**Bayes' Theorem Examples**

### Example 1: Cavity and Toothache

Given historical data:
- P(cavity) = 0.1 (10%)
- P(toothache) = 0.05 (5%)
- P(Cavity | toothache) = 0.8 (80%)

We need to find P(toothache | cavity) using Bayes' law:
P(B|A) = {P(A|B)P(B)}/P(A)

Substituting the given values:
P(toothache | cavity) = {P(cavity | toothache) P(toothache) } / P(cavity)
= (0.8 * 0.05) / 0.1
= 0.4

### Example 2: HIV Test

Given:
- Prevalence of HIV in San Francisco = 3% (P(hiv+))
- False positive rate of the HIV test = 0.001 (P(t+|hiv-))
- False negative rate of the HIV test = 0.01 (P(t-|hiv+))

We need to find the probability that a random person who tests positive is actually infected, P(hiv+|t+), also known as the positive predictive value.

Using Bayes' theorem:
P(hiv+|t+) = P(t+|hiv+)P(hiv+) / P(t+)

First, calculate:
- P(t-|hiv-) = 1 - P(t+|hiv-) = 1 - 0.001 = 0.999
- P(t+|hiv+) = 1 - P(t-|hiv+) = 1 - 0.01 = 0.99

Then, find P(t+):
P(t+) = P(t+&hiv+) + P(t+&hiv-)
= P(t+|hiv+)P(hiv+) + P(t+|hiv-)P(hiv-)
= 0.99 * 0.03 + 0.001 * 0.97

Now, substitute these values into the equation for P(hiv+|t+):
P(hiv+|t+) = (.99 * 0.03) / {0.99 * 0.03 + 0.001 * 0.97}
= 0.968, or approximately 96.8%


Cross-tab analysis, also known as chi-square or contingency table analysis, is used to evaluate the association between categorical variables. In this case, we analyze data from a sample of 580 newspaper readers, focusing on two variables: income level (Low or High) and primary newspaper read (USA Today or Wall Street Journal).

We test the null hypothesis (H0) that there is no relationship between income level and newspaper choice, against the alternative hypothesis (Ha) that there is a relationship. To do this, we use the Pearson chi-squared test, which compares observed frequencies to expected frequencies calculated under the assumption of no association.

The expected frequencies are calculated as (Row total x Column Total) / Overall Total. The chi-square value for each cell is calculated as `(o - e)^2 / e`, where `o` is the observed frequency and `e` is the expected frequency. The overall chi-square value is the sum of these values.

To determine if the chi-square value is significant, we calculate the degrees of freedom (df) as `(# rows - 1) x (# columns - 1)`, which in this 2x2 table is 1. The output shows the chi-square statistic, df, and p-value. We reject the null hypothesis if the p-value is less than 0.05.

In this case, the p-value is very small (< 0.001), indicating a significant association between newspaper readership and income. We can also calculate the critical chi-squared value using a probability calculator, which is 3.841 for a 95% confidence level. Since our calculated chi-square value (187.783) is larger than this critical value, we reject the null hypothesis.

The standardized deviations, calculated as `(o-e)/sqrt(e)`, can be used to evaluate the hypothesis visually. A score greater than 1.96 in absolute value indicates a significant deviation from the model of independence. In this plot, all cells contribute to the association between income and readership, with all standardized deviations larger than 1.96 in absolute value.

The results suggest that there are fewer low-income respondents reading the Wall Street Journal and more high-income respondents reading the Wall Street Journal than expected. Conversely, there are more low-income respondents reading USA Today and fewer high-income respondents reading USA Today than expected.

It's essential to check the expected values, as small expected counts (less than 5) may require simulation methods to calculate the p-value. If some cells have an expected count below 1, it may be necessary to collapse rows and/or columns. In this case, all expected frequencies are larger than 5, so the p-value is unlikely to be biased.


### Computing the Set

To determine if a complex number **c = a + bi** belongs to the Mandelbrot set, there is no formula. However, there are conditions that can be applied:
- Exclusion condition: If **a^2 + b^2 ≥ 4**, then **c** belongs to the set.
- Inclusion condition: If there exists a **z** where **z_j = z_k** for **j ≠ k**, then **c** belongs to the set.

For other **c** values, the defining sequence can be computed until **z_n** satisfies **a_n^2 + b_n^2 ≥ 4**, indicating **c** does not belong to the set. If **c** is a member, this condition will never be met unless the inclusion condition is satisfied.

In practice, a ceiling **Max** is set for **n** (e.g., 256, 512, 100000), and **c** is considered to belong to the set if **a_n^2 + b_n^2 < 4** for all **n ≤ Max**. An approximation of the inclusion rule can be used by checking if the absolute difference between **z_n** and **z_(n-1)** is less than a predefined constant **z_small**.

### Drawing the Set

Drawing the Mandelbrot set on a medium with a fixed number of points (e.g., a computer screen with **x** horizontal and **y** vertical resolution) requires mapping complex plane coordinates to medium coordinates. Since the complex plane has an infinite number of points, a subset of points corresponding to the medium's resolution must be selected and mapped.

For example, a 2000x1000 drawing medium can be used to draw the complex plane area from -1.0 to 0.5 on the horizontal axis and -0.25 to 0.5 on the vertical axis. Dividing the horizontal range (1.5) by 2000 gives a step size of 0.00075, allowing for the selection of points along both axes.

To draw the set, one must decide on the ceiling value **Max** and optionally **z_small** for the inclusion rule. The simplest strategy is to color points as members or non-members of the set. A more interesting approach is to color points based on the number of iterations **n** it takes for **a_n^2 + b_n^2 ≥ 4**, using a color scale with **Max** values.


This course introduces fundamental principles of highway and road design, including safety, speed, terrain, and operating volumes, as they apply to roadway width, side slopes, curvature, and gradient. Key topics include horizontal curves, compound curves, cross-section areas and volumes, vertical curves, and alignments.

Upon completion of this course, students will be able to solve fundamental problems dealing with superelevation, friction, sight distance, stopping distance, and response-initiation time. They will also be able to calculate volumes of earthwork related to route design and construction, interpret field data associated with preliminary route location and slope staking, and perform a preliminary route survey and slope staking calculation.

Specific learning objectives include:

* Solving problems dealing with superelevation, friction, sight distance, stopping distance, and response-initiation time
* Calculating volumes of earthwork related to route design and construction
* Interpreting field data associated with preliminary route location and slope staking
* Performing a preliminary route survey and slope staking calculation
* Calculating the principal lengths of parts of a circular curve
* Planning and producing a circular curve in the field
* Solving route surveying and design problems using a computer and software
* Calculating tangent and offset elevations on vertical curves
* Planning and producing a compound or reverse curve
* Demonstrating vertical and compound curve design
* Solving fundamental problems dealing with highway drainage
* Identifying the standard design elements of a railroad track
* Demonstrating professional and ethical conduct as expected in industry

The course will cover various topics, including:

* Highway design principles
* Route surveying and design
* Earthwork calculations
* Circular curve design
* Vertical curve design
* Compound curve design
* Highway drainage
* Railroad track design

Students will use various tools and software, including:

* Civil Engineers Scale
* Surveying field book
* Computer-aided design (CAD) software
* Data recorders

The grade will be determined using the Delaware Tech grading system, with the following weighted average:

* Formative assignments (20%)
* Summative labs (20%)
* Summative tests (60%)

The course will emphasize the development of skills in:

* Clear and effective communication
* Critical thinking
* Collaboration
* Professional and ethical conduct
* Information literacy
* Quantitative reasoning and scientific inquiry

The course is designed to meet the Core Curriculum Competencies and Program Graduate Competencies, including:

* CETAASCET: Apply the knowledge, techniques, skills, and applicable tools of the discipline to engineering activities
* CETAASSET: Apply the knowledge, techniques, skills, and applicable tools of the discipline to engineering and surveying activities

The College is committed to providing reasonable accommodations for students with disabilities. Students are encouraged to schedule an appointment with the campus Disabilities Support Counselor to request an accommodation needed due to a disability.


Q: How many centiliters are in 100 liters?

To solve this, recall that 1 liter is equal to 100 centiliters. Therefore, 100 liters would be 100 * 100 = 10,000 centiliters.

Key conversion facts:
- 1 liter = 100 centiliters
- 100 centiliters = 1 liter
- To convert liters to centiliters, multiply by 100.
- To convert centiliters to liters, divide by 100.

Examples:
- 5 liters = 5 * 100 = 500 centiliters
- 0.07 liters = 0.07 * 100 = 7 centiliters
- 400 centiliters = 400 / 100 = 4 liters
- 857,213 centiliters = 857,213 / 100 = 8,572.13 liters
- 6500 centiliters = 6500 / 100 = 65 liters

Understanding the prefix "centi" means one hundredth, which is why there are 100 centiliters in a liter. Similarly, knowing that there are 1,000 liters in a kiloliter, we can deduce there are 100,000 centiliters in a kiloliter because 1,000 * 100 = 100,000.

Given these relationships, it's straightforward to convert between liters and centiliters by either multiplying or dividing by 100, depending on the direction of the conversion.


The SEQUENCE function in Google Sheets generates numbered lists. The function is defined as `=SEQUENCE(rows, columns, start, step)`, where arguments 2, 3, and 4 are optional. To use them, the previous arguments must be specified.

1. **Ascending list of numbers**: `=SEQUENCE(5)` generates a list of numbers from 1 to 5.
2. **Horizontal list of numbers**: `=SEQUENCE(1,5)` creates a horizontal list of numbers from 1 to 5.
3. **Two-dimensional array of numbers**: `=SEQUENCE(5,5)` generates a 5x5 array of numbers.
4. **Start from a specific value**: `=SEQUENCE(5,1,100)` starts the list from 100.
5. **Use a custom step**: `=SEQUENCE(5,1,1,10)` increments the list by 10.
6. **Descending numbers**: `=SEQUENCE(5,1,5,-1)` counts down from 5.
7. **Negative numbers**: `=SEQUENCE(5,1,-1,-1)` starts from -1 and counts down.
8. **Dates**: `=SEQUENCE(5,1,TODAY(),1)` generates a list of dates, which are stored as numbers in spreadsheets.
9. **Decimal numbers**: Combining SEQUENCE with an Array Formula, e.g., `=ArrayFormula(SEQUENCE(5,1,10,1) / 10)`, allows for decimal counts.
10. **Constant numbers**: Setting the step value to 0, e.g., `=SEQUENCE(5,1,10,0)`, generates an array of constant numbers.

Advanced uses include:
- **Monthly sequences**: `=ArrayFormula(DATE(2021,SEQUENCE(12),1))` generates the first day of each month.
- **Text and emoji sequences**: Using the TEXT function, e.g., `=ArrayFormula(TEXT(SEQUENCE(100,1,1,1),"")&"Ben Collins")`, repeats a text string.
- **Lists of grouped numbers**: `=ArrayFormula(ROUNDUP(SEQUENCE(20)/5))` groups numbers into sets of 5.
- **Repeating sequences**: Combining SEQUENCE with the MOD function, e.g., `=ArrayFormula(MOD(SEQUENCE(100,1,0),3)+1)`, creates a repeating sequence.

Examples of using SEQUENCE in complex formulas include generating Fibonacci numbers and creating grids of combinations. SEQUENCE can also be used to create non-linear and looping sequences, such as repeating a sequence of numbers or generating a list of dates with a specific pattern.

To solve specific problems, such as generating a list that starts with a constant or creating a sequence of years from abbreviated inputs, SEQUENCE can be combined with other functions like TEXT, ARRAYFORMULA, and DATE. For instance, `=ARRAYFORMULA(TEXT(SEQUENCE(10;1;1;1);"\A\B\-0"))` generates a list starting with "AB-" followed by a sequence of numbers, and `=SEQUENCE(11,1,1950,1)` generates a list of years from 1950 to 1960.


To change a decimal to a fraction, analyze the decimal places. For example, 2.60 can be shortened to 2.6. The last decimal, 6, is in the tenths place. Reading the fraction, you would say "two and six tenths," which translates to 2 6/10. Simplifying this fraction gives 2 3/5.

For 0.653, the decimal can be written as 653/1000. To simplify, find the greatest common divisor (GCD) of 653 and 1000.

When converting decimals to fractions, consider the place value of the last decimal. For instance, 2.6 has the last decimal in the tenths place, making it 2 6/10 or 2 3/5 in simplest form.

Converting mixed fractions to improper fractions is also essential. For example, 5 3/5 can be converted to an improper fraction by multiplying the whole number part (5) by the denominator (5) and adding the numerator (3), resulting in 28/5.

To change a fraction to a decimal, divide the numerator by the denominator. For example, 3/4 as a decimal is 0.75.

Key steps to change decimals to fractions:
- Identify the place value of the last decimal.
- Write the decimal as a fraction with the decimal part as the numerator and the place value as the denominator.
- Simplify the fraction by finding the greatest common divisor (GCD) of the numerator and denominator.

Key steps to change fractions to decimals:
- Divide the numerator by the denominator.
- Express the result as a decimal.

Examples:
- 2.60 = 2 6/10 = 2 3/5
- 0.653 = 653/1000
- 2.5 = 5/2
- 3 1/2 as a percent in fraction form: convert 3 1/2 to an improper fraction (7/2), then to a decimal (3.5), and finally to a percentage (350%). As a fraction, 3 1/2 is 7/2, and as a percentage in fraction form, it's 350% or 7/2 * 100%.

To convert a percentage to a fraction, divide by 100. For example, 66 2/3% = 66.67/100 = 2/3. To find the number that 66 2/3% represents 74, set up the equation: (2/3)*x = 74, solve for x.

To change a mixed fraction to an improper fraction, multiply the whole number part by the denominator and add the numerator. For example, 5 3/5 = (5*5) + 3 = 28/5.

To simplify a fraction, find the greatest common divisor (GCD) of the numerator and denominator and divide both by the GCD. For example, 6/8 can be simplified to 3/4 by dividing both the numerator and denominator by 2.

To convert a fraction to a decimal, divide the numerator by the denominator. For example, 1/2 as a decimal is 0.5.

To change a decimal to a fraction when the fraction looks like 0.43 3/4, first convert the decimal part (0.43) to a fraction, then add the fraction part (3/4). For 0.43, it's approximately 43/100, and adding 3/4 gives a mixed fraction.

To convert 6 2/3% to a fraction, divide by 100: (6 2/3)/100 = (20/3)/100 = 1/15. To convert .6 1/3 to a decimal or fraction, first convert the fraction part (1/3) to a decimal (approximately 0.33), then add the decimal part (0.6), resulting in approximately 0.93. As a fraction, it remains 6 1/3/10 or 19/30 when simplified.


**Kirchhoff’s Current Law (KCL)**

Kirchhoff’s Current Law states that at any moment, the algebraic sum of flowing currents through a point in a network is zero. This means that the sum of incoming currents to a point is equal to the sum of outgoing currents from that point.

For example, in a network with currents **I**_{1}, **I**_{2}, **I**_{3}, **I**_{4}, and **I**_{5}, the law can be expressed as: **I**_{1} + **I**_{5} = **I**_{2} + **I**_{3} + **I**_{4}. In a simpler case, **I**_{1} = **I**_{2} + **I**_{3}, which can be illustrated with a specific example where 6A = 2A + 4A.

**Kirchhoff’s Voltage Law (KVL)**

Kirchhoff’s Voltage Law states that the algebraic sum of the EMF applied in a circuit is equal to the algebraic sum of the voltage drops in the elements. This law is also known as the Voltage Law or Mesh Law. KCL and KVL are used together to simplify circuits and find missing data.

To illustrate the application of KCL and KVL, consider a circuit with meshes ABC, CA, and CDA. Assume currents flow in the directions indicated by arrows. Applying KCL to Junctions C and A yields:

- Current in mesh ABC = *i*_{1}
- Current in Mesh CA = *i*_{2}
- Current in Mesh CDA = *i*_{1} – *i*_{2}

Applying KVL to Mesh ABC, with 20V acting in the clockwise direction, gives the equation: **10***i*_{1}** + 4***i*_{2}** = 20. Note that only **R**_{1}, **R**_{2}, and **E**_{1} are involved in mesh ABC.

For mesh ACD, with 12 volts acting in the clockwise direction, the equation becomes: **8****i**_{1}** – 12****i**_{2}** = 12. Here, only **R**_{2}, **R**_{3}, and **E**_{2} are involved in mesh ACD.

Solving these two equations yields: *i*_{1} = 1.895 A and *i*_{2} = 0.263 A. Using different circuit-solving techniques can simplify the process of solving circuits.


There are six properties of multiplication of whole numbers that help solve problems easily. These properties are: Closure Property, Commutative Property, Zero Property, Identity Property, Associativity Property, and Distributive Property.

**Closure Property of Whole Numbers**: The product of two whole numbers is always a whole number. For example, 8 × 9 = 72, 0 × 16 = 0, 11 × 15 = 165, and 20 × 1 = 20.

**Commutativity of Whole Numbers**: The multiplication of whole numbers is commutative, meaning the order of numbers does not change the product. For example, 7 × 4 = 28 and 4 × 7 = 28. This property can be verified with various examples, such as 7 × 6 = 42 and 6 × 7 = 42.

**Multiplication By Zero/Zero Property of Multiplication of Whole Numbers**: When a number is multiplied by 0, the product is always 0. For example, 3 × 0 = 0, 9 × 0 = 0, and 115 × 0 = 0. This property can be verified with various examples, such as 20 × 0 = 0 and 1 × 0 = 0.

**Multiplicative Identity of Whole Numbers**: When a number is multiplied by 1, the product is the number itself. For example, 1 × 2 = 2, 1 × 6 = 6, and 13 × 1 = 13. This property can be verified with various examples, such as 25 × 1 = 25 and 117 × 1 = 117.

**Associativity Property of Multiplication of Whole Numbers**: The multiplication of whole numbers is associative, meaning the product of three or more numbers does not change when their order is changed. For example, (6 × 5) × 3 = 90, 6 × (5 × 3) = 90, and (6 × 3) × 5 = 90. This property can be verified with various examples, such as (2 × 3) × 5 = 30 and 2 × (3 × 5) = 30.

**Distributive Property of Multiplication of Whole Numbers**: The product of a number and the sum of two or more numbers is equal to the sum of the products of that number and each of the numbers separately. For example, 3 × (2 + 5) = 21 and 3 × 2 + 3 × 5 = 21. This property can be verified with various examples, such as 1 × (5 + 9) = 14 and 1 × 5 + 1 × 9 = 14.

Key properties of multiplication include:
1. The product of two numbers does not change when the order of the numbers is changed.
2. The product of three numbers does not change when the groupings of the numbers are changed.
3. The product of a number and 1 is the number itself.
4. The product of a number and 0 is always 0.
5. The product of a number and the sum of two numbers is the same as the sum of the products of that number and each of the numbers separately.

In a multiplication question, the number to be multiplied is called the multiplicand, the number by which we multiply is called the multiplier, and the result of multiplication is called the product.

**Worksheets on Properties of Multiplication**:

1. Fill in the blanks:
(i) Number × 0 = 0
(ii) 54 × 1000 = 54000
(iii) Number × 1 = Number itself
(iv) 8 × (5 × 7) = (8 × 5) × 7
(v) 7 × 9 = 9 × 7
(vi) 5 × 6 × 12 = 12 × 5 × 6
(vii) 62 × 10 = 620
(viii) 6 × 32 × 100 = 6 × 100 × 32

2. Fill in the blanks using properties of multiplication:
(i) 62 × 5 = 5 × 62
(ii) 31 × 0 = 0
(iii) 332 × 9 = 332 × 9
(iv) 134 × 1 = 134
(v) 26 × 16 × 78 = 26 × 78 × 16
(vi) 43 × 34 = 34 × 43
(vii) 540 × 0 = 0
(viii) 29 × 4 × 15 = 4 × 15 × 29
(ix) 47 × 1 = 47

3. Fill in the blanks using properties of multiplication:
(i) 629 × 9 = 9 × 629
(ii) 999 × 38 = 38 × 999
(iii) 99,125 × 1 = 99,125
(iv) 4,875 × 69 = 69 × 4,875
(v) 1 × 242 = 242
(vi) 24,210 × 0 = 0
(vii) 12,982 × 0 = 0
(viii) 43,104 × 1 = 43,104
(ix) 0 × 48,452 = 0
(x) 35 × 43,194 = 43,194 × 35
(xi) 4,589 × 1 = 4,589
(xii) 57,130 × 1,576 = 1,576 × 57,130
(xiii) 700 × 516 = 516 × 700
(xiv) 2,942 × 0 = 0


**American Chances**
The bet odds for favorite wagers start with a negative number and indicate how much you must wager to win $100. If the odds are -110, you would have to wager $110 to win $100. If the odds are -200, you would have to wager $200 to win $100.

**American Money Line**
Odds are denoted by a + or – and a number. The number +200 represents how much a $100 stake might reward the bettor. If the wager is successful, the player will be paid a total of $300 ($100 original wager + $200 net profit).

**Plus or Minus in Betting Odds**
A “minus” (-) before the number denotes the team’s favored status. A “plus” (+) before the number denotes an underdog team.

**Example: Bet Against**
If you bet $100 on a team with +200 odds, you would win $200 on that play, and you would get your $100 wager back, bringing the total to $300.

**Equivalent Odds**
A 200 moneyline price is equivalent to fractional 2/1 odds, or 3.00 in decimal form.

**Money Line Bet**
If you bet the favorite, the chances tell you how much you must wager to win $100. If you bet the underdog, the odds tell you how much you would win on a $100 bet.

**Related Questions and Answers**

## What does +7 spread mean?
If a game’s spread is seven points, the underdog will get seven points, which will be shown on the odds as +7. The favored squad is laying seven points and is set at -7.

## What does minus 160 mean in betting?
If the odds are negative (-), that much money must be bet to win $100. If the odds are positive (+), $100 would profit if the bet were successful.

## How much do I win if I bet $100 on odds?
Your wins are represented by the leftmost number, while your stake is shown by the rightmost number. For instance, 2/1 odds suggest that for every $1 wagered, you will win $2.

## What do odds of +350 mean?
A +350 underdog entails a $100 risk for a $350 reward. You may either push, lose $100, or win $350.

## What does +450 mean in odds?
If the number is -450, you would have to wager $450 to earn back $100 in profit. On the other side, if there is a “plus” in front of the odds number, it denotes an underdog team or person.

## What does 20 to 1 odds pay?
Interpreting the Win Odds: 20-1 means that you would receive $20 in profit for every $1 wagered.

## What does 7 to 2 odds pay?
Odds of 7-2 indicate that the bettor will benefit $7 for every $2 staked.

## How often does the moneyline favorite win NFL?
NFL moneyline favorites have prevailed 66.5 percent of the time since 1985.

## Is it better to bet moneyline or spread?
Better rewards are obtained by betting on favorites against the spread as opposed to the moneyline.

## What is a 2 leg parlay?
A normal parlay is the initial kind of parlay. A wager requires at least two legs to be deemed a parlay.

## What does +3 spread mean?
Depending on the sportsbook and the state, the odds on both sides of a spread bet are often set at -110.

## What does +1.5 spread mean?
The puck line is another name for the point spread in hockey betting. In the NHL, the puck line is often placed at 1.5, which indicates that the favorite must win by two goals or more.

## How do you read odds?
The payout for a successful wager on a $1 wager is represented by the decimal odds as one number.

## What does +5000 mean in betting?
+5000 is an underdog.

## Is sport betting profitable?
Although sports betting might be rewarding, most participants end up broke, which is why sportsbooks are a thing.

## What does +115 mean in betting?
Chances with a Plus (Underdog): If the outcome is +150, a $100 wager will result in a profit of $150.

## What does +225 mean in betting?
The Jets are +225 while the Buffalo Bills are -278 to win. This implies that if you wanted to wager $100 on the Bills winning, you would have to risk $278.

## Do you lose money on negative odds?
No, you do not lose money on negative odds. Your profit on those odds is lower than it would be if they were positive.

## What does minus 140 mean in betting?
The more you must risk to recover your initial bet, the bigger the favorite is. Consider a squad that has a -140 moneyline rating. This indicates that for every $1 of profit, you must risk $1.40.

## How much do you win if you bet $10 on?
Odds in decimals: If you wagered $10 on 4.0 odds, you would get a total payout of $40 if you won.

## What is a good gambling percentage?
A long-term winning percentage of around 55% is excellent for professional sports bettors.

## What does 3+ made threes mean?
The “player threes made” line is extremely straightforward; you are just placing a wager on whether or not a player will score more than or less than the specified number of three points.

## What does a negative spread mean?
A negative spread means the preferred team.

## What are 9 to 4 odds?
9/4: If you win, you will earn 9 units for every 4 units you bet (plus your stake).

**Conclusion**
The “what does the + and – mean in sports betting” is a question that has been asked before. The answer to the question is that it means you are getting an increase or decrease of your bet.


## A rectangle and a square are equal in area. The side of the square is 24 m. If the length of the rectangle is 36 m, find the breadth of the rectangle.

To find the breadth of the rectangle, we first calculate the area of the square, which is side squared. So, the area of the square is 24 x 24 = 576 m². Since the area of the square is equal to the area of the rectangle, we can set up the equation: 576 m² = length x breadth. Given the length of the rectangle is 36 m, we substitute it into the equation: 576 m² = 36 x breadth. To find the breadth, we divide both sides of the equation by 36: 576 / 36 = breadth, which gives us a breadth of 16 m.

Additionally, we can calculate the perimeter of the rectangle using the formula 2(length + breadth) = 2(36 + 16) = 2(52) = 104 m, and the perimeter of the square using the formula side x 4 = 24 x 4 = 96 m.

The breadth of the rectangle is 16 m.


**3.1 Introduction to Integers**

This section covers the following topics:
- Locating positive and negative numbers on the number line
- Ordering positive and negative numbers
- Finding opposites
- Simplifying expressions with absolute value
- Translating word phrases to expressions with integers

**3.1.1 Locate Positive and Negative Numbers on the Number Line**

A negative number is a number that is less than 0. Negative numbers can be used to describe temperatures below zero, bank overdrafts, elevations below sea level, and depths below the ocean surface.

The number line is used to represent both positive and negative numbers. The counting numbers (1, 2, 3, …) on the number line are all positive. A plus sign (+) is usually omitted before a positive number. If there is no sign, the number is assumed to be positive.

To extend the number line to include negative numbers, mark several units to the left of zero, keeping the intervals the same width as those on the positive side. Label the marks with negative numbers, starting with -1 at the first mark to the left of 0, -2 at the next mark, and so on.

**Example 1**

Plot the numbers on a number line:
- 3
- -3
- -2

**Solution**

Draw a number line. Mark 0 in the center and label several units to the left and right.
- To plot 3, start at 0 and count three units to the right.
- To plot -3, start at 0 and count three units to the left.
- To plot -2, start at 0 and count two units to the left.

**3.1.2 Order Positive and Negative Numbers**

The number line can be used to compare and order positive and negative numbers. Going from left to right, numbers increase in value. Going from right to left, numbers decrease in value.

Inequality symbols can be used to show the ordering of positive and negative numbers. The notation a < b (read "a is less than b") is used when a is to the left of b on the number line. The notation a > b (read "a is greater than b") is used when a is to the right of b on the number line.

**Example 2**

Order each of the following pairs of numbers using < or >:
- 14 ____ 6
- -1 ____ 9
- -1 ____ -4
- 2 ____ -20

**Solution**

Begin by plotting the numbers on a number line.
- Compare 14 and 6: 14 is to the right of 6, so 14 > 6.
- Compare -1 and 9: -1 is to the left of 9, so -1 < 9.
- Compare -1 and -4: -1 is to the right of -4, so -1 > -4.
- Compare 2 and -20: 2 is to the right of -20, so 2 > -20.

**3.1.3 Find Opposites**

On the number line, the negative numbers are a mirror image of the positive numbers with zero in the middle. The opposite of a number is the number that is the same distance from zero on the number line, but on the opposite side of zero.

**OPPOSITE**

The opposite of a number is the number that is the same distance from zero on the number line, but on the opposite side of zero.

**Example 3**

Find the opposite of each number:
- 7
- -10

**Solution**

- The number -7 is the same distance from 0 as 7, but on the opposite side of 0. So -7 is the opposite of 7.
- The number 10 is the same distance from 0 as -10, but on the opposite side of 0. So 10 is the opposite of -10.

**Opposite Notation**

The symbol "-" can have different meanings.
- Between two numbers, the symbol indicates the operation of subtraction.
- In front of a number, the symbol indicates a negative number.
- In front of a variable or a number, it indicates the opposite.

**Opposite Notation**

- a means the opposite of the number a
The notation -a is read "the opposite of a".

**Example 4**

Simplify: -(-6)

**Solution**

The opposite of -6 is 6.

**Integers**

The set of counting numbers, their opposites, and 0 is the set of integers.

**Integers**

Integers are counting numbers, their opposites, and zero.
…-3, -2, -1, 0, 1, 2, 3…

**Example 5**

Evaluate -x:
- when x = 8
- when x = -8

**Solution**

- To evaluate -x when x = 8, substitute 8 for x: -(8) = -8
- To evaluate -x when x = -8, substitute -8 for x: -(-8) = 8

**3.1.4 Simplify Expressions with Absolute Value**

The distance between 0 and any number on the number line is called the absolute value of that number. The absolute value of a number is never negative.

The symbol for absolute value is two vertical lines on either side of a number.

**Absolute Value**

The absolute value of a number is its distance from 0 on the number line.
The absolute value of a number n is written as |n|.
|n| ≥ 0 for all numbers

**Example 6**

Simplify:
- |3|
- |-44|
- |0|

**Solution**

- |3| = 3 because 3 is 3 units from zero
- |-44| = 44 because -44 is 44 units from zero
- |0| = 0 because 0 is already at zero

**Example 7**

Evaluate:
- |x| when x = -35
- |-y| when y = -20
- |-u| when u = 12
- |-p| when p = -14

**Solution**

- To find |x| when x = -35, substitute -35 for x: |-35| = 35
- To find |-y| when y = -20, substitute -20 for y: |-(-20)| = |20| = 20
- To find |-u| when u = 12, substitute 12 for u: -|12| = -12
- To find |-p| when p = -14, substitute -14 for p: -|-14| = -14

**Example 8**

Simplify:
- |9-3|
- 4|-2|

**Solution**

- |9-3| = |6| = 6
- 4|-2| = 4 * 2 = 8

**Example 9**

Simplify: |8+7|-|5+6|

**Solution**

|8+7|-|5+6| = |15|-|11| = 15 - 11 = 4

**Example 10**

Simplify: 24-|19-3(6-2)|

**Solution**

24-|19-3(6-2)| = 24-|19-3(4)| = 24-|19-12| = 24-|7| = 24 - 7 = 17

**3.1.5 Translate Word Phrases into Expressions with Integers**

Look for words that indicate a negative sign, such as "negative" or "opposite".

**Example 11**

Translate each phrase into an expression with integers:
- the opposite of positive fourteen
- the opposite of -11
- negative sixteen
- two minus negative seven

**Solution**

- the opposite of positive fourteen = -14
- the opposite of -11 = -(-11) = 11
- negative sixteen = -16
- two minus negative seven = 2 - (-7)

**Example 12**

Translate into an expression with integers:
- The temperature is 12 degrees Fahrenheit below zero.
- The football team had a gain of 3 yards.
- The elevation of the Dead Sea is 1,302 feet below sea level.
- A checking account is overdrawn by $40.

**Solution**

- The temperature is 12 degrees Fahrenheit below zero = -12°F
- The football team had a gain of 3 yards = 3 yards
- The elevation of the Dead Sea is 1,302 feet below sea level = -1,302 feet
- A checking account is overdrawn by $40 = -$40


To sketch polynomials that start out in standard form, we'll use the **Rational Root Theorem**. This method helps find rational roots, which are numbers that can be written as a fraction of whole numbers.

## All Possible Rational Roots

The Rational Root Theorem generates a finite list of rational numbers that could be roots of the equation. Although the list may be large, it's a better starting point than randomly trying numbers.

## Rational Root Theorem: Step-by-Step

1. List all factors of the constant term, including itself and 1.
2. List all factors of the leading coefficient.
3. Create all possible fractions with the numerator as a factor of the constant term and the denominator as a factor of the leading coefficient.
4. These fractions, including their positive and negative versions, are the possible rational roots.
5. Test each possible root by plugging it into the polynomial or using synthetic division. If the result is 0, it's a root.
6. Cross off possibilities that don't work and synthetically divide by the root that works to get a lower-degree polynomial.
7. Apply the Rational Root Theorem to the new polynomial, which may have fewer possibilities.

## Example 1

Given the polynomial with a constant term of 6 (factors: 1, 2, 3, 6) and a leading coefficient of 2 (factors: 1, 2), the possible rational roots are:
±1, ±2, ±3, ±6, ±1/2, ±3/2
After simplification and removing duplicates, the final list includes 1, 2, 3, 6, 1/2, 3/2. Testing each root reveals that 2 is a root. Using synthetic division, we factor out the root and get a quadratic equation. Factoring the quadratic equation yields roots of 2, 1/2, and -3.

## Example 2

For a polynomial with a constant term of -4 (factors: 1, 2, 4) and a leading coefficient of 1 (factors: 1), the possible rational roots are:
±1, ±2, ±4, ±1/2, ±2/2, ±4/2
After simplification, the possible roots are ±1, ±2, ±4, ±1/2. Testing these roots using synthetic division will reveal the actual roots of the polynomial.

By following the Rational Root Theorem, you can turn a polynomial from standard form into factored form.


**Number of Highly Composite Numbers Not Divisible by n**

The sequence A199337 represents the number of highly composite numbers not divisible by n. A highly composite number is a positive integer with more divisors than any smaller positive integer.

**Key Facts:**

* The sequence is well-defined, as for any n, there exists an m such that n divides A002182(k) for all k >= m.
* This follows from Ramanujan's equation (1915): [log_p P] <= e_p <= 2*[log_p P'], where P is the greatest prime factor of N, e_p is the exponent of p in the prime factorization of N, and P' is the next prime after P.
* The right inequality gives N <= Product_{prime p} p^(2*[log_p P']) = A003418(P')^2, implying P > oo as N > oo.
* The left inequality implies e_p > oo for any p, as N > oo.

**Important Numbers:**

* a(6) = 3, as only 1, 2, and 4 are not divisible by 6 among highly composite numbers.
* a(12) = 5 - 1 = 4, and a(30) = 9 - 1 = 8, as 12 and 60 divide all A002182(k) >= n.

**Example:**

To illustrate, we prove that n = 12 = A002182(5) and n = 60 = A002182(9) divide all A002182(k) >= n. Using eq.(54), we get e_2 >= 2 and e_3 >= 1 when [log_2 P] >= 2 and [log_3 P] >= 1, which is the case for P >= 5. This implies all N = a(n) > 2^4*3^2*5^2 = 3600 must have gpf(N) > 3, i.e., gpf(N) >= 5.

**Mathematica Code:**

Table[Length[Select[t, Mod[#, n] > 0 &]], {n, 100}] (* T. D. Noe, Mar 18 2012 *)

**Cross-References:**

Cf. A002182, A106037, A329570, A329571 (bounds from the Ramanujan formula).
Sequence in context: A294643 A029614 A143615 * A279789 A249389 A016606
Adjacent sequences: A199334 A199335 A199336 * A199338 A199339 A199340

**Keyword:**

nonn

**Author:**

J. Lowell, Nov 05 2011

**Extensions:**

Extended by T. D. Noe, Mar 18 2012

**Status:**

approved


**The RSA Algorithm**

The RSA algorithm was developed by Rivest, Shamir, and Adleman. It uses an expression with exponentials to encrypt plaintext in blocks, with each block having a binary value less than some number n. The block size must be less than or equal to log2(n), and in practice, it is k-bits, where 2^k < n < 2^(k+1).

Encryption and decryption are performed using the following formulas:

C = M^e mod n
M = C^d mod n
M = (M^e)^d mod n
M = M^(ed) mod n

Both the sender and receiver know the value of n. The sender knows the value of e, and only the receiver knows the value of d. This makes it a public key encryption algorithm with a public key of KU = {e, n} and a private key of KR = {d, n}.

For the RSA algorithm to be satisfactory for public key encryption, the following requirements must be met:

* It is possible to find values of e, d, and n such that M^(ed) = M mod n for all M < n.
* It is relatively easy to calculate M^e and C^d for all values of M < n.
* It is infeasible to determine d given e and n.

To meet the first requirement, a relationship of the form M^(ed) = M mod n is needed. This can be achieved using Euler's theorem, which states that for two prime numbers p and q, and two integers n and m, such that n = pq and 0 < m < n, the following relationship holds:

m^kΦ(n) + 1 = m mod n

where Φ(n) is the Euler totient function, which is the number of positive integers less than n and relatively prime to n.

The desired relationship can be achieved if ed = kΦ(n) + 1, which is equivalent to saying that ed ≡ 1 mod Φ(n) and d = e^(-1) mod Φ(n). This means that e and d are multiplicative inverses mod Φ(n), and according to the rule of modular arithmetic, this is true only if d (and therefore e) is relatively prime to Φ(n).

The steps involved in the RSA algorithm for generating keys are:

1. Select two prime numbers, p and q.
2. Calculate n = p*q.
3. Calculate Φ(n) = (p-1)(q-1).
4. Select e such that e is relatively prime to Φ(n) and less than Φ(n).
5. Determine d such that ed ≡ 1 mod Φ(n) and d < Φ(n).

**Security of RSA**

There are three approaches to attack the RSA:

1. Brute force key search (infeasible given the size of numbers).
2. Mathematical attacks (based on the difficulty of computing Φ(N), by factoring modulus N).
3. Timing attacks (on the running time of decryption).

The mathematical approach takes three forms:

1. Factor n = p*q, hence find Φ(n) and then d.
2. Determine Φ(n) directly without determining p and q, and find d.
3. Find d directly, without first determining Φ(n).

Timing attacks can be used to determine a private key by keeping track of how long a computer takes to decipher messages. However, there are simple countermeasures that can be used, such as:

* Constant exponentiation time: ensures that all exponentiations take the same amount of time before returning a result.
* Random delay: adds a random delay to the exponentiation algorithm to confuse the timing attack.
* Blinding: multiplies the ciphertext by a random number before performing exponentiation.


If the open circuit voltage of a circuit containing ideal sources and resistors is measured at 10V, while the current through the short circuit across the circuit is 600mA, what would be the power absorbed by a 60Ω resistor placed across the terminals?

To find the power absorbed, we first need to calculate the Thevenin resistance. The open circuit voltage (Voc) is 10V, and the short circuit current (Isc) is 600mA or 0.6A.

The Thevenin resistance (Rth) is calculated as Voc/Isc = 10V/0.6A = 16.66Ω. The load resistance (Rl) is given as 60Ω.

The current flowing through the terminals (I) can be calculated using the formula I = Voc / (Rth + Rl) = 10V / (16.66Ω + 60Ω) = 10V / 76.66Ω = 0.13A.

The power absorbed by the 60Ω resistor (P) is calculated using the formula P = I^2 * Rl = (0.13A)^2 * 60Ω = 1.02W.

Therefore, the power absorbed by the 60Ω resistor is 1.02W.


The number 1001 has a unique property that makes it useful for math tricks. When multiplied by any number, it creates a special pattern. For example, 123 x 1001 = 123123, where the original number 123 repeats itself.

This property can be utilized to perform a math trick. First, ask someone to come up with a 3-digit number, such as 135. Then, ask them to repeat the number beside the original, resulting in 135135. Next, ask them to divide this number by 13, then by 11, and finally by 7. Without knowing the result, you can reveal the original number, 135, which will likely surprise them.

The trick works because repeating the number beside the original is equivalent to multiplying it by 1001. When you divide the result by 13, 11, and 7 (the prime factors of 1001), you are essentially dividing by 1001, which returns the original number. This is based on the mathematical principle that abc(1001) = abcabc, and abc(1001) / 1001 = abc.

The number 1001 can also be represented in binary, which can be used for other tricks. Additionally, it can be converted to a different base, such as from binary (base 2) to decimal (base 10), resulting in a different value. For instance, 1001 in binary is equal to 9 in decimal. This highlights the importance of creativity and thinking in math, as simple operations and number theory can be used to create interesting games and tricks.


**Calculus Derivative Questions and Problems**

1. **Range of a Relation and Function**: Find the range of the relation {(1,4),(2,5),(3,5)} and determine whether it is a function. The range is {4,5}, and since each input maps to exactly one output, it is a function.

2. **Evaluating a Function**: Given f(x) = x^2 - 6x/x+2, find f(1). Substitute x = 1 into the function: f(1) = (1)^2 - 6(1)/(1+2) = 1 - 6/3 = 1 - 2 = -1.

3. **Derivative of a Function**: Find the derivative of f(x) = cos(x) at the point (2, cos(2)). The derivative of cos(x) is -sin(x), so f'(2) = -sin(2).

4. **Derivative of a Polynomial**: Find f'(x) for f(x) = 2x^3 + 3x^2 - x + 15. Using the power rule, f'(x) = 6x^2 + 6x - 1.

5. **Differential Equation**: Given dy/dx = (xy)/(2), sketch a slope field and find the function f that satisfies the differential equation for the tangent line to the curve y = f(x) through a given point.

6. **Tangent Line to a Curve**: Find the equation of the line tangent to the curve x = sec(t), y = tan(t), at t = π/6. First, find the slope by taking the derivative of y with respect to x, which is csc(t). At t = π/6, the slope is 2. The point of tangency is (sec(π/6), tan(π/6)) = (2/√3, 1/√3). The equation of the tangent line is y - 1/√3 = 2(x - 2/√3).

7. **Cubic Polynomial Function**: Given a cubic polynomial f(x) = 4x^3 + ax^2 + bx + k with a local minimum at x = 1 and a point of inflection at x = 2, find the values of a, b, and k.

8. **Secant Lines**: For the function f(x) = √x and the point P(4,2), find the slope of the secant line passing through P and Q(x, f(x)) for x values 1, 3, and 5.

9. **Derivative and Tangent Line**: For f(x) = 4x^2 + 2x, find f'(1) and the equation of the tangent line at (1, f(1)). f'(1) = 10, and the equation of the tangent line is y - 6 = 10(x - 1).

10. **Critical Values**: Given g'(x) = (x^2 - 16)/(x - 2) and g(3) = 4, find all values of x where the graph of g has a critical value.

11. **Tangent Line**: For f(x) = 7x - 5x^2 at (1, 9), find the slope m of the tangent line and its equation. The slope m = f'(1) = 7 - 10(1) = -3. The equation of the tangent line is y - 9 = -3(x - 1).

12. **Tangent Line**: For f(x) = 7x - 5x^2 at (2, 34), find the slope m of the tangent line and its equation. The slope m = f'(2) = 7 - 10(2) = -13. The equation of the tangent line is y - 34 = -13(x - 2).

13. **Differential Equation and Tangent Line**: Given dy/dx = 6x^2/y, find the equation of the tangent line to f(x) at (1,2) and write the function f(x).

14. **Secant Line and Mean Value Theorem**: For f(x) = x^2 - x - 12, find the equation of the secant line joining (2,6) and (4,0), and use the Mean Value Theorem to determine a point c in the interval [2,4] where f'(c) equals the slope of the secant line. The equation of the secant line is y = x - 4. The slope of the secant line is 1, so we need to find c where f'(c) = 1. Since f'(x) = 2x - 1, solving 2c - 1 = 1 gives c = 1.


This article discusses experiments and results for minimizing the Rastrigin function, a mathematical function used in optimization techniques evaluation. The experiments were performed using Genetic Algorithm (GA) and Particle Swarm Optimization (PSO) on a specific setup, parameters, and system. The fitness function is the given function to be optimized.

The Rastrigin function is defined as:
f(x) = ∑[x(i)^2 - 10cos(2πx(i))] + 10n, where x(i) ∈ [-5.12, 5.12] for all i = 1, …, n. The function was evaluated on 5 different values of n between 10 and 100.

The Matlab implementation of the fitness function is:
```matlab
function z = testfunction(x)
    [M, N] = size(x);
    z = [];
    sum = 0;
    for j = 1:M
        for i = 1:N
            sum = sum + (x(j, i)*x(j, i) - 10*cos(2*pi*x(j, i)));
        end
        z = [z; sum + 10*N];
    end
```

The experimental results are presented in the following table:

| N | GA Final Minimum Value | PSO Global Minimum Value (C1=1, C2=2) | PSO Global Minimum Value (C1=3, C2=3) | PSO Global Minimum Value (C1=3, C2=2) | PSO Global Minimum Value (C1=2, C2=3) |
| --- | --- | --- | --- | --- | --- |
| 10 | 27.8663 (500 epochs) / 9.1 (1000 epochs) | 12.723 | 4.0002 | 8.4609 | 7.0004 |
| 20 | 113.9635 (500 epochs) | 74.584 | 13.001 | 31.879 | 29.679 |
| 40 | 399.6707 (500 epochs) / 235 (1000 epochs) | 191.02 | 43.23 | 127.1 | 115.58 |
| 60 | 918.9643 (500 epochs) / 619 (1000 epochs) | 372.2 | 96.065 | 193.74 | 189.75 |
| 80 | 1392.1043 (500 epochs) | 489.16 | 220.28 | 314.04 | 286.96 |

The general analysis of the results shows that PSO performs better than GA, given the same number of iterations. Within PSO, the parameters C1=3 and C2=3 seem to perform the best, while C1=1 and C2=2 perform the worst in achieving the global minimum.

The experiments used the following parameters:
- Number of chromosomes: 10
- One-point crossover GA with a 5% mutation rate
- Maximum number of iterations: 500

Graphical visualizations of the results are presented for different values of N, showing the convergence of GA and PSO with different parameters. The plots show that PSO1 with C1=1 and C2=2 converges slower than the other parameters, while PSO2 with C1=3 and C2=3 reaches the minimum faster.

The results are not intended to be benchmarks but rather to illustrate experimental setups in labs and to be used by mentors teaching Machine Learning Lab Work. Changing parameters, mutation functions, and population sizes can drastically affect the results obtained. For benchmark results, peer-reviewed research papers should be consulted.


# Introduction to Fisher's Exact Test and Bayesian Models

A recent lecture by Bob Carpenter on Bayesian models sparked interest in exploring Fisher's exact test and its Bayesian approach. This post delves into the problem statement, classical approach using Fisher's exact test, and creating a Stan model for Bayesian analysis.

## Problem Statement

The data consists of the number of left-handed and right-handed individuals in male and female groups.

| sex | n left handed | n right handed |
| --- | --- | --- |
| male | 9 | 43 |
| female | 4 | 44 |

The question is whether the rate of left-handedness differs between the male and female groups.

## Classical Approach

In frequentist statistics, Fisher's exact test is used to determine if there is a significant association between two categorical variables. The test is applied to the given data.

```r
m <- matrix(c(9, 43, 4, 44), nrow = 2, byrow = TRUE,
           dimnames = list(sex = c("male", "female"),
                           handedness = c("left", "right")))
m
fisher.test(m)
```

The output shows a p-value of 0.2392, indicating that the null hypothesis cannot be rejected, suggesting no significant difference in left-handedness between the male and female groups.

## Creating a Stan Model

A Stan model is created to fit the data using a Bayesian approach. The model assumes that the counts of left-handed individuals in each group follow a binomial distribution with a probability of being left-handed (θ).

```r
model_code <- "
data {
  int<lower=0> beta_a;
  int<lower=0> beta_b;
  int<lower=0> n_total_1;
  int<lower=0> n_total_2;
  int<lower=0> n_hits_1;
  int<lower=0> n_hits_2;
  int<lower=0, upper=1> sample_prior_only;
}
parameters {
  real<lower=0, upper=1> theta_1;
  real<lower=0, upper=1> theta_2;
}
model {
  theta_1 ~ beta(beta_a, beta_b);
  theta_2 ~ beta(beta_a, beta_b);
  if (sample_prior_only != 1) {
    n_hits_1 ~ binomial(n_total_1, theta_1);
    n_hits_2 ~ binomial(n_total_2, theta_2);
  }
}
generated quantities {
  real diff;
  diff = theta_1 - theta_2;
}
"
```

The model is compiled and run using the `stan_model` and `sampling` functions from the `rstan` package.

## Prior Information

The prior distribution for the model is chosen to be a Beta distribution, which generates values between 0 and 1. Two priors are considered: a flat prior (beta(1, 1)) and an informative prior (beta(5, 40)).

```r
steps <- seq(from = 0, to = 1, by = .01)
ggplot(data.frame(x = steps, y = dbeta(steps, shape1 = 1, shape2 = 1))) +
  geom_line(aes(x, y)) +
  scale_x_continuous(breaks = (0:10) / 10) +
  ylim(0, 2) +
  labs(x = "p(left-handed)", y = "density", title = "beta(1,1)")

ggplot(data.frame(x = steps, y = dbeta(steps, shape1 = 5, shape2 = 40))) +
  geom_line(aes(x, y)) +
  scale_x_continuous(breaks = (0:10) / 10) +
  labs(x = "p(left-handed)", y = "density", title = "beta(5,40)")
```

## Sampling the Posterior

The model is run using the `sampling` function to generate posterior samples.

```r
m_informative <- run_model(beta_a = 5, beta_b = 40, sample_prior_only = 0)
m_flat <- run_model(beta_a = 1, beta_b = 1, sample_prior_only = 0)
```

The posterior samples are summarized and visualized using the `ggmcmc` package.

```r
ggs_density(ggs_flat) +
  facet_grid(Parameter ~ ., labeller = label_parsed) +
  ggtitle("flat prior: beta(1, 1)") +
  theme_grey(base_size = 14) +
  theme(legend.position = "bottom") +
  xlim(-.2, .5)

ggs_density(ggs_informative) +
  facet_grid(Parameter ~ ., labeller = label_parsed) +
  ggtitle("informative prior: beta(5, 40)") +
  theme_grey(base_size = 14) +
  theme(legend.position = "bottom") +
  xlim(-.2, .5)
```

The results show that the informative prior model is more skeptical of high left-handedness rates, while the flat prior model assigns more probability to these values. The models demonstrate the regularizing effect of prior information, where the informative prior pulls the male probabilities towards more plausible values.

## Conclusion

The Bayesian approach using Stan provides a flexible and interpretable way to analyze the data. The results show that there is no significant difference in left-handedness between the male and female groups. The informative prior model provides a more realistic estimate of the left-handedness rates, while the flat prior model demonstrates the importance of prior information in Bayesian analysis.


To solve the given equations, follow these steps:

The first equation, 7x + 2y = -28, cannot be solved without another expression, as it requires simultaneous equations to find the values of x and y.

The second equation is 21d^2 + 85d - 26 = 0. To solve this quadratic equation, factor it as follows:
21d^2 + 85d - 26 = (3d + 13)(7d - 2) = 0

This gives two possible solutions:
1. 3d + 13 = 0, which simplifies to d = -13/3
2. 7d - 2 = 0, which simplifies to d = 2/7

Therefore, the solutions to the second equation are d = -13/3 or d = 2/7.


## Problem of the Day #6: Tiling a 2 by 2 by n Box with Rectangular Prisms
### Introduction
The problem asks for the number of distinct ways to place $2n$ identical $1\times1\times2$ rectangular prisms inside a $2\times2\times n$ box.

### Solution
Let $a_n$ be the number of ways to completely fill a $2\times2\times n$ box with $1\times1\times2$ blocks, and let $b_n$ be the number of ways to fill a $2\times2\times n$ box with $1\times1\times2$ blocks such that there is a $1$-block gap at the top that is not filled and the part at the top that is filled has $2$ vertical blocks.

### Initial Conditions
By casework for $n\leq2$, we can see that $a_0=1$, $b_0=0$, $a_1=2$, $b_1=0$, $a_2=9$, and $b_2=4$.

### Recurrence Relations
We have two recurrence relations:
- $b_n=4a_{n-2}+b_{n-1}$
- $a_n=2a_{n-1}+a_{n-2}+b_n$

### Simplifying the Recurrence Relations
To eliminate $b$, we consider the difference $a_{n+1}-a_n$ and simplify it to obtain $a_{n+1}-3a_n-3a_{n-1}+a_{n-2}=0$.

### Characteristic Polynomial
The characteristic polynomial of this relation is $x^3-3x^2-3x+1$, which has roots $x=-1$, $x=2+\sqrt{3}$, and $x=2-\sqrt{3}$.

### General Solution
The general solution for $a_n$ is $a_n=\alpha(-1)^n+\beta(2+\sqrt{3})^n+\gamma(2-\sqrt{3})^n$.

### Solving for $\alpha$, $\beta$, and $\gamma$
Using the initial conditions $a_0=1$, $a_1=2$, and $a_2=9$, we can solve for $\alpha$, $\beta$, and $\gamma$:
- $\alpha = \frac{1}{3}$
- $\beta = \frac{1}{6}(2+\sqrt{3})$
- $\gamma = \frac{1}{6}(2-\sqrt{3})$

### Final Solution
Substituting the values of $\alpha$, $\beta$, and $\gamma$ into the general solution, we get $a_n=\frac{1}{3}(-1)^n+\frac{1}{6}(2+\sqrt{3})^{n+1}+\frac{1}{6}(2-\sqrt{3})^{n+1}$.

### Bonus Challenge
Write a program to verify that this solution is correct.


The "Triangles and its Properties Class 7 Worksheet" is a guide for learning about the properties of triangles, including descriptions and properties based on interior angles, length, and other characteristics. This worksheet aims to enhance skills in identifying triangles and their properties.

A triangle is a three-sided closed polygon with 3 sides and 3 vertices, and it has different types based on interior angles and measurements of lengths. The worksheet will help learners improve their skills in identifying types of triangles and understanding their properties.

The "Triangles and its Properties Class 7 Worksheet" will evaluate learners' ability to identify types of triangles and their properties. To use this worksheet, learners will first discuss the topic, then complete a 5-item true or false quiz on triangle properties, followed by a math drill to enumerate triangle types. Finally, learners will identify triangle figures based on length or interior angles.

Mastering triangle properties and types is essential for higher mathematics topics and has practical applications in real-life situations. By completing this worksheet, learners will develop a solid understanding of triangles and their properties, preparing them for more advanced mathematical concepts.


# What defines an obtuse triangle?

## Definition of an Obtuse Triangle

An obtuse triangle is a triangle with one angle greater than 90 degrees. A triangle can be either obtuse, acute, or right, and only one angle in a triangle can be obtuse.

## Definition of a Scalene Triangle

A scalene triangle is a triangle with all three sides of unequal length and all three angles of different measures. This type of triangle is distinguished based on the properties of its sides, where none of the sides are equal.

## Types of Triangles

A right triangle can be isosceles or scalene. An obtuse triangle, which has one angle greater than 90 degrees, can also be isosceles or scalene. This means that an obtuse triangle is not necessarily scalene, and a scalene triangle is not necessarily obtuse.

## Properties of Scalene Triangles

A scalene triangle has all three sides of different lengths and all three angles of different measures. However, the sum of all interior angles in a scalene triangle is always 180 degrees, meeting the angle sum property of a triangle.

## Obtuse-Angled Triangles

An obtuse-angled triangle has one interior angle measuring more than 90 degrees. If one angle measures more than 90 degrees, the sum of the remaining two angles is less than 90 degrees.

## Relationship between Scalene and Obtuse Triangles

A scalene triangle can be acute-angled, obtuse-angled, or right-angled. This means that a scalene triangle can also be an obtuse triangle, but not all obtuse triangles are scalene.

## Key Characteristics

- An obtuse triangle has one angle greater than 90 degrees.
- A scalene triangle has all three sides of unequal length and all three angles of different measures.
- The sum of all interior angles in any triangle is 180 degrees.
- A triangle can be classified based on its angles (obtuse, acute, right) and its sides (scalene, isosceles, equilateral).


Shannon's theory of information, also known as information theory, is a mathematical framework for understanding the concept of information. However, it is essential to evaluate whether this theory completely captures every possible meaning of the word "information." Information may be key to understanding the rules of the universe and creating artificial intelligence.

To comprehend existing theories of information, it is crucial to understand Shannon's information theory. This theory is a framework of thought with a specific vocabulary, which is necessary to appreciate the whole concept. The theory defines information indirectly by defining the quantity of information contained in a message or event.

The self-information of a message x, denoted as h(x), is the amount of information gained by receiving x. This concept is analogous to how physics defines mass and energy in terms of their quantities. The self-information of x is a function of the probability of occurrence of x, and it acts like the quantity of bits when applied to computer memory.

The theory is based on two principles: the quantity of information is a function only of the probability of occurrence, and the quantity of information acts like the quantity of bits when applied to computer memory. These principles constrain the self-information function h to the form h(x) = f(p_X(x)), where f is a function that maps probabilities to information.

The function f can be derived as f(p) = -log_2 p, where p is a probability value. This derivation is based on the concept of computer memory, where the number of bits of memory determines the number of distinguishable states. The self-information of x is then defined as h(x) = -log_2 p_X(x), where p_X(x) is the probability of x.

The unit of self-information is called a bit, which is a unit quantifying information, like Joules for energy or kilograms for mass. This unit can be regarded as undefined in absolute terms, but its usage gives it semantic meaning. The bit is an analytic continuation of computer bits, extending the concept to non-uniform distributions over a non-power-of-2 number of counterfactuals.

The properties of self-information include the fact that an event with a probability of 1 contains no information, and an event with a probability of 0 contains infinite information. The less likely an event, the more information it contains. The self-information of x goes up as the probability of x goes down.

The concept of entropy, denoted as H, is the expected self-information of a random variable. It is a measure of uncertainty, quantifying what is unknown. Entropy is a type of statistical dispersion, and it can be used as an alternative to statistical variance. The maximum value of entropy is achieved when the distribution is uniform, and the minimum value is 0, which indicates a deterministic distribution.

Conditional entropy, denoted as H[Y|X], is the expected conditional self-information of Y given X. It measures the average uncertainty about Y given an observation from X. The conditional entropy is useful because it takes all x into consideration, and it can be used to think of H[Y|X] as a property of a stochastic function from X to Y.

Mutual information, denoted as I[X,Y], is a measure of the strength of the statistical dependency between X and Y. It is always non-negative and can be thought of as a measure of information content. Mutual information is symmetric, and it measures how much information can flow through a stochastic function in either direction.

The channel capacity of a stochastic function g is defined as the supremum of mutual information over all possible input distributions. It measures the maximum amount of information that can be transmitted through the function.

Shannon information has several limitations and criticisms. One of the main issues is that it is blind to the meaning of messages, depending only on the probability of the message. Another criticism is that it ignores the scrambling of information, such as encryption. Additionally, Shannon information is not well-defined for continuous distributions, and it can be infinite for deterministic information.

To address these limitations, alternative approaches have been proposed, such as algorithmic information theory, which uses the length of the shortest program that can output a message as the information measure. Mutual information can also be used to address some of these limitations, as it takes into account the relationship between the message and the observer.

In conclusion, Shannon's information theory provides a mathematical framework for understanding the concept of information. However, it has several limitations and criticisms, and alternative approaches have been proposed to address these issues. Mutual information is a powerful tool for understanding the relationship between random variables, and it can be used to address some of the limitations of Shannon information.

The properties of conditional entropy include the fact that H[Y|X] = H[X,Y] - H[X], and the minimum value of H[Y|X] is 0, which occurs when p(y|x) is always deterministic. The maximum value of H[Y|X] is H[Y], which occurs when X and Y are independent.

Bayes' rule can be rewritten in terms of self-information as h(y|x) = h(x|y) + h(y) - h(x). This shows that the information contained in y given x is proportional to the information contained in x given y plus the information contained in y.

Cross-entropy and KL-divergence are measures of the difference between two probability distributions. Mutual information can be written in terms of KL-divergence as I[X,Y] = KL[p_X,Y || p_X * p_Y].


In an upcoming geometry course, the topic of rotations in $\mathbb{R}^3$ will be covered, including the classification of all finite rotation groups. To introduce groups, GAP will be utilized, and the symmetry groups of the Platonic solids will be explored. The transformation group of the Rubik's cube will be computed, as well as the megaminx, a version of the Rubik's cube on a dodecahedron.

The megaminx group can be defined using the following generators:
f1:=(1,3,5,7,9)(2,4,6,8,10)(20,31,42,53,64)(19,30,41,52,63)(18,29,40,51,62);
f2:=(12,14,16,18,20)(13,15,17,19,21)(1,60,73,84,31)(3,62,75,86,23)(2,61,74,85,32);
f3:=(23,25,27,29,31)(24,26,28,30,32)(82,95,42,3,16)(83,96,43,4,17)(84,97,34,5,18);
f4:=(34,36,38,40,42)(35,37,39,41,43)(27,93,106,53,5)(28,94,107,54,6)(29,95,108,45,7);
f5:=(45,47,49,51,53)(46,48,50,52,54)(38,104,117,64,7)(39,105,118,65,8)(40,106,119,56,9);
f6:=(56,58,60,62,64)(57,59,61,63,65)(49,115,75,20,9)(50,116,76,21,10)(51,117,67,12,1);
f7:=(67,69,71,73,75)(68,70,72,74,76)(58,113,126,86,12)(59,114,127,7,13)(60,115,128,78,14);
f8:=(78,80,82,84,86)(79,81,83,85,87)(71,124,97,23,14)(72,125,98,24,15)(73,126,89,25,16);
f9:=(89,91,93,95,97)(90,92,94,96,98)(80,122,108,34,25)(81,123,109,35,26)(82,124,100,36,27);
f10:=(100,102,104,106,108)(101,103,105,107,109)(91,130,119,45,36)(92,131,120,46,37)(93,122,111,47,38);
f11:=(111,113,115,117,119)(112,114,116,118,120)(102,128,67,56,47)(103,129,68,57,48)(104,130,69,58,49);
f12:=(122,124,126,128,130)(123,125,127,129,131)(100,89,78,69,111)(101,90,79,70,112)(102,91,80,71,113);

The megaminx group is defined as megaminx:=Group(f1,f2,f3,f4,f5,f6,f7,f8,f9,f10,f11,f12). The order of the megaminx group is $2^{115} 3^{58} 5^{28} 7^{19} 11^{10} 13^9 17^7 19^6 23^5 29^4 31^3 37^3 41^2 43^2 47^2 53^2 59^2 61 \cdot 67 \cdot 71 \cdot 73 \cdot 79 \cdot 83 \cdot 89 \cdot 97 \cdot 101 \cdot 103 \cdot 107 \cdot 109 \cdot 113$, which can be computed using GAP.


Given the polynomial $Q(x) = x^4 - 3x^3 + 2x^2 + x - 3$, evaluate $Q(-3)$ by substitution and then using the remainder theorem and synthetic division.

First, substitute $x = -3$ into the polynomial:
$Q(-3) = (-3)^4 - 3(-3)^3 + 2(-3)^2 + (-3) - 3$

Simplifying the expression:
$Q(-3) = 81 - 3(-27) + 2(9) - 3 - 3$
$Q(-3) = 81 + 81 + 18 - 3 - 3$
$Q(-3) = 174$

Next, apply the remainder theorem and synthetic division to evaluate $Q(-3)$. The remainder theorem states that the remainder of the division of a polynomial $Q(x)$ by $x - c$ is $Q(c)$. Synthetic division is a method for dividing a polynomial by $x - c$.

To use synthetic division for $Q(-3)$, divide $Q(x)$ by $x + 3$:
- The coefficients of $Q(x)$ are $1, -3, 2, 1, -3$.
- The divisor is $x + 3$, so $c = -3$.

Perform synthetic division:
1. Bring down the leading coefficient, which is $1$.
2. Multiply $-3$ by $1$ and add to $-3$, getting $-3 - 3 = -6$.
3. Multiply $-3$ by $-6$ and add to $2$, getting $2 + 18 = 20$.
4. Multiply $-3$ by $20$ and add to $1$, getting $1 - 60 = -59$.
5. Multiply $-3$ by $-59$ and add to $-3$, getting $-3 + 177 = 174$.

The result of the synthetic division is $174$, which confirms that $Q(-3) = 174$. This demonstrates the equivalence of evaluating a polynomial by direct substitution and by using the remainder theorem with synthetic division.


# Key Concepts in Triangle Geometry

A triangle is a polygon with three edges and three vertices, and the sum of its interior angles is always 180 degrees. An exterior angle of a triangle is supplementary to an interior angle. The area of a triangle can be calculated using the formula T = a*h /2, where a is the length of the side and h is the height or altitude.

## Problem Sets

1. **Isosceles Triangle**: In an isosceles triangle ABC with base BC, given | ∢BAC | = α and | CABCA | = 4α, calculate all interior angles.
2. **Function Analysis**: For the function f(x) = x*tanx, find vertical and horizontal asymptotes, intervals of decrease and increase, local maxima and minima, and intervals of concavity and inflection. Sketch the graph.
3. **Angle Calculation**: Determine the angle ϕ given ϕ + 71° 15' = 90°.
4. **Roof Consumption**: Calculate the consumption of sheet metal to cover a cone-shaped roof with a base diameter of 4.3m and a deviation of 36° from the base plane, assuming 8% waste.
5. **Instantaneous Power**: Describe how instantaneous power changes in an AC circuit during one period.
6. **Angle Ratio**: Calculate the interior angles of a triangle with sizes in the ratio 3:4:5.
7. **River Width**: Calculate the width of a river seen from a 15m high building 30m away from the river bank at an angle of 15°.
8. **Isosceles Triangles**: Determine how many isosceles triangles can be constructed from six lines of lengths 3cm, 4cm, 5cm, 7cm, 8cm, and 9cm, with two lines of each length.
9. **Triangular Numbers**: Find the 10th triangular number, where triangular numbers count objects arranged in an equilateral triangle.
10. **Cable Car Climb**: Calculate the climb of a cable car given the lower station altitude (1025m), upper station altitude (1272m), and horizontal distance (1921m).
11. **Perimeter and Side Lengths**: Given a triangle perimeter of 162dm and side length ratios a:b = 2:3 and a:c = 8:7, determine the side lengths.
12. **Isosceles Triangle Angles**: Calculate the angle between the arms of an isosceles triangle with a base angle of 78° 20'.
13. **Circumference and Side Lengths**: Given a triangle circumference of 125cm, where the shortest side is 12cm shorter than the longest side and the longest side is 7cm longer than the middle side, find the length of the middle side.
14. **Triangular Plot Dimensions**: Determine the dimensions of a triangular plot on a map with a scale of 1:2000, given the actual dimensions are 90m, 110m, and 130m.
15. **Third Side Length**: Given two sides of a triangle (a = 6cm, b = 13cm), determine the possible length of the third side.
16. **Triangle Formation**: Calculate the ratio a:c given a:b = 3:2 and α:β = 2:1 in triangle ABC.
17. **Non-congruent Triangles**: Determine how many non-congruent triangles can be formed with sticks of lengths 3, 5, and 7.
18. **Stick Set Property**: Analyze a set of at least 14 sticks with the property that no 3 sticks can form a triangle.
19. **Observatory Distance**: Calculate the distance between points A and C given the distance between observatories A and B (975m), angle BAC (63°), and angle ABC (48°).
20. **Ladder Length**: Calculate the length of a ladder leaning against a wall, touching at 340cm high, with its lower end 160cm away from the wall.

## Multiple Choice Questions

1. What is the sum of the interior angles of a triangle?
A) 180 degrees
B) 360 degrees
C) 90 degrees
D) 270 degrees

2. Which of the following is a property of an isosceles triangle?
A) Two sides are equal
B) Two angles are equal
C) The base is always the longest side
D) All of the above

3. What is the formula for the area of a triangle?
A) T = a*h /2
B) T = a*h
C) T = a + h
D) T = a - h

4. What type of triangle has all sides equal?
A) Isosceles
B) Equilateral
C) Right
D) Scalene

5. What is the name of the number sequence that counts objects arranged in an equilateral triangle?
A) Triangular numbers
B) Square numbers
C) Circle numbers
D) Rectangle numbers


**Introduction to the Mole Concept**

When asking students to compare the weight of a dozen pizzas and a dozen skyscrapers, or the number of items in each, they can easily provide answers. However, when introducing the concept of the mole, chemistry students often become baffled.

**Teaching Goals**

The primary objectives in teaching the mole concept in a first-year general chemistry course are two-fold:

1. Students should develop a solid understanding of the mole concept, enabling them to answer questions with ease, similar to their understanding of the concept of a dozen.
2. Students should be able to efficiently use dimensional analysis to convert between grams, moles, and particles, laying the foundation for a seamless introduction to stoichiometry later in the course.

**Common Problems**

Two main issues arise when teaching the mole concept:

1. Some students struggle with the idea of conversion factors, often creating incorrect and creative factors, such as "1 atom = 58.44 mol" or "1 g = 6.02 x 10^23 atoms." Despite emphasizing the importance of only two conversion factors, students may still produce random factors on tests and quizzes.
2. Other students focus solely on solving math problems, lacking a visual understanding of the relationship between a mole, a particle, and mass. When asked relational questions, they may spend too much time setting up math problems instead of using logical reasoning.

**Solution**

To address these issues, incorporating competition and self-guided lessons can increase student engagement and eagerness to practice. A mole concept game can help students think through relationships without relying on calculators. By recognizing the relationship between mass, number of items, and a familiar quantity (the dozen), students can develop a deeper understanding of the mole concept.

**Self-Guided Lesson**

The self-guided lesson includes:

* Recognizing the relationship between mass, number of items, and a familiar quantity (the dozen)
* Using dimensional analysis to perform conversions
* Applying these skills to tackle the concept of the mole

**Practice Materials**

Available worksheets cover both the mole concept and mole conversion calculations. Answer keys are provided for students to check their work, learn from mistakes, and practice without the stress of a failing grade. This approach encourages students to try, make mistakes, and learn from them, increasing their excitement to practice, even among lower-level students.