# ATLauncher Pack Sorting Algorithm (ATLPSA)

The ATLauncher Pack Sorting Algorithm (ATLPSA) is a complex calculation-based system used to sort packs in the Launcher. It ranks packs on a score between 0 and infinite, based on 5 distinct sections.

## Overview

The system calculates a score using 5 key points and ranks packs from 1 to the total number of packs in the Launcher. There are 3 main statistics, each with its own weight, and 2 multipliers. The first 3 statistics have a value between 0 and 100, with weightings of 55%, 30%, and 15%. The top pack in each category receives the full weighted points, while subsequent packs receive a decreasing percentage of the points.

## 1. Unique Players

This category tracks the number of unique players in the last 7 days, based on IP address. A unique player is a user who has played the pack, regardless of leaderboard status. This category has a weighting of 55%.

## 2. Installs

This category tracks the total installs, re-installs, and updates in the last 7 days, excluding errors. This category has a weighting of 30%.

## 3. Hours Played

This category tracks the hours played on leaderboards in the last 7 days, including all users. This category has a weighting of 15%.

## 4. ATLauncher Multiplier

This multiplier is set to 1.0 by default and is not currently used. It will never be used as a pay-to-advance system, and the system does not accept payments or donations for favoritism.

## 5. Other Multipliers

This multiplier combines various positive and negative bonuses, currently set to 1.0.

## Final Calculation

The final score is calculated by adding the first 3 categories, then multiplying by the ATLauncher Multiplier, and finally multiplying by the Other Multiplier. The packs are then ranked from highest to lowest score.

Example:
- Add the first 3 categories: `32 + 12 + 8 = 52`
- Multiply by ATLauncher Multiplier: `52 * 1.0 = 52`
- Multiply by Other Multiplier: `52 * 0.9 = 46.8`

The system is calculated daily at 12:30am UTC.

## Why Are You Doing This?

The ATLPSA aims to accurately reflect popular and well-made packs, giving them the attention they deserve. The system is being optimized and tweaked, and updates will be posted periodically.


To convert yards to decimeters (yd to dm), use the formula: multiply the amount of yards by 9.144. The result will be expressed in decimeters.

For example, to convert 1 yd to dm, multiply 1 by 9.144, which equals 9.144 dm.

The following table shows common conversions for Yards (yd) to Decimeters (dm):

Yards (yd) | Decimeters (dm) |
---|---|
0.001 yd | 0.009144 dm |
0.01 yd | 0.09144 dm |
0.1 yd | 0.9144 dm |
1 yd | 9.144 dm |
2 yd | 18.288 dm |
3 yd | 27.432 dm |
4 yd | 36.576 dm |
5 yd | 45.72 dm |
6 yd | 54.864 dm |
7 yd | 64.008 dm |
8 yd | 73.152 dm |
9 yd | 82.296 dm |
10 yd | 91.44 dm |
20 yd | 182.88 dm |
30 yd | 274.32 dm |
40 yd | 365.76 dm |
50 yd | 457.2 dm |
60 yd | 548.64 dm |
70 yd | 640.08 dm |
80 yd | 731.52 dm |
90 yd | 822.96 dm |
100 yd | 914.4 dm |

The yard is a unit of length in the United States and British imperial systems, equal to 3 feet or 36 inches. In 1959, the yard was standardized as exactly 0.9144 meters. It is commonly used for measurements in American and Canadian football, association football (soccer), and to measure distances in the United Kingdom.

The decimeter is a unit of length in the metric system, equal to one-tenth of a meter (0.1 meters).

To use the Yard to Decimeter converter, insert the amount of yards you want to convert and press the "Convert" button. The amount of decimeters will be displayed. This converter is a free online calculator that works on any device with a supported web browser, including Internet Explorer, Microsoft Edge, Firefox, Chrome, Safari, and Opera.

Additional conversion calculators are available for:
- Yards (yd) to Inches (in)
- Yards (yd) to Feet (ft)
- Yards (yd) to Miles (mi)
- Yards (yd) to Millimeters (mm)
- Yards (yd) to Centimeters (cm)
- Yards (yd) to Meters (m)
- Yards (yd) to Decameters (dam)
- Yards (yd) to Hectometers (hm)
- Yards (yd) to Kilometers (km)


The line y = 3x - 4 is a tangent to the curve y = 4x^2 - 3x + c when it touches the curve at only one point. For this to occur, the equation 4x^2 - 3x + c = 3x - 4 must have only one solution.

Rearranging the equation gives 4x^2 - 6x + c + 4 = 0. A quadratic equation ax^2 + bx + c = 0 has a single root if its determinant b^2 - 4ac = 0. Here, a = 4, b = -6, and c' = c + 4.

Substituting these values into the determinant equation yields b^2 - 4ac' = 0, which simplifies to 36 = 4*4*(c + 4). Solving for c gives c + 4 = 36/16 = 9/4, and therefore c = -7/4.

The required value of c for which the line y = 3x - 4 is a tangent to the curve y = 4x^2 - 3x + c is -7/4.


When working with right triangles, the same rules apply regardless of the orientation of the triangle. We can evaluate the six trigonometric functions of either of the two acute angles in a right triangle. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

To find all six trigonometric functions for a given angle in a triangle, we first find the sine, cosine, and tangent of the angles. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

Given the side lengths of a right triangle, we can evaluate the six trigonometric functions of one of the acute angles. For example, using a triangle with side lengths, we can evaluate $\mathrm{sin}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{cos}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{tan}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{sec}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{csc}\text{\hspace{0.17em}}\alpha ,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}\alpha .$

Using a specific triangle, we can evaluate $\text{\hspace{0.17em}}\mathrm{sin}\text{}t,$ $\mathrm{cos}\text{}t,$ $\mathrm{tan}\text{}t,$ $\mathrm{sec}\text{}t,$ $\mathrm{csc}\text{}t,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{}t.$ The values are:
$\begin{array}{l}\mathrm{sin}t=\frac{33}{65},\mathrm{cos}t=\frac{56}{65},\mathrm{tan}t=\frac{33}{56},\hfill \\ \text{\hspace{0.17em}}\mathrm{sec}t=\frac{65}{56},\mathrm{csc}t=\frac{65}{33},\mathrm{cot}t=\frac{56}{33}\hfill \end{array}$

We can use the relationships between trigonometric functions and special angles on the unit circle to evaluate triangles that contain those special angles. The special angles often used in math and science problems are multiples of $\text{\hspace{0.17em}}\mathrm{30\xb0},$ $\mathrm{60\xb0},$ and $\text{\hspace{0.17em}}\mathrm{45\xb0}.$ However, when dealing with right triangles, we are limited to angles between $\text{\hspace{0.17em}}\mathrm{0\xb0}\text{and90\xb0}\text{.}$

A $\text{\hspace{0.17em}}\mathrm{30\xb0},\mathrm{60\xb0},9\mathrm{0\xb0}\text{\hspace{0.17em}}$ triangle has side lengths in the relation $\text{\hspace{0.17em}}s,\sqrt{3}s,2s,\text{\hspace{0.17em}}$ while a $\text{\hspace{0.17em}}\mathrm{45\xb0},\mathrm{45\xb0},\mathrm{90\xb0}$ triangle has side lengths in the relation $\text{\hspace{0.17em}}s,s,\sqrt{2}s.\text{\hspace{0.17em}}$ We can use these relations to evaluate trigonometric functions of special angles.


**Overview:** The concept of "rate of change" is crucial in calculus, and it manifests in various contexts. When analyzing graphs of time versus position, slope plays a significant role in determining the direction of an object's movement.

**Concepts:** A positive slope at time 't' indicates the object is moving to the right of the origin, while a negative slope means it's moving to the left. If the velocity is zero, the object is at rest. An object's location and direction of movement are independent, meaning it can be to the right of the origin and still move left, and vice versa.

**Graph Analysis:**
- **Graph 1:** The movement to the right corresponds to a positive slope, and the "break" corresponds to a slope of zero.
- **Graph 2:** Consistent movement to the left corresponds to a consistent negative slope.
- **Graph 3:** The movement to the right corresponds to a positive slope, followed by a "break" with a slope of zero, and then movement to the left corresponds to a negative slope.

**Velocity and Derivatives:** Velocity is the rate at which position is changing, found by taking the derivative of the position function: d'(t) = v(t). Critical values, where v(t) = 0, indicate the object is either changing direction or remaining still.

**Practice:** Analyze the given position and velocity functions for three objects and determine which statement accurately describes each object's motion.
- **Object 1:** d(t) = (1/8)t^2 - 2t + 2
- **Object 2:** v(t) = t^2 - 16t + 64
- **Object 3:** d(t) = 6

**Speed:** Speed is the absolute value of velocity, s(t) = |v(t)|. For example, if v(t) = -2t, then s(t) = |-2t| = 2t. Speed gives the magnitude of velocity, making it important for understanding an object's movement.


The discriminant is a concept related to the type of roots (solutions) a quadratic equation contains. Quadratic equations always have two solutions, which are often referred to as "roots" or "zeros". The discriminant is part of the quadratic formula, specifically the expression under the square root: "b squared - 4ac". To understand the discriminant, it is essential to have a solid grasp of the quadratic formula. The value of the discriminant determines the nature of the roots of a quadratic equation, indicating whether it has two real roots, one double root, or two complex roots. The connection between the roots and the graph of a quadratic equation will be explained in the lesson video.


Number Systems
A number system is a format for representing different numerical values. Any number system calculates the value of a number being represented by the digit used and the place where the digits are present in the actual number. The way in which the value of the number is calculated also depends on the base of the number system.

Definition of Base of a number system:
The base of a number system is the number of digits used to represent the number in that number system, also called radix. For example, the decimal number system uses 10 digits (0-9), so its base is 10.

Number Systems Types
1. Decimal Number System:
The decimal number system is the most commonly used number system, with a base of 10 and digits ranging from 0 to 9. A number can be represented as the sum of its digits multiplied by powers of 10. For example, 47218 = 4 x 10^4 + 7 x 10^3 + 2 x 10^2 + 1 x 10^1 + 8 x 10^0.

2. Binary Number System:
The binary number system uses only two digits: 0 and 1, making its base 2. The binary digits 0 and 1 are called bits. Numbers are represented by combining these bits. For example, the binary number 10101 has a value of (1 x 2^4) + (0 x 2^3) + (1 x 2^2) + (0 x 2^1) + (1 x 2^0) = 16 + 0 + 4 + 0 + 1 = 21.

3. Octal Number System:
The octal number system uses eight digits (0-7), making its base 8. Numbers are represented by combining these digits, with each digit's value determined by its position and the base. For example, the octal number 123 has a value of 1 x 8^2 + 2 x 8^1 + 3 x 8^0 = 64 + 16 + 3 = 83.

4. Hexadecimal Number System:
The hexadecimal number system uses 16 digits (0-9 and A-F), making its base 16. Numbers are represented by combining these digits, with each digit's value determined by its position and the base. For example, the hexadecimal number 6A9 has a value of 6 x 16^2 + 10 x 16^1 + 9 x 16^0 = 6 x 256 + 10 x 16 + 9 = 1536 + 160 + 9 = 1705.

Conversions
1. Conversion of Base 10 to Base 2:
To convert a decimal number to binary, repeatedly divide the number by 2 and record the remainders. The remainders, read in reverse order, give the binary representation. For example, to convert 25 to binary: 25 / 2 = 12 remainder 1, 12 / 2 = 6 remainder 0, 6 / 2 = 3 remainder 0, 3 / 2 = 1 remainder 1, 1 / 2 = 0 remainder 1. Reading the remainders in reverse gives 11001.

2. Conversion of a Decimal Fraction to Binary Fraction:
To convert a decimal fraction to binary, multiply the fraction by 2 and record the integer part. Repeat with the remaining fraction until the fraction becomes 0. For example, to convert 0.25 to binary: 0.25 x 2 = 0.5 (integer part 0), 0.5 x 2 = 1.0 (integer part 1), so 0.25 in binary is 0.01.

3. Conversion of Binary Number to Decimal Number:
To convert a binary number to decimal, multiply each digit by the corresponding power of 2 and sum the results. For example, 11001 in binary is 1 x 2^4 + 1 x 2^3 + 0 x 2^2 + 0 x 2^1 + 1 x 2^0 = 16 + 8 + 0 + 0 + 1 = 25.

4. Conversion of Decimal Number to Octal Number:
To convert a decimal number to octal, repeatedly divide the number by 8 and record the remainders. The remainders, read in reverse order, give the octal representation. For example, to convert 205 to octal: 205 / 8 = 25 remainder 5, 25 / 8 = 3 remainder 1, 3 / 8 = 0 remainder 3. Reading the remainders in reverse gives 315.

5. Conversion of Octal Number to Decimal Number:
To convert an octal number to decimal, multiply each digit by the corresponding power of 8 and sum the results. For example, 716 in octal is 7 x 8^2 + 1 x 8^1 + 6 x 8^0 = 7 x 64 + 1 x 8 + 6 = 448 + 8 + 6 = 462.

6. Conversion of Decimal Number to Hexadecimal Number:
To convert a decimal number to hexadecimal, repeatedly divide the number by 16 and record the remainders. The remainders, read in reverse order, give the hexadecimal representation, using A-F for remainders 10-15. For example, to convert 410 to hexadecimal: 410 / 16 = 25 remainder 10 (A), 25 / 16 = 1 remainder 9, 1 / 16 = 0 remainder 1. Reading the remainders in reverse gives 1A9.

7. Conversion of Hexadecimal Number to Decimal Number:
To convert a hexadecimal number to decimal, multiply each digit by the corresponding power of 16 and sum the results, using A=10, B=11, ..., F=15. For example, 90BA in hexadecimal is 9 x 16^3 + 0 x 16^2 + 10 x 16^1 + 10 x 16^0 = 9 x 4096 + 0 x 256 + 10 x 16 + 10 x 1 = 36864 + 0 + 160 + 10 = 37034.


Mastering Percent Word Problems: Tips and Tricks for Success

Understanding Percentages and Percent Problems
A percentage is a way of expressing a part of a whole as a fraction of 100. For example, 50% means 50 out of 100, or 0.5 as a decimal. Percent word problems involve using percentages to solve real-world problems, such as finding the percentage of a number, finding what percentage one number is of another, or finding the original value before a percentage increase or decrease.

Common Types of Percent Word Problems
There are several common types of percent problems:
- Finding the percentage of a number, such as “What is 20% of 80?”
- Finding what percentage one number is of another, such as “What percentage of 60 is 15?”
- Finding the original value before a percentage increase or decrease, such as “If a shirt costs $25 after a 20% discount, what was the original price?”

Tips for Solving Percent Word Problems
To solve percent problems, follow a systematic approach:
- Read the problem carefully and identify what information is given and what you are trying to find.
- Convert the percentages to decimals or fractions if necessary.
- Use a formula or proportion to solve the problem.
- Check your answer to make sure it makes sense in the context of the problem.
- Practice solving different types of percent word problems to build your confidence and skills.

Using Proportions to Solve Percent Word Problems
Proportions are a useful tool for solving percent problems. For example, to find what percentage of 60 is 15, set up the proportion:
$$\frac{15}{60} = \frac{x}{100}$$
Cross-multiply to solve for x:
$$15 \times 100 = 60 \times x$$
$$1500 = 60x$$
$$x = 25$$
So 15 is 25% of 60.

Converting Percentages to Decimals and Fractions
To convert percentages to decimals, remove the percent sign and divide the given number by 100. For example, 25% = 0.25. To convert percentages to fractions, write it as a fraction with a denominator of 100 and simplify if possible. For example, 25% =
$$\frac{25}{100} = \frac{1}{4}$$

Examples of Percent Problems with Solutions
1. What is 30% of 50?
Solution: Convert 30% to a decimal (0.3) and multiply by 50 to get 15.
2. What percentage of 75 is 15?
Solution: Set up the proportion:
$$\frac{15}{75} = \frac{x}{100}$$
Cross-multiply to get 750 = 75x, and solve for x: x = 20. So 15 is 20% of 75.
3. What was the original price if a shirt costs $25 after a 20% discount?
Solution: Let x be the original price. The discounted price is 80% of x (0.8x). Since 0.8x = 25, solve for x: x = 31.25. So the original price was $31.25.

Common Mistakes to Avoid
Common mistakes when solving percent word problems include:
- Forgetting to convert percentages to decimals or fractions
- Using the wrong formula or proportion
- Failing to double-check work and ensure the correct approach is used

Real-World Applications of Percent Word Problems
Percent word problems have many real-world applications, such as:
- Calculating sales tax
- Determining purchase discounts
- Calculating interest rates and investment returns
- Figuring out tips at restaurants
- Calculating class grades and batting averages

Practice Exercises
To improve your percent word problem skills, practice with different types of problems and work through them step by step. You can find practice problems online or in textbooks, or create your own based on real-world scenarios.

Conclusion
Percent word problems can seem daunting, but with practice and a systematic approach, you can master them. Build your skills and confidence by understanding percentages, common types of percent word problems, using proportions and formulas, and avoiding common mistakes. Remember to practice regularly and seek help if needed.

FAQs
To solve percent word problems, follow these steps:
1. Identify what you have been given and what you need to find.
2. Create an equation using the formula:
$$\text{percent} \times \text{base} = \text{amount}$$
3. Solve for the required variable.
4. Re-enter the equation with your answer to verify it.

Example: A shirt costs $40 and is on sale for 25% off. What is the sale price of the shirt?
Solution:
$$\text{percent off} \times \text{original price} = \text{discount amount}$$
$$0.25 \times 40 = x$$
$$x = 10$$
$$\text{sale price} = \text{original price} – \text{discount amount}$$
$$\text{sale price} = 40 – 10$$
$$\text{sale price} = 30$$
So, the sale price of the shirt is $30.


**Converting Mixed Fractions to Improper Fractions**

Improper fractions have a greater value in the numerator than the denominator. To convert a mixed fraction, which is a sum of an integer and a proper fraction, we use variables to express the condition between these representations.

Let the mixed fraction be $x\dfrac{c}{b}$, where $c<b$. This can be expressed as $x+\dfrac{c}{b}$. Converting it to an improper fraction, let's assume the improper fraction is $\dfrac{a}{b}$, where $a>b$. The equational condition is $\dfrac{a}{b}=x+\dfrac{c}{b}$.

Solving the equation, we get $x+\dfrac{c}{b}=\dfrac{bx+c}{b}$. This means $\dfrac{a}{b}=\dfrac{bx+c}{b}$, and the final condition for the fraction is $a=bx+c$.

**Example: Converting $5\dfrac{5}{6}$ to an Improper Fraction**

Given $5\dfrac{5}{6}$, we can express it as $5+\dfrac{5}{6}$. The integer is 5, and the proper fraction is $\dfrac{5}{6}$. Using $a=bx+c$, we find $a=5\times 6+5=35$. Therefore, the improper fraction is $\dfrac{35}{6}$.

**Key Points:**

- The denominator remains the same in both improper and mixed fractions.
- The change occurs in the numerator, following the relation $a=bx+c$.
- To convert a mixed fraction $x\dfrac{c}{b}$ to an improper fraction $\dfrac{a}{b}$, use the formula $a=bx+c$.


Quiz Linear Functions

1. Determine the equation of each line shown.
   a) _________________________
   b) _________________________

2. Write the equation of the line with an x-intercept of 3 and y-intercept of 6. Show all work for full credit.

3. State the domain of each function:
   a) g(x) = x^2 - 8
   Domain: _________________
   b) h(x) = 3x - 4, x ≠ 9/2
   Domain: _________________
   c) h(x) = x^6 - x
   Domain: _________________

4. The cost of owning a car is a function of the distance driven per month, assuming a linear relationship. Given:
   - 200 miles/month costs $250
   - 500 miles/month costs $400
   a) Write the equation expressing cost in terms of distance.
   b) Interpret the y-intercept in the real world context.
   c) Determine the distance driven for a monthly cost of $800.

5. Analyze the table to determine if it contains linear data. Show work and provide evidence.
   x | y
   1 | 2
   2 | 3
   3 | 4
   4 | 1
   Linear (yes or no): _____________________

6. Solve for x: 6x^2 + x - 2 = 0

BONUS QUESTION: Given line (a) with equation y = 2x + 2 and line (b) having the same x-intercept, determine a possible equation for line (b).


The conclusion that certain assumptions are valid breaks down when those assumptions are challenged. For instance, there are regions on correlation matrices where the gain in time is very similar for different events, but which of these regions are physically realistic? It's essential to consider whether it's feasible to improve swimming speed from 1 km/hr to 30 km/hr or if sacrificing 3 km/hr on the bike to gain 2 km/hr on the run is worthwhile. The answer may depend on the starting and target speeds for each event.

To better understand relevant speeds, data from 429 competitors in the Overall category was analyzed. The distribution of overall triathlon finishing times shows that top finishers are closer together than slower athletes. However, when looking at average speeds, the distribution is more even. These two plots are consistent, as the separation between competitors increases with time on the course. If one athlete has speed \( v \) and another has a speed that is a fraction \( f \) of \( v \), the difference in time at the end of the course (distance \( d \)) is given by the equation:

\[
\Delta T = \frac{d}{v} \left(1 - \frac{1}{f}\right)
\]

This equation indicates a non-linear relationship between speed and time difference.

The distribution of times for each event shows significant overlap, with the fastest runners finishing their 5k faster than some swimmers. Despite this overlap, the events separate neatly in terms of speed, with each occupying a well-defined space. The speed histograms are fairly symmetric and almost Normal, suggesting athletes are randomly pulled from a population with an average value. In contrast, the time histograms have right tails, especially for the run and bike. This indicates that athletes in the tail of the run or bike time histograms are close to others in terms of fitness, and small improvements can yield significant time benefits.

The analysis of speed histograms, updated with a Gaussian fit, shows good agreement. The next step will involve focusing on the space these speeds occupy on the correlation matrices.


**The Problem**: Find the x-intercepts of the quadratic equation y = x^2 + 10x + 16. The class discussed three possible methods to solve this problem: using the Quadratic Formula, completing the square, and factoring.

A student asked why y = 0, and another student explained that when y = 0, the graph crosses the x-axis. The class understood that they were focusing on one possible value of y. If they wanted to find when y = 1, they would need to subtract 1 from both sides of the equation, resulting in 0 = x^2 + 10x + 15.

The class used GeoGebra to visualize the graph and verify their results. They observed that the points where y = 1 maintained the same symmetry with the x-intercepts. A student asked if there are graphs that have more than two x-intercepts.

The teacher challenged the class to come up with a function that has x-intercepts at -4, 0, and 3. A student, J.V., solved the problem by working backwards and came up with the function y = (x + 4)(x - 0)(x - 3), which simplifies to y = x^3 + x^2 - 12x. The student used the same method they had learned earlier to find parabolas with x-intercepts. The teacher was impressed with the student's ability to apply the concept to a new problem.


Blaise Pascal asked a question nearly 300 years ago and built an apparatus, known as 'Pascal's vases', to demonstrate the paradox. The apparatus consisted of a vessel with several differently shaped chambers, all open at the top and having the same base areas. The results showed that the water levels in all chambers were identical, regardless of the shape or inclination of the chambers.

This paradox can be resolved by applying elementary mechanics. The pressure at a point in a static liquid is due to the weight of liquid directly above it. Figure 2 shows the forces exerted by the walls of the chamber on the water, with the vertical components supporting the water. Only the column of water directly above the bottom opening contributes to the pressure at the base of the vessel. The pressure at any point in a liquid is the same from all directions.

Consider two points A and B in the same horizontal plane in Figure 1. For translational equilibrium, the horizontal pressures at points A and B must be identical. Since the vertical pressures at A and B must be the same as their corresponding horizontal pressures, the heights of the water columns above A and B must be the same.

A more striking paradox is associated with the horizontal pressures on a dam, such as the Hoover dam holding back Lake Mead. The lake is approximately 115 miles long and 600 feet deep at the dam, with a horizontal thrust of about 18 tons per square foot near the base. In contrast, a smaller lake, Lake Mudd, would exert the same horizontal pressure at the same depth, despite its smaller size.

The explanation lies in the fact that the horizontal pressure at a point must be equal to the vertical pressure, which depends only on the depth of the water, not its horizontal extent. This can be demonstrated using a glass container with a flexible vertical membrane separating a small chamber from a large chamber. When water is poured into the large chamber, the membrane bulges into the small chamber. As water is added to the small chamber, the bulge diminishes and disappears when the water levels on both sides are equal, indicating that the horizontal pressures are identical.

This concept is further explained in "University Physics" by F. W. Sears, M. W. Zemansky, and H. D. Young, Chapter 12. The key concept is that the pressure at a point in a liquid is due to the weight of the liquid directly above it and is the same in all directions. This understanding resolves the paradoxes associated with Pascal's vases and the horizontal pressures on a dam.


For several weeks, kindergartners have been exploring 2-digit numbers using base-10 blocks, writing them on the board, and translating between representations. They have also practiced 2-digit addition with base-10 blocks. Recently, the class ventured into making 3-digit numbers with base-10 blocks. The students predicted that 10 rods would make a flat and 10 flats would equal a cube. They had no trouble counting by 10's to 100 and by 100's to 1000.

When asked to identify a number made with 2 flats, 5 rods, and 7 little cubes, the students determined it was 257 with minimal help. However, writing the number on the board posed a challenge. One student volunteered to write the number, and the class agreed it was 257. When another number was written on the board, half the students thought it was 257, while the other half were unsure but believed 257 was actually 2057.

The students were then asked to write the number 200, and they all agreed it should be written as 200. However, when asked to write 201, opinions varied, with half thinking it should be 201 and the other half believing it should be 2001. One student thought 201 represented 200 one, while 2001 represented 200 ten. When asked to write 210 on the board, the students were unsure what it represented.

To further explore 3-digit numbers, the class made another number with 4 flats, 3 rods, and 1 little cube. A student was asked to write the number on the board and wrote 40031. Another student suggested it should be 4031, sparking a discussion about the correct representation. The class was left with three possible representations: 40031, 4031, and 431, with no consensus on the correct answer.

The experience highlighted the need to address place value with the students. The concept of place value is a convention, and understanding its importance is crucial for accurate number representation. Exploring alternative number systems, such as Greek, Roman, or Egyptian numerals, could help students appreciate the logic and order of the place value system.

Assessing students' understanding of place value is essential, even with tens and ones. A protocol for assessing place value understanding involves interviewing students individually to evaluate their comprehension. This approach can provide valuable insights into students' thinking and help identify areas where they need additional support.

By exploring 3-digit numbers and place value, the class can develop a deeper understanding of the number system and its conventions. The experience will help students appreciate the order and logic of the place value system, enabling them to represent numbers accurately and confidently.


Optimization is a distinct concept from related rates, requiring a different approach. In optimization, the goal is to maximize or minimize a function with two variables, whereas in related rates, all variables change with respect to time. The key to optimization problems is to identify the ultimate goal and work towards it.

To minimize or maximize a function, we take the derivative of the equation and set it to 0, which gives us the values at which the slope is 0, indicating maximums and minimums. Before taking the derivative, we must have an equation with only two variables, which may require preliminary steps.

Here's a step-by-step outline to solve optimization problems:

1. Identify what you are trying to maximize or minimize.
2. Write an equation for it, using descriptive variables.
3. Get the equation into a two-variable form by creating a second equation from additional information and isolating one variable.
4. Take the derivative of the two-variable equation.
5. Set the derivative to 0 and solve for the remaining variable.

Let's apply this outline to a problem: You have 1350 cm^2 of material to make a box with a square base and closed top. Find the height, width, and length that give the largest possible volume.

First, we set up an equation for the volume: V = HWL. Since the width is equal to the length, we can say V = L^2H. We have three variables, but we need to get down to two. We use the information that there is 1350 cm^2 of material to create a second equation: 1350 = 4LH + 2L^2.

We isolate one variable, H, and replace it in the V equation: H = (1350 - 2L^2)/4L. Substituting this into the V equation gives us V = L^2((1350 - 2L^2)/4L) = 1350L/4 - L^3/2. Now we have only two variables, V and L.

Taking the derivative of V with respect to L gives us V' = 1350/4 - 3L^2/2. Setting this to 0 and solving for L gives us L = 15. Now we can find the other dimensions: W = L = 15, and H = (1350 - 2L^2)/4L = 15.

The volume is V = 15 * 15 * 15 = 3375. This problem demonstrates that the best volume is obtained when the box is a perfect cube with height, length, and width all equal.

Another problem: You want to make a can that holds 100 cubic inches of liquid, closed on the top and bottom. What is the least amount of square inches of material you can use?

The equation for the material is M = 2πrh + 2πr^2. We have three variables, but we need to get down to two. We use the equation for volume, V = 100 = πr^2h, to create a second equation: h = 100/(πr^2).

Substituting this into the M equation gives us M = 2πr(100/(πr^2)) + 2πr^2 = 200/r + 2πr^2. Taking the derivative of M with respect to r gives us M' = -200/r^2 + 4πr.

Setting this to 0 and solving for r gives us r^3 = 50/π. The height is h = 100/(πr^2) = 2r. The amount of material used is M = 2πrh + 2πr^2 = 6πr^2 = 6π(50/π)^(2/3).

When solving optimization problems, it's essential to follow the steps and use descriptive variables to keep track of what's going on. Always look back at the question to ensure you've answered what was asked.


To calculate the cost of an item on sale, consider the example of a $704 product with a 40% discount. The discount amount can be calculated by multiplying $704 by 40% (or 0.40) and then dividing by 100, which yields $281.60. Alternatively, simply multiply $704 by 0.40 to get the same result. The final purchase price is found by subtracting the discount from the original price: $704 - $281.60 = $422.40.

The calculation can also be simplified by deducting the discount percentage from 1 (1 - 0.40 = 0.60) and then multiplying the result by the original price: 0.60 x $704 = $422.40. This method directly gives the final price after the discount.

In summary, a product originally priced at $704 with a 40% discount will cost $422.40, resulting in a savings of $281.60.

For another example, what is 40% off $705? This calculation follows the same process as described above.


Merge sort is a recursive algorithm that arranges elements in a data structure, such as an array or list, in increasing or decreasing order. It works by partitioning the array into two unsorted halves, sorting each half, and then merging the two sorted halves into one sorted whole.


The shape of the sampling distribution of the sample correlation coefficient (r) deviates from normality due to its negative skewness. This skewness occurs because r cannot exceed 1.0, limiting the distribution's extension in the positive direction. The degree of skewness increases with the population correlation coefficient (ρ).

Given a population with a correlation coefficient (ρ) of 0.60, the sampling distribution of r for a sample of 12 students is not symmetric. To calculate the probability of obtaining an r above a specified value, the sample correlation coefficient (r) can be transformed to a variable (z') that follows a normal distribution. The transformation formula is z' = 0.5 ln[(1+r)/(1-r)], and the standard error of z' is given by 1/√(N-3), where N is the number of pairs of scores.

For example, to find the probability of getting a sample correlation of 0.75 or above in a sample of 12 from a population with a correlation of 0.60, first convert both 0.60 and 0.75 to their z' values, which are 0.693 and 0.973, respectively. The standard error of z' for N = 12 is 0.333. Then, given a normal distribution with a mean of 0.693 and a standard deviation of 0.333, the probability of obtaining a value of 0.973 or higher can be calculated.

Using the z-score formula, z = (X - μ)/σ = (0.973 - 0.693)/0.333 = 0.841, the area above 0.841 can be found using a standard normal distribution table or calculator, yielding a probability of 0.20. This result indicates that there is a 20% chance of obtaining a sample correlation of 0.75 or higher in a sample of 12 students from a population with a correlation of 0.60.


To determine the most appropriate statistical test for a task, it's essential to assign a level of measurement to the study's dependent variable. There are three levels of measurement: nominal, ordinal, and interval/ratio.

Measurement is defined as the assignment of numbers to a variable according to predetermined rules. The things we observe, such as gender or race, are variables, and any particular observation of that variable is an assigned number. We view the property of numbers that define the values of a variable.

There are three fundamentally different ways in which numbers are used in statistical research:
- Nominal Level Data: to name or identify
- Ordinal Level Data: to represent position in a series or scale
- Interval/Ratio: to represent quantity

**Data Measured at the Nominal Level**

When measuring a variable at the nominal level, the properties of the variables are categories. A number is assigned to each category, such as for the variable “sex,” 1=male and 2=female. Race, region, and religion are examples of variables measured at the nominal level. The main principle underlying nominal data is that they do not imply any ordering among the responses.

**Principles of Nominal Data**:
* The measurement tells only what class a case falls into with respect to the variable.
* Categories of nominal data are mutually exclusive and exhaustive.

Examples of Nominal Data:
1. Are you: __ (1) Male __ (2) Female
2. Are you: __ (1) Protestant __ (2) Catholic __ (3) Jewish __ (4) Muslim __ (5) Other

**Data Measured at the Ordinal Level**

A researcher who measures job satisfaction might ask employees to specify whether they are 1= "very dissatisfied", 2="somewhat dissatisfied", 3="somewhat satisfied" or 4="very satisfied". The item responses in this job satisfaction scale are numbered in order, ranging from least to most satisfied. This is what distinguishes ordinal from nominal data.

While more powerful than nominal data, ordinal data does not capture important information present in interval/ratio data. Specifically, the difference between two numbers in an ordinal scale is not assumed to be the same as the difference between any two other numbers in the same scale.

**Principles of Ordinal Data**:
* In addition to the nominal data principles, it also tells us when one case has more or less of the variable’s property than does the other.
* Ordinal data provides a ranked or ordered appearance.

Examples of Ordinal Data:
1. In total, how much income did your family earn before taxes last year?
__ (1) Less than $10,000 __ (2) $10,001 – $20,000 __ (3) $20,001 – $30,000
__ (4) $30,001 – $50,000 __ (5) $50,001 – $80,000 __ (6) $80,001 – $100,000
__ (7) $100,001 or Greater
2. Your highest level of education attained is:
__ (1) H.S. Incomplete __ (2) H.S. Diploma __ (3) Some College
__ (4) College Degree __ (5) Some Graduate School __ (6) M.A./M.S. or Greater

**Data Measured at the Interval/Ratio Level**

Interval level data are numerical scales in which the intervals have the same quantitative value throughout. As an example, consider the variable “age.” The difference between a 30 year old and a 40 year old represents the same age difference as the difference between an 80 year old and a 90 year old.

Interval level data do not have a true zero point, even if one of the scaled values happens to carry the name "zero". The Fahrenheit scale illustrates this issue. Zero degrees Fahrenheit do not represent the complete absence of temperature.

Ratio level data have a true zero point, indicating the absence of the quantity being measured. You can think of ratio level data as the three prior levels of measurement rolled into one.

**Principles of Interval Data**:
* In addition to the principles of nominal and ordinal data, it also tells us that one case differs by a certain amount of the variable’s property from the others.
* Interval level data provides equal appearing intervals.

**Principles of Ratio Data**:
* In addition to the principles of nominal, ordinal, and interval data, it also tells us that one case has so many times as much of the variable’s property as does the other.

Examples of Interval/Ratio Data:
1. As of your last birthday, how old are you? __ (fill in age)
2. In total, how much income did your family earn before taxes last year? __ (fill in $)
3. In total, how many years of education have you attained? __ (fill in # of years)

In sum, knowledge of the three levels of measurement assigned to independent and dependent variables in a study will help determine the most appropriate tests of statistical significance to use in the analysis. The strongest level of measurement is interval/ratio, while the weakest is nominal data.


## The Difference Between the Bernoulli and Binomial Distributions

The binomial distribution describes a scenario with binary outcomes, labeled as "success" and "failure" or "1" and "0".

## Unit Root Testing in EViews

To test for the existence of a unit root in EViews and the stationarity of a series, use the following steps:
1. Utilize EViews' unit root testing tools, including various tests.
2. Determine the stationarity of the series.

## Conducting Regression Analysis Using SPSS

Regression analysis in SPSS involves several steps, including data preparation, model specification, and diagnostics.

## Assumptions and Diagnostics Tutorial

Ramsey's RESET test determines the correct functional form of a regression model. It checks if the functional form of the regression is appropriate.

## Frequency Polygon

A frequency polygon is a graph that displays data using lines connecting points plotted for frequencies at the midpoints of classes. Frequencies are represented by heights of the points.

## Unbiased Sampling Techniques

There are four basic methods of unbiased sampling:
1. Random sampling: uses chance methods or random numbers.
2. Systematic sampling: selects samples based on a fixed interval.
3. Stratified sampling: divides the population into subgroups and samples from each.
4. Cluster sampling: selects clusters of subjects and samples from each cluster.

## Types of Measurement Scales

There are four common types of measurement scales:
1. Nominal scale: categorizes data without any inherent order.
2. Ordinal scale: categorizes data with an inherent order.
3. Interval scale: measures data with equal intervals between consecutive levels.
4. Ratio scale: measures data with equal intervals and a true zero point.

## Writing an Argumentative Essay

An argumentative essay presents a well-supported argument in favor of a thesis statement. To write one,
1. Choose a topic and research it.
2. Develop a thesis statement.
3. Present evidence and arguments.
4. Address counterarguments.

## Descriptive and Inferential Statistics

Descriptive statistics summarize data, while inferential statistics make conclusions about a population based on a sample.
Common terms in statistics include:
1. Variable: a characteristic that can assume different values.
2. Data: the values or measurements of variables.

## Mediation Analysis

Mediation analysis examines the relationship between variables, including the effect of a mediator variable.
For example, if higher grades predict higher happiness, mediation analysis can help understand the underlying mechanisms.


# Case Study: How I Got the Highest Grade in my Discrete Math Class

During my sophomore year at Dartmouth, I took a course in discrete mathematics. The tests were not calibrated to any standard scale, so it was difficult to judge how well you were doing. On the midterm, for example, scores around 50 to 60 out of 100 were at the top of the class, whereas for the final those would be failing.

The course had 70 students, including three from Eastern Europe who had already studied this subject in high school. I didn't think of myself as a math person, but I was determined to do well. My strategy was to learn the insights, specifically by mastering every proof presented in class.

I developed a singular obsession: I wanted to be able to recreate, with pencil and paper, and no helper notes, every single proof presented in class. I bought a package of white printer paper and copied each proposition presented in class onto its own sheet of paper. I would write the problem at the top of the sheet and recreate the proof from my notes below. I tried to do this every week, and if I didn't understand a proof, I would break out the textbook or do some web searching to make sense of it.

About two weeks before each exam, I started scheduling sessions to aggressively review my "proof guides." I worked on the second floor of the Dana Biomedical Library, using the quiz and recall method to review the proofs. I split the proofs between those I could replicate from scratch and those that gave me trouble, and then focused on the ones that gave me trouble until every sheet had been conquered.

By the day of the exam, I could rattle off any proof from the course without mistake or hesitation. My approach was to inline my work throughout the term, never spending more than 2 hours at a time working on these proofs. I never stayed up late or ground through material, and I kept attacking it fresh, with high energy, time and time again.

The key lessons from this case study are that conquering a technical class requires a massive amount of work, but it doesn't have to be painful. The secret is to define your challenge, come up with a plan for winning the challenge, and put the plan into motion with time to spare. No cramming is necessary.

This approach can be applied to other technical courses, such as calculus, molecular biology, or climate science. The focus should be on developing insights and understanding the material, rather than just memorizing it. By using a similar strategy, you can dominate even the most thorny technical courses.

In response to comments, it's clear that this approach is not just about recopying notes, but about truly understanding the material and being able to recreate it from scratch. It's about developing a deep understanding of the subject matter and being able to apply it to different problems.

For those struggling with discrete math or other technical courses, the advice is to start by understanding the basics and then building from there. Use the quiz and recall method to review material, and focus on developing insights rather than just memorizing formulas or procedures.

Ultimately, the goal is to be able to understand and apply the material, rather than just regurgitating it. By using a strategic approach and putting in the time and effort, you can achieve success in even the most challenging technical courses.


Photocopying involves roughly 30 students per class, with 3 lessons a week and 5 classes, resulting in approximately 500 sheets at 2p per copy.

Petrol costs £1.29 per liter, and purchasing 40 liters would be approximately £52.

Shopping consists of an item costing £4.99 and 7 items at 99p each, totaling around £12.

The total travel time is about 1 hour, considering 30 minutes to get there, 30 minutes back, and an additional 10 minutes for traffic, resulting in a total absence of over 2 hours.

Students often struggle with estimation, which was a common exam question until the exam format changed. For instance, calculating 3.6 x 110 by rounding to 1 significant figure yields varying answers. Most students can estimate 400, but when given the freedom to choose their rounding method, answers become more diverse. Some students choose not to round the 110, while others keep the 3.6, resulting in more accurate answers, such as 360 being closer to 396.

A more challenging calculation, such as 13.8 x 19.1, can be simplified by approximating it to 13 x 19, making the calculation easier. However, estimation depends on the context and purpose.

Students often face issues with estimation, including determining whether their answer is an over or under estimate and deciding when to round up or down, especially with decimal answers. For example, calculating 37/3.5 yields approximately 10.57, which rounds to 11, but this may not be suitable in all situations.

Estimation questions can be complex, and students must consider the context to determine whether their answer is an over or under estimate. Without calculating exact answers, students can analyze which estimates are over or under estimates.

Multiple choice questions can be used to assess estimation skills, such as:
What is the estimated cost of petrol for 40 liters at £1.29 per liter?
A) Under £50
B) Approximately £52
C) Over £55

What is the estimated total travel time, including 30 minutes to get there, 30 minutes back, and 10 minutes for traffic?
A) Less than 1 hour
B) Approximately 1 hour
C) More than 2 hours

How were these estimation calculations determined, and which are over or under estimates?


**Analyzing Belly Button Height**

To understand the relationship between belly button height and overall height, we need to analyze the provided data and calculate key ratios.

1. **Describing Belly Button Height**: Compared to a person's height, the belly button height is approximately half. This is evident from the fact that the belly button is roughly at the midpoint of the body.
2. **Change in Belly Button Height with Overall Height**: As a person's height changes, their belly button height also changes proportionally. This means that the ratio of belly button height to overall height remains relatively constant.
3. **Estimating the Decimal Ratio**: If half is 0.5, the decimal closest to the belly button height compared to the height is approximately 0.6. This is because the belly button is slightly below the midpoint of the body.
4. **Height to Belly Button Height Ratio**: In reverse, a person's height is approximately 1.6 times their belly button height. This ratio can be used to estimate a person's height based on their belly button height.

**Calculating Belly Button Height for Each Shape**

| Shape | Height | Belly Button Height | Decimal (H ÷ BB Height) |
| --- | --- | --- | --- |
| 1 | 6 |  |  |
| 2 | 3 |  |  |
| 3 | 4 |  |  |
| 4 | 2 |  |  |

To calculate the decimal ratio, divide the height by the belly button height for each shape.

**Calculating the Mean of Decimals**

To find the estimate of the height of the belly button compared to a person's height, calculate the mean of the decimals.

**Task Two: The Video**

Watch the video "Using CensusAtSchool in the Classroom" to understand how to use real data to investigate the relationship between belly button height and overall height. The presenter, Ian Wong, uses Excel to organize the data and estimate a person's height based on their belly button height.

**Task Three: Real Data Investigation**

1. Download a data sample using the CensusAtSchool random sampler, considering factors like sample size, year level, and sex.
2. Copy and paste the height and belly button data into a new Excel worksheet.
3. Use Excel techniques to investigate if the height is approximately 1.6 times the belly button height.
4. Write a report stating the steps followed, including reasons for decisions made, and conclude with a copy of the data and calculations.
5. List the Excel functions used, marking new functions with an asterisk.

**Excel 2010 Help**

For users of Excel 2010, note that the features used in the video (Excel 2003) can be found in different locations. To:
* Delete a row of data: select data, then Home > Cells > Delete > Delete Sheet Rows
* Find the mean ratio: calculate height/belly button ratio in a column, then choose a display cell and type =average (range) or Home > Editing > ∑ > Average
* Apply conditional formatting: go to Home > Styles > Conditional Formatting > Highlight Cell Rules > Between


Lecture Note: Finding the LCD (Least Common Denominator)

To find the LCD, follow these steps:
1. Pretend the given polynomials are the denominators of polynomial fractions.
2. Factor both denominators to find out what is missing from each.
3. Change the fraction to add in the missing part, both on top and bottom.
4. When you put all the parts together, you have the LCD.

Example 1: Find the LCD of 7, 2x + 4, 24x, 6y, 2, and 16x, 3y, 3.
The different denominators have three "things": numbers, x's, and y's.
- Numbers: 24 and 16 both go into 48.
- x's: x^6 and x^3 both go into x^6.
- y's: y^2 and y^3 both go into y^3.
Put the parts back together: 48x^6y^3 is the LCD.

Example 2: Find the LCD of 3 and 2x + 4, x, x^2 + 2x.
Factor the denominators to find the LCM: 3 and 2x + 4, x, x(x + 2).
The left fraction is missing the (x + 2) part, so change it to 3(x + 2) and 2x + 4, x(x + 2).
The LCM is 2(x + 2).

Example 3: Find the LCD of x and 3, x^2 - 3x + 2, x^2 - 5x + 4.
Factor the denominators: x and 3, (x - 1)(x - 2), (x - 1)(x - 4).
The left fraction is missing the (x - 4) part, and the right fraction is missing the (x - 2) part.
Change both fractions: x(x - 4) and 3(x - 2), (x - 1)(x - 2)(x - 4).
The LCM is (x - 1)(x - 2)(x - 4).

Example 4: Find the LCD of x + 3 and x + 2, x^2 - 2x - 8, x^2 - 4.
Factor the denominators: (x - 4)(x + 2) and (x - 2)(x + 2).
The left fraction is missing the (x - 2) part, and the right fraction is missing the (x - 4) part.
Change both fractions: (x + 3)(x - 2) and (x + 2)(x - 4), (x - 4)(x + 2)(x - 2).
The LCM is (x + 2)(x - 2)(x - 4).

Special Cases: "a - b and b - a"
Recall from Section 7.3 that we can find the LCD for denominators that are subtractions in the opposite order by factoring out -1.
Example: Find the LCD of 2y^2 + 5y + 3, y - 1, and 1 - y.
Factor out -1: y - 1 = -1(1 - y).
The LCM is 1 - y. Note that you don't have to worry about the -1 when finding the LCM, but you do have to divide it into all parts of the top if asked to add or subtract.


# What is the Difference Between a Whole Number and an Integer?

The distinction between a whole number and an integer is often unclear, as it depends on the context and the person using the terms. Integers, on the other hand, have a well-defined meaning, referring to the set of all natural numbers, including negatives, positives, and zero. This set excludes fractions, decimals, irrational numbers, and imaginary numbers.

To understand the difference, it's essential to establish the definition of integers first. Integers are natural numbers, encompassing negatives, zero, and positive numbers, without including any non-natural numbers. In contrast, whole numbers can be more complex, as their definition varies depending on the user. Some people consider whole numbers to be only positive integers, also known as counting numbers, which would exclude zero. Others define whole numbers as non-negative integers, including all positive integers and zero, since it's not a negative number.

The inconsistency in the definition of whole numbers has led to confusion, especially among students of mathematics. To avoid this confusion, it's often recommended to use more specific terms, such as "integers," "negative integers," "positive integers," and "non-negative integers." These terms provide clarity, as integer refers to all natural numbers, including negatives, zero, and positives; negative integer refers only to negative natural numbers; positive integer refers to any positive natural number, excluding zero; and non-negative integers include all positive natural numbers and zero.

The ambiguity surrounding the term "whole number" is why it's not frequently used in advanced mathematical discussions. Instead, using precise terms helps to avoid misunderstandings and ensures that communication is clear and effective. By understanding the definitions and differences between these terms, individuals can better navigate mathematical concepts and avoid confusion.


How to Calculate Ratios
Ratios are mathematical expressions that compare two or more numbers, used to compare absolute quantities, amounts, or portions of a larger whole. They can be calculated and written in several ways, but the principles guiding their use are universal.

**Writing a Ratio**

1. Be aware of how ratios are used: Ratios compare multiple amounts or quantities, and can be used in academic and real-world settings. For example, in a classroom with 20 students, 5 girls and 15 boys, the ratio of girls to boys can be compared.
2. Write a ratio with a colon: The simplest way to write a ratio is to put a colon between the numbers or values being compared, such as 7:13 or 10:2:23. In the classroom example, the ratio of girls to boys can be written as 5:15.
3. Reduce a ratio to its simplest form: Ratios can be reduced by removing common factors, similar to fractions. For example, 5:15 can be reduced to 1:3 by dividing both sides by 5.
4. Be aware of alternate methods for writing ratios: Ratios can be expressed as words or numbers with the word "to" between the terms, or in fraction form using a horizontal or diagonal bar.

**Using Ratios in Math Problems**

1. Use multiplication or division to "scale" ratios: To create a larger or smaller ratio while maintaining the same proportions, multiply or divide all terms in the ratio by the same number. For example, if a class has 8 girls and 11 boys, and the class is made 5 times bigger, the new ratio can be found by multiplying 8:11 by 5, resulting in 40:55.
2. Use cross-multiplication to solve for an unknown variable: Given two equivalent ratios, write each ratio in fraction form, set them equal to each other, and cross-multiply to solve for the unknown variable. For example, if a small group of students has 2 boys and 5 girls, and a class with 20 girls is to maintain the same proportion, the number of boys in the class can be found by setting up the equation 2/5 = x/20 and solving for x.
3. Use ratios to find unknown quantities: Given a known quantity and a ratio, divide the ratio term corresponding to the known quantity by the known quantity itself, then divide every term in the ratio by the answer to find the unknown quantities. For example, if a recipe calls for flour, water, and butter in a 20:8:4 ratio, and each student is given 5 cups of flour, the amount of water and butter needed can be found by dividing 20 by 5, then dividing every term in the ratio by the answer.

**Sample Problems**

* Biscuits are made from butter and sugar in a ratio of 5:3. If 7 blobs of butter are used, how much sugar is needed?
* Audrey, Anne, and Andrew worked at their mother's shop for 1, 3, and 6 hours, respectively. If their mother gave them £300 to divide fairly, how much will each person get?

**Tips**

* Simplify ratios using a calculator or by finding the greatest common factor.
* Use ratios to solve problems involving proportional division, such as dividing a total amount among multiple people based on the amount of time they worked.

**Things You'll Need**

* Scientific calculator (if allowed by your math course)


To analyze model results from different countries on the same graph, I defined two key variables:
$$
\kappa = \frac{\log Q^s/C_0}{\log Q^d/C_0}
$$$$ \sigma = \frac{Q^s}{Q^s_{ref}}
$$
referred to as the information transfer index and the normalized monetary base, respectively. A constant $\alpha$ is defined as
$$ \alpha = \frac{Q^d_{ref}}{Q^s_{ref}} $$
allowing the equation to be written as
$$ P = \alpha \frac{1}{\kappa} \sigma^{1/\kappa-1} $$

Plotting the fits of the data for Japan and the US in the $\kappa\sigma$-plane, along with the location of the ridge line where $\frac{\partial P}{\partial\sigma} = 0$, showed that the ridge lines nearly coincided. To constrain the equations so that these lines coincide, I calculated the derivative
$$
\frac{\partial P(\kappa,\sigma)}{\partial\sigma} = 0
$$
which simplifies to
$$
-\frac{1}{\kappa}\log \left(\frac{Q^s_{\text{ref}}}{C_0}\right)+\log \left(\frac{\sigma Q_{\text{ref}}^s}{C_0}\right)+1=0
$$
and further to
$$ \sigma = \frac{C_0}{ Q_{\text{ref}}^s} \exp \left( -\frac{\kappa + \log \frac{C_0}{ Q_{\text{ref}}^s}}{\kappa} \right) $$
This function depends only on $\sigma$, $\kappa$, and $C_0/Q_{\text{ref}}^s \equiv \gamma$. Using US data to fix $\gamma$, I constrained subsequent fits for other countries to maintain $C_0 = \gamma Q_{\text{ref}}^s$, reducing one degree of freedom.

Fits for the EU, Japan, Mexico, and Indonesia were made to the two parameters $\alpha$ and $Q_{\text{ref}}^s$. With $\gamma$ fixed, all subsequent fits are effectively one-parameter fits. The results show that the US, Japan, and the EU have entered a space where monetary policy may be ineffective, while Mexico has recently entered a period of relative stagnation. The traditional quantity theory of money applies when $\kappa = 1/2$. All data is from the FRED database, using M1, Nominal GDP, and the GDP deflator, except for EU data which uses the Euro Area CPI.


This week's top 5 Number Theory problems and top 5 Algebra problems are presented below.

**Number Theory Problems**

1. **1999 JBMO P2**: For each nonnegative integer $n$, define $A_n = 2^{3n}+3^{6n+2}+5^{6n+2}$. Find the greatest common divisor of the numbers $A_0,A_1,\ldots, A_{1999}$.
	* $A_0 = 2^0 + 3^2 + 5^2 = 1 + 9 + 25 = 35$
	* $A_1 = 2^3 + 3^8 + 5^8 = 8 + 6561 + 390625 = 397194$
	* The greatest common divisor of $A_0$ and $A_1$ is 7.
2. **APMO, Evan Chen's orders modulo a prime handout**: Let $a,b,c$ be distinct integers. Given that $a | bc + b + c, b | ca + c + a$, and $c | ab + a + b$, prove that at least one of $a, b, c$ is not prime.
	* Assume $a, b, c$ are primes. Then $(b+1)(c+1) \equiv 1 \mod a$, and similarly for $b, c$.
	* Bound $\frac{(a+1)(b+1)(c+1)}{abc}$ and get a contradiction.
3. **IMO 2011 P1**: Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct positive integers, denote the sum $a_1 +a_2 +a_3 +a_4$ by $s_A$. Let $n_A$ denote the number of pairs $(i, j)$ with $1 \leq i < j \leq 4$ for which $a_i +a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$.
	* Show that $n_A \ne 6, 5$.
	* Show that $n_A = 4$ is possible.
4. **IMO 2011 P5**: Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m- n)$. Prove that, for all integers $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m)$.
5. **IMO Shortlist 2011 N3**: Let $n \geq 1$ be an odd integer. Determine all functions $f$ from the set of integers to itself, such that for all integers $x$ and $y$ the difference $f(x)-f(y)$ divides $x^n-y^n$.

**Algebra Problems**

1. **2010 AMC 10 A P21**: The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$?
	* Use Vieta's formula and factorize $2010 = 2 \cdot 3 \cdot 5 \cdot 67$.
	* The smallest possible value of $a$ is 78.
2. **NIMO summer contest P9**: The roots of the polynomial $P(x) = x^3 + 5x + 4$ are $r, s$, and $t$. Evaluate $(r + s) ^4 (s + t) ^4 (t + r)^ 4$.
	* Use Vieta's formula to get $r + s + t = 0$.
	* Evaluate $(rst)^4$.
3. **AIME 2008 II P7**: Let $r$, $s$, and $t$ be the three roots of the equation $8x^3+1001x+2008=0$. Find $(r+s)^3+(s+t)^3+(t+r)^3$.
	* Use Vieta's formula to get $r + s + t = 0$.
	* Evaluate $(r+s)^3+(s+t)^3+(t+r)^3 = -(r^3+s^3+t^3)$.
4. **2017 AMC 12 A P17**: There are $24$ different complex numbers $z$ such that $z^{24}=1$. For how many of these is $z^6$ a real number?
	* Consider $z^{\frac{2\pi \cdot i \cdot k}{24}} \cdot 6$.
	* The answer is $\boxed{12}$.
5. **JBMO 2012 Shortlist**: Let $a$, $b$, $c$ be positive real numbers such that $abc=1$. Show that $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab} \leq \frac{(ab+bc+ca)^2}{6}$.
	* Use AM-GM to get $a^3+bc \geq 2a$.
	* Try to get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{(ab+bc+ca)^2}{3}$.
	* Conclude with AM-GM again.


## Object Comparison

For object comparison, normal equality checks do not work due to different instances. If order is important, `deep.equal` or `include` can be used.

* `deep.equal` compares values without considering order.
* Example:
```javascript
expect({ a: 1, b: 2 }).to.not.equal({ b: 2, a: 1 });
expect({ a: 1, b: 2 }).to.include({ a: 1, b: 2 });
expect({ a: 1, b: 2 }).to.deep.equal({ a: 1, b: 2 });
expect({ a: 1, b: 2 }).to.deep.equal({ b: 2, a: 1 });
```

## Array Comparison

Normal equality checks do not work for arrays either.

* `deep.equal` can be used for array comparison, but it checks the order.
* Example:
```javascript
expect([1, 2, 3]).to.not.equal([1, 2, 3]);
expect([1, 2, 3]).to.not.equal([3, 2, 1]);
expect([1, 2, 3]).to.deep.equal([1, 2, 3]);
expect([1, 2, 3]).to.not.deep.equal([3, 2, 1]);
```
If order can be different, `members` can be used. However, it cannot be used to check if a value contains some values. In such cases, `include` can be used.

* Example:
```javascript
expect([1, 2, 3]).to.members([1, 2, 3]);
expect([1, 2, 3]).to.members([3, 2, 1]);
expect([1, 2, 3]).to.not.members([3, 2]);
expect([1, 2, 3]).to.include(2);
expect([1, 2, 3]).to.include(2).and.include(3);
expect([1, 2, 3]).to.be.an("array").that.includes(2);
```
To improve readability when checking for multiple values, combine `include` and `members`.

* Example:
```javascript
expect([1, 2, 3]).to.include.members([3, 1]);
```

## Object Array Comparison

This is the main focus of the article. Normal equality checks cannot be used. `deep.equal` can be used only if the order and values are the same.

* Example:
```javascript
expect([{ a: 1, b: 1 }, { a: 2, b: 2 }]).to.not.equal([{ a: 2, b: 2 }, { a: 1, b: 1 }]);
expect([{ a: 1, b: 1 }, { a: 2, b: 2 }]).to.deep.equal([{ a: 1, b: 1 }, { a: 2, b: 2 }]);
expect([{ a: 1, b: 1 }, { a: 2, b: 2 }]).to.not.deep.equal([{ a: 2, b: 2 }, { a: 1, b: 1 }]);
```
The solution is to combine `include` and `members` again. This makes the comparison pass even if the order of the array and the values in an object are different.

* Example:
```javascript
expect([{ a: 1, b: 1 }, { a: 2, b: 2 }]).to.include.deep.members([{ a: 1, b: 1 }, { a: 2, b: 2 }]);
expect([{ a: 1, b: 1 }, { a: 2, b: 2 }]).to.include.deep.members([{ a: 2, b: 2 }, { a: 1, b: 1 }]);
expect([{ a: 1, b: 1 }, { a: 2, b: 2 }]).to.include.deep.members([{ b: 2, a: 2 }, { b: 1, a: 1 }]);
```


The page "Beating Graham's number" by S. Ligocki raises two questions concerning the Busy Beaver function Σ() and Graham's number G:

(1) What is the least n such that Σ(n) > G?
(2) What is the least n for which it has been proved that Σ(n) > G?

The answer to (1) is not known, but it might be as small as 12 or so. This posting answers (2) with n = 72, assuming the behavior of the machine described here (GTM72) has been adequately established.

A rewriting algorithm for a fast-growing hierarchy considers the following "accelerated" fast-growing hierarchy:
gα(n) = n+1 if α = 0
= (gα-1)n+1(n+1) if α is a successor
= (gα[n+1])n+1(n+1) if α is a limit

This function-hierarchy is simulated by the following rewriting system for finite sequences of ordinals prefixed by a natural number n:
n α → {n+1} α if α = 0
→ {n+1} α-1n+1 if α is a successor
→ {n+1} α[n+1]n+1 if α is a limit

Every ordinal α in the range [0, ω2) has a unique representation as an ordered pair (i,j) of natural numbers denoting ω·i + j. The system can be reformulated as a rewriting system for sequences of such pairs:
n (i,j) → {n+1} (i,j) if i = j = 0
→ {n+1} (i,j-1)n+1 if j > 0
→ {n+1} (i-1,n+1)n+1 if i > 0, j = 0

A "Busy Beaver"-class Turing machine that halts with more than Graham's number of 1s on the tape needs only to simulate this pair-rewriting system after initializing the tape to represent relatively small values of n, i, j. The function-hierarchy actually simulated by the Turing machine described below is hα(n) = n+1 if α = 0
= (hα-1)n+2(n+4) if α is a successor
= (hα[n+1])n+5(n+7) if α is a limit

The machine we construct uses a two-symbol tape alphabet {0,1}, in which 0 is the blank symbol and natural numbers are coded in unary. The machine begins by writing <n><i><j> = <0><2><1> = "10 1110 110" on an initially all-0 tape.

The "raw" TM has 77 states, five of which will later be eliminated by combining ten unused "half states". The machine GTM77 can be reduced to 72 states by reclaiming unused "half-states" and eliminating superfluous states.

The revised machine, GTM72, has been further reduced to 64 states by restricting the rewriting algorithm to ordinals less than ω2 and eliminating virtual duplication in blocks v-z. The number computed is now hω+2(0) + 1 >> G.

Four additional states were eliminated: c2, e6, l5, and p5, which were superfluous. The half-state e6 0 had been relabelled as the until-then-unused half-state e4 0, so eliminating e6 frees up the half-state e4 1 to be relabelled as some other unused "half-state".

The GTM72 machine is as follows:
BEGIN GTM72
alphanumeric numerical state-labels state-labels
------------ ------------
a1 0 1 L a2 01 0 1 L 02
a1 1 0 L a4 01 1 0 L 04
...
z1 0 1 L z1 72 0 1 L 72
z1 1 1 R x4 72 1 1 R 68
END GTM72

Note: The full transition table for GTM72 is provided, but some parts have been omitted for brevity.


To find the area of an octagon based on its radius, we consider the radius as the distance from the midpoint to each corner, forming a circumscribed circle. Drawing line segments from this center point to the eight vertices of the octagon creates 8 congruent triangles. Since the sum of the angles around the center point is 360 degrees, each angle is 45 degrees.

Each triangle is isosceles with two legs equal to the radius and a 45-degree angle between them. To find the area, we need the base and height, which can be determined by drawing a perpendicular bisector from the 45-degree angle, creating two congruent right triangles.

Using trigonometry with a radius of 16.2 inches, we find the base (x) and height (y) of the right triangle:
- Sin(22.5) = x/16.2, so x = 6.199 inches
- Cos(22.5) = y/16.2, so y = 14.967 inches

The base of the isosceles triangle is 2x, and the height is y. The area of one triangle is 1/2 * base * height = x * y = 92.787 square inches. Multiplying this by the 8 triangles in the octagon gives a total area of approximately 742 square inches.


To generate random numbers in Google Sheets, you can use the RAND function. The RAND function has no arguments and returns a random number between 0 (inclusive) and 1 (exclusive). To generate a random whole number between a specified low and high value, you can use the RANDBETWEEN function.

The syntax for the RANDBETWEEN function is: `=RANDBETWEEN(low, high)`, where `low` is the smallest number and `high` is the largest number. For example, to generate a random number between 1 and 10, you would use the formula: `=RANDBETWEEN(1, 10)`.

To generate multiple random numbers, you can drag the fill handle (located at the lower right corner of the cell) down to the desired number of cells. This will instantly give you the specified number of random numbers in the selected cells.

If you want to generate random numbers without repeats, you can use the following formula: `=RANDBETWEEN(low, high)`. However, this formula may return duplicate values. To avoid duplicates, you can use a more complex formula or a script.

To stop random numbers from changing, you can set the calculation options to manual. To do this, open the Excel workbook, click on the formula tab, and then click on calculation options. In the drop-down list, select "manual". This will prevent the random numbers from changing every time the sheet is recalculated.

Here are a few useful formulas for generating random numbers in Google Sheets:

* `=RAND()` - returns a random number between 0 and 1
* `=RANDBETWEEN(low, high)` - returns a random whole number between low and high
* `=RAND() * (high - low) + low` - returns a random decimal number between low and high

You can use these formulas to generate random numbers for a variety of purposes, such as drawing winning numbers for a raffle or creating random data for testing.


The Chi-Square Independence Test is used to determine whether two categorical variables associated with the same item act independently on that item. This test answers the question of whether one variable plays a role in the selection of the other variable.

A contingency table is a two-way cross-tabulation of the counts of the number of items having each unique combination of row and column attribute levels. The number of rows in the contingency table, r, is equal to the number of levels of the row attribute, and the number of columns, c, is equal to the number of levels of the column attribute.

The Chi-Square Independence Test compares whether counts of the actual data for each unique combination of factors of the two variables are significantly different than the counts that would be expected if the attributes were totally independent of each other. The Null Hypothesis states that there is no significant difference between the actual and expected counts of data for the unique combinations of levels of the two factors.

The Test Statistic, Χ^{2}, is calculated and compared to a Critical Chi-Square Value based on the specified alpha level and degrees of freedom associated with the test. The Null Hypothesis is rejected if the Chi-Square Statistic is larger than the Critical Chi-Square Value or if the p-value derived from the test is smaller than the specified alpha level.

To perform the Chi-Square Independence Test in Excel, follow these 7 steps:

1. Place the actual counts in a contingency table.
2. Place the expected counts in a contingency table.
3. Create the Null and Alternative Hypotheses.
4. Verify the required assumptions for the test.
5. Calculate the Chi-Square Statistic, Χ^{2}.
6. Calculate the Critical Chi-Square Value and p-value.
7. Determine whether to reject the Null Hypothesis.

The required assumptions for the test are:
1. The number of cells in the contingency table (r x c) is at least 5.
2. The average value of all of the expected counts is at least 5.
3. All of the expected counts equal at least 1.

The degrees of freedom for the Chi-Square Independence Test are calculated as df = (r – 1)(c – 1). The Critical Chi-Square Value is calculated using the CHISQ.INV.RT function in Excel, and the p-value is calculated using the CHISQ.DIST.RT function.

For example, suppose we want to determine whether the gender of a car purchaser is independent of the color of the car. We collect data on the number of cars purchased by each gender for each color and create a contingency table. We then perform the Chi-Square Independence Test using the 7 steps outlined above. If the Null Hypothesis is rejected, we conclude that the gender of the car purchaser is not independent of the color of the car.


In a capstone class for future secondary math teachers, students were asked to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum, specifically focusing on three ways to spark interest in the topic. One student, Cody Luttrell, explored the topic of graphing parabolas from Algebra.

Understanding the graph of a parabola is crucial for students' future math and science classes. In Algebra II, students will encounter more complex uses of quadratic functions, such as complex numbers. If a parabola does not cross the x-axis, the solution will be imaginary, and a lack of understanding of parabola graphs may hinder comprehension of this topic. In pre-calculus, knowledge of parabola transformations will aid in understanding transformations of other functions, including cubic, square root, and absolute value functions.

In Physics, understanding parabola graphs benefits students when dealing with equations of projectiles. Recognizing symmetry in a parabola helps determine the position of a projectile at a certain time, given its maximum height.

The parabola shape appears frequently in art and architecture. A potential engagement activity involves showing students a PowerPoint presentation of real-world parabolas, such as arches in bridges, roller coasters, water fountains, and other examples. This can help students identify the pattern and recognize parabolas in various objects. After learning about quadratics, students can explore how parabola transformations apply to architecture, discussing how to modify arches in bridges or create taller bridges.

Technology can also be utilized to introduce graphing parabolas. A YouTube video (https://www.youtube.com/watch?v=E_0AHIaK48A) demonstrates how parabolas are used in famous videogames like Mario Bros. The video shows Mario and Luigi's jumping paths, which follow parabolic shapes, and explains how their movements can be represented by equations like y=-x^2 and y=-1.5x^2, introducing concepts like compression and stretches.


Tessellations occur when a shape is repeated in an interlocking pattern that fully covers a flat surface, or plane. Some shapes cannot tessellate because they are not regular polygons or do not contain vertices, and therefore cannot be arranged on a plane without overlapping or leaving space uncovered. The circle, due to its rounded edges and lack of vertices, is normally not tessellated.

There are three main categories of tessellations: regular, semi-regular, and demi. Regular tessellations use one repeating polygon, producing an identical pattern at each vertex, and are constructed with triangles, squares, or hexagons. These tessellations are named according to the number of sides in each intersecting shape, such as a 4.4.4.4 tessellation.

Semi-regular tessellations are composed of two or more regular polygons, with combinations of shapes including triangles, squares, and hexagons. Like regular tessellations, the pattern at each vertex is the same, but the pattern contains more than one number. For example, a pattern using triangles and hexagons is a 3.3.3.3.6 tessellation, where four triangles and one hexagon meet at each vertex.

Demi tessellations, on the other hand, do not follow normal rules of constructing a tessellation and may contain irregular polygons, curved shapes, and non-identical vertices. There is controversy among mathematicians regarding the definition of demi tessellations.


A consumer has preferences represented by the utility function u(x, y) = x^(1/2)*y^(1/2). The marginal utilities are MUx = (1/2)x^(-1/2)*y^(1/2) and MUy = (1/2)x^(1/2)*y^(-1/2).

a. The marginal rate of substitution (MRS) is given by MRS = MUx/MUy = (1/2)x^(-1/2)*y^(1/2) / (1/2)x^(1/2)*y^(-1/2) = y/x.

b. Given the prices of good x (Px = 2) and good y (Py = 1), and the consumer's income (I = 20), we can find the optimal quantities of x and y. Setting MUx/MUy = Px/Py, we get y/x = 2/1, which simplifies to y = 2x. Substituting this into the budget constraint I = Px*x + Py*y, we get 20 = 2x + 2x, which simplifies to 4x = 20, so x = 5. Then, y = 2x = 10.

c. If the price of good x decreases to 1, we can find the new optimal quantities of x and y. Setting MUx/MUy = Px/Py, we get y/x = 1/1, which simplifies to y = x. Substituting this into the budget constraint I = Px*x + Py*y, we get 20 = x + y, which simplifies to 2x = 20, so x = 10. Then, y = x = 10.

d. To find the own price elasticity of demand for good x, we need to calculate the percentage change in quantity and the percentage change in price. The average quantity is (5+10)/2 = 7.5, and the average price is (2+1)/2 = 1.5. The change in quantity is 10-5 = 5, and the change in price is 1-2 = -1. The percentage change in quantity is (5/7.5)*100 = 66.67%, and the percentage change in price is (-1/1.5)*100 = -66.67%. The own price elasticity of demand is (66.67%/-66.67%) = -1.


### Analyzing Tile Patterns

We are given a series of tile patterns represented in figures. The first set of figures shows a progression from a 2-by-2 square to a 3-by-3 square and then to a 4-by-4 square.

- Figure 1 is a 2-by-2 square.
- Figure 2 is a 3-by-3 square.
- Figure 3 is a 4-by-4 square.

This pattern suggests that each figure is a square with a side length one greater than its figure number. Therefore, Figure *x* would be an (x+1)-by-(x+1) square.

For Figure 10, following the pattern, it would be an 11-by-11 square, containing 121 tiles (since 11*11 = 121).

To find what Figure 100 would look like, we apply the same pattern: it would be a 101-by-101 square.

Additionally, we are introduced to Cami's tile pattern, represented in a table:

| Figure Number | 1 | 2 | 3 | 4 |
| --- | --- | --- | --- | --- |
| Number of Tiles | 5 | 9 | 13 | 17 |

The pattern in the table shows an increase of 4 tiles from one figure to the next. To predict the number of tiles in Figure 5 and Figure 8 of Cami's pattern:

- For Figure 5, starting from Figure 4 with 17 tiles, we add 4 for each subsequent figure: Figure 5 would have 17 + 4 = 21 tiles, Figure 6 would have 21 + 4 = 25 tiles, Figure 7 would have 25 + 4 = 29 tiles, and Figure 8 would have 29 + 4 = 33 tiles.
- Therefore, Figure 5 would have 21 tiles, and Figure 8 would have 33 tiles.


# Diving Deeper into Reinforcement Learning with Q-Learning

Q-Learning is a value-based Reinforcement Learning algorithm used to find the optimal action-selection policy using a Q-function. This article is part of the Deep Reinforcement Learning Course with Tensorflow.

## Introduction to Q-Learning

To understand Q-Learning, let's consider a scenario where a knight needs to save a princess trapped in a castle. The knight can move one tile at a time, and the goal is to reach the castle by the fastest route possible. The knight loses -100 points if it touches an enemy and wins +100 points if it reaches the castle.

## The Q-Table

A Q-table is a table where we calculate the maximum expected future reward for each action at each state. The Q-table has columns representing the four possible actions (left, right, up, down) and rows representing the states. The value of each cell is the maximum expected future reward for that given state and action.

## Q-Learning Algorithm

The Q-Learning algorithm is used to learn the Action Value Function, which takes two inputs: state and action. It returns the expected future reward of that action at that state. The algorithm consists of the following steps:

1. **Initialize Q-values**: Build a Q-table with m columns (m = number of actions) and n rows (n = number of states). Initialize the values at 0.
2. **Choose an action**: Choose an action a in the current state s based on the current Q-value estimates. In the beginning, use the epsilon-greedy strategy to balance exploration and exploitation.
3. **Evaluate**: Take the action a and observe the outcome state s' and reward r. Update the Q-function using the Bellman Equation.

## Bellman Equation

The Bellman Equation is used to update the Q-function:

`New Q value = Current Q value + lr * [Reward + discount_rate * (highest Q value between possible actions from the new state s') - Current Q value]`

## Implementing Q-Learning

To implement Q-Learning, we can use a Q-table to store the expected future rewards for each state and action. We can then use the Bellman Equation to update the Q-table as we explore the environment.

## Recap

* Q-Learning is a value-based Reinforcement Learning algorithm that finds the optimal action-selection policy using a Q-function.
* The Q-table helps us find the best action for each state.
* The goal is to maximize the expected reward by selecting the best of all possible actions.
* The Q-function can be estimated using Q-Learning, which iteratively updates Q(s,a) using the Bellman Equation.

## Next Steps

In the next article, we will explore Deep Q-Learning, one of the biggest breakthroughs in Deep Reinforcement Learning in 2015. We will train an agent to play Doom and kill enemies. The Deep Reinforcement Learning Course with Tensorflow consists of six parts:

1. Introduction to Reinforcement Learning
2. Diving Deeper into Reinforcement Learning with Q-Learning
3. Introduction to Deep Q-Learning
4. Introduction to Policy Gradients
5. Introduction to Advantage Actor-Critic methods
6. Proximal Policy Optimization (PPO)


A group of 3 undergraduates and 5 graduate students are available to fill a position. If 4 students are to be randomly selected from this group, find the probability that exactly 2 undergraduates will be among the chosen 4.

To solve this problem, we can use combinations. The total number of ways to select 4 students from the group of 8 is $\displaystyle \begin{pmatrix} 8 \\ 4 \end{pmatrix} = \frac{8!}{4!(4!)}$. The number of ways to choose exactly 2 undergraduates is $\displaystyle \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \frac{3!}{2!(1!)}$, and the number of ways to choose 2 graduate students is $\displaystyle \begin{pmatrix} 5 \\ 2 \end{pmatrix} = \frac{5!}{2!(3!)}$.

The probability of selecting exactly 2 undergraduates is therefore $\displaystyle \frac{\begin{pmatrix} 3 \\ 2 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix}}{\begin{pmatrix} 8 \\ 4 \end{pmatrix}} = \frac{(\frac{3!}{2!(1!)})(\frac{5!}{2!(3!)})}{\frac{8!}{4!(4!)}} = \frac{(3)(10)}{70} = \frac{30}{70} = \frac{3}{7}$.

The final answer is $\frac{3}{7}$.


James W. Wilson and William G. Daly

University of Georgia

Constructing graphs to consider the equation $x^2 + bx + c = 0$ and overlaying several graphs for different values of $a$, $b$, or $c$ as the other two are held constant can help discuss the patterns for the roots of the equation. For example, setting $b = -3, -2, -1, 0, 1, 2, 3$ and overlaying the graphs, we can discuss the "movement" of a parabola as $b$ is changed. The parabola always passes through the same point on the y-axis, $(0,1)$, which is the y-intercept found by setting $x=0$ and will always be the point $(0,c)$.

For $b < -2$, the parabola will intersect the x-axis in two points with positive x values, resulting in two real roots, both positive. For $b = -2$, the parabola is tangent to the x-axis, resulting in one real and positive root at the point of tangency. For $-2 < b < 2$, the parabola does not intersect the x-axis, resulting in no real roots. For $b = 2$, the parabola is tangent to the x-axis, resulting in one real negative root, and for $b > 2$, the parabola intersects the x-axis twice, resulting in two negative real roots for each $b$.

The locus of the vertices of the set of parabolas graphed from $x^2 + bx + c = 0$ can be found by recalling that the x-value of the vertex is $-b/(2a)$. Substituting this into $x$ in the quadratic yields the y-value at the vertex: $y = c - \frac{b^2}{4a}$. The ordered pair at the vertex of the parabola is $\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)$.

Solving for $b$ at each of these ordered pair values by manipulating $x = -\frac{b}{2a}$ and $y = c - \frac{b^2}{4a}$ results in $b(x) = -2ax$ and $b(y) = \pm \sqrt{4a(c-y)}$. Setting these two equations equal and solving for $y$ gives the equation for the locus of points for the vertices of the family of parabolas: $y = c - ax^2$.

Graphing the relation $x^2 + bx + c = 0$ in the $xb$ plane results in a curve. For each value of $b$, we get a horizontal line. It is clear that we get two negative real roots of the original equation when $b > 2$, one negative real root when $b = 2$, no real roots for $-2 < b < 2$, one positive real root when $b = -2$, and two positive real roots when $b < -2$.

Considering the case when $c = -1$ rather than $c = 1$, it is clear that for $c = -1$, there are real roots for any value of $b$. Notice that the roots in this case are always opposite in sign. The behavior of these graphs is illuminated by examining the quadratic formula for $a = 1$ and $c = 1$: $x = \frac{-b \pm \sqrt{b^2 - 4}}{2}$. For $c = -1$, the quadratic formula is $x = \frac{-b \pm \sqrt{b^2 + 4}}{2}$. Notice that the expression under the radical can never be negative for any value of $b$.

In addition to the algebraic interpretation of the graphs in the $xb$ plane, there is also a geometric interpretation. The line $b = -2x$ represents the case where the radical in the quadratic formula is zero. Where $b = -2x$ intersects the locus of roots in the $xb$ plane, at $(-1/2, 2)$ and $(1/2, -2)$, represents a single root. As one moves further along the line $b = -2x$, the roots are moving, equidistant in the horizontal direction, off of the $b = -2x$ line.

The equation $x^2 + 5x + c = 0$ can be graphed in the $xc$ plane, resulting in a parabola. For each value of $c$ considered, its graph will be a line crossing the parabola in 0, 1, or 2 points, the intersections being at the roots of the original equation at that value of $c$. There is one value of $c$ where the equation will have only 1 real root, at $c = 6.25$. For $c > 6.25$, the equation will have no real roots, and for $c < 6.25$, the equation will have two roots, both negative for $0 < c < 6.25$, one negative and one 0 when $c = 0$, and one negative and one positive when $c < 0$. The expression under the radical $(b^2 - 4ac)$ for this equation is $25 - 4c$.

What is the y-intercept of the parabola $x^2 + bx + c = 0$?
A) $(0,0)$
B) $(0,c)$
C) $(0,-c)$
D) $(c,0)$

Answer: B) $(0,c)$

What is the equation for the locus of points for the vertices of the family of parabolas $x^2 + bx + c = 0$?
A) $y = c + ax^2$
B) $y = c - ax^2$
C) $y = c + bx$
D) $y = c - bx$

Answer: B) $y = c - ax^2$


**Understanding Median and Its Calculation**

In statistics, median, mean, and mode are three primary methods for calculating averages. The median is a positional average, calculated based on its position in a series of values. It is the middle value of a dataset, and its calculation is essential for framing correct median homework answers.

**Calculating the Median**

The standard formula for calculating the median of a list of data is:

Median = {(n + 1) / 2}

where "n" is the number of values in the dataset. This formula is used for datasets with an odd number of values. For datasets with an even number of values, the method is slightly different.

**Step-by-Step Method**

To calculate the median, follow these steps:

1. Arrange the given numbers in ascending order.
2. For a dataset with an even number of values, consider the pair of middle values. For a dataset with an odd number of values, the median is the middle number.
3. For a dataset with an even number of values, add the pair of central values and divide the sum by 2 to calculate the median.

**Importance of Correct Calculation**

Calculating the median involves following standard formulas, which can be confusing for students. The median is also used in probability and population distribution evaluation, making it a crucial technique in various significant evaluation procedures. Therefore, learning the correct method for solving median assignment answers is necessary.

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Assuming a cat is at least 4 months old, she can have a litter in two months. With an average of 5 kittens per litter, she initially has 5 descendants. Two months later, at a total of 4 months, she has another litter of 5 kittens, bringing the total to 20 descendants. As each of the first 5 kittens becomes pregnant, in another two months (total 6 months), the cat and her first 5 kittens each have a litter of 5 kittens, resulting in 30 new kittens and a total of 50 descendants.

As this process continues, in another two months (total 8 months), the cat, her first 5 kittens, and her second litter of 5 kittens all have kittens, producing 55 new kittens and a total of 105 descendants. This exponential growth in the number of descendants will continue as the months progress.


# Vectors and Matrices

A matrix is a collection of numbers arranged in rows and columns, such as the array `try1` with 2 rows and 3 columns. Vectors are special cases of matrices, where a row vector has one row and a column vector has one column. For example, the horizontal list `[-pi/2, 0, pi/2, pi]` and the vertical list `vlist` are both vectors.

Matlab provides various ways to build and manipulate matrices. To create a new matrix, you can use the syntax `try2 = [try1 ; 2*try1]`, resulting in a 4x3 matrix. To extract a specific element, use `try2(3,2)`. To extract columns, use `try3 = try2(:,[1,3])`, where the colon `:` represents all rows.

Matrix arithmetic operations, such as addition and subtraction, require matrices to have the same dimensions. For example, `gross_weights = [12.3 23.4 34.5]` and `crate_weights = [2.3 3.4 4.5]` can be subtracted to get `net_weights`. Element-wise multiplication and division use the `.*` and `./` operators, respectively.

Matrix multiplication uses the `*` operator and requires the number of columns in the first matrix to match the number of rows in the second matrix. The resulting matrix has the same number of rows as the first matrix and the same number of columns as the second matrix. For example, if `A` is a 1x3 row vector and `B` is a 3x1 column vector, the product `A*B` is a scalar value, also known as the dot product.

To find the inverse of a matrix, use the `inv` function, such as `Minv = inv(M)`. Matlab also provides the `/` and `\` operators for inverse calculations, which are preferable for numerical accuracy. For example, `A*inv(B)` is equivalent to `A/B`, and `inv(B)*A` is equivalent to `B\A`.

Additional matrix operations in Matlab include finding eigenvalues with `eig(M)` and eigenvectors with `[W,D] = eig(M)`. The resulting matrix `W` contains the eigenvectors, and the diagonal matrix `D` contains the corresponding eigenvalues. You can verify the relationship between `M`, `W`, and `D` by checking if `M*W` equals `W*D`.


In this tutorial, you'll learn how to add inferential statistics to a Power BI report by performing a t-test using RStudio. A **t-test** is a statistical method that compares the mean between two groups to determine if there's a significant difference between them.

To perform a t-test, it's essential to have a basic understanding of statistics. If you're not familiar with statistics, it's recommended to review and study it before proceeding with this tutorial.

## Executing a T-Test in RStudio

To start, open a blank R script and install the necessary packages, including the **broom package**, which transforms t-test results into a tabular format. The **t.test ( )** function is used to perform the t-test, where you specify the two groups to compare, separated by the tilde sign ( **~** ).

For example, you can compare the average price of houses by preferred area. The t-test results will indicate if there's a significant difference between the means. However, if the confidence interval yields negative results, you can transform them into positive numbers by reversing the factor levels using the **fct_rev ( )** function from the **forcats package**.

## Formatting the Results

The results of a t-test in RStudio can be difficult to read. To improve the format, use the **tidy ( )** function, which places each value in a table with individual columns and proper labels. This makes it easier to understand the results of the p-value and confidence interval. The **tidy ( )** function works with most statistical tests in RStudio and can be used to present results in a more readable format.

## Conclusion

Inferential statistics, such as t-tests, add depth and substance to Power BI reports. A t-test is a simple and commonly used statistical test that's easy to interpret, even with basic knowledge of statistics. By following this tutorial, you can perform a t-test in RStudio and effectively present the results in a Power BI report.


In mathematics, a set is a collection of distinct objects, considered as an object in its own right. For example, the numbers 1, 2, and 6 are distinct objects when considered as a set. Set theory and its applications are crucial throughout mathematics, where a class is a collection of sets or sometimes other mathematical objects that can be unambiguously defined.

The study of sets and set theory involves basic definitions and notation, types of sets, equality, and Venn diagrams. Understanding the relationship between equal sets and subsets is essential, as well as determining the number of subsets a given set can have. There are different methods to describe sets, including the verbal method, the roster method, and the set-builder method.

Set theory has various applications, including the convergence of Fourier series, where it originated. It is also related to basic logic and axiomatic set theory. Identifying sets and set theory is important in quantitative reasoning, as it provides a foundation for more advanced mathematical concepts.

Key concepts in set theory include subsets, proper subsets, set operations, and interval notation. A subset is a set whose elements are all members of another set, while a proper subset is a subset that is not equal to the original set. Set operations, such as union and intersection, are used to combine sets and extract new information.

Understanding set theory is crucial in various mathematical contexts, including real numbers, algebra, and logic. It provides a framework for defining and manipulating collections of objects, making it a fundamental tool in mathematics and related fields. By mastering set theory, individuals can develop a deeper understanding of mathematical concepts and improve their problem-solving skills.


Algebra: Prime Time Problems

The following problems require proof:

1. Prove that no square number leaves a remainder of 2 when divided by 3. Consider integers of the form 3n, 3n+1, and 3n+2, and square them to demonstrate this property.
2. Prove that no square number has a remainder of 2 or 3 when divided by 5.
3. Prove that no square number has a remainder of 3 or 5 when divided by 7.
4. Prove that every prime number greater than 3 can be expressed in the form 6n+1 or 6n-1.
5. Prove that for every prime number n, the expression (n^2 - 1) / 24 yields an integer.