text:
# (Crash) Dummies Doing Physics

1. Nov 13, 2014

### MissBisson

1. The problem statement, all variables and given/known data
a. A set of crash tests consists of running a test car moving at a speed of 11.2 m/s (25 mi/hr) into a solid wall. Strapped securely in an advanced seat belt system, a 57.0 kg (126lbs) dummy is found to move a distance of 0.750 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time.

b. Using the direction of motion as the positive direction, calculate the average acceleration of the dummy during that time (in g's, with 1g=9.81m/s2).

c.In a different car, the distance the dummy moves while being stopped is reduced from 0.750 m to 0.250 m, calculate the average force on the dummy as that car stops.

2. Relevant equations
Ke=1/2mv^2
w=force x distance
p=mv
Impulse=Favg x t

3. The attempt at a solution
i tried solving the first part using momentum:
pf-pi=Favg x t
but that didnt work

I also tried using teh kinetic energy equation
Ke=1/2mv^2
w=Ke
w=1/2mv^2
f x d = 1/2mv^2
But this also didnt work..

I dont know hat im doing wrong and i can seem to find anyother way of solving this problem.

2. Nov 13, 2014

### Simon Bridge

Last edited by a moderator: May 7, 2017
3. Nov 13, 2014

### sam400

The only thing you need to calculate the force is time. I think you might have to use some kinematic equations for that part, assuming that the dummy underwent constant acceleration.

4. Nov 14, 2014

### haruspex

Soapbox: This is a flawed question. Average acceleration is defined as change in velocity divided by change in time: $\Delta v/\Delta t$. Ergo, average force is change in momentum divided by change in time: $\Delta p/\Delta t$. There is insufficient information to calculate that. You can calculate an "average over distance" from change in energy and displacement, but that's different from the standard meaning of average force.
As sam400 writes, to answer the question as posed you need to make some assumption about the pattern of deceleration. This will depend on the crumple characteristics of the vehicle. If it is constant deceleration then the average over distance happens to give the same answer, but more likely the deceleration will increase from 0 steadily up to a maximum before becoming approximately constant.

5. Nov 14, 2014

### MissBisson

Okay so i had done is this:
Try 1:
1/2mv^2 = F x d
1/2(57)(11.2)^2 = F x (0.75)
This gave me F = 4766.72N and since it acts on the dummie it is -4766.72N but this was wrong

Try 2:
Pf=0
Pi=mv = 57 x 11.2 = 638.4kgm/s
i found that it takes 0.0669seconds for the dummie to travel 0.75m using proportions with the velocity
Impulse = Favg x t
638.4 = Favg x 0.0669 Favg = 9533.44 and again i put negative so -9533.44N but this was also wrong.

Try 3:
a = dv/dt
a= 0-11.2/0.0669-0
a=-167.4m/s^2
F=ma
F=(57)(-167.4)
F=-9542N This didnt work

Try 4:
vf^2=vi^2 +2a(xf-xi)
0=(11.2)^2 + 2a(0.75) a = 86.63m/s^2
F=ma

6. Nov 14, 2014

### sam400

The impulse of the dummy will be a negative one, so I feel it might be a sign related problem. If it's something online, it could be significant figure related too, unless the system lets you know about how close the value is from it.

7. Nov 14, 2014

### MissBisson

It tells me the amount of significant figures i need and i always enter the negative answer. when i input my answer it tell me this:

The force that acts on the dummy has to do (negative) work on the dummy. It must slow the dummy and stop it. Thus the amount of work equals the kinetic energy of the dummy. From the equation of work done by a force we can calculate that force.

Maybe this will help..

8. Nov 15, 2014

### Staff: Mentor

How does the question specify the number of sig figs required of your answer?

9. Nov 15, 2014

### haruspex

In try 1 you entered -4766.72N, but now you have the positive version. Did you try both?
I see no reason to enter a negative value here. You are not asked for the work done. The force could be positive or negative depending on which direction you take as positive, and the question does not specify that.
If you take the direction of movement of the car as positive then the displacement is positive and the force negative, giving negative work.
If you take the direction of movement of the car as negative then the displacement is negative and the force positive, giving negative work
And it specifies 6? But your conversions from lbs and miles were not that accurate. Only the first two digits of your answer are correct.
.


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# 9 Generalized linear models

Linear regression is suitable for outcomes which are continuous numerical scores. In practice this requirement is often relaxed slightly, for example for data which are slightly skewed, or where scores are somewhat censored ( e.g. questionnaire scores which have a minium or maximum).

However, for some types of outcomes standard linear models are unsuitable. Examples here include binary (zero or one) or count data (i.e. positive integers representing frequencies), or proportions (e.g. proportion of product failures per batch). This section is primarily concerned with binary outcomes, but many of the same principles apply to these other types of outcome.

### Logistic regression

In R we fit logistic regression with the glm() function which is built into R, or if we have a multilevel model with a binary outcome we use glmer() from the lme4:: package.

Fitting the model is very similar to linear regression, except we need to specify the family="binomial" parameter to let R know what type of data we are using.

Here we use the titanic dataset (you can download this from Kaggle, although you need to sign up for an account).

Before we start fitting models, it’s best to plot the data to give us a feel for what is happening.

Figure 1 reveals that, across all fare categories, women were more likely to survive the disaster than men. Ticket class also appears to be related to outcome: those with third class tickets were less likely to survive than those with first or second class tickets. However, differences in survival rates for men and women differed across ticket classes: women with third class tickets appear to have been less advantaged (compared to men) than women with first or second class tickets.

titanic <- read.csv('data/titanic.csv')
titanic %>%
ggplot(aes(factor(Pclass), Survived,
group=Sex, color=Sex)) +
stat_summary() +
stat_summary(geom="line") +
xlab("Ticket class")

Given the plot above, it seems reasonable to predict survival from Sex and Pclass, and also to include the interaction between these variables.

To run a logistic regression we specify the model as we would with lm(), but instead use glm() and specify the family parameter:

m <- glm(Survived ~ Sex * factor(Pclass),
data=titanic, family = binomial(link="logit"))

Because it can become repetitive to write out the family parameter in full each time, I usually write a ‘helper function’ called logistic() which simply calls glm with the right settings. For example:

# define a helper function for logistic regression the '...'
# means 'all arguments', so this function passes all it's
# arguments on to the glm function, but sets the family correctly
logistic <- function(...) {
}

Which you can use like so:

logistic(Survived ~ Sex * factor(Pclass), data=titanic)

Call:  glm(formula = ..1, family = binomial(link = "logit"), data = ..2)

Coefficients:
(Intercept)                  Sexmale          factor(Pclass)2
3.4122                  -3.9494                  -0.9555
factor(Pclass)3  Sexmale:factor(Pclass)2  Sexmale:factor(Pclass)3
-3.4122                  -0.1850                   2.0958

Degrees of Freedom: 890 Total (i.e. Null);  885 Residual
Null Deviance:      1187
Residual Deviance: 798.1    AIC: 810.1

#### Tests of parameters

As with lm() models, we can use the summary() function to get p values for parameters in glm objects:

titanic.model <- logistic(Survived ~ Sex * factor(Pclass), data=titanic)
summary(titanic.model)

Call:
glm(formula = ..1, family = binomial(link = "logit"), data = ..2)

Deviance Residuals:
Min       1Q   Median       3Q      Max
-2.6248  -0.5853  -0.5395   0.4056   1.9996

Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept)               3.4122     0.5868   5.815 6.06e-09 ***
Sexmale                  -3.9494     0.6161  -6.411 1.45e-10 ***
factor(Pclass)2          -0.9555     0.7248  -1.318  0.18737
factor(Pclass)3          -3.4122     0.6100  -5.594 2.22e-08 ***
Sexmale:factor(Pclass)2  -0.1850     0.7939  -0.233  0.81575
Sexmale:factor(Pclass)3   2.0958     0.6572   3.189  0.00143 **
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

Null deviance: 1186.7  on 890  degrees of freedom
Residual deviance:  798.1  on 885  degrees of freedom
AIC: 810.1

Number of Fisher Scoring iterations: 6

You might have spotted in this table that summary reports z tests rather than t tests for parameters in the glm model. These can be interepreted as you would the t-test in a linear model, however.

Where there are categorical predictors we can also reuse the car::Anova function to get the equivalent of the F test from a linear model (with type 3 sums of squares; remember not to use the built in anova function unless you want type 1 sums of squares):

car::Anova(titanic.model, type=3)
Analysis of Deviance Table (Type III tests)

Response: Survived
LR Chisq Df Pr(>Chisq)
Sex                  97.547  1  < 2.2e-16 ***
factor(Pclass)       90.355  2  < 2.2e-16 ***
Sex:factor(Pclass)   28.791  2  5.598e-07 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Note that the Anova table for a glm model provides $$\chi^2$$ tests in place of F tests. Although they are calculated differently, you can interpret these $$\chi^2$$ tests and p values as you would for F tests in a regular Anova.

#### Predictions after glm

As with linear models, we can make predictions from glm models for our current or new data.

One twist here though is that we have to choose whether to make predictions in units of the response (i.e. probability of survival), or of the transformed response (logit) that is actually the ‘outcome’ in a glm (see the explainer on transformations and links functions).

You will almost always want predictions in the units of your response, which means you need to add type="response" to the predict() function call. Here we predict the chance of survival for a new female passenger with a first class ticket:

new.passenger = expand.grid(Pclass=1, Sex=c("female"))
predict.glm(titanic.model, newdata=new.passenger, type="response")
1
0.9680851

And we could plot probabilities for each gender and class with a standard error for this prediction if desired:

new.passengers = expand.grid(Pclass=1:3, Sex=c("female", "male"))

# this creates two vectors: $fit, which contains # predicted probabilities and$se.fit
preds <- predict.glm(titanic.model,
newdata=new.passengers,
type="response",
se.fit=T)

new.passengers %>%
mutate(fit = preds$fit, lower=fit - preds$se.fit,
upper=fit + preds\$se.fit) %>%
ggplot(aes(factor(Pclass), fit,
ymin=lower, ymax=upper,
group=Sex, color=Sex)) +
geom_pointrange() +
geom_line() +
xlab("Ticket class") +
ylab("Probability of survival")

#### Evaluating logistic regression models

glm models don’t provide an R2 statistic, but it is possible to evaluate how well the model fits the data in other ways.

Although there are various pseudo-R2 statistics available for glm; see https://www.r-bloggers.com/evaluating-logistic-regression-models/

One common technique, however, is to build a model using a ‘training’ dataset (sometimes a subset of your data) and evaluate how well this model predicts new observations in a ‘test’ dataset. See http://r4ds.had.co.nz/model-assess.html for an introduction.

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text:
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# M06 #11

Author Message
Senior Manager
Joined: 16 Jan 2009
Posts: 355
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

### Show Tags

14 Jul 2009, 23:14
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

90
105
168
420
2520

= $$(2C8*2C6*2C4*2C2)/4!$$
Could someone please explain why we are dividing the number of combinations by 4!?
_________________

Lahoosaher

CIO
Joined: 02 Oct 2007
Posts: 1216

### Show Tags

14 Jul 2009, 23:23

m06q11-80504.html
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Re: M06 #11   [#permalink] 14 Jul 2009, 23:23
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# M06 #11

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text:
Students can Download Maths Chapter 2 Measurements Ex 2.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

8th Maths Exercise 2.1 Question 1.
Fill in the blanks:
(i) The ratio between the circumference and diameter of any circle is _________.
(ii) A line segment which joins any two points on a circle is a ______.
(iii) The longest chord of a circle is _______.
(iv) The radius of a circle of diameter 24 cm is ______.
(v) A part of circumference of a circle is called as _____.
Solution:
(i) π
(ii) chord
(iii) diameter
(iv) 12 cm
(v) an arc

Samacheer Kalvi 8th Maths Book Solutions Question 2.
Match the following

Solution:
(i) 3
(ii) 4
(iii) 5
(iv) 2
(v) 1

Class 8 Maths Exercise 2.1 Solution Question 3.
Find the central angle of the shaded sectors (each circle is divided into equal sectors)

Solution:

8th Maths 2.1 Question 4.
For the sectors with given measures, find the length of the arc, area and perimeter, (π = 3.14)
(i) central angle 45°, r = 16 cm
(ii) central angle 120°, d = 12.6 cm
(iii) central angle 60°, r = 36 cm
(iv) central angle 72°, d = 10 cm
Solution:
(i) Central angle 45°, r = 16 cm
Length of the arc l = $$\frac{\theta^{\circ}}{360^{\circ}}$$ × 2πr units
l = $$\frac{45^{\circ}}{360^{\circ}}$$ × 2 × 3.14 × 16 cm
l = $$\frac{1}{8}$$ × 2 × 3.14 × 16 cm
l = 12.56 cm
Area of the sector = $$\frac{\theta^{\circ}}{360^{\circ}}$$ × πr2
A = $$\frac{45^{\circ}}{360^{\circ}}$$ × 3.14 × 16 × 16
A = 100.48 cm2
Perimeter of the sector
P = l + 2r units
P = 12.56 + 2(16) cm
P = 44.56 cm

(ii) Central angle 120°, d = 12.6 cm
∴ r = $$\frac{12.6}{2}$$ cm
r = 6.3 cm
Length of the arc

Area of the sector missing
Perimeter of the sector
P = l + 2r units
P = 6.28 + 2(5) cm
P = 6.28 + 10 cm
P = 16.28 cm

Question 5.
From the measures given below, find the area of the sectors.

Solution:
(i) Area of the sector
A = $$\frac{l r}{2}$$ sq. units
l = 48 m
r = 10 m
= $$\frac{48 \times 10}{2}$$ m2
= 24 × 10 m2
= 240 m2
Area of the sector = 240 m2

(ii) length of the arc l = 12.5 cm
Area of the sector
A = $$\frac{l r}{2}$$ sq. units
= $$\frac{12.5 \times 6}{2}$$
= 12.5 × 3 cm2 cm2
= 37.5 cm2
Area of the sector = 37.5 cm2

(iii) length of the arc l = 50 cm
Area of the sector
A = $$\frac{l r}{2}$$ sq. units
= $$\frac{50 \times 13.5}{2}$$
= 25 × 13.5 cm2 cm2
= 337.5 cm2
Area of the sector = 337.5 cm2

Question 6.
Find the central angle of each of the sectors whose measures are given below (π = $$\frac{22}{7}$$)

Solution:
(i) Radius of the sector = 21 cm
Area of the sector = 462 cm2
$$\frac{l r}{2}$$ = 462
$$\frac{l \times 21}{2}$$ = 462

∴ Central angle of the sector = 120°

(ii) Radius of the sector = 8.4 cm
Area of the sector = 18.48 cm2

(iii) Radius of the sector = 35 m
Length of the arc l = 44 m

Question 7.
(i) A circle of radius 120 m is divided into 8 equal sectors. Find the length of the arc of each of the sectors.
(ii) A circle of radius 70 cm is divided into 5 equal sectors. Find the area of each of the sectors.
Solution:
(i) Radius of the circle r = 120 m
Number of equal sectors = 8

(ii) Radius of the sector r = 70 cm
Number of equal sectors = 5

Note: We can solve this problem using A = $$\frac{1}{n}$$ πr2 sq. units also.

Question 8.
Find the area of a sector whose length of the arc is 50 mm and radius is 14 mm.
Solution:
Length of the arc of the sector l = 50 mm
Area of the sector = $$\frac{l r}{2}$$ sq. units = $$\frac{50 \times 14}{2}$$ mm2 = 50 × 7 mm2 = 350 mm2
Area of the sector = 350 mm2

Question 9.
Find the area of a sector whose perimeter is 64 cm and length of the arc is 44 cm.
Solution:
Length of the arc of the sector l = 44 cm
Perimeter of the sector P = 64 cm
l + 2r = 64 cm
44 + 2 r = 64 .
2 r = 64 – 44
2 r = 20
r = $$\frac{20}{2}$$ = 10 cm2
Area of the sector = $$\frac{l r}{2}$$ sq. units = $$\frac{44 \times 10}{2}$$ cm2 = 22 × 10 cm2 = 220 cm2
Area of the sector = 220 cm2

Question 10.
A sector of radius 4.2 cm has an area 9.24 cm2. Find its perimeter
Solution:
Radius of the sector r = 4.2 cm
‘ Area of the sector = 9.24 cm2
$$\frac{l r}{2}$$ = 9.24
$$\frac{l \times 4.2}{2}$$ = 9.24
l × 2.1 = 9.24
l = $$\frac{9.24}{2.1}$$
l = 4.4 cm
Perimeter of the sector = 1 + 2r units = 4.4 + 2(4.2) cm
= 4.4 + 8.4 cm = 12. 8 cm
Perimeter of the sector = 12.8 cm

Question 11.
Infront of a house, flower plants are grown in a circular quadrant shaped pot whose radius is 2 feet. Find the area of the pot in which the plants grow. (π = 3.14)

Solution:
Central angle of the quadrant = 90°
Radius of the circle = 2 feet

Area of the quadrant = 3.14 sq. feet (approximately)

Question 12.
Dhamu fixes a square tile of 30 cm on the floor. The tile has a sector design on it as shown in the figure. Find the area of the sector, (π = 3.14).

Solution:
Side of the square = 30 cm
∴ Radius of the sector design = 30 cm
Area of the quadrant = $$\frac{1}{4}$$ πr2 sq. units
= $$\frac{1}{4}$$ × 3.14 × 30 × 30 cm2
= 3.14 × 15 × 15 cm2
∴ Area of the sector design = 706.5 cm2 (approximately)

Question 13.
A circle is formed with 8 equal granite stones as shown in the figure each of radius 56 cm and whose central angle is 45°. Find the area of each of the granite. (π = $$\frac{22}{7}$$)

Solution:
Number of equal sectors ‘n’ = 8
Radius of the sector ‘r’ = 56 cm
Area of the each sector = $$\frac{1}{n}$$ πr2 sq. units
= $$\frac{1}{8} \times \frac{22}{7}$$ × 56 × 56 cm2 = 1232 cm2
Area of each sector = 1232 cm2 (approximately)


text:
# Counting Queen's paths on a rectangular chessboard

Given two positive integers $$\m,n\$$, imagine a chessboard of size $$\m \times n\$$. A chess queen is on the upper-left corner. In how many ways can it reach the lower-right corner, by moving only right, down, or diagonally right-down (possibly moving many steps at once, because it's a queen)?

The resulting 2D sequence is A132439.

Standard rules apply. The shortest code in bytes wins.

## Test cases

m n result
1 1      1
1 2      1
1 3      2
1 4      4
1 5      8
2 2      3
2 3      7
2 4     17
2 5     40
3 3     22
3 4     60
3 5    158
4 4    188
4 5    543
5 5   1712

1!1=1;m!n=sum[m!(n-x)+(m-x)!n+(m-x)!(n-x)|x<-[1..m+n]]

Try it online!

Initially a boring port of the comment on OEIS

-27 from some tricks off dingledooper's answer

-10 from Christian Sievers

• I think you can replace max m n with m+n – Christian Sievers Apr 13 '20 at 8:45
• @ChristianSievers seems like it, and with that the m*n>0 check was also unnecessary, ty! – clapp Apr 13 '20 at 9:02

# Python 2, 75 bytes

@xnor's idea which is 2 bytes smaller than the one below. Returns True for 1 1.

f=lambda m,n:sum(f(m-i,n)+f(m,n-i)+f(m-i,n-i)for i in range(1,n|m))or m>0<n

Try it online!

# Python 2, 79 77 bytes

f=lambda m,n:(m==n==1)+sum(f(m-i,n)+f(m,n-i)+f(m-i,n-i)for i in range(1,n+m))

Try it online!

• I think the 77-byter is surely correct. The correctness is clear and it will end in finite time because n+m is decreasing. – newbie Apr 13 '20 at 3:59
• Yes, you're definitely right. It probably just takes longer for larger inputs. – dingledooper Apr 13 '20 at 17:52
• Here's a way to shave two bytes off the base case for the first one, with True for 1: Try it online! – xnor Apr 13 '20 at 21:36
• @xnor Clever idea; thanks for the improvement! – dingledooper Apr 13 '20 at 21:48

# K (ngn/k), 33 bytes

{(x~1 1)+/o'x-/:{|/~x,-/x}#1_+!x}

Try it online!

!x odometer

+ transpose

1_ drop the first

{ }# filter

|/~x,-/x either element or their difference is 0

x-/: subtract each right

o' recur each

+/ sum

(x~1 1) does x match 1 1? use as initial value for the sum

# Erlang (escript), 92 bytes

f(1,1)->1;f(M,N)->lists:sum([f(M,N-X)+f(M-X,N)+f(M-X,N-X)||X<-lists:seq(1,max(M,N)),M*N>0]).

Try it online!

# Charcoal, 40 bytes

ＮθＦＮＦθ⊞υ∨¬υΣΦυ∨∨⁼ι÷μθ⁼κ﹪μθ⁼⁻ικ⁻÷μθ﹪μθＩ⊟υ

Try it online! Link is to verbose version of code. Explanation:

Ｎθ

Input the number of columns.

ＦＮ

Loop over the rows.

Ｆθ

Loop over each column.

⊞υ∨¬υ

The first result is always 1...

ΣΦυ∨∨⁼ι÷μθ⁼κ﹪μθ⁼⁻ικ⁻÷μθ﹪μθ

... otherwise find the existing results that are a Queen's move away, take the total, and push that to the list of results. The list therefore simulates a matrix, requiring its index to be divided or taken modulo the number of columns as appropriate to compare with the loop variables.

Ｉ⊟υ

Output the last result calculated.

# J, 60 50 bytes

_2{(],1#.]#~(#:#)(0=*/*-/)@:-"1(#:i.@#))^:(*/@[)&1

Try it online!


text:
# Basic trig question for physics

I have to calculate the phase constant for a mass and spring system for simple harmonic motion. If I find that phi equals the arctan of some negative over another negative number, do I have to add pi to the answer for it to be in quadrant III?

I know this is more of a precalculus question but I don't know if the rules are different for physics.

yes, you have to add $$\pi$$.


text:
Displacement with magnitude of deceleration

1. Sep 25, 2007

Jacque77

You are driving your car and the traffic light ahead turns red. You apply the brakes for 3.59 s, and the velocity of the car decreases to +4.99 m/s. If the car's deceleration has a magnitude of 2.53 m/s2, what is the car's displacement during this time?

I know Δd=Vi(Δt)+1/2at2. I am not sure how to handle the deceleration magnitude and if this is the best formula to use.

2. Sep 25, 2007

learningphysics

You can use this formula (for some reason it is less commonly mentioned or used):

d = vf*t - (1/2)at^2

where vf is the final velocity. since it is decelerating a = -2.53m/s^2

or you can you use formula with vi... so first get vi using

vf = vi + at

3. Sep 25, 2007

Jacque77

so if I understand you properly:
d= 4.99m/s * 3.59s - (.5)(-2.53m/s^2)(3.59)^2

I get 17.914--16.30 = 34.21

Am I using the deceleration properly? I haven't had physics in 30+ yrs

4. Sep 25, 2007

learningphysics

Yup. That's right.

Also try calculating vi and then using

d = vi*t + 0.5at^2

you should get the same result that way also. that's a good way to double check your answer.

5. Sep 25, 2007

Jacque77

I don't understand how to use the second formula with the numbers given - the Vf=vi+at.

6. Sep 25, 2007

learningphysics

you can get vi from that equation:

4.99 = vi + (-2.53)(3.59)

vi = 14.0727

then you can use d = vi*t + (1/2)at^2

7. Sep 25, 2007

learningphysics

Then using

d = vi*t + (1/2)at^2 = 14.0727*3.59 + (1/2)(-2.53)(3.59)^2 = 34.2175m

So it's the same answer (only a slight difference due to rounding)

The best way to solve this problem, is using d = vf*t - (1/2)at^2 as you did... but for some strange reason this formula isn't mentioned in physics texts very often... that's why I mentioned this other method also...

8. Sep 25, 2007

Jacque77

OMG! It worked!! (14.0727)(3.59)+(.5)(-2.53)(3.59)^2 is
50.52-16.30 is 34.21!!!!!! yipee.


text:
Definition:Left Normal Element of Relation

Definition

Let $A$ be a class.

Let $\RR$ be a relation on $A$.

An element $x$ of $A$ is left normal with respect to $\RR$ if and only if:

$\forall y \in A: \map \RR {x, y}$ holds.


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# Why are allenes chiral?

How is 1-bromo-3-chloropropa-1,2-diene ($\ce{BrHC=C=CHCl}$) chiral? It doesn't have any carbon that is connected to four different groups.

• You probably will never realize how useful your questions were for students in the future. Aug 9, 2021 at 17:28

Asked and answered, but I think one thing that's missing is that allenes are not planar like alkenes or alkynes are. You can refer to this question for an explanation.

The dihedral angle between the two halogens is 90 degrees (ideally). Here's an animation hopefully providing a better view of the 3D structure:

This results in the two mirror images being non-superimposable upon each other, which means the molecule satisfies the criterion for chirality. If you rotate the right-hand molecule by $90^\circ$ such that the blue and white atoms at the top coincide, the green and white atoms will not match each other.

In general, any allene with the formula $\ce{R^1R^2C=C=CR^3R^4}$ will be chiral as long as $\ce{R^1}\neq\ce{R^2}$ and $\ce{R^3}\neq\ce{R^4}$. In this case we have $\ce{H} \neq \ce{Br}$ and $\ce{H} \neq \ce{Cl}$, so the compound in the question is chiral.

If $\ce{R^1} = \ce{R^2}$ or $\ce{R^3} = \ce{R^4}$, then the compound will possess a mirror plane and will therefore be achiral.

Chirality is a property of objects in which they lack certain symmetry operations, specifically improper rotations, including the mirror plane and inversion operations. For example, 3-dimensional chiral objects lack mirror symmetry. According to Wikipedia:

The feature that is most often the cause of chirality in molecules is the presence of an asymmetric carbon atom.

This is not the only cause of chirality. Any molecule of low enough symmetry to lack a mirror plane is chiral.

Similarly, stereogenic centers need not be carbon atoms (or any atoms), they just need to be a point in the molecule's space which generates chirality for the molecule. The central carbon atom in low symmetry allenes serves this purpose. This is also an example of axial chirality, which is chirality about an axis (the $\ce{C=C=C}$ axis) instead of about a point (which is point chirality — the more familiar kind).

To expand on the already touched points: The central carbon of an allene is sp-hybridized meaning it has 2 half-empty p orbitals for pi-bonding purposes. Regardless of which 2 we choose to leave half-empty (px, py, or pz) they are going to be perpendicular to each other, giving rise to the non-planar allene.


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# Show that affine varieties are quasi-compact.

A topological space $X$ is called quasi-compact if whenever ${(U_i)}_{i∈S}$ are a family of open subsets such that $∪_{i∈S}(U_i) = X$ then there are a finite number of $(U_i)$’s which actually cover X.

I'm thinking first we show that if $(U_{fi})$, $i ∈ S$ (where ${f_i}$, $i ∈ S$ a collection of elements of $R[X]$) is a family of principal open subsets which cover an affine variety $X$, then there is a finite number which cover $X$.
Then we use the fact that principal open subsets are a basis for the Zariski topology and above statement to show that a finite number of the $(U_j)$ are sufficient to cover $X$.

• What is the definition of affine variety you are using? – user363120 Mar 7 '17 at 20:27
• Is an affine variety the set of zeros in $\Bbb A^n$ of a polynomial, or is it a reduced scheme of finite type? – Alex Mathers Mar 7 '17 at 20:45
• Its the set of zeros – user286826 Mar 7 '17 at 20:49
• You can see also here! – Armando j18eos Mar 8 '17 at 9:17

Every descending chain of Zariski closed subsets in $X$, $$X_1 \supseteq X_2,...$$

corresponds by inclusion reversal to $$I(X_1) \subseteq I(X_2)...$$

However, we know that Polynomial rings are Noetherian, so the ascending chain of ideals stabilizes.

Hence, every descending chain of closed sets does as well. By taking complements, we see that every ascending chain of open subsets stabilizes.

Let $\{B_i\}$ be some open covering of $X$. Consider the open sets that can be written as the finite union of elements in our open cover:$$\mathcal{A}:=\{ U \mid U=\bigcup_{i=1}^{k}B_i\}.$$

By the axiom of dependent choice, we know that if $\mathcal{A}$ does not contain $X$, then there exists some infinite ascending chain of open sets, which is a contradiction.

Affine schemes are Noetherian as topological spaces, meaning every descending chain of closed subspaces $Z_1\supseteq Z_2\supseteq\cdots$ stabilizes. This is the same as saying that each increasing chain of open subspaces $U_1\subseteq U_2\subseteq\cdots$ stabilizes.

Now try to show that if $X$ is not quasicompact, you can get an infinite strictly increasing chain of open subsets, giving a contradiction.


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About neutral kaons and similar systems; from quantum field theory to effective mass matrices

Comment: 29 pages, LaTeX, 1 postscript logo

Higgs Pair-Production in the Standard Model at Next Generation Linear $e^+e^-$ Colliders

Comment: 8 pages, 4 figures

Support of Evidence for Neutrinoless Double Beta Decay

Comment: 15 pages, latex2e, 13 figures

R-parity violation effect on the top-quark pair production at linear colliders

Comment: Accepted by Phys.Rev.D

Three heavy jet events at hadron colliders as a sensitive probe of the Higgs sector

Comment: Revtex, 11 pages, 4 figures embedded in the text. Main changes with respect to Version 1: Numerical results re-calculated using the CTEQ5L pdf, improved discussion on the experimental consequences, new references added. Conclusions remain unchanged. As will appear in Phys. Rev. D

The Lorentz and CPT violating effects on H^0\to f^+ f^- (ZZ, W^+ W^-) decays

Comment: 10 Pages, 3 Figures

Chargino Contributions in $B \to \phi K_S$ Asymmetry

CP asymmetry in $B \to \phi K_S$ decay is studied in a special context of supersymmetry theories, in which the charginos play an important role. We find that in addition to the gluino, chargino can also make large contributions to CP asymmetry in $B \to \phi K_S$ decay. After considering the constraints from $B \to J/psi K_S$ decay, we study three special scenarios: (a). Large mixing on left-handed charm and top squarks (LL mixing); (b). Large mixing on right-handed charm and top squarks (RR mixing); (c). Large mixing on left-handed charm and top squarks plus right-handed charm and top squarks (LL + RR mixing). We show quantitatively that because of large squark mixing within second and third generations, an $\mathcal O$(1) effect on CP violation in $B \to \phi K_S$ is possible.

Mass spectrum of heavy hybrid mesons in the QCD string model

Comment: withdrawn

Upper Bound On Gluino Mass From Thermal Leptogenesis

Comment: 11 pages, 3 figures. Minor corrections, new references added, to appear in PLB

Where are the missing members of the baryon antidecuplet?

Comment: 5 pages, 1 figure. Missing reference added


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# Number System: Learn definition, types like Binary, Octal, Decimal, Hexadecimal and conversion formula here!

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A number system is a way of expressing numbers through writing. The number system is a mathematical language for consistently representing numbers in a particular set using digits or other symbols. It represents the arithmetic and algebraic structure of the figures and gives a unique representation for each number.

Numbers are used for counting/ checking/labelling objects and performing various arithmetic calculations. Examples of numbers in mathematics are natural numbers, whole numbers, rational and irrational numbers, prime and composite numbers, etc.

In this article, we will learn about the different types of number systems like binary, octal, decimal, hexadecimal, and their interconversion with solved examples.

## What is a Number System?

The number system can be understood as a system of writing or expressing numbers in different forms and vice versa. For example binary, decimal, etc. The Hindu–Arabic numeral system is the most widely used, allowing any number to be represented using a combination of 10 fundamental numeric symbols known as digits. It is used in linear algebra, calculus, general mathematics, physics, etc.

## History of Numbers and Number Systems

The Egyptians were the first to establish a ciphered number system, and the Greeks were the next to map their counting numerals onto the Ionian and Doric alphabets. Roman numerals, which used combinations of letters from the Roman alphabet, were popular in Europe until the late 14th century when the superior Hindu–Arabic numeral system expanded, and the Hindu–Arabic numeral system is still the most widely used system for representing numbers today.

## Number System Definition

A number system presents a unique representation of numbers. It also allows users to perform arithmetic operations like addition, subtraction, and division which play an essential role in computer applications and digital domains. Learn about the basics of the number system, conversion, and implementation.

## Types of Number Systems

The four common types of Number systems are

• Decimal Number System: Has base 10.
• Binary Number System: Has base 2.
• Octal Number System: Has base 8.
• Hexadecimal Number System: Has base 16.

### Decimal Number System

The decimal number system, which we use every day, uses numbers 0 through 9, i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Because there are 10 numbers in this number system, the decimal number system’s base number is 10. Any number that is represented without a base has a base of 10. Each value has a power of ten place value, meaning that the digit in the tens place is 10 times bigger than the digit in the one’s place. When we get to 9, we make the rightmost number 0 and add 1 to the left, resulting in 10. We continue to move ahead and so on.

For example, the decimal number 8379 consists of the digit 9 in the unit’s or one’s position, 7 in the tens position, 3 in the hundreds position, and 8 in the thousands position. Its value can be formulated as

$$(8\times1000)+ (3\times100)+ (7\times10)+ (9\times1)$$

$$(8\times10^3)+ (3\times10^2)+ (7\times10^1)+ (9\times10^0)$$

$$8000 + 300 + 70 + 9$$

$$8379$$

### Binary Number System

The binary number system is a Base-2 numbering system that follows the same mathematical rules as the widely used decimal or base-10 system. Instead of using powers of ten ($$10^n$$), such as 1, 10, 100, 1000, binary numbers use powers of two ($$2^n$$), effectively doubling the value of each succeeding bit as it advances, such as 1, 2, 4, 8, 16, 32, and so on.

The binary number system is used by computers to modify and store all of their data, including numbers, words, films, photos, and music. 1 + 1 = 10 since the number “2” does not exist in this system. In 1854, British mathematician George Boole published important work outlining an algebraic logic system based on the binary system, which became known as Boolean algebra.

### Octal Number System

The Octal Number System is a number system with an eight-digit base with digits ranging from 0 to 7 i.e., 0, 1, 2, 3, 4, 5, 6, and 7. The term octal is used to describe numbers with an eight-digit base. Octal numbers have a wide range of uses and significance, including in computers and digital numbering systems.

Octal numbers have a wide range of applications and significance, including use in computers and digital numbering systems. When queried by ground radar, aeroplane transponders broadcast a “squawk” code, which is expressed as a four-octal-digit number. On the radar screen, this code is used to identify different aircraft.

Learn the various concepts of the Binomial Theorem here.

### Hexadecimal Number System

Hexadecimal is made up of two words: ‘Hexa’ and ‘deci’, where ‘Hexa’ means six and ‘deci’ indicates ten. The hexadecimal number system is a 16-digit representation of integers 0 through 9 and letters A through F. To put it another way, the first 9 digits or numerals are represented as numbers, while the next 6 digits are represented as symbols from A to F. The hexadecimal number system is extremely similar to the decimal number system, which has a base of nine. After nine numbers, the tenth digit is represented by a symbol: 10 as A, 11 as B, 12 as C, 13 as D, 14 as E, and 15 as F.

The Base16 transfer encoding uses the hexadecimal system to divide each byte of plaintext into two 4-bit values and two hexadecimal digits. Because they give a human-friendly representation of binary-coded data, hexadecimal numbers are widely employed by software engineers and system designers. Four bits (binary digits) are represented by each hexadecimal digit, also known as a nibble (or nybble).

### Number System Chart

Here’s a handy table that shows the values of various numbers in the four number systems.

 Decimal Binary Octal Hexa 1 1 1 1 2 10 2 0 3 11 3 2 4 100 4 3 5 101 5 4 6 110 6 5 7 111 7 6 8 1000 10 7 9 1001 11 8 10 1010 12 9 11 1011 13 A 12 1100 14 B 13 1101 15 C 14 1110 16 D 15 1111 17 E 16 10000 20 F 17 10001 21 10 18 10010 22 11 19 10011 23 12 20 10100 24 13

### Summary

Here’s the summary of what we just learned about the four number systems

 S.No. Number System with Description 1 Binary Number SystemThe base is 2. Digits used : 0, 1 2 Octal Number SystemThe base is 8. Digits used : 0 to 7 3 Hexadecimal Number SystemThe Base is 16. Digits used: 0 to 9, Letters used: A- F

With a clear knowledge of the numbers system, let’s proceed towards the conversion of one form to another.

## Conversion in Number System

The number systems decimal, binary, octal, and hexadecimal are all positional value number systems. To convert a binary, octal, or hexadecimal number to a decimal number, just multiply each digit by its positional value. We’ll learn how to convert between various number systems in this section.

### Decimal to Binary

By repeatedly dividing a number by two and recording the result, decimal values can be transformed into binary. Steps to be followed:

#### Conversion of Integer Decimal Numbers to Binary

1. Divide the number by 2.
2. Get the integer quotient for the next iteration.
3. Get the remainder for the binary digit.
4. Repeat the above steps until the quotient is equal to 0.

Solved Example

Let’s convert 29 to Binary

We repeatedly divide 29 by 2 till we get 0 reminders.

 Dividend Divisor Reminder 2 29 1 2 14 0 2 7 1 2 3 1 2 1 1 2 0 —

We read the reminder in reverse order. So $$29_{10} = 10111_{2}$$

Next is the conversion of the fraction part.

Conversion of Fractional Decimal Numbers to Binary

A fraction is made of a whole part and a decimal part. For the conversion of the whole part to binary follow the same steps as mentioned above. For the fractional part, follow the steps below:

1. Multiply two to the fractional part of the decimal number.
2. The whole part of the product is our binary equivalent.
3. Step 1 is repeated using only the fractional part of the decimal value, followed by step 2.
4. The binary number is found for some significant digits.

Also, read about Relations and Functions here.

Solved Example

Let’s convert 29.188 in Binary to 5 significant digits. We have already converted 29 to binary above. Let’s work on the fractional part.

 — Binary 0.188 x 2 = 0.376 0 0.376 x 2 = 0.752 0 0.752 x 2 = 1.504 1 0.504 x 2 = 1.008 1 0.008 x 2 = 0.016 0

We have reached 5 significant digits. So we stop here. So $$43.188_{10} = 101011.00110_{2}$$

### Decimal to Octal

By repeatedly dividing a number by 8 and recording the result, decimal values can be transformed into octal. Steps to be followed:

Conversion of Integer Decimal Numbers to Octal

1. Divide the number by 2.
2. Get the integer quotient for the next iteration.
3. Get the remainder for the octal digit.
4. Repeat the above steps until the quotient is equal to 0.

Solved Example

 — Quotient Reminder 43/8 5 3 5/8 0 5

We read the reminder in reverse order. So $$43_{10} = 53_{8}$$

Next is the conversion of the fraction part.

Note: The dividend (here given decimal number) is the number to be divided, the divisor (here base of octal, i.e., 8) is the number to be divided by, and the quotient (remaining divided decimal number) is the outcome of the division.

Conversion of Fractional Decimal Numbers to Octal

If the decimal fractional component is M, multiply it by 8 because the octal number system’s base is 8. For the fractional part, follow the steps below:

1. Multiply 8 to the fractional part of the decimal number.
2. The whole part of the product is our octal equivalent.
3. Step 1 is repeated using only the fractional part of the decimal value, followed by step 2.
4. The octal number is found for some significant digits.

Solved Example

Let’s convert 43.188 in Octal to 5 significant digits. We have already converted 43 to octal above. Let’s work on the fractional part.

 — Octal 0.188 x 8 = 1.504 1 0.504 x 8 = 4.032 4 0.32 x 8 = 2.56 2 0.56 x 8 = 4.48 4 0.48 x 8 = 3.84 3

We have reached 5 significant digits. So we stop here. So $$43_{10} = 53.14243_{8}$$

### Decimal to Hexadecimal

By repeatedly dividing a number by 16 and recording the result, decimal values can be transformed into hexadecimal. Steps to be followed:

Conversion of Integer Decimal Numbers to Hexadecimal

1. Divide the number by 16.
2. Get the integer quotient for the next iteration.
3. Get the remainder for the hexadecimal digit.
4. Repeat the above steps until the quotient is equal to 0.

Solved Example

Let’s convert 765 to Hexadecimal

We repeatedly divide 765 by 16 till we get 0 reminders.

 — — Reminder 16 765 E 16 56 8 16 3 3 16 0 —

We read the reminder in reverse order. So $$765_{10} = 38E_{2}$$

Next is the conversion of the fraction part.

Conversion of Fractional Decimal Numbers to Hexadecimal

If the decimal fractional component is M, multiply it by 16 because the hexadecimal number system’s base is 16. For the fractional part, follow the steps below:

1. Multiply 16 to the fractional part of the decimal number.
2. The whole part of the product is our hexadecimal equivalent.
3. Step 1 is repeated using only the fractional part of the decimal value, followed by step 2.
4. The hexadecimal number is found for some significant digits.

Solved Example

Let’s convert 765.565 in Hexadecimal to 5 significant digits. We have already converted 765 to Hexadecimal above. Let’s work on the fractional part.

 — Hexadecimal 0.565 x 16 = 9.04 9 0.04 x 16 = 0.64 0 0.64 x 16 = 10.24 10 = A 0.24 x 16 = 3.84 3 0.84 x 16 = 13.44 13 = D

We have reached 5 significant digits. So we stop here. So $$765.565_{10} = 90A3D_{2}$$

Know more about Sequences and Series here.

### Binary to Decimal Conversion

To convert a binary number to a decimal number, follow the steps below:

1. Count the number of digits in the binary number.
2. Starting from the left start multiply the first digit of the binary number by 2 raised to the power of the number of digits.
3. Decrease the exponent as you move to the left.
4. Add all the digits together.

Here’s the general formula to convert a binary number to a decimal number.

decimal=$$b_n\times2^n+b_{n-1}\times2^{n-1}+b_{n-2}\times2^{n-2}+…$$

Solved Example

Let’s convert $$111001_2$$ to decimal.

$$111001_2 = 1\times2^5+1\times2^4+1\times2^3+0\times2^2+0\times2^1+1\times2^0 = 5710$$

### Binary to Hexadecimal Conversion

There is no direct method to convert Binary to Hexadecimal. We first convert the binary number to decimal and then convert the decimal number to binary by using the steps we learned above.

The second method to convert Binary to Hexadecimal is by grouping the numbers in a series of 4 from left to right and using the conversion table to convert them to hexadecimal.

Solved Example

Let’s convert $$(11111111)_2$$ to hexadecimal.

$$= (11111111)_2$$

$$= (1111)(1111)$$

$$= (15)(15)$$

$$= (ff)_16$$

### Binary to Octal Conversion

Just like binary to hexadecimal conversion, there is no direct method to convert from binary to octal. We again rely on binary to decimal and decimal to octal. Both of these methods have been already discussed.

The second method is the grouping method. Much like we did in the case of hexadecimal we group the binary numbers in a series of three from left to right. Then we use the conversion chart to convert it into an octal number.

Solved Example

Let’s convert $$(101010101)_2$$ to octal

$$= (101010101)_2$$

$$= (101)(010)(101)$$

$$= (525)_8$$

### Octal to Decimal Conversion

To convert a number from octal to decimal, follow these steps:

1. Count the number of digits.
2. Starting from the left multiply the octal number by 8 raised to the power of the number of digits in the octal number.
3. Decrease the exponent as you move to the left.
4. Add all the digits together.

Here’s the general formula to convert an octal number to a decimal number.

decimal=$$o_n\times8^n+o_{n-1}\times8^{n-1}+o_{n-2}\times8^{n-2}+…$$

Solved Example

Let’s convert $$16512_8$$ to decimal.

$$16512_8 = 1\times8^4+6\times8^3+5\times8^2+1\times8^1+0\times8^0 = 7498$$

### Octal to Binary Conversion

It takes two steps to convert an octal number to a binary number. We must first convert the given octal number to its decimal counterpart, then convert the decimal to binary. We have already learnt about those processes.

### Octal to Hexadecimal Conversion

We can convert octal to hexadecimal by two indirect methods. The first method involves converting the octal number to a decimal number and then we convert the decimal number to hexadecimal. The second is to convert the octal number to binary and then convert it to hexadecimal. Since there is no direct method to convert octal to binary we avoid the second method. Both of these methods are discussed in the sections above.

### Hexadecimal to Decimal Conversion

To convert a number from hexadecimal to decimal, follow these steps:

1. Count the number of digits.
2. Convert the alphabets into equivalent numbers as shown in the Number System Table above.
3. Starting from the left multiply the hexadecimal number by 16 raised to the power of the number of digits in the hexadecimal number.
4. Decrease the exponent as you move to the left.
5. Add all the digits together.

Here’s the general formula to convert an octal number to a decimal number.

decimal=$$h_n\times16^n+h_{n-1}\times16^{n-1}+h_{n-2}\times16^{n-2}+…$$

Solved Example

Let’s convert $$A37E_{16}$$ to decimal.

$$A37E_{16} = A\times16^3+3\times16^2+7\times16^1+E\times16^0$$

$$A37E_{16} = 10\times16^3+3\times16^2+7\times16^1+14\times16^0$$

$$= 41854$$

### Hexadecimal to Binary Conversion

To convert the hexadecimal number into binary no direct method exists. We use the intermediate of the decimal number system by converting hexadecimal to decimal as shown above and then converting the decimal to binary as discussed in previous sections.

### Hexadecimal to Octal Conversion

We again do not have a direct method to convert from hexadecimal to octal and resolve to the use of decimal system by converting hexadecimal to decimal as shown above and then convert the decimal to octal as discussed in previous sections.

## Solved Examples on Number System

Here are some solved examples on number system to help you for practice.

Solved Example 1: Convert 43.01 to binary till 7 significant digits.

Solution: First lets consider the whole number part. We repeatedly divide 43 by 2 till we get 0 remainder.

 Dividend Divisor Reminder 2 43 1 2 21 0 2 10 1 2 5 0 2 2 1 2 1 1 2 0 —

Now for the fractional part of .01 we start multiplying it by 2.

 – Binary 0.1 x 2 = 0.2 0 0.2 x 2 = 0.4 0 0.4 x 2 = 0.8 0 0.8 x 2 = 1.6 1 0.6 x 2 = 1.2 1 0.2 x 2 = 0.4 0 0.4 x 2 = 0.8 0

So $$43.01_{10} = 101011.0001100_{2}$$

Solved Example 2: Convert 473.05 to binary till 5 significant digits.

Solution:

First let’s consider the whole number part. We repeatedly divide 473 by 8 till we get a quotient of 0.

 — Quotient Reminder 473/8 5 1 59/8 7 3 7/8 0 7

Now for the fractional part of .05 we start multiplying it by 8.

 — Octal 0.5 x 8 = 0.2 0 0.40 x 8 = 3.20 3 0.20 x 8 = 1.60 1 0.60 x 8 = 4.80 4 0.80 x 8= 6.4 6

So $$473.05_{10} = 731.03146_{2}$$

## Number System FAQs

Q.1 What are the 4 number systems?
Ans.1 The four common types of Number systems are.Decimal Number System: Has base 10.
Binary Number System: Has base 2.
Octal Number System: Has base 8.
Hexadecimal Number System: Has base 16.

Q.2 What is number system?
Ans.2 The number system can be understood as a system of writing or expressing numbers in different forms and vice versa. For example binary, decimal, etc. The Hindu–Arabic numeral system is the most widely used, allowing any number to be represented using a combination of 10 fundamental numeric symbols known as digits.

Q.3 How many number systems are there?
Ans.3 There are four number systems that we commonly use: Decimal (Base 10), Binary (Base 2), Hexadecimal (Base 16) and Octal (Base 8).

Q.4 What is the importance of numbers?
Ans.4 We all use numbers in our day-to-day lives. In school and at work, we utilise numbers for counting money, measurements, phone numbers, phone passwords, locks, reading, page numbers, and TV channels. Engineers utilise numbers in their calculations when creating structures and roadways.

Q.5 What number system do we use?
Ans.5 The decimal positional numeral system is the most widely used numeral system, with the decimal referring to the usage of ten symbols to form all numbers (0, 1, 2, 3, 4, 5, 6, 7, 8, 9).


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## anonymous 4 years ago -7+n/22=-1 Show steps? answers:-15 (its hard for me to show my work; cause i do it in my head.)

1. anonymous

n/22 = 6 n = 6*22 = 132

2. anonymous

3. anonymous

4. anonymous

it's not right

5. anonymous

correct me if im wrong.

6. anonymous

the answer should be 132 not -15

7. anonymous

thats not even an option on my paper.

8. anonymous

So the problem is wrong

9. Mertsj

$\frac{-7+n}{22}\times22=-1\times22$

10. anonymous

@smr_kml You're reading the question wrong. The question is as Mertsj has written it. The answer is correct.

11. Mertsj

$-7+n=-22$

12. anonymous

The steps are as follows: Multiply both sides by 22, then add seven to both sides.

13. anonymous

$-7+\frac{n}{22}=-1$ n/22 = -1 +7 n/22 = 6 n = 132

14. Mertsj

$-7+n+7=-22+7$

15. Mertsj

n=-15

16. anonymous

The takeaway lesson from this, @staceyyyy1, is to use parentheses.

17. anonymous

I got the same answer @JayDS

18. anonymous

Of course you got the same answer, smr, you're both solving the same incorrect problem.

19. anonymous

she has to put ( )

20. anonymous

Ok I'm sorry I didn't know that it's seperate

21. Mertsj

22. Mertsj

@smr_kml There is A RAT in sepARATe

23. anonymous

To me it was plenty clarified by the fact that she said the answer was -15. However, she should in the future use parentheses for clarity.

24. anonymous

right! I'm sorry again everyone!

25. anonymous

thanks everyone. and thanks especially to Mertsj and nbouscal(:


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# Revision history [back]

In the meantime I have learned that these things can be done with the aid of the GAP-package QPA. This also works in SAGE via letting GAP be the intermediator.

In the meantime I have learned that these things can be done for admissible quotients of path algebras with the aid of the GAP-package QPA. This also works in SAGE via letting GAP be the intermediator.

So, if we make the additional restriction that our algebra is commutative, it works as follows:

gap> LoadPackage("qpa");;
gap> k:=Rationals;
Rationals
gap> Q:=Quiver(1,[[1,1,"a"],[1,1,"b"]]);
<quiver with 1 vertices and 2 arrows>
gap> kQ:=PathAlgebra(k,Q);
<Rationals[<quiver with 1 vertices and 2 arrows>]>
gap> AssignGeneratorVariables(kQ);
\#I  Assigned the global variables [ v1, a, b ]
gap> rels:=[a^2,b^2, a*b-b*a];
[ (1)*a^2, (1)*b^2, (1)*a*b+(-1)*b*a ]
gap> A:=kQ/rels;
<Rationals[<quiver with 1 vertices and 2 arrows>]/
<two-sided ideal in <Rationals[<quiver with 1 vertices and 2 arrows>]>,
(3 generators)>>
true
<algebra of dimension 3 over Rationals>
Algebra( Rationals, [ [(1)*a], [(1)*b], [(1)*a*b] ] )
3
gap>

I have not yet a solution for the non-commutative case, sorry.

In the meantime I have learned that these things can be done for admissible quotients of path algebras with the aid of the GAP-package QPA. This also works in SAGE via letting GAP be the intermediator.

So, if we make the additional restriction that our algebra is commutative, it works as follows:

gap> LoadPackage("qpa");;
gap> k:=Rationals;
Rationals
gap> Q:=Quiver(1,[[1,1,"a"],[1,1,"b"]]);
<quiver with 1 vertices and 2 arrows>
gap> kQ:=PathAlgebra(k,Q);
<Rationals[<quiver with 1 vertices and 2 arrows>]>
gap> AssignGeneratorVariables(kQ);
\#I  Assigned the global variables [ v1, a, b ]
gap> rels:=[a^2,b^2, a*b-b*a];
[ (1)*a^2, (1)*b^2, (1)*a*b+(-1)*b*a ]
gap> A:=kQ/rels;
<Rationals[<quiver with 1 vertices and 2 arrows>]/
<two-sided ideal in <Rationals[<quiver with 1 vertices and 2 arrows>]>,
(3 generators)>>
true
<algebra of dimension 3 over Rationals>
Algebra( Rationals, [ [(1)*a], [(1)*b], [(1)*a*b] ] )
3
gap>

If we let $k$ be a finite field, then the command IdempotentsForDecomposition(A); works, but I don't have a computational solution for the case $k=\mathbb{Q}$. I have not yet a solution for the non-commutative case, sorry.case.

In the meantime I have learned that these things can be done for admissible quotients of path algebras with the aid of the GAP-package QPA. This also works in SAGE via letting GAP be the intermediator.

So, if we make the additional restriction that our algebra is commutative, it It works as follows:

gap> LoadPackage("qpa");;
gap> k:=Rationals;
Rationals
gap> Q:=Quiver(1,[[1,1,"a"],[1,1,"b"]]);
<quiver with 1 vertices and 2 arrows>
gap> kQ:=PathAlgebra(k,Q);
<Rationals[<quiver with 1 vertices and 2 arrows>]>
gap> AssignGeneratorVariables(kQ);
\#I  Assigned the global variables [ v1, a, b ]
gap> rels:=[a^2,b^2, a*b-b*a];
[ (1)*a^2, (1)*b^2, (1)*a*b+(-1)*b*a ]
gap> A:=kQ/rels;
<Rationals[<quiver with 1 vertices and 2 arrows>]/
<two-sided ideal in <Rationals[<quiver with 1 vertices and 2 arrows>]>,
(3 generators)>>
true
<algebra of dimension 3 over Rationals>
Algebra( Rationals, [ [(1)*a], [(1)*b], [(1)*a*b] ] )
3
gap>

If we let $k$ be a finite field, then the command IdempotentsForDecomposition(A); works, but I don't have a computational solution for the case $k=\mathbb{Q}$. I have not yet a solution for the non-commutative case.

 5 No.5 Revision slelievre 12586 ●11 ●122 ●250 http://carva.org/samue...

In the meantime I have learned that these things can be done for admissible quotients of path algebras with the aid of the GAP-package GAP package QPA. This also works in SAGE Sage via letting GAP be the intermediator.

It works as follows:

gap> LoadPackage("qpa");;
gap> k:=Rationals;
Rationals
gap> Q:=Quiver(1,[[1,1,"a"],[1,1,"b"]]);
<quiver with 1 vertices and 2 arrows>
gap> kQ:=PathAlgebra(k,Q);
<Rationals[<quiver with 1 vertices and 2 arrows>]>
gap> AssignGeneratorVariables(kQ);
\#I  Assigned the global variables [ v1, a, b ]
gap> rels:=[a^2,b^2, a*b-b*a];
[ (1)*a^2, (1)*b^2, (1)*a*b+(-1)*b*a ]
gap> A:=kQ/rels;
<Rationals[<quiver with 1 vertices and 2 arrows>]/
<two-sided ideal in <Rationals[<quiver with 1 vertices and 2 arrows>]>,
(3 generators)>>
true
<algebra of dimension 3 over Rationals>

If we let $k$ be a finite field, then the command IdempotentsForDecomposition(A); command
IdempotentsForDecomposition(A);

works, but I don't have a computational solution for the case $k=\mathbb{Q}$. $k=\mathbb{Q}$. I have do not yet have a solution for the non-commutative case.


text:
# ACP Seminar (Astronomy - Cosmology - Particle Physics)

Speaker: Serguey Todorov Petcov (SISSA/INFN, Kavli IPMU) Predictions for the Dirac CP Violation in the Lepton Sector Wed, Mar 12, 2014, 13:30 - 14:30 Seminar Room A 1161.pdf After the successful determination of the reactor neutrino mixing angle \mbox{$\theta_{13} \cong 0.16 \neq 0$}, a new feature suggested by the current neutrino oscillation data is a sizeable deviation of the atmospheric neutrino mixing angle $\theta_{23}$ from $\pi/4$. Using the fact that the neutrino mixing matrix $U = U^\dagger_{e}U_{\nu}$, where $U_{e}$ and $U_{\nu}$ result from the diagonalisation of the charged lepton and neutrino mass matrices, and assuming that $U_{\nu}$ has a i) bimaximal (BM), ii) tri-bimaximal (TBM) form, or else iii) corresponds to the conservation of the lepton charge $L' = L_e - L_\mu - L_{\tau}$ (LC), we investigate quantitatively what are the minimal forms of $U_e$, in terms of angles and phases it contains, that can provide the requisite corrections to $U_{\nu}$ so that $\theta_{13}$, $\theta_{23}$ and the solar neutrino mixing angle $\theta_{12}$ have values compatible with the current data. In the case of the standard'' ordering of the 12 and the 23 rotations in $U_e$, the Dirac CP violation phase $\delta$, present in the PMNS matrix $U$, is predicted to have a value in a narrow interval around i) $\delta \cong \pi$ in the BM (or LC) case, ii) $\delta \cong 3\pi/2$ or $\pi/2$ in the TBM case, the CP conserving values $\delta = 0, \pi, 2\pi$ being excluded in the TBM case at more than $4\sigma$.


text:
# Emergency stretch hook

Is there a way to automatically set parameters when emergency stretch is required?

Here is my situation. After I implemented the solution chosen for unbreakable spaces in Bible verse references, I get some overfull boxes due to verse references at the end of lines. To fix these overfull lines, I can allow hyphenation between the book name and the chapter, as an exception (I don't want to allow it unless it is really necessary).

I've made myself two macros to make this easier. For example, I'll use:

(\BRallowhypbch\ibibleverse{Ac}(4:13)\BRforbidhypbch).

To allow hyphenation between Actes and 4.13 in Actes 4.13 as an exception.

I'm wondering if, when LaTeX gets down to use \emergencystretch (because other methods of justifications have failed), there is a way to trigger a hook that would call my two macros before and after the problematic paragraph (or line if possible) to fix it.

-
Would you please make the question more clear? Do you want \BRallowhypbch activated when you set \emergencystretch? –  egreg Jun 14 '11 at 15:10
@egreg: I fixed the explanation a bit... The question is whether, when LaTeX has to employ some "tougher" justification method, it is possible to trigger hooks to "help" it. –  ℝaphink Jun 14 '11 at 15:13
@Raphink Are you using microtype? –  Yiannis Lazarides Jun 14 '11 at 15:32
@Yiannis: yes, the beta version for XeTeX. –  ℝaphink Jun 14 '11 at 17:18
@Raphink normally this should help. Can you post a small image with a sample paragraph? I think it will be best to rather box and unbox the full paragraph, or otherwise use egreg's solution or something similar. TeX goe through a paragraph three times and then if everything fails and emergencystretch>0 kicks in the last round. To hook into it, you will need to catch the whole paragraph, check for badness, change parameters and retry. –  Yiannis Lazarides Jun 14 '11 at 18:49

I would define a particular environment:

\newenvironment{specialpar}[1]{\par#1}{\par}

This allows to enclose the problematic paragraph and do experiments with the various methods, for example

\begin{specialpar}{\emergencystretch=3em \BRallowhypbch}
Problematic paragraph ...
\end{specialpar}

The settings in the argument will be in force only for this paragraph; if the argument is left empty, there will be no particular setting and the paragraph will be typeset with the default values.

-


text:
# United States congressional apportionment

(Redirected from Congressional apportionment)

United States congressional apportionment is the process by which seats in the United States House of Representatives are redistributed amongst the 50 states following each constitutionally mandated decennial census. Each state is apportioned a number of seats which approximately corresponds to its share of the aggregate population of the 50 states.[1] However, every state is constitutionally guaranteed at least one seat.

## Reapportionment

The 435 seats of the House grouped by state (post-2010 Census reapportionment).
While citizens of each state are given nearly equal voting power through representation in the House, the situation in the Senate is significantly different where each state gets two votes no matter how large its population may be. A Wyoming voter, therefore, has roughly 66 times more representation in the Senate than a California voter. (Citizens in DC and territories have no voting power in Congress.)

The number of seats in the House of Representatives is currently set to 435, and has been since 1913, except for a temporary increase to 437 after the admissions of Alaska and Hawaii. Though the actual reapportionment will normally occur in respect of a decennial census, the law that governs the total number of representatives and the method of apportionment to be carried into force at that time can be created prior to the census.

The decennial apportionment also determines the size of each state's representation in the U.S. Electoral College—that is, any state's number of electors equals the size of its total congressional delegation (i.e., House seat(s) plus Senate seats).[2]

Federal law requires the Clerk of the House of Representatives to notify each state government of its entitled number of seats no later than January 25 of the year immediately following the census. After seats have been reapportioned, each state determines the boundaries of congressional districts—geographical areas within the state of approximately equal population—in a process called redistricting.[3] Any citizen of the State can challenge the constitutionality of the redistricting in their US district court.[4]

Because the deadline for the House Clerk to report the results does not occur until the following January, and the states need sufficient time to perform the redistricting, the decennial census does not affect the elections that are held during that same year. For example, the electoral college apportionment during 2000 presidential election was still based on the 1990 census results. Likewise, the congressional districts and the electoral college during the 2020 general elections will still be based on the 2010 census.

## Constitutional texts

The subject of Congressional apportionment is addressed twice in the U.S. Constitution. Initially, the apportionment of House seats was guided by two sentences in Article I, Section 2, clause 3:

Representatives and direct Taxes shall be apportioned among the several States which may be included within this Union, according to their respective Numbers, which shall be determined by adding to the whole Number of free Persons, including those bound to Service for a Term of Years, and excluding Indians not taxed, three fifths of all other Persons.

The Number of Representatives shall not exceed one for every thirty Thousand, but each State shall have at least one Representative;…

Following the end of the Civil War, the first of those provisions was superseded by Section 2 of the Fourteenth Amendment:

Representatives shall be apportioned among the several States according to their respective numbers, counting the whole number of persons in each State, excluding Indians not taxed,[5] But when the right to vote at any election for the choice of electors for President and Vice-President of the United States, Representatives in Congress, the Executive and Judicial officers of a State, or the members of the Legislature thereof, is denied to any of the male[6] inhabitants of such State, being twenty-one years of age,[7] and citizens of the United States, or in any way abridged, except for participation in rebellion, or other crime, the basis of representation therein shall be reduced in the proportion which the number of such male citizens shall bear to the whole number of male citizens twenty-one years of age in such State.

## House size

Ratio of representation in the House, 1789–1913
Years Source Constituents
per Rep.
1789 U.S. Const. ≥30,000
1793–1803 1790 Census 34,436
1803–1813 1800 Census 34,609
1813–1823 1810 Census 36,377
1823–1833 1820 Census 42,124
1833–1843 1830 Census 49,712
1843–1853 1840 Census 71,338
1853–1863 1850 Census 93,020
1863–1873 1860 Census 122,614
1873–1883 1870 Census 130,533
1883–1893 1880 Census 151,912
1893–1903 1890 Census 173,901
1903–1913 1900 Census 193,167
1913–1923 1910 Census 210,583
The U.S. population has increased more rapidly than the membership of the House of Representatives.

The size of the U.S. House of Representatives refers to total number of congressional districts (or seats) into which the land area of the United States proper has been divided. The number of voting representatives is currently set at 435. There are an additional five delegates to the House of Representatives. They represent the District of Columbia and the territories of American Samoa, Guam, the Northern Mariana Islands, which first elected a representative in 2008,[8] and the U.S. Virgin Islands. Puerto Rico also elects a resident commissioner every four years.

### Controversy and history

Since 1789, when the Federal Government began operating under the Constitution, the number of citizens per congressional district has risen from an average of 33,000 in 1790 to almost 700,000 as of 2008. Prior to the 20th century, the number of representatives increased every decade as more states joined the union, and the population increased. In 1911, Public Law 62-5 raised the membership of the U.S. House to 433 with a provision to add one permanent seat each upon the admissions of Arizona and New Mexico as states. As provided, membership increased to 435 in 1912.

In 1921, Congress failed to reapportion the House membership as required by the United States Constitution. This failure to reapportion may have been politically motivated, as the newly elected Republican majority may have feared the effect such a reapportionment would have on their future electoral prospects.[9][10] Then in 1929 Congress (Republican control of both houses of congress and the presidency) passed the Reapportionment Act of 1929 which capped the size of the House at 435 (the then current number). This cap has remained unchanged for more than eight decades. Three states – Wyoming, Vermont, and North Dakota – have populations smaller than the average for a single district.

The "ideal" number of members has been a contentious issue since the country's founding. George Washington agreed that the original representation proposed during the Constitutional Convention (one representative for every 40,000) was inadequate and supported an alteration to reduce that number to 30,000.[11] This was the only time that Washington pronounced an opinion on any of the actual issues debated during the entire convention.[12]

In Federalist No. 55, James Madison argued that the size of the House of Representatives has to balance the ability of the body to legislate with the need for legislators to have a relationship close enough to the people to understand their local circumstances, that such representatives' social class be low enough to sympathize with the feelings of the mass of the people, and that their power be diluted enough to limit their abuse of the public trust and interests.

"... first, that so small a number of representatives will be an unsafe depositary of the public interests; secondly, that they will not possess a proper knowledge of the local circumstances of their numerous constituents; thirdly, that they will be taken from that class of citizens which will sympathize least with the feelings of the mass of the people, and be most likely to aim at a permanent elevation of the few on the depression of the many;..."[13]

Madison also addressed Anti-Federalist claims that the representation would be inadequate, arguing that the major inadequacies are of minimal inconvenience since these will be cured rather quickly by virtue of decennial reapportionment. He noted, however,

"I take for granted here what I shall, in answering the fourth objection, hereinafter show, that the number of representatives will be augmented from time to time in the manner provided by the Constitution. On a contrary supposition, I should admit the objection to have very great weight indeed."

Madison argued against the assumption that more is better:

"Sixty or seventy men may be more properly trusted with a given degree of power than six or seven. But it does not follow that six or seven hundred would be proportionally a better depositary. And if we carry on the supposition to six or seven thousand, the whole reasoning ought to be reversed. ... In all very numerous assemblies, of whatever character composed, passion never fails to wrest the scepter from reason."[13]

#### Clemons v. Department of Commerce

A 2009 lawsuit, Clemons v. Department of Commerce, sought a court order for Congress to increase the size of the House's voting membership and then reapportion the seats in accordance with the population figures of the 2010 Census. The intent of the plaintiff was to rectify the disparity of congressional district population sizes among the states that result from the present method of apportionment. Upon reaching the U.S. Supreme Court in December 2010, the holdings of the lower district and appellate courts were vacated and the case remanded to the U.S. District Court from which the case originated with instructions that the district court dismiss the case for lack of jurisdiction.

### Proposed expansion

The first proposed amendment to the Constitution within the Bill of Rights attempted to set a pattern for growth of the House along with the population, but has not been ratified.

Article the first... After the first enumeration required by the first article of the Constitution, there shall be one Representative for every thirty thousand, until the number shall amount to one hundred, after which the proportion shall be so regulated by Congress, that there shall be not less than one hundred Representatives, nor less than one Representative for every forty thousand persons, until the number of Representatives shall amount to two hundred; after which the proportion shall be so regulated by Congress, that there shall not be less than two hundred Representatives, nor more than one Representative for every fifty thousand persons.[14]

The proposed Wyoming Rule calls for expanding the House until the standard Representative-to-population ratio equals that of the smallest entitled unit (currently the state of Wyoming). This proposal is primarily designed to address the fact that some House districts are currently nearly twice the size of others; for instance, there are nearly 1 million residents in Montana's single district, compared to about 570,000 in Wyoming's. See List of U.S. states by population. It should be noted that, while a larger House size will generally result in the smallest and largest districts being proportionally closer in size, this is not always the case. Therefore in some cases, the Wyoming Rule may actually result in an increase in the ratio of the sizes of the largest and smallest districts. For instance, after the 1990 Census and with a House size of 435, the largest district (Montana's At-large district) had 799,065 residents, 76% larger than the smallest district (Wyoming's At-large district with 453,588 residents). The Wyoming Rule would have given a House size of 547 in 1990. Using that size, the largest district (North Dakota's At-large district) would have had 638,800 residents, 92% larger than the smallest districts (Delaware's two districts at approximately 333,084 residents each), which is larger than the 76% figure mentioned above.

On May 21, 2001, Rep. Alcee Hastings sent a dear colleague letter pointing out that U.S. expansion of its legislature had not kept pace with other countries.[15]

In 2007, during the 110th Congress, Representative Tom Davis introduced a bill in the House of Representatives that would add two seats to the House, one for Utah and one for the District of Columbia. It was passed by the House, but was tripped up by procedural hurdles in Senate and withdrawn from consideration. An identical bill was reintroduced during the 111th Congress. In February 2009 the Senate adopted the measure 61-37. In April 2010, however, House leaders decided to shelve the proposal.[16]

## Apportionment methods

Apart from the requirement that the number of delegates for each state is at least one, a state's number of representatives is in principle proportional to population, thus assuring reasonably consistent representation to the people regardless of the state boundaries and populations. No method of calculating a fair distribution of voting power across the various states was known until recently and five distinct apportionment methods have been used since the adoption of the Constitution, none of them producing fully proportional distribution of power among the states.

The current method, known as the method of equal proportions, has been used since the 1940 Census. The Vinton or Hamilton method, used from 1850 to 1900, was susceptible to what is known as the apportionment paradox or Alabama paradox.[17]

In 2008, a so-called One-Person-One-Vote model was introduced by J. Toplak in Temple Law Review,[18] which distributes the states' power in the House of Representatives exactly 'according to their Numbers'. Under this system, however, members of the House of Representatives do not have equal voting power. The method would be constitutional since the U.S. Constitution does not require Congressmen to have equal voting powers but does require the voters to have votes of equal weight.[citation needed][19]

### The Method of Equal Proportions

Further information: Huntington–Hill method

The apportionment methodology currently used is the method of equal proportions,[20][21][22] so called because it guarantees that no additional transfer of a seat (from one state to another) will reduce the ratio between the numbers of persons per representative in any two states.[23] According to NationalAtlas.gov, the method of equal proportions minimizes the percentage differences in the populations of the congressional districts.[24]

In this method, as a first step, each of the 50 states is given its one guaranteed seat in the House of Representatives, leaving 385 seats to assign.

The remaining seats are allocated one at a time, to the state with the highest priority number. Thus, the 51st seat would go to the most populous state (currently California). The priority number is determined by a formula that is mathematically computed to be the ratio of the state population to the geometric mean of the number of seats it currently holds in the assignment process, n (initially 1), and the number of seats it would hold if the seat were assigned to it, n+1.

The formula for determining the priority of a state to be apportioned the next available seat defined by the method of equal proportions is

$A_{n} = \frac{P}{\sqrt{n(n+1)}}$

where P is the population of the state, and n is the number of seats it currently holds before the possible allocation of the next seat. An equivalent, recursive definition is

$A_{n+1} = \sqrt{\frac{n}{n+2}} \ A_{n}$

where n is still the number of seats the state has before allocation of the next, and for n = 1, the initial A1 is explicitly defined as

$A_{1} = \frac{P}{\sqrt{2}}$

Consider the reapportionment following the 2010 U.S. Census: beginning with all states initially being allocated one seat, the largest value of A1 corresponds to the largest state, California, which is allocated seat 51. After being allocated its 2nd seat, its priority value decreases to its A2 value, which is reordered to a position back in line. The 52nd seat goes to Texas, the 2nd largest state, because its A1 priority value is larger than the An of any other state. However, the 53rd seat goes back to California because its A2 priority value is larger than the An of any other state. The 54th seat goes to New York because its A1 priority value is larger than the An of any other state at this point. This process continues until all remaining seats are assigned. Each time a state is assigned a seat, n is incremented by 1, causing its priority value to be reduced and reordered among the states, whereupon another state normally rises to the top of the list.

The Census 2010 Ranking of Priority Values[25] shows the order in which seats 51–435 were apportioned after the 2010 Census, with additional listings for the next five priorities. Minnesota was allocated the final (435th) seat. North Carolina, which gained the 435th seat in the 2000 census,[26] missed its 14th seat by 15,754 residents as the 436th seat to be allocated.

## Past apportionments

Note: The first apportionment was established by the Constitution based on population estimates made by the Philadelphia Convention, and was not based on any census or enumeration.

## Changes following the 2010 census

On December 21, 2010 the U.S. Census Bureau released its official apportionment results for congressional representation. The changes were in effect for the U.S. elections in 2012.[27]

Gain four Gain two Gain one Lose one Lose two
Texas Florida Arizona
Georgia
South Carolina
Utah
Washington
Illinois
Iowa
Louisiana
Massachusetts
Michigan
Missouri
New Jersey
Pennsylvania
New York
Ohio

## Past increases

The size of the U.S. House has increased as follows:[28]

### 1789–1800

March 4, 1789 59 n/a Const. Art. I, § 2, cl. 3 Seats apportioned by the Constitution
November 21, 1789 64 5 North Carolina ratified the Constitution with the seats apportioned by the Constitution
May 29, 1790 65 1 Rhode Island ratified the Constitution with the seats apportioned by the Constitution
March 4, 1791 67 2 Stat. 191 Vermont admitted
June 1, 1792 69 2 Kentucky admitted
March 4, 1793 105 36 Apportionment Act of 1792, 1 Stat. 253 Apportionment of the First Census
June 1, 1796 106 1 Stat. 492 Tennessee admitted

### 1801–1820

March 1, 1803 107 1 Stat. 175 Ohio admitted.
March 4, 1803 142 35 Stat. 128 Apportionment of the Second Census.
April 30, 1812 143 1 Stat. 703 Louisiana admitted.
March 4, 1813 182 39 Stat. 669 Apportionment of the Third Census.
December 11, 1816 183 1 Stat. 290 Indiana admitted.
December 10, 1817 184 1 Stat. 349 Mississippi admitted.
December 3, 1818 185 1 Stat. 430 Illinois admitted.
December 14, 1819 186 1 Stat. 492 Alabama admitted.
March 15, 1820 Stat. 555 Maine admitted, 7 seats transferred from Massachusetts

### 1821–1840

August 10, 1821 187 1 Stat. 547 Missouri admitted
March 4, 1823 213 26 Stat. 651 Apportionment of the Fourth Census
March 4, 1833 240 27 Stat. 516 Apportionment of the Fifth Census
June 15, 1836 241 1 Stat. 51 Arkansas admitted
January 26, 1837 242 1 Stat. 50 Michigan admitted

### 1841–1853

March 4, 1843 223 19 Stat. 491 Apportionment of the Sixth Census, the only time the size of the House was reduced, except for the minor readjustment in 1963.
March 3, 1845 224 1 Stat. 743 Florida admitted.
December 29, 1845 226 2 Stat. 798 Texas annexed and admitted.
December 28, 1846 228 2 Stat. 743
Stat. 52
May 29, 1848 230 2 Stat. 58
Stat. 235
March 4, 1849 231 1 Stat. 235 Wisconsin given another seat.
September 9, 1850 233 2 Stat. 452 California admitted.
March 4, 1853 Stat. 432 Apportionment of the Seventh Census.
234 1 10 Stat. 25 Additional seat apportioned to California

The 1850 Apportionment bill provided a method to be used in future reapportionments, as well as establishing the then-current 233 as the number of seats to be apportioned after future censuses. Due to census returns being incomplete in California, an additional act provided that California retain the same representation it had when admitted, until a new census could be taken. California would otherwise have lost one seat, and so the total number of seats was increased by one to 234.

### 1854–1880

May 11, 1858 236 2 11 Stat. 166 Minnesota admitted.
February 14, 1859 237 1 11 Stat. 383 Oregon admitted.
January 29, 1861 238 1 11 Stat. 269 Kansas admitted
June 2, 1862 239 1 12 Stat. 411 California apportioned an extra seat
March 4, 1863 233 6 Stat. 432 Apportionment of the Eighth Census, in accordance with the 1850 act, which provided for an apportionment of 233 seats
241 8 12 Stat. 353 Supplemental apportionment of 8 seats, for an overall increase of 2 seats in the 38th Congress
June 20, 1863 12 Stat. 633 West Virginia admitted, three seats transferred from Virginia
March 4, 1873 283 40 17 Stat. 28 Apportionment of the Ninth Census, replacing the 1850 act
292 9 17 Stat. 192 Supplemental apportionment added one seat each for nine states

### 1881–1900

March 4, 1883 325 32 47th Congress, ch. 20, 22 Stat. 5 Apportionment of the Tenth Census.
November 2, 1889 328 3 North Dakota and South Dakota admitted. One seat goes to the former, two to the latter.
November 8, 1889 329 1 Montana admitted.
November 11, 1889 330 1 Washington admitted.
July 3, 1890 331 1 Idaho admitted.
July 10, 1890 332 1 Wyoming admitted.
March 4, 1893 356 24 Apportionment Act of 1891 Apportionment of the Eleventh Census.
January 4, 1896 357 1 Utah admitted.

### 1901–present

March 4, 1903 386 29 Apportionment Act of 1901, 31 Stat. 733 Apportionment of the Twelfth Census
November 16, 1907 391 5 Oklahoma Enabling Act Oklahoma admitted
January 6, 1912 393 2 Apportionment Act of 1911, Sec. 2; New Mexico Enabling Act New Mexico admitted
February 14, 1912 394 1 Apportionment Act of 1911, Sec. 2 Arizona admitted
March 4, 1913 435 41 Apportionment Act of 1911, Pub.L. 62–5, 37 Stat. 13 Apportionment of the Thirteenth Census
January 3, 1963 435 2 Reapportionment Act of 1929, ch. 28, 46 Stat. 26, 2 U.S.C. § 2a Apportionment of the Eighteenth Census
Note: The Reapportionment Act of 1929 stated that the "then existing number of Representatives" would be apportioned after each census, which would have dictated an apportionment of 437 seats, but the Alaska Statehood Act and Hawaii Admission Act explicitly stated that the new seats were temporary increases[29]

## Notes

• Delegate counts in italics represent temporary counts assigned by Congress until the next decennial census or by the U.S. Constitution in 1789 until the first U.S. Census.
• Elections held in the year of a census use the apportionment determined by the previous census.
1. ^ The populations of Washington, D.C. and federal territories are not included in this figure.
2. ^
3. ^
4. ^
5. ^ Rendered moot by the Revenue Act of 1924 and Indian Citizenship Act of 1924.
6. ^ Modified by section 1 of the Nineteenth amendment.
7. ^ Modified by section 1 of the Twenty-sixth amendment.
8. ^ Bush signs federalization bill, Agnes E. Donato, Saipan Tribune, May 10, 2008.
9. ^ "Fair Representation, Meeting The Ideal of One Man One vote" - Michel Balinski and H. Peyton Young -- Page 51
10. ^ [1]
11. ^ George Will Called Me An Idiot, Jonah Golderg, National Review, January 15, 2001.
12. ^ Madison's notes on the Constitutional Convention - Tuesday September 17, 1787
13. ^ a b The Federalist #55
14. ^ "Constitutional Amendments Not Ratified". United States House of Representatives. Archived from the original on 27 September 2007. Retrieved 2007-09-30.
15. ^ House of Representatives? Hardly., Alcee Hastings, May 21, 2001.
16. ^ Marimow, Ann E.; Pershing, Ben (April 21, 2010). "Congressional leaders shelve D.C. voting rights bill". The Washington Post.
17. ^ "Congressional Apportionment-Historical Perspective". U.S. Census Bureau. Retrieved 27 October 2013..
18. ^ Toplak, Jurij (2008). "Equal Voting Weight of All: Finally 'One Person, One Vote' from Hawaii to Maine?". Temple Law Review (Temple University) 81 (1): 123–176.
19. ^ Jurij Toplak (January 4, 2011). "Equal Voting Weight of All: Finally 'One Person, One Vote' from Hawaii to Maine?". pp. 158–175. Retrieved 2013-10-17.
20. ^ "The History of Apportionment in America". American Mathematical Society. Retrieved 2009-02-15.
21. ^ "2 USC §2a". Cornell University Law School, Legal Information Institute. Retrieved 2008-05-13.
22. ^ "Computing Apportionment". U.S. Census Bureau. Retrieved 2009-02-14.
23. ^ Edward V Huntington (1921). "The Mathematical Theory of the Apportionment of Representatives". Proceedings of the National Academy of Sciences U.S.A. 7 (4): 123–7. doi:10.1073/pnas.7.4.123. PMC 1084767. PMID 16576591.
24. ^ "Congressional Apportionment". NationalAtlas.gov. Retrieved 2009-02-14.
25. ^ "PRIORITY VALUES FOR 2010 CENSUS". U.S. Bureau of the Census. Retrieved 2012-06-07.
26. ^ "Census 2000 Ranking of Priority Values". U.S. Bureau of the Census. 2001-02-21. Retrieved 2008-05-13.
27. ^ "APPORTIONMENT POPULATION AND NUMBER OF REPRESENTATIVES, BY STATE: 2010 CENSUS". US Census. 2010-12-21. Retrieved 2013-02-23.
28. ^ The Size of the U. S. House of Representatives and its Constituent State Delegations, thirty-thousand.org.
29. ^ Both acts included the phrasing That such temporary increase in the membership shall not operate to either increase or decrease the permanent membership of the House of Representatives as prescribed in the Act of August 8, 1911 (37 Stat. 13) nor shall such temporary increase affect the basis of apportionment established by the Act of November 15, 1941 (55 Stat. 761; 2 U. S. C., sec. 2a), for the Eighty-third Congress and each Congress thereafter.


text:
mersenneforum.org > Math Estimating the number of prime factors a number has
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 2012-05-15, 18:28 #1 henryzz Just call me Henry     "David" Sep 2007 Liverpool (GMT/BST) 2·5·599 Posts Estimating the number of prime factors a number has Has anyone found a formula that gives the probability of a number with a given size having n prime factors? I haven't ever seen one and a google search didn't turn anything up.
 2012-05-15, 19:17 #2 mart_r     Dec 2008 you know...around... 76110 Posts Maybe this post is relevant to your question: http://www.mersenneforum.org/showthread.php?t=13737 Last fiddled with by mart_r on 2012-05-15 at 19:21
2012-05-15, 22:00   #3
CRGreathouse

Aug 2006

135338 Posts

Quote:
 Originally Posted by henryzz Has anyone found a formula that gives the probability of a number with a given size having n prime factors? I haven't ever seen one and a google search didn't turn anything up.
The Erdős-Kac theorem with continuity correction gives a decent estimate, but it's not very useful unless n is large.

 2012-05-16, 10:06 #4 henryzz Just call me Henry     "David" Sep 2007 Liverpool (GMT/BST) 10111011001102 Posts My aim with this information is to estimate how likely it is for an aliquot sequence with for example 2^4*3*c100 to lose/keep the 3 on the next iteration. $\pi_k(n)=\frac{n(\log\log n)^{k-1}}{(k-1)!\log n}$ should be good enough especially if I can estimate the error. Since I am using this for aliquot sequences I know certain factors do not divide the composite. In the case of 2^4*3*c100 this is 2 and 3(Note I could also do with doing things like 2^4*7 not just the first n primes. The exponent 4 isn't crucial as well.)
 2012-05-20, 21:38 #5 henryzz Just call me Henry     "David" Sep 2007 Liverpool (GMT/BST) 2×5×599 Posts Is there a way of removeing a factor from this formula? For example if I know that 2 is not a factor of the composite. How much difference should this make? I imagine with 2 it would make quite a bit of difference. With much larger numbers e. g. 31 it wouldn't make nearly so much difference.
 2012-05-22, 04:22 #6 CRGreathouse     Aug 2006 3·1,993 Posts Use a weighted sum of the appropriate pi_k. For example, if you wanted the number of odd 3-almost-primes up to x, that's pi_3(x) - pi_2(x/2).
2012-05-22, 09:16   #7
henryzz
Just call me Henry

"David"
Sep 2007
Liverpool (GMT/BST)

2·5·599 Posts

Quote:
 Originally Posted by CRGreathouse Use a weighted sum of the appropriate pi_k. For example, if you wanted the number of odd 3-almost-primes up to x, that's pi_3(x) - pi_2(x/2).
Brilliant thanks. Wierdly I actually thought of something similar while going to sleep last night.
Does this sort of start a chain where pi_2(x/2) needs correcting by subtracting from it pi_1(x/4) etc.?
This would lead to pi_k(x)-pi_(k-1)(x/2)+pi_(k-2)(x/4)-pi_(k-3)(x/8) ... which you would continue until you have the necessary precision.

2012-05-23, 01:13   #8
CRGreathouse

Aug 2006

597910 Posts

Quote:
 Originally Posted by henryzz Brilliant thanks. Wierdly I actually thought of something similar while going to sleep last night. Does this sort of start a chain where pi_2(x/2) needs correcting by subtracting from it pi_1(x/4) etc.? This would lead to pi_k(x)-pi_(k-1)(x/2)+pi_(k-2)(x/4)-pi_(k-3)(x/8) ... which you would continue until you have the necessary precision.
No, one subtraction is good enough for that problem. But for more complicated problems you might need a lot of terms, yes.

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# Magic rotated checkerboard function made in python!

Here's a puzzle for all you python programmers out there:

I defined a function that goes like this:

def checkers(num):
magic = [_____________________________________________________________]
print('\n'.join(magic + magic[num-2::-1]))

Here are the results from calling the function with different numbers:

checkers(1)

Output:

_|
_|

checkers(2)

Output:

  _|
_|_|_|
_|

checkers(3)

Output:

    _|
_|_|_|
_|_|_|_|_|
_|_|_|
_|

checkers(4)

Output:

      _|
_|_|_|
_|_|_|_|_|
_|_|_|_|_|_|_|
_|_|_|_|_|
_|_|_|
_|

checkers(5)

Output:

        _|
_|_|_|
_|_|_|_|_|
_|_|_|_|_|_|_|
_|_|_|_|_|_|_|_|_|
_|_|_|_|_|_|_|
_|_|_|_|_|
_|_|_|
_|

I hope you get the gist as to what the function does, but if not:
Whatever number we pass into the brackets, it will print out a checker board rotated 45 degrees, with the number we passed in as each of its dimensions.

For your challenge, find out how magic is defined.

You don't have to define it the same way I did, but just for a little hint on one way to define it, each underscore between the two square-brackets represents a character, with no unnecessary whitespaces (I do love PEP-8, don't get me wrong).

• Are the trailing whitespaces on each row significant? Oct 21 '20 at 1:34
• The 1x1 checkerboard usually has fewer than two squares though.
– Bass
Oct 21 '20 at 4:06
• @Bubbler The trailing spaces were conclusive, but I guess I accidentally removed some from the last one. I'll add them in. Oct 21 '20 at 15:32
• @Bass It is one square if you look at it sideways and squint really hard. Oct 21 '20 at 17:04

From a fellow abuser of listcomps, magic is

" "*(num-i-1)+"_|"*(2*i+1)+" "*(num-i-1)for i in range(num)
It's a list of the rows in the upper half.

The hint is

magic(1) not being square: this occurs because num-2 is -1 and so what would be all but the last row in reverse order becomes the last (and only) row

49 bytes, respecting the spaces on the right side:

('_|'*(2*i+1)).center(num*4-2)for i in range(num)
Try it online!

That said, it isn't a very interesting puzzle since string formatting tasks are common in regular programming, except the bit of inferring what the magic should evaluate to. I hope you come up with more interesting ones later :)