cost for remaining = q S1: Cost of two loaves = p+q Price /loaf = (p+q)/2 = 0.9p (10% discount) p+q = 1.8p or q = 0.8p or 4p -5q = 0 Not sufficient. S2: 10 = p+5q Not sufficient. S1 & S2: p+5q = 10 4p-5q=0 or 5p = 10 p = 2 q = 4x2/5 = 1.6 Sufficient w/ 2 equations. Retired Moderator Joined: 05 Jul 2006 Posts: 1317 Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 20 Sep 2006, 05:44 i cant get the warding of st one...........Help guys Director Joined: 02 Jun 2006 Posts: 850 Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 20 Sep 2006, 05:58 Statement 1 says that when you buy 2 loaves of bread instead of 1, you get a discount of 10% per loaf. If you buy 2, then the cost is (p+q) Cost per loaf = (p+q)/2 = 0.5p+0.5q If you buy 1, the cost is p. Cost per loaf = p/1 = p You get a 10% discount per loaf.. i.e. 0.5p+0.5q = 0.9p or 0.5q = 0.4p or 5q = 4p yezz wrote: i cant get the warding of st one...........Help guys Retired Moderator Joined: 05 Jul 2006 Posts: 1317 Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 20 Sep 2006, 07:58 Statement 1 says that when you buy 2 loaves of bread instead of 1, you get a discount of 10% per loaf. If you buy 2, then the cost is (p+q) Cost per loaf = (p+q)/2 = 0.5p+0.5q If you buy 1, the cost is p. Cost per loaf = p/1 = p You get a 10% discount per loaf.. i.e. 0.5p+0.5q = 0.9p or 0.5q = 0.4p or 5q = 4p but dont you think that the part in read is the average cost per loaf not cost per loaf Retired Moderator Joined: 05 Jul 2006 Posts: 1317 Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 20 Sep 2006, 08:04 For each customer, a bakery charges p dollars for the first loaf of bread bought by the customer and charges q dollars for each add'l loaf bought by the customer. What's the value of p? (1) A customer who buys 2 loaves is charged 10% less per loaf than a customer who buys a single loaf. (2) A customer who buys 6 loaves of bread is charged 10 dollars. general formula to calculate price x = p+nq where n is number of loafs in excess of one from one original price of two loafs is = p+q fro one i think it means p+q = 2(0.9)p = 1.8p thus 0.8p=q from two p+5q = 10 both together p+5(0.8p) = 10 thus 5p = 10 and p = 2 and q = 1.6 what is my mistake here Manager Joined: 13 Sep 2006 Posts: 194 Location: New York Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 20 Sep 2006, 10:11 I used a combination of math and some guess work. I have seen these type of problems before and made the mistake of thinking it was E in the past. However this time I knew it was D ...From the two stems it looks like you are going to get two equations with two variables which one can solve. I just made the sure the equations weren't equal to each other when I chose D...I didn't actually work all the way through the problem. Director Joined: 29 Nov 2012 Posts: 672 Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 13 Aug 2013, 00:21 What would be the equations for this question? Director Joined: 26 Oct 2016 Posts: 588 Location: United States Concentration: Marketing, International Business Schools: HBS '19 GMAT 1: 770 Q51 V44 GPA: 4 WE: Education (Education) Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 26 Jan 2017, 05:51 2 Question stem tells you that p$ for 1st loaf and q $for additional loaf. 1. if someone buys 2 loaf , he will be charged p+q$.
price per laof = (p+q) / 2 $if one buys only 1 loaf then one pays only p$

St1 tells that

(p+q)/2 = (1-10%) p
p/2+q/2 = 9/10p
q=0.8 p

Cant figure out p insufficient.

2. 6 loaves price is p+5q given as 10$p+5q=10 Insufficient. Together 1 & 2 p+5*0.8p = 10 p =2$
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Re: For each customer, a bakery charges p dollars for the first loaf of  [#permalink]

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08 Oct 2017, 08:41
Hi

In this why cant the answer be B.

We have the equation, p+5q=10

The only possible value that q can take is 1. Thus, we can get p as 5.

Bunuel wrote:
fozzzy wrote:
What would be the equations for this question?

For each customer, a bakery charges p dollars for the first loaf of bread bought by the customer and charges q dollars for each additional loaf bought by the customer. What is the value of p ?

(1) A customer who buys 2 loaves is charged 10 percent less per loaf than a customer who buys a single loaf:

Price of 2 loaves = $(p+q). Price per loaf =$(p+q)/2

Price of a single loaf = $p. Given that (p+q)/2=0.9p. Two unknowns. Not sufficient. (2) A customer who buys 6 loaves of bread is charged 10 dollars --> p+5q=10. Not sufficient. (1)+(2) We have two distinct linear equations with two unknowns: (p+q)/2=0.9p and p+5q=10, thus we can solve for both p and q. Sufficient.. Answer: C. Hope it's clear. Manhattan Prep Instructor Joined: 04 Dec 2015 Posts: 948 GMAT 1: 790 Q51 V49 GRE 1: Q170 V170 Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 08 Oct 2017, 12:55 sinhap07 wrote: Hi In this why cant the answer be B. We have the equation, p+5q=10 The only possible value that q can take is 1. Thus, we can get p as 5. You don't necessarily know that q is an integer. Dollar amounts can be non-integers. For instance, the first loaf could cost 3.2 dollars, and the remaining 5 loaves could cost 1.36 dollars each. 3.2 + 5(1.36) = 10, which fits statement 2. _________________ Chelsey Cooley | Manhattan Prep | Online tutoring and curriculum My latest GMAT blog posts | My latest GRE blog posts | Suggestions for blog articles are always welcome! GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 938 Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 16 Sep 2018, 11:46 dwag wrote: For each customer, a bakery charges p dollars for the first loaf of bread bought by the customer and charges q dollars for each additional loaf bought by the customer. What is the value of p ? (1) A customer who buys 2 loaves is charged 10 percent less per loaf than a customer who buys a single loaf. (2) A customer who buys 6 loaves of bread is charged 10 dollars. $${\text{\ }}p\,\,\,:\,\,{\text{first}}\,\,{\text{loaf}}$$ $${\text{\ }}q\,\,\,{\text{:}}\,\,\,{\text{any}}\,\,{\text{additional}}\,\,{\text{loaf}}$$ $$? = q$$ $$\left( 1 \right)\,\,{\left( {\frac{{\,{\text{charge}}\,}}{{{\text{loaf}}}}} \right)_{\,2\,\,{\text{loaves}}}} = \frac{{p + q}}{2}\,\,\,\,\mathop = \limits^{{\text{given}}} \,\,\frac{9}{{10}}p\,\,\,\,\, \Rightarrow \,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\frac{q}{p} = \frac{4}{5}$$ $$\left\{ \begin{gathered} \,{\text{Take}}\,\,\left( {p,q} \right) = \left( {0.5,0.4} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0.4 \hfill \\ \,{\text{Take}}\,\,\left( {p,q} \right) = \left( {1,0.8} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0.8 \hfill \\ \end{gathered} \right.$$ $$\left( 2 \right)\,\,6\,\,{\text{loaves}}\,\,{\text{for}}\,\,\ 10\,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,\left( {p\,;q} \right) = \left( {2\,;\frac{8}{5}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = \frac{8}{5} \hfill \\ \,{\text{Take}}\,\,\left( {p\,;q} \right) = \left( {3\,;\frac{7}{5}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = \frac{7}{5} \hfill \\ \end{gathered} \right.$$ $$\left( {1 + 2} \right)\,\,\,\left\{ \begin{gathered} p + 5q = 10 \hfill \\ \frac{q}{p} = \frac{4}{5} \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{k}}\,\,{\text{technique}}} \,\,\,\,\left( {5k} \right) + 5\left( {4k} \right) = 10\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 4k\,\,{\text{unique}}\,$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our high-level "quant" preparation starts here: https://gmath.net Intern Status: Applying Affiliations: test Joined: 16 Apr 2012 Posts: 32 Location: India Yawer: Yawer Concentration: Marketing, Technology GMAT 1: 700 Q48 V38 WE: Consulting (Internet and New Media) Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 16 Sep 2018, 12:32 IMO C. Both will give equations of 2 variables, and then we can solve to get p. Manager Joined: 10 Dec 2019 Posts: 52 Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 16 Feb 2020, 05:25 Bunuel wrote: fozzzy wrote: What would be the equations for this question? For each customer, a bakery charges p dollars for the first loaf of bread bought by the customer and charges q dollars for each additional loaf bought by the customer. What is the value of p ? (1) A customer who buys 2 loaves is charged 10 percent less per loaf than a customer who buys a single loaf: Price of 2 loaves =$(p+q).
Price per loaf = $(p+q)/2 Price of a single loaf =$p.

Given that (p+q)/2=0.9p.

Two unknowns. Not sufficient.

(2) A customer who buys 6 loaves of bread is charged 10 dollars --> p+5q=10. Not sufficient.

(1)+(2) We have two distinct linear equations with two unknowns: (p+q)/2=0.9p and p+5q=10, thus we can solve for both p and q. Sufficient..

Hope it's clear.

Hi Bunuel

(1) A customer who buys 2 loaves is charged 10 percent less per loaf than a customer who buys a single loaf:

Price of 2 loaves = $(p+q). Price per loaf =$(p+q)/2

Price of a single loaf = $p. As per my understanding p+q = 1p -10%p which will be p+q = 0.9p But unable to understand where is (p+q)/2 coming from? Math Expert Joined: 02 Sep 2009 Posts: 65771 Re: For each customer, a bakery charges p dollars for the first loaf of [#permalink] ### Show Tags 16 Feb 2020, 05:43 1 Anurag06 wrote: Bunuel wrote: fozzzy wrote: What would be the equations for this question? For each customer, a bakery charges p dollars for the first loaf of bread bought by the customer and charges q dollars for each additional loaf bought by the customer. What is the value of p ? (1) A customer who buys 2 loaves is charged 10 percent less per loaf than a customer who buys a single loaf: Price of 2 loaves =$(p+q).
Price per loaf = $(p+q)/2 Price of a single loaf =$p.

Given that (p+q)/2=0.9p.

Two unknowns. Not sufficient.

(2) A customer who buys 6 loaves of bread is charged 10 dollars --> p+5q=10. Not sufficient.

(1)+(2) We have two distinct linear equations with two unknowns: (p+q)/2=0.9p and p+5q=10, thus we can solve for both p and q. Sufficient..

Hope it's clear.

Hi Bunuel

(1) A customer who buys 2 loaves is charged 10 percent less per loaf than a customer who buys a single loaf:

Price of 2 loaves = $(p+q). Price per loaf =$(p+q)/2

Price of a single loaf = $p. As per my understanding p+q = 1p -10%p which will be p+q = 0.9p But unable to understand where is (p+q)/2 coming from? Say bakery charges 2 dollars for the first loaf of bread bought by the customer and charges 1.6 dollars for each additional loaf bought by the customer. If you buy 2 loaves, you pa$2 + $1.6 =$3.6, so ($2 +$1.6)/2 = $3.6/2 =$1.8 per loaf, which is 10 percent less per loaf than a customer who buys a single loaf for $2 ($1.8 = 0.9*\$2).
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Re: For each customer, a bakery charges p dollars for the first loaf of  [#permalink]

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16 Feb 2020, 08:03
Thank you Bunuel. I was getting confused because of consideration about avg and single loaf price as same.

Re: For each customer, a bakery charges p dollars for the first loaf of   [#permalink] 16 Feb 2020, 08:03


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