A fact family is a group of three numbers where two of them add up to the third, introducing students to the relationship between addition and subtraction in primary arithmetic. Fact families illustrate number relationships and the conceptual relationship between addition and subtraction through four different equations.

For example, using the numbers 3, 5, and 8, the equations demonstrate this relationship. Given two numbers, two fact families can be built. Using 5 and 7 as an example, the third number can be found by either adding the two numbers together or subtracting the smaller number from the larger one. In subtraction, it is essential to subtract the smaller number from the larger to avoid resulting in a negative number.


Find the center C and the radius a for the spheres in Exercises 51–52.

51. (x + 2)2 + y2 + (z - 2)2 = 8
52. x2 + y2 + z2 + 4x - 4z = 0

To find the center and radius, we need to rewrite the equations in standard form.

For equation 51:
(x + 2)2 + y2 + (z - 2)2 = 8
(x - (-2))2 + (y - 0)2 + (z - 2)2 = 8
The center is (-2, 0, 2) and the radius is √8 = 2√2.

For equation 52:
x2 + y2 + z2 + 4x - 4z = 0
(x2 + 4x) + y2 + (z2 - 4z) = 0
(x2 + 4x + 4) + y2 + (z2 - 4z + 4) = 4 + 4
(x + 2)2 + y2 + (z - 2)2 = 8
The center is (-2, 0, 2) and the radius is √8 = 2√2.


**Mathematical Functions**

The following mathematical functions are defined:

1. **ABS**: Computes the absolute value of a number.
2. **ACOS**: Computes the arc cosine of a number in radians.
3. **ACOSH**: Computes the inverse hyperbolic cosine of a number.
4. **ASIN**: Computes the arc sine of a number in radians.
5. **ASINH**: Computes the inverse hyperbolic sine of a number.
6. **ATAN2**: Computes the arc tangent of an x and an y coordinate.
7. **ATANH**: Computes the inverse hyperbolic tangent of a number.
8. **CEILING**: Rounds a number up, away from zero, to the nearest multiple of significance.
9. **COMBIN**: Computes the possible number of groups of size number-chosen that can be formed from number objects.
10. **COS**: Computes the cosine of a number.
11. **COSH**: Computes the hyperbolic cosine of a number.
12. **DEGREES**: Converts an angle in radians into degrees.
13. **ECMA**: Rounds a number up, away from zero, to the nearest multiple of a given significance.
14. **EVEN**: Rounds a number to the nearest even integer, away from zero.
15. **EXP**: Computes e^x, where the constant e is the base of the natural logarithm.
16. **FACT**: Computes the factorial of a number.
17. **FACTDOUBLE**: Computes the double factorial of a number.
18. **FLOOR**: Rounds a number down, toward zero, to the nearest multiple of a given significance.
19. **GCD**: Computes the greatest common divisor of one or more numbers.
20. **INT**: Rounds a number down to an integer.
21. **ISO**: Rounds a number up, to the nearest multiple of a given significance.
22. **LCM**: Computes the least common multiple of one or more arguments.
23. **LN**: Computes the natural logarithm of a number.
24. **LOG**: Computes the logarithm of x to the base base.
25. **LOG10**: Computes the base-10 logarithm of a number.
26. **MOD**: Computes the remainder when a number is divided by another number.
27. **MROUND**: Rounds a number to the nearest multiple of another number, away from zero.
28. **ODD**: Rounds a number to the nearest odd integer, away from zero.
29. **PI**: Returns the value π.
30. **POWER**: Computes a number raised to the power of a second number.
31. **PRODUCT**: Multiplies the numeric values of the arguments.
32. **QUOTIENT**: Computes the integer portion of the division of two given numbers.
33. **RADIANS**: Converts an angle in degrees into radians.
34. **RAND**: Computes an evenly distributed random real number greater than or equal to 0 and less than 1.
35. **RANDBETWEEN**: Computes a random integer number in a given range.
36. **ROUND**: Rounds a number to a specified number of digits.
37. **ROUNDDOWN**: Rounds a number down, toward zero, to a specified number of digits.
38. **ROUNDUP**: Rounds a number up, away from zero, to a specified number of digits.
39. **SIGN**: Determines the sign of a number.
40. **SIN**: Computes the sine of a number.
41. **SINH**: Computes the hyperbolic sine of a number.
42. **SQRT**: Computes the positive square root of a number.
43. **SQRTPI**: Computes the positive square root of a number × π.
44. **SUM**: Adds the numeric values of the arguments.
45. **SUMIF**: Applies selection criteria on the values in one range of cells and sums the values of the cells in a corresponding range.
46. **TAN**: Computes the tangent of a number.
47. **TANH**: Computes the hyperbolic tangent of a number.

**Statistical Functions**

1. **AVERAGE**: Computes the arithmetic mean of the numeric values of its arguments.
2. **AREAS**: Finds the number of areas designated by reference.
3. **MAX**: Computes the largest of a set of numbers.
4. **MAXA**: Computes the largest of a set of numbers.
5. **MIN**: Computes the smallest of a set of numbers.
6. **MINA**: Computes the smallest of a set of numbers.

**Reference Functions**

1. **ADDRESS**: Creates a cell address, given the specified row and column numbers.
2. **ROW**: Finds the number of the row(s) corresponding to a given reference.


The chain rule is a method of differentiating a function within a function. We know from basic differentiation that $\frac{d}{dx}(g(x)) = \lim_{h\to 0}\left(\frac{g(x+h) - g(x)}{h}\right)$. Let $v = \frac{g(x+h) - g(x)}{h} - g'(x)$, clearly $v\to 0$ as $h\to 0$. This idea can be extended to a function of a function. Let $w = \frac{f(y+k)-f(y)}{k} - f'(y)$, clearly $w\to 0$ as $k\to 0$. Rearranging these equations gives $g(x+h) = g(x) + \{g'(x)+v\}h$ and $f(y+k) = f(y) + \{f'(y)+w\}k$.

If we let $k = \{g'(x)+v\}h$ and $y=g(x)$, clearly $k\to0$ as $h\to0$. This reduces the equation to $f(g(x) + \{g'(x)+v\}h) = fg(x) + \{f'g(x)+w\}\{g'(x)+v\}h$. The left-hand side of this statement is equivalent to $fg(x+h)$, and we are now in a position to simplify $\frac{fg(x+h)-fg(x)}{h}$.

$\frac{fg(x+h)-fg(x)}{h} \equiv \frac{fg(x)+\{f'g(x)+w\}\{g'(x)+v\}h - fg(x)}{h} \equiv \{f'g(x)+w\}\{g'(x)+v\}$. As $h\to0$, $k,v\to0$, and $w\to0$ as $k\to0$. $\Rightarrow \lim_{h\to0}\{f'g(x)+w\}\{g'(x)+v\} = g'(x)f'g(x)$. $\frac{d}{dx}(fg(x)) = g'(x)f'g(x)$, which is the chain rule.

To prove the product rule using logarithmic differentiation, consider $y = uv$ for some functions of $x$: $u, v$. This means that $\ln y = \ln uv$ and $\ln y = \ln u + \ln v$. Differentiating both sides implicitly gives $\frac{1}{y} \frac{dy}{dx} = \frac{u'}{u} + \frac{v'}{v}$ and $\frac{1}{y} \frac{dy}{dx} = \frac{u'v + v'u}{uv}$. Since $y = uv$, we can multiply through by $uv$ to get $\frac{dy}{dx} = u'v + v'u$, which is the product rule.

Logarithmic differentiation also has useful applications to more complicated derivatives. For example, given $y = xe^x\cos (1+x^2)$, find $\frac{dy}{dx}$. Taking logs of both sides and rearranging gives $\ln y = \ln x + x\ln e + \ln \cos (1+x^2)$. Differentiating both sides implicitly gives $\frac{1}{y}\frac{dy}{dx} = \frac{1}{x} + 1 - \frac{2x\sin(1+x^2)}{\cos(1+x^2)}$. Multiplying through by $y$ gives $\frac{dy}{dx} = e^x\left((x+1)\cos(1+x^2) - 2x^2\sin(1+x^2)\right)$.

The quotient rule can also be proven using logarithmic differentiation. Consider $y = \frac{u}{v}$ for some functions of $x$: $u, v$. This means that $\ln y = \ln \frac{u}{v}$ and $\ln y = \ln u - \ln v$. Differentiating both sides implicitly gives $\frac{1}{y} \frac{dy}{dx} = \frac{u'}{u} - \frac{v'}{v}$ and $\frac{1}{y} \frac{dy}{dx} = \frac{u'v - v'u}{uv}$. Since $y = \frac{u}{v}$, we can multiply through by $uv$ to get $\frac{dy}{dx} = \frac{u'v - v'u}{v^2}$, which is the quotient rule.

To evaluate $\frac{d}{dx}\left(\frac{3x^2\sin x}{\sqrt{x^2+1}}\right)$, we can use logarithmic differentiation. Taking logs of both sides and rearranging gives $\ln y = \ln 3 + 2\ln x + \ln \sin x - \frac{1}{2}\ln (x^2+1)$. Differentiating both sides implicitly gives $\frac{1}{y}\frac{dy}{dx} = \frac{2}{x} + \frac{\cos x}{\sin x} - \frac{x}{x^2+1}$. Multiplying through by $y$ gives $\frac{dy}{dx} = \frac{3x^2\sin x}{\sqrt{x^2+1}}\left(\frac{2}{x} + \frac{\cos x}{\sin x} - \frac{x}{x^2+1}\right)$.

In a triangle $ABC$, if we connect the vertices to the midpoint of the opposite side, they all meet at a common point (they are concurrent), this point is called the centroid. The line through a vertex which is perpendicular to the opposite side is called an altitude. The three altitudes of a triangle meet at a common point (they are concurrent) if this point is called the orthocentre.

Given a triangle $ABC$, with midpoints of $BC$, $AC$, and $AB$ denoted as $F$, $E$, and $D$ respectively, and $Q$ as the point of intersection of $AF$ and $BE$, we can show that $Q$ lies on the line connecting $C$ to the midpoint of $AB$. Let $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$, and $\overrightarrow{OC} = \mathbf{c}$. Then $\overrightarrow{OD} = \mathbf{\frac{1}{2}(a+b)}$, $\overrightarrow{OE} = \mathbf{\frac{1}{2}(a+c)}$, and $\overrightarrow{OF} = \mathbf{\frac{1}{2}(b+c)}$. The vector equations for the lines $AF$ and $BE$ are $\mathbf{r_{AF}} = \mathbf{a} + \lambda (\mathbf{b + c - 2a})$ and $\mathbf{r_{BE}} = \mathbf{b} + \mu (\mathbf{a+c-2b})$. They intersect when $\mathbf{r_{AF}} = \mathbf{r_{BE}}$.

Solving for $\lambda$ and $\mu$, we get $\lambda = \frac{1}{3}$ and $\mu = \frac{1}{3}$. The point of intersection $Q$ lies on $\mathbf{r_{AF}}$ with $\lambda = \frac{1}{3}$, and its position vector is $\overrightarrow{OQ} = \mathbf{a} + \frac{1}{3}(\mathbf{b} + \mathbf{c} - 2\mathbf{a}) = \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$. We can show that $Q$ lies on the line connecting $C$ to the midpoint of $AB$ by showing that $\overrightarrow{OQ} = \mathbf{r_{CD}}$, where $\mathbf{r_{CD}} = \mathbf{c} + \omega (\mathbf{a + b - 2c})$.

To show that $AH \perp BC$, we can use the vector equation $\mathbf{r_{AH}} = \mathbf{a} + \lambda (\mathbf{h} - \mathbf{a}) = \mathbf{a} + \lambda (\mathbf{b} + \mathbf{c})$ and $\mathbf{r_{BC}} = \mathbf{b} + \mu (\mathbf{c} - \mathbf{b})$. If $AH \perp BC$, then $(\mathbf{b} + \mathbf{c}) \cdot (\mathbf{c} - \mathbf{b}) = 0$, which implies $|\mathbf{b}| = |\mathbf{c}|$. Similarly, we can show that $BH \perp AC$ and $AB \perp CH$.

The final part of the question can be easily deduced by subtracting the two statements we have already shown to be true, eliminating $p_3$ and $q_3$. This gives $p_1^2 + q_1^2 = p_2^2 + q_2^2$, which implies $AC \perp BH$ and $BC \perp AH$, and therefore $AB \perp CH$.

In conclusion, we have shown that the centroid and orthocentre of a triangle are concurrent, and we have proven the product rule and quotient rule using logarithmic differentiation. We have also evaluated a complex derivative using logarithmic differentiation and shown that $AH \perp BC$, $BH \perp AC$, and $AB \perp CH$.


### What is Norton’s Theorem?

Norton’s Theorem states that any linear circuit can be simplified to an equivalent circuit with a single current source and parallel resistance connected to a load. The circuit must be linear, meaning all underlying equations are linear with no exponents or roots.

### Simplifying Linear Circuits

The Norton equivalent circuit consists of a current source and parallel resistance connected to a load. A current source provides a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current.

### Thevenin’s Theorem vs. Norton’s Theorem

Norton’s Theorem is similar to Thevenin’s Theorem, as it reduces the original circuit to a simpler equivalent circuit. The steps to calculate the Norton source current and Norton resistance are also similar to those in Thevenin’s Theorem.

### Identify The Load Resistance

To apply Norton’s Theorem, first identify the load resistance and remove it from the original circuit.

### Find The Norton Current

Next, find the Norton current by placing a direct wire connection between the load points and determining the resultant current. This is the opposite of Thevenin’s Theorem, where the load resistor is replaced with a break. The total current through the short between the load connection points becomes the Norton source current.

### Find Norton Resistance

To calculate the Norton resistance, remove the power sources from the original circuit, replace voltage sources with wires, and replace current sources with breaks. Then, calculate the total resistance from one load connection point to the other.

### Determine The Voltage Across The Load Resistor

Re-connect the original load resistance to the Norton equivalent circuit and analyze it as a simple parallel arrangement. The voltage and current values for the load resistor can be determined using simple parallel circuit analysis.

**REVIEW:**

* Norton’s Theorem reduces a network to an equivalent circuit composed of a single current source, parallel resistance, and parallel load.
* Steps to follow for Norton’s Theorem:
 1. Find the Norton source current by removing the load resistor and calculating the current through a short.
 2. Find the Norton resistance by removing all power sources and calculating total resistance between the open connection points.
 3. Draw the Norton equivalent circuit and analyze voltage and current for the load resistor following parallel circuit rules.

**RELATED WORKSHEET:**


## Related questions with answers

Consider the differential equation $y ^ { \prime \prime } + 4 y = \sin ^ { 2 } x$. The objective is to find a particular solution $y _ {p}$ of the given equation using the method of variation of parameters.

To solve the equation, we first note that the differential equation is $y ^ { \prime \prime } + 4 y = \sin ^ { 2 } x$. The method of variation of parameters will be used to find $y _ {p}$.

The given differential equation is a second-order linear non-homogeneous equation with constant coefficients. The right-hand side of the equation is $\sin ^ { 2 } x$, which can be rewritten using the trigonometric identity $\sin ^ { 2 } x = \frac{1 - \cos(2x)}{2}$.

The solution to the homogeneous equation $y ^ { \prime \prime } + 4 y = 0$ is $y _ {h} = c _ {1} \cos(2x) + c _ {2} \sin(2x)$, where $c _ {1}$ and $c _ {2}$ are arbitrary constants.

Using the method of variation of parameters, we assume that the particular solution has the form $y _ {p} = u _ {1} \cos(2x) + u _ {2} \sin(2x)$, where $u _ {1}$ and $u _ {2}$ are functions of $x$ to be determined.

The method of variation of parameters involves finding $u _ {1}$ and $u _ {2}$ such that $y _ {p}$ satisfies the given differential equation. This requires computing the derivatives of $y _ {p}$ and substituting them into the differential equation.

By applying the method of variation of parameters, we can determine the particular solution $y _ {p}$ of the given differential equation $y ^ { \prime \prime } + 4 y = \sin ^ { 2 } x$.


To prove Heron's Area Formula, we start with the formula for the area of a triangle: area = (1/2)bc.sin(A). We can express sin(A) as 2 sin(A/2).cos(A/2), so the area formula becomes area = bc.sin(A/2).cos(A/2).

We need to find expressions for cos(A/2) and sin(A/2) in terms of the semi-perimeter s and the sides a, b, and c of the triangle. The semi-perimeter s is given by s = (a+b+c)/2.

Using the Law of Cosines, we have cos(A) = (b^2 + c^2 - a^2) / (2bc). We can express cos(A) in terms of cos(A/2) as 2cos^2(A/2) - 1 = (b^2 + c^2 - a^2) / (2bc), which simplifies to 2cos^2(A/2) = 1 + (b^2 + c^2 - a^2) / (2bc).

Simplifying further, we get cos^2(A/2) = (b^2 + c^2 + a^2 + 2bc - 2a^2 - 2ab - 2ac + 2a^2) / (4bc), which reduces to cos^2(A/2) = (b+c+a)(b+c-a) / (4bc). Then, cos^2(A/2) = (2s-2a)2s / (4bc), and finally, cos^2(A/2) = s(s-a) / (bc).

Similarly, we can express sin(A) in terms of sin(A/2) as 1 - 2 sin^2(A/2) = (b^2 + c^2 - a^2) / (2bc), which simplifies to 2sin^2(A/2) = 1 - (b^2 + c^2 - a^2) / (2bc).

Simplifying further, we get 2sin^2(A/2) = (2bc - b^2 - c^2 + a^2) / (2bc), which reduces to sin^2(A/2) = (a^2 - (b-c)^2) / (4bc). Then, sin^2(A/2) = (a-b+c)(a+b-c) / (4bc), which simplifies to sin^2(A/2) = (2s-2b)(2s-2c) / (4bc), and finally, sin^2(A/2) = (s-b)(s-c) / (bc).

Now, we can substitute the expressions for cos(A/2) and sin(A/2) into the area formula: area = bc.sin(A/2).cos(A/2) = bc.sqrt[s(s-a)(s-b)(s-c)] / (bc), which simplifies to area = sqrt[s(s-a)(s-b)(s-c)]. This is Heron's Area Formula.


**Algebraic Representation of Numbers**

Representing numbers in algebraic form is useful. Key concepts include:

* Consecutive numbers: n, n+1, n+2 …
* Even number: 2n
* Consecutive even numbers: 2n, 2n+2, 2n+4 …
* Odd number: 2n+1
* Consecutive odd numbers: 2n+1, 2n+3, 2n+5 …
* 2-digit number (say ab) = 10a + b
* 3-digit number (say abc) = 100a + 10b + c
* n-digit number = 10^(n-1)digit_1 + 10^(n-2)digit_2 + … + digit_n

**Algebraic Identities**

Important identities include:

* (a + b)^2 = a^2 + 2ab + b^2
* (a - b)^2 = a^2 - 2ab + b^2
* (a + b)(a - b) = a^2 - b^2
* a^3 + b^3 = (a + b)(a^2 - ab + b^2)
* a^3 - b^3 = (a - b)(a^2 + ab + b^2)
* (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
* (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
* (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
* (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc

**Exponents**

Key concepts include:

* a^0 = 1
* a^1 = a
* a^m × a^n = a^(m + n)
* a^m / a^n = a^(m - n)
* (a^m)^n = a^(m × n)
* a^(-m) = 1 / a^m
* (ab)^m = a^m × b^m

**Divisibility Rules**

Rules for divisibility include:

* Divisible by 2: last digit is divisible by 2
* Divisible by 3: sum of digits is divisible by 3
* Divisible by 4: last 2 digits are divisible by 4
* Divisible by 5: last digit is 5 or 0
* Divisible by 7: subtract twice the unit digit from the remaining number
* Divisible by 8: last 3 digits are divisible by 8
* Divisible by 9: sum of digits is divisible by 9
* Divisible by 11: difference between sum of digits at odd and even places is 0 or divisible by 11
* Divisible by 13: difference between number of thousands and remainder of division by 1000 is divisible by 13

**Approximation**

Approximation techniques include:

* Writing numerator with respect to denominator
* Using simple methods that are easy to understand and apply

**Using Options**

Using options to solve problems can be an effective strategy, including:

* Substitution: plugging in options to see which one satisfies the given condition
* Deduction: using options to deduce the correct answer

**Example Problems**

Example problems include:

* Finding two numbers given their sum and product
* Maximizing the product of two numbers given their sum
* Finding distinct values of [x^2/n] where x can be from 1 to n

**Kaprekar's Constant**

6174 is known as Kaprekar's constant, which can be reached by:

* Taking any four-digit number with at least two different digits
* Arranging the digits in descending and ascending order
* Subtracting the smaller number from the bigger number
* Repeating the process until 6174 is reached

**Formulas**

Formulas include:

* N = floor(Sum/e) for dividing a number into equal parts to maximize the continued product
* X = S * a / (a + b) and Y = S * b / (a + b) for finding two numbers given their sum and product.


One tenth as a percentage is 10%. To understand this, we can break it down: 1/10 = 0.1, and when converted to a percentage, it equals 10%. This is because 1/10 as a percentage is the same as 10/100, which simplifies to 10%. The percentage rate can be calculated by multiplying 1/10 by 100%, which gives 0.10 * 100% = 10%. Essentially, one tenth is equal to ten percent.

Related concepts include understanding that 9 tenths as a percentage equals 90%, following the same conversion logic. The key concept here is converting fractions to percentages, where the fraction 1/10 directly translates to 10% when considering 10/100.

In summary, the conversion of one tenth to a percentage is a straightforward process that results in 10%. This fundamental concept in mathematics and arithmetic is crucial for more complex calculations involving percentages and fractions.


4 Multivariable and logistic models

This chapter explores multiple regression, which introduces the possibility of more than one predictor in a linear model, and logistic regression, a technique for predicting categorical outcomes with two levels.

4.1 Regression with multiple predictors

Multiple regression extends simple two-variable regression to the case that still has one response but many predictors. The method is motivated by scenarios where many variables may be simultaneously connected to an output.

We consider data about loans from the peer-to-peer lender, Lending Club, which includes terms of the loan as well as information about the borrower. The outcome variable is the interest rate assigned to the loan.

The data set includes results from 10,000 loans, and we look at a subset of the available variables. The first six observations in the data set are shown in Table 4.1, and descriptions for each variable are shown in Table 4.2.

Notice that the past bankruptcy variable is an indicator variable, where it takes the value 1 if the borrower had a past bankruptcy in their record and 0 if not.

Using an indicator variable in place of a category name allows for these variables to be directly used in regression.

Two of the other variables are categorical, each of which can take one of a few different non-numerical values; we discuss how these are handled in the model in Section 4.1.1.

The data can be found in the openintro package: loans_full_schema.

Based on the data in this dataset, we created two new variables: credit_util, which is calculated as the total credit utilized divided by the total credit limit, and bankruptcy, which turns the number of bankruptcies to an indicator variable.

We refer to this modified dataset as loans.

4.1.1 Indicator and categorical predictors

Let's start by fitting a linear regression model for interest rate with a single predictor indicating whether or not a person has a bankruptcy in their record.

Results of this model are shown in Table 4.3.

Interpret the coefficient for the past bankruptcy variable in the model. Is this coefficient significantly different from 0?

The variable takes one of two values: 1 when the borrower has a bankruptcy in their history and 0 otherwise. A slope of 0.74 means that the model predicts a 0.74% higher interest rate for those borrowers with a bankruptcy in their record.

Suppose we had fit a model using a 3-level categorical variable, such as verified_income.

The output from software is shown in Table 4.4.

This regression output provides multiple rows for the variable. Each row represents the relative difference for each level of verified_income.

However, we are missing one of the levels: Not Verified. The missing level is called the reference level and it represents the default level that other levels are measured against.

4.1.2 Many predictors in a model

The world is complex, and it can be helpful to consider many factors at once in statistical modeling.

We want to construct a model that accounts for not only for any past bankruptcy or whether the borrower had their income source or amount verified, but simultaneously accounts for all the variables in the loans data set.

4.1.3 Adjusted R-squared

We first used R^2 in Section 3.2.5 to determine the amount of variability in the response that was explained by the model.

Adjusted R-squared is computed as R_adj^2 = 1 - (s_residuals^2 / (n-k-1)) / (s_outcome^2 / (n-1)).

Because k is never negative, the adjusted R-squared will be smaller – often times just a little smaller – than the unadjusted R-squared.

4.2 Model selection

The best model is not always the most complicated. Sometimes including variables that are not evidently important can actually reduce the accuracy of predictions.

In this section, we discuss model selection strategies, which will help us eliminate variables from the model that are found to be less important.

4.2.1 Model selection strategies

Two common strategies for adding or removing variables in a multiple regression model are called backward elimination and forward selection.

Backward elimination starts with the model that includes all potential predictor variables. Variables are eliminated one-at-a-time from the model until we cannot improve the adjusted R-squared.

4.3 Model diagnostics

4.5 Logistic regression

In this section, we introduce logistic regression as a tool for building models when there is a categorical response variable with two levels, e.g., yes and no.

Logistic regression is a type of generalized linear model (GLM) for response variables where regular multiple regression does not work very well.

4.5.1 Resume data

We consider experiment data from a study that sought to understand the effect of race and sex on job application callback rates.

The researchers enumerated important characteristics, such as years of experience and education details, and they used these characteristics to randomly generate the resumes.

Finally, they randomly assigned a name to each resume, where the name would imply the applicant's sex and race.

4.5.2 Modeling the probability of an event

Logistic regression is a generalized linear model where the outcome is a two-level categorical variable.

The outcome, Y_i, takes the value 1 with probability p_i and the value 0 with probability 1-p_i.

4.5.3 Logistic model with many variables

We used statistical software to fit the logistic regression model with all 8 predictors described in Table 4.9.

Like multiple regression, the result may be presented in a summary table, which is shown in Table 4.10.

The structure of this table is almost identical to that of multiple regression; the only notable difference is that the p-values are calculated using the normal distribution rather than the t-distribution.

4.5.4 Model diagnostics

Logistic regression conditions.

There are two key conditions for fitting a logistic regression model:

1. Each outcome Y_i is independent of the other outcomes.
2. Each predictor x_i is linearly related to logit(p_i) if all other predictors are held constant.

4.5.5 Groups of different sizes

Any form of discrimination is concerning, and this is why we decided it was so important to discuss this topic using data.

The resume study also only examined discrimination in a single aspect: whether a prospective employer would call a candidate who submitted their resume.

4.5.6 Exercises

4.6 Chapter review

4.6.1 Terms

We introduced the following terms in the chapter:

* Adjusted R-squared
* Akaike information criterion
* Backward elimination
* Full model
* Generalized linear model
* Logistic regression
* Multiple regression
* Parsimonious
* Reference level

4.6.2 Chapter exercises

Logistic regression fact checking: Determine which of the following statements are true and false. For each statement that is false, explain why it is false.

Multiple regression fact checking: Determine which of the following statements are true and false. For each statement that is false, explain why it is false.

4.6.3 Interactive R tutorials

Navigate the concepts you've learned in this chapter in R using the following self-paced tutorials. All you need is your browser to get started!


Newton's law of cooling is a physical law that can be applied to both cooling and heating, as it is symmetrical. This law can be used to predict the time of death from the temperature of a corpse and the ambient temperature, but a more interesting and useful example is cooking chicken.

The law states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature, or dT(t)/dt = -k(T(t)-Tambient), where T is the temperature of the object, t is time, k is a constant, and Tambient is the ambient temperature. This differential equation solves to T(t) = Tambient + (T(0) - Tambient)e^{-kt}.

To calculate the temperature needed to cook chicken, we need to know the constant k, which is specific to the particular chicken and filling in the oven. If the chicken took 50 minutes to go from 10°C to 53°C with the oven at 150°C, we can calculate the constant k as k = -1/t ln((T(t)-Tambient) / (T(0)-Tambient)). Plugging in the values, we get k = -1/50 * ln((53-150)/(10-150)) = 0.0073 min^{-1}.

Using this constant, we can calculate the temperature needed for the chicken to be done in another 30 minutes as Tambient = (T(t) - T(0)e^{-kt}) / (1 - e^{-kt}). Plugging in the values, we get Tambient = (80 - 53*e^{-0.0073*30}) / (1 - e^{-0.0073*30}) = 200°C.

Note that this calculation can be done using a web-app, eliminating the need for manual calculations. The application of Newton's law of cooling to cooking demonstrates the effectiveness of science in everyday life. However, it is essential to consider the potential consequences of using a higher temperature, such as burning the skin, and the benefits of cooking at a consistent temperature, like 150°C, for a juicier result.


Integers can be positive or negative numbers. Adding positive integers is straightforward, but adding negative integers requires a different approach.

## Adding With Negative Integers

Consider a situation where one person has $6 and the other owes $2. The total amount of money they have is represented by -2 + 6 = 4. Owing $2 is equivalent to -2.

### Adding To a Negative Integer

Adding a positive integer to a negative integer involves moving to the right on a number line. For example, -2 + 6 = 4.

### Adding a Negative Integer

Adding a negative integer to a positive integer involves moving to the left on a number line. Think of adding a negative number as receiving a fine, resulting in a decrease.

### More Examples of Adding With Negative Integers

## Subtracting With Negative Numbers

### Subtracting From Negative Integers

Subtracting positive numbers on a number line involves moving to the left. For instance, -2 - 3 = -5.

### Subtracting Negative Integers

Subtracting negative numbers involves moving to the right. This is equivalent to moving to the left by the opposite positive number. For example, walking -3 miles to the north is the same as walking 3 miles to the south.

When explaining subtraction of negative numbers, relate it to finding the difference between two values. For example, the temperature difference between 8 degrees and -4 degrees is 8 - (-4) = 12.

### Changes in Temperature

Integers are useful for showing temperature and changes in temperature. If the temperature is 16 degrees during the day and falls to -9 degrees overnight, the change in temperature can be calculated as -9 - 16 = -25. Alternatively, calculate the distance from zero for each temperature (-9 is 9 units from zero, and 16 is 16 units from zero) and add them together to get a difference of 25. Since the temperature is falling, the result is -25.


Giving students an insight into mathematical savants and other mathematical geniuses is a good way of invoking a sense of wonder about the subject. Calendar savants are able to correctly name the day of the week from any given date in history – almost instantly. Whilst it is still not clear how they do this, it is possible to achieve the same feat using a relatively straightforward algorithm. To make things easier, limit the prediction to the first decade of the 21st century.

**Step 1 – Memorise a twelve digit number: 622 – 503 – 514 – 624.** Each digit corresponds to a month of the year. January is 6, February is 2, March is 2, etc.

**Step 2 – Learn to work in mod 7.** In mod 7, we look at the remainder when dividing by 7. For example, 36 divided by 7 is 5 remainder 1, so 36 is 1 (mod 7). A number like 14 would be 0 (mod 7).

To calculate the day of the week for any date in the 21st century, follow these steps:

**Step 1 – Work out the day of the month mod 7.** For example, March 10, 2004: 10 is 3 mod 7.

**Step 2 – Add the month value.** The month value is 2, so 3 + 2 = 5.

**Step 3 – Count how many years since 2000.** For March 10, 2004, this is 4 years. Add 4 to 5 = 9, which is 2 (mod 7).

**Step 4 – Adjust for leap years.** There was a leap year in 2000, then in 2004, etc. For March 10, 2004, there were 1 leap years (if the date was before March 1st, we wouldn’t count the 2004 leap year). Add 1 to 2 = 3.

**Step 5 – Determine the day of the week.** Sunday = 0, Monday = 1, Tuesday = 2, etc. So, March 10, 2004, was a Wednesday.

**Lesson Plan:**

Learning objectives: This activity can be used as a starter or as part of a larger lesson on mental arithmetic or mathematical savants.

**5 minutes:** Watch a video on autistic savants and discuss their abilities.

**5 minutes:** Demonstrate the algorithm using a student-chosen date from 2000 to 2009.

**10 minutes:** Discuss the algorithm and have students practice in pairs using the "How to Calculate like a Mathematical Savant" sheet.

Note: The original multiple-choice questions and answers were not provided in the text. If you provide them, I can refine them for you.


A colleague is experiencing an issue with their HP 48gII calculator, which is displaying division results as fractions instead of decimal numbers. For example, when entering "2 Enter 5 /", the display shows "2/5" as a fraction. The calculator is currently in Fix 3 mode, and addition, multiplication, and subtraction operations are working as expected. The goal is to change the division display back to decimal numbers.


**Math Challenges and Problems**

1. **Magic Square**: In a Magic Square, all rows, columns, and diagonals add up to the 'Magic Constant'. How can you change the magic constant of a given square?

2. **Number Patterns**: Investigate the number patterns that emerge when you square a number and subtract your answer. What kind of number do you get? Can you prove it?

3. **Calculator Challenge**: Start with 1,000,000 on your calculator and reduce it to 7 using only the 7 key and basic arithmetic operations.

4. **Division Problems**: Use problems from the NRICH site to teach division, exploring spatial awareness, addition, and multiplication.

5. **Pied Piper Problem**: Investigate the different numbers of people and rats that could be present if you know the total number of legs.

6. **Napier's Bones**: Learn how to use Napier's bones to calculate multiplication and division, invented by John Napier 400 years ago.

7. **Calculator Division**: If you divide one whole number by another and get 3.125, what are the two numbers if they are both under 50?

8. **Stamp Prices**: Determine which stamps are needed to post a parcel weighing 825g using unlimited numbers of 1st and 2nd class stamps.

9. **Lowest Number**: Find the lowest number that always leaves a remainder of 1 when divided by numbers from 2 to 10.

10. **Clock Face**: Draw a line on a clock face so that the numbers on both sides have the same total.

11. **Prime Number Code**: Decipher a code using two different prime numbers less than 10, multiplying them together and shifting the alphabet forwards by the result.

12. **Route Products**: Find the product of numbers on routes from A to B, determining which route has the smallest and largest product.

13. **Spectacle Lenses**: Calculate the number of lenses needed for 9 families with either 3 or 4 eyes on the planet Zargon.

14. **Rod Lengths**: Determine which rod is twice the length of the first rod made using different numbers of cubes.

15. **Puzzles and Books**: Solve a riddle to find out how many puzzles and books Santa left for girls and boys.

16. **Height and Tea**: Estimate the height and tea consumption of a small man using a picture of him next to a mug.

17. **Gift Piles**: Calculate the number of gifts in each pile using given information.

18. **Function Machine**: Determine the output numbers when the numbers 5, 7, and 4 are input into a function machine.

19. **Mathematics Representation**: Explore the differences between 'recording mathematics' and 'representing mathematical thinking' in an article for teachers.

20. **Locked Room**: Help Skippy and Anna escape a locked room by solving a problem to find the key number.

21. **Jump Heights**: Calculate the length or height of two children's first jumps after they improved their results.

22. **Times Tables**: Read an article about learning times tables, featuring an interview with Liz in November.

23. **Number Guessing**: Find a number that, when divided by 5, leaves a remainder of 4, and when divided by 3, leaves a remainder of 2.

24. **Countdown Game**: Play a version of the classic Countdown Game.

25. **Domino Pieces**: Determine which domino pieces are missing from a set of 24 pieces with 125 spots.

26. **Deca Tree**: Calculate the number of leaves on a Deca Tree after a woodcutter cuts off a trunk, branch, twig, and leaf.

27. **Array Modelling**: Use an array of objects to model number properties involving multiplication, allowing children to represent their thinking with concrete materials.

28. **Rabbit Pens**: Calculate the number of each type of rabbit in pens using given statements.

29. **Cows and Sheep**: Determine the number of cows and sheep in each field using logical reasoning.

30. **Picnic Sharing**: Calculate what each person will have when a picnic is shared equally between Petros and Michael.

31. **Cross Arrangement**: Arrange 5 different digits in a cross according to given rules.

32. **Dinner Party Seating**: Seat 44 people at 15 square tables with 4 seats each, using all tables with no empty places.

33. **Product Variations**: Calculate the number of different products that can be obtained by multiplying three numbers less than 10 together.

34. **Smallest Whole Number**: Find the smallest whole number that can be made using the numbers 1, 2, 3, 4, and 5, and the operations x and ÷, each used only once.

35. **Dot Arrangements**: Calculate the total number of dots in a 3 by 5 arrangement, and explore what happens for other sizes.

36. **Clock Face Cake**: Determine who received each piece of a birthday cake cut into a clock face shape.

37. **Multiplication Reconstruction**: Reconstruct the original multiplication calculation where some digits have been replaced by letters and others by asterisks.

38. **Date Calculations**: Add the dates of all days in a week, then multiply the first date by 7 and add 21. Explain the result.

39. **Bucket Liquid**: Calculate the amount of liquid in each of three buckets, each holding a maximum of 5 liters, using given clues.

40. **Family Size**: Determine the number of children in the Brown family using information about Sally and her brother.

41. **Consecutive Numbers**: Find two consecutive numbers whose product is 56,406.

42. **Multiplication Methods**: Explore unconventional multiplication methods and watch videos to understand them.

**Multiple Choice Questions**

1. What is the magic constant of a Magic Square?
a) The sum of all rows
b) The sum of all columns
c) The sum of all diagonals
d) The sum of all rows, columns, and diagonals

2. Which operation can be used to reduce 1,000,000 to 7 on a calculator?
a) Addition
b) Subtraction
c) Multiplication
d) All of the above

3. What is the purpose of Napier's bones?
a) To calculate multiplication and division
b) To calculate addition and subtraction
c) To calculate fractions and decimals
d) To calculate algebraic expressions

4. What is the lowest number that always leaves a remainder of 1 when divided by numbers from 2 to 10?
a) 11
b) 12
c) 13
d) 251

5. How many different products can be obtained by multiplying three numbers less than 10 together?
a) 10
b) 20
c) 30
d) 120

**Answers**

1. d) The sum of all rows, columns, and diagonals
2. d) All of the above
3. a) To calculate multiplication and division
4. d) 251
5. d) 120


#### Grand Piano

The problem involves finding the largest area rectangle that fits in an isosceles triangle with a base twice its height. Let the base be 2 and the height be 1. If the height of the rectangle is \(h\), then its width is \(2(1-h)\), and its area is \(2h-2h^2\). Through differentiation, the maximum area occurs when \(h=\frac{1}{2}\), resulting in a ratio of the rectangle's length to its width of **2:1**.

#### Extension

Consider a scenario where the corner is not a 90° angle. What would be the largest area rectangle that could fit around it?

#### Cycling Digits

Examples of cycling digits include:
$$105263157894736842 \times 2 = 210526315789473684$$
$$210526315789473684 \times 2 = 421052631578947368$$
$$421052631578947368 \times 2 = 842105263157894736$$

#### Extension

There exists a number such that removing its units digit and placing it at the front results in the same value as multiplying the original number by 3. Is this claim true?

#### The Mutilated Chessboard

A standard chessboard has 32 white and 32 black squares. After removing two diagonally opposite corners, there are either 30 white and 32 black squares or 32 white and 30 black squares. Since each domino covers one white and one black square, a combination of dominoes will always cover an equal number of black and white squares. Therefore, it is impossible to cover all the squares on the mutilated chessboard.

#### Extension

Is it possible to perform a knight's tour on the mutilated chessboard?

#### Tennis

A set can be won by hitting the ball 12 times, for example, by winning three service games with four aces and the opponent losing their service games with four double faults.


## What is RST
Rough set theory (RST) is a mathematical framework that deals with uncertainty.

## Mathematical Model
An information system (IS) or S is presented mathematically as a pair S = (U, A), where U is a finite, nonempty set called the universe, and A is a finite, nonempty set called the set of attributes. We associate Va with its values, called the domain of a. A decision system (DS) or S is a triple DS = (U, C, D), where U is a universal set, C is the condition set, and D is the decision set. DS can also be represented as DS = (U, A), where A = C U D, and C and D are disjoint, non-empty attribute sets.

## Model
To present a decision system to a computer system, we need to present two main objects: the universal object and the attribute.

## Decision System Attribute
A decision system contains attributes that can be classified as either condition or decision attributes. An attribute has values that present the domain of attributes.

## Decision Matrix
A decision matrix is a combination of universe objects and attributes. For each universe object, a value of each attribute must be decided, which can be either a condition or a decision.

## Discernibility Matrix
According to the universe objects and their attribute values, the decision attribute value is first decided for comparison. If not specified, the comparison will be between all universe objects. The comparison is between each pair of universe objects that differ in the decision value. The reducts present the main condition attributes that represent the decision system, and other condition attributes' differentiation will not affect the decision value.

## System Requirements
The system requires .Net Framework 2 and SQL Server Express or a higher version. To set up the system, create a new user on SQL Server named "Permanent", create a database with Permanent as the owner, and restore the RoughBack backup file. When running the application for the first time, it will require server connection data. From the Main Data menu, set the attributes, then set the universe from the main form, and finally open the discernibility matrix from the main menu.


**The Task**

The original task involves a math problem with a pseudocontext that needs to be corrected. The task starts at an abstract level, constructing linear equations, and then proceeds to a concrete level by asking students to evaluate plans for 20 hours of internet use each month. However, this approach has several issues, including a lack of intellectual need for the solution and an unrealistic scenario.

**Revised Approach**

To improve the task, several changes can be made:

1. **Change the context**: Instead of using a pseudocontext, use a real-world scenario that students can relate to.
2. **Put students in the shoes of the person experiencing the problem**: Ask students to consider the problem from the perspective of someone who is actually experiencing it.
3. **Lower the floor on the task**: Start with a more concrete and accessible scenario, rather than an abstract one.
4. **Raise the ceiling on the task**: Allow students to develop the concept further and explore more complex ideas.
5. **Provoke an intellectual need for the solution**: Create a scenario where students need to find a solution to a real problem, rather than just following a procedure.

**Example**

One example of a revised task is to ask students to consider a scenario where they need to choose between two internet plans. The plans have different pricing structures, and students need to determine which plan is the best value. To make the task more concrete, students can be asked to write down a number between 1 and 25, representing the number of months they plan to use the internet. They can then evaluate the plans based on this number and determine which plan is the best value.

**Assessment and Extension**

The assessment for this task can be straightforward, with students presenting their findings and explaining their reasoning. To extend the task, students can be asked to come up with their own internet plans and evaluate them based on different scenarios.

**Comments and Feedback**

Several comments and feedback from educators highlight the importance of considering real-world scenarios and complexities in math problems. Some suggestions include:

* Using piecewise linear equations to model real-world scenarios
* Considering the value of time and convenience in financial decisions
* Using authentic and messy real-world problems to develop mathematical modeling skills
* Emphasizing the importance of communication and persuasion in mathematical modeling

**Call for Submissions**

Educators are encouraged to submit their own math tasks and makeovers for consideration. The goal is to create a community of educators who can share and learn from each other's experiences in developing more effective and engaging math tasks.


The "Number 9906" challenge is a brain teaser that tests quick thinking and sharp observation. The goal is to find the hidden "Number 9906" among "Number 9909" within 14 seconds. This challenge is a workout for the brain, making you think and figure things out.

To solve the challenge, try breaking it down into smaller parts. It's a puzzle where you need to fill in the missing "Number 9906" among a bunch of "Number 9909". At first glance, the task might seem easy, but it can be challenging.

The challenge is designed to sharpen visual acumen and elevate IQ and visual problem-solving abilities. To participate, observe the image and locate the "Number 9906" within 14 seconds. The solution is provided after the time is up.

The challenge is as follows:
Observe the image and find the "Number 9906" among "Number 9909" within 14 seconds.
Time starts now:
14…
9…
8…
7…
6…
5…
4…
3…
2…
1…
Time's up!

Note: The solution is provided in the image below the challenge. Regular updates on new brain teasers are available on the website Fresherslive.


Two towns, A and B, are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling at 20 km/h from A to B notices that a bus passes him every 18 minutes in the direction of his motion and every 6 minutes in the opposite direction. The distance between two buses when one takes off is V.T, where V is the velocity of the bus and T is the frequency of the bus.

Let's denote the speed of the cyclist as U = 20 km/h. When the cyclist and bus move in the same direction, the relative velocity is V - U. The time to cover the distance between two buses is t = 18 minutes = 18/60 hours. The distance covered is t(V - U), which equals V.T. Therefore, t(V - U) = V.T, which gives VT = 3/10(V - U).

When the cyclist and bus move in opposite directions, the relative velocity is V + U. The time to cover the distance between two buses is t = 6 minutes = 6/60 hours. Equating this to V.T gives t(V + U) = V.T, which yields VT = 1/10(V + U).

Equating the two expressions for VT, we get 3/10(V - U) = 1/10(V + U). Simplifying this equation gives 3V - 3U = V + U, which further simplifies to 2V = 4U. Therefore, V = 2U = 40 km/h.

Substituting V = 40 km/h into one of the equations, we get 40.T = 1/10(40 + 20). Solving for T gives T = 3/20 hours, which is equivalent to T = 9 minutes. Thus, the period of the bus service is 9 minutes, and the buses travel at a constant speed of 40 km/h.


A null hypothesis is a statistical hypothesis that is tested for possible rejection under the assumption that it is true. The null hypothesis is denoted by H0 and is typically stated as a statement of no effect or no difference. For example, "the correlation between frustration and aggression is zero" or "the average income for men is similar to that for women."

In hypothesis testing, the null hypothesis is assumed to be true until there is enough evidence to suggest otherwise. The alternative hypothesis, denoted by H1 or Ha, is the hypothesis that is being tested against the null hypothesis. The alternative hypothesis is typically the hypothesis that there is an effect or a difference.

To test a null hypothesis, a significance level, typically denoted by α, is chosen. The significance level is the probability of rejecting the null hypothesis when it is true. Common significance levels are 0.05, 0.01, and 0.001.

The test statistic is calculated from the sample data, and its distribution is compared to the distribution of the test statistic under the null hypothesis. If the test statistic is unlikely under the null hypothesis, the null hypothesis is rejected.

For example, suppose we want to test the hypothesis that the mean return from a portfolio over a 200-day period is greater than zero. The null hypothesis would be that the mean return is equal to zero, and the alternative hypothesis would be that the mean return is greater than zero.

The formula for the test statistic is:

z = (x - μ) / σ

where x is the sample mean, μ is the population mean under the null hypothesis, and σ is the standard deviation of the sample mean.

If the test statistic is greater than the critical value from the standard normal distribution, the null hypothesis is rejected.

In a two-tailed test, we are observing whether the data is statistically different or statistically the same. The typical levels of significance are P<0.001, P<0.01, P<0.05, and P<0.10.

The reasons for using a two-tailed test is that even though the experimenters expect cloud seeding to increase rainfall, it is possible that the reverse occurs and, in fact, a significant decrease in rainfall results.

The null hypothesis can be written mathematically as: H0: p = p0, where p is the sample proportion and p0 is the population proportion under the null hypothesis.

The alternative hypothesis can be written mathematically as: H1: p ≠ p0.

The test statistic for the null hypothesis is:

z = (p - p0) / sqrt(p0(1-p0)/n)

where p is the sample proportion, p0 is the population proportion under the null hypothesis, and n is the sample size.

If the test statistic is greater than the critical value from the standard normal distribution, the null hypothesis is rejected.

In conclusion, the null hypothesis is a statistical hypothesis that is tested for possible rejection under the assumption that it is true. The null hypothesis is denoted by H0 and is typically stated as a statement of no effect or no difference. The alternative hypothesis is the hypothesis that is being tested against the null hypothesis. The test statistic is calculated from the sample data, and its distribution is compared to the distribution of the test statistic under the null hypothesis. If the test statistic is unlikely under the null hypothesis, the null hypothesis is rejected.


# Find the common difference of the arithmetic sequence with a4=10 and a11=45.

To find the common difference, we can use the formula for the nth term of an arithmetic sequence: an = a1 + (n-1)d, where an is the nth term, a1 is the first term, n is the term number, and d is the common difference.

Given a4 = 10 and a11 = 45, we can set up two equations:

a4 = a1 + (4-1)d = 10
a11 = a1 + (11-1)d = 45

Simplifying the equations:

a1 + 3d = 10
a1 + 10d = 45

Subtracting the first equation from the second equation:

7d = 35

Dividing both sides by 7:

d = 5

Therefore, the common difference of the arithmetic sequence is 5.

---

Solve the absolute value equation: |6w - 2| = |6w + 18|

Since the absolute values are equal, we can set up two equations:

6w - 2 = 6w + 18
6w - 2 = -(6w + 18)

Simplifying the first equation:

-2 = 18 (no solution)

Simplifying the second equation:

6w - 2 = -6w - 18

Adding 6w to both sides:

12w - 2 = -18

Adding 2 to both sides:

12w = -16

Dividing both sides by 12:

w = -16/12
w = -4/3

Therefore, the solution to the absolute value equation is w = -4/3.

---

Stanley has $1.45 in dimes and quarters. If he has 10 coins in total, how many of each does he have?

Let x be the number of dimes and y be the number of quarters.

Since the total number of coins is 10:

x + y = 10

Since the total value is $1.45:

0.10x + 0.25y = 1.45

We can solve this system of equations by substitution or elimination.

Multiplying the first equation by 0.10:

0.10x + 0.10y = 1.00

Subtracting this equation from the second equation:

0.15y = 0.45

Dividing both sides by 0.15:

y = 3

Substituting y = 3 into the first equation:

x + 3 = 10

Subtracting 3 from both sides:

x = 7

Therefore, Stanley has 7 dimes and 3 quarters.


# What Is Yield?

Yield is a percentage that measures the return on an investment over a fixed period, including price increases and dividends paid. It is calculated as the net realized return divided by the invested amount. Higher yields are often associated with lower risk and higher income, but this is not always the case. For example, a high yield may be due to a falling stock price and a rising dividend yield.

## Basics

Yield refers to the returns accrued from an investment during a specified timeframe, manifesting as a percentage derived from the invested capital, market value, or nominal value of the security. Yields can be classified as recognized or anticipated, depending on the security's valuation stability.

## Yield Computation Formula

The yield formula represents the cash flow generated by an investor based on the invested amount in a security. It is typically calculated annually, but quarterly and monthly yields are also prevalent. The yield equation is:

Yield = Net Realized Return / Principal Amount

For example, if an investor buys a stock for $100 and sells it for $120 after a year, with a $2 dividend, the yield is:

($20 + $2) / $100 = 0.22, or 22%

## Analyzing Yield Insights

Yield reveals an investor's ability to recover substantial cash flows from investments. A higher yield may indicate lower risk and increased income potential, but it is crucial to understand the nuances involved. Elevated yield values might stem from a declining market value, reducing the formula's denominator and inflating the yield.

## Diverse Yields in Finance

Investments present various yield types, dependent on the security, investment duration, and return amount.

### Yield on Stocks

Cost Yield = (Price Increase + Dividends Paid) / Purchase Price
Current Yield = (Price Increase + Dividend Paid) / Current Price

For instance, if an investor made a profit of $20 and received a dividend of $2, the total return on investment is ($20 + $2) / $100 = 0.22, or 22%. The current yield would be ($20 + $2) / $120 = 0.1833, or 18.33%.

### Yield on Bonds

Nominal Yield = (Annual Interest Earned / Face Value of Bond)

For example, if a Treasury bond with a face value of $1,000 pays 5% annual interest, its yield is $50 / $1,000 = 0.05, or 5%.

## Bond Yield Metrics

- Yield to Maturity (YTM): forecasts the annualized total return if a bond is held until maturity.
- Yield to Worst (YTW): gauges the lowest potential bond yield without default risk.
- Yield to Call (YTC): indicates the bond's yield at the call date.
- Tax-Equivalent Yield (TEY) for Municipal Bonds: adjusts taxable bond yields to match tax-free municipal bond yields.
- Regulatory Standard (SEC Yield): standardizes bond fund comparisons, factoring in associated fees.
- Mutual Fund Yield: represents a mutual fund's net income return, computed by dividing the annual income distribution payment by the fund's share value.

## Conclusion

Yield is a pivotal metric, quantifying the return on investment over a specified period. Whether gauging stock or bond yields, intricate calculations and varied metrics offer insights into investment performance. The association of higher yields with lower risk and increased income demands nuanced scrutiny, considering market fluctuations. Regulatory standards and the adaptability of companies, issuers, and fund managers in yield reporting emphasize the multifaceted nature of this financial measure.


To make an object move in a circular path, a force must be exerted towards the center of the circle, known as the centripetal force. In contrast, any object within a rotating container experiences a force that pushes it away from the center, referred to as the centrifugal force. This force is given by the equation *F* = *m**ω*^{2}*r*, where *m* is the mass of the object, *ω* is its angular speed, and *r* is the distance from the center of the circle.

The centrifugal force is utilized in a device called a centrifuge to separate particles from a liquid. For instance, blood cells can be separated from the surrounding liquid when blood is contained in a rotating tube, a process known as centrifugation. However, when particles are separated from a liquid, the equation must be modified to account for Archimedes' principle, resulting in *F* = *m*[1 – (*ρ*_{0}/*ρ*)]*ω*^{2}*r*, where *ρ*_{0} is the density of the liquid and *ρ* is the density of the particles.

The centrifugal force causes particles to move to the bottom of the tube, analogous to gravity moving an object towards the Earth's surface. This allows for the definition of a centrifugal potential energy, similar to gravitational potential energy. The difference in potential energy between two radii, *r* and *r*_{0}, is given by Δ*E* = *m*[1 – (*ρ*_{0}/*ρ*)]*ω*^{2}[(*r*_{0}^{2} – *r*^{2})/2].

According to the Boltzmann distribution, the number of particles at a given radius *r* is *n _{r}* = *n*_{0}exp{-(*m*/2*kT*)[1 – (*ρ*_{0}/*ρ*)]*ω*^{2}(*r*_{0}^{2}–*r*^{2})}, where *n*_{0} is the number of particles at the bottom of the tube, *k* is Boltzmann's constant, and *T* is the temperature in Kelvin.

To minimize the number of particles not at the bottom of the tube, *mω*^{2} must be maximized. For smaller particles, such as protein molecules, a higher angular speed *ω* is required to achieve separation. Centrifuge tubes typically rotate at 2,000 revolutions per minute for blood cells, while ultracentrifuges rotate at 30,000-60,000 revolutions per minute for protein molecules.

A technique called density gradient ultracentrifugation is used to separate molecules based on their mass. By introducing a solution with a density gradient into an ultracentrifuge, larger molecules can be separated and detected due to their high concentration at a specific point, causing an increase in the refractive index of the solution.

The force acting on a large molecule in a density gradient is given by *F* = *m*[1 – (*ρ*_{r}/*ρ*)]*ω*^{2}*r*, where *ρ*_{r} is the density at radius *r*. When *ρ* = *ρ*_{r}, there is no force acting on the molecule, and it does not move in the density gradient. The molecule can be detected due to its high concentration at this point, causing an increase in the refractive index of the solution.

Appendix 1 derives equation 2, which accounts for Archimedes' principle in a centrifugal force field. The total force acting on an object immersed in a liquid is given by *F _{A}* = *m*[(*ρ*/*ρ*) – (*ρ*/*ρ*_{0})]*ω*^{2}*r*, which simplifies to equation 2.

Appendix 2 derives equation 3, which describes the difference in potential energy between two radii. The work done in moving a particle between *r*_{0} and *r* is given by the integral of the force with respect to distance, and this work is stored as potential energy. The resulting equation 3 describes the difference in potential energy between two radii.


# A Beginner's Guide to Regression Trees using Sklearn | Decision Trees

## Introduction to Decision Trees

Decision Trees are a type of machine learning algorithm that uses a greedy approach to make predictions. At every node generation, the algorithm focuses on making the best split at that given time, without considering the final tree's overall performance.

## Equation of Regression Tree

The equation of a Regression Tree is given by:

$f(x) = \displaystyle \sum_{m=1}^{M} C_m.1(X \epsilon R_m)$

where $R_1, R_2...R_m$ are the different regions.

## Predictions in Regression Trees

There are two main ways to predict a regression tree:

1. **Prediction using Stratification of feature space**: This method involves dividing the feature space into smaller regions and predicting the target variable based on the mean value of the training observations in each region.
2. **Prediction using Tree Pruning**: This method involves cutting the tree into smaller parts to prevent overfitting.

## Prediction using Stratification of Feature Space

The goal of this method is to reduce the Residual Sum of Squares (RSS) at each node. The RSS is calculated as:

$RSS = \displaystyle \sum_{j=1}^{J} \sum_{i \epsilon R_j} (y_i - \widehat{y}_{R_j})^ {2}$

where $\widehat{y}_{R_j}$ is the mean value of the training observations in the $Jth$ box.

## Disadvantages of Predicting using Stratification

This algorithm can produce good results on the training data but may overfit the data, leading to poor test set performance.

## Predicting using Tree Pruning

Tree Pruning is a technique used to prevent overfitting by cutting the tree into smaller parts. There are two types of pruning:

1. **Pre-pruning or early stopping**: This involves stopping the tree growth early to prevent overfitting.
2. **Post Pruning**: This involves removing branches from the tree after it has been grown.

## Regression Tree Analysis using Sklearn

To analyze a regression tree using Sklearn, we can use the following code:
```python
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
from sklearn.model_selection import train_test_split
from sklearn.datasets import load_boston
from sklearn.tree import DecisionTreeRegressor
from sklearn.metrics import mean_squared_error

X, y = load_boston(return_X_y=True)
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.33, random_state=42)

clf = DecisionTreeRegressor()
clf.fit(X_train, y_train)
predictions = clf.predict(X_test)
print(mean_squared_error(y_test, predictions))
print(np.sqrt(mean_squared_error(y_test, predictions)))
```

## Finding the Relation between Tree Depth and Mean Square Error

We can train the model with different depth values and plot the relation between them using the following code:
```python
mses = []
for depth in range(1, (clf.tree_.max_depth + 1)):
    d_tree_reg = DecisionTreeRegressor(max_depth=depth)
    d_tree_reg.fit(X_train, y_train)
    tree_predictions = d_tree_reg.predict(X_test)
    mses.append(mean_squared_error(y_test, tree_predictions))
tree_depths = [depth for depth in range(1, (clf.tree_.max_depth + 1))]
plt.figure(figsize=(10, 6))
plt.grid()
plt.plot(tree_depths, mses)
plt.xlabel("Tree Depth")
plt.ylabel("Mean Square Error")
```
This will help us find the optimal tree depth that results in the minimum Mean Square Error.


# Problem A

Shopping

You have just moved into a new apartment and need to buy items from multiple stores, minimizing driving distance. Your city is a set of intersections connected by roads, with your house and stores located at intersections. You want to find the shortest route that starts at your house, visits all stores, and returns home.

## Input

The first line contains two integers $N$ and $M$, representing the number of intersections (1-100,000) and roads, respectively. Intersections are numbered from 0 to $N-1$, with your house at intersection 0. The next $M$ lines contain three integers $X$, $Y$, and $D$, indicating a bidirectional road of length $D$ (1-1,000,000) between intersections $X$ and $Y$. The following line contains a single integer $S$ (1-10), representing the number of stores to visit. The subsequent $S$ lines contain one integer each, indicating the intersection where each store is located. No two stores are at the same intersection, and none are at intersection 0. All stores are reachable from your house.

## Output

Output the length of the shortest possible shopping trip from your house, visiting all stores, and returning home.


To simplify a fraction, divide both the numerator and the denominator by the greatest common factor, reducing it to its lowest form. Start by making a list of the factors of the numerator and a separate list of the factors of the denominator. A factor is a number that can be divided into another number.

For example, in the fraction 6/18, the numerator's factors are 1, 2, 3, and 6, while the denominator's factors are 1, 2, 3, 6, 9, and 18. The greatest common factor is the highest common number between the two lists, which in this case is 6.

To simplify, divide the numerator by the greatest common factor: 6 divided by 6 equals 1. Then, divide the denominator by the greatest common factor: 18 divided by 6 equals 3. The new values of the numerator and the denominator are then used to rewrite the fraction. In this example, 6/18 simplifies to 1/3, which is its lowest form. This process ensures that the value of the fraction remains the same while expressing it in the simplest terms possible.


**Cluster Analysis: Basic Concepts and Algorithms**

**Introduction to Cluster Analysis**

Cluster analysis is the process of finding groups of objects that are similar to each other and different from objects in other groups. It has various applications, including understanding, summarization, and reducing the size of large data sets.

**What is Cluster Analysis?**

Cluster analysis is a grouping of objects based on the data, where the objects in a group are similar to each other and different from objects in other groups. It is not simple segmentation, dividing students into different registration groups alphabetically, or results of a query.

**Types of Clusterings**

There are two main types of clusterings: hierarchical and partitional. Hierarchical clustering produces a set of nested clusters organized as a hierarchical tree, while partitional clustering divides data objects into non-overlapping subsets.

**Types of Clusters**

There are several types of clusters, including:

1. Well-separated clusters: a cluster is a set of points such that any point in a cluster is closer to every other point in the cluster than to any point not in the cluster.
2. Center-based clusters: a cluster is a set of objects such that an object in a cluster is closer to the center of a cluster than to the center of any other cluster.
3. Contiguous clusters: a cluster is a set of points such that a point in a cluster is closer to one or more other points in the cluster than to any point not in the cluster.
4. Density-based clusters: a cluster is a dense region of points separated by low-density regions from other regions of high density.
5. Property or conceptual clusters: finds clusters that share some common property or represent a particular concept.

**Clustering Algorithms**

There are several clustering algorithms, including:

1. K-means clustering: a partitional clustering approach that requires the number of clusters to be specified.
2. Hierarchical clustering: produces a set of nested clusters organized as a hierarchical tree.
3. Density-based clustering: clusters data points into dense regions separated by low-density regions.

**K-means Clustering**

K-means clustering is a partitional clustering approach that requires the number of clusters to be specified. The algorithm works by assigning each point to the cluster with the closest centroid and updating the centroids until convergence.

**Evaluating K-means Clusters**

The most common measure for evaluating K-means clusters is the Sum of Squared Error (SSE). SSE measures the distance between each point and its nearest cluster centroid.

**Limitations of K-means**

K-means has several limitations, including:

1. Difficulty handling clusters of differing sizes, densities, and non-globular shapes.
2. Sensitivity to outliers.
3. Requirement for the number of clusters to be specified.

**Hierarchical Clustering**

Hierarchical clustering produces a set of nested clusters organized as a hierarchical tree. It can be visualized as a dendrogram, a tree-like diagram that records the sequences of merges or splits.

**Strengths of Hierarchical Clustering**

Hierarchical clustering has several strengths, including:

1. No requirement for the number of clusters to be specified.
2. Ability to handle clusters of varying sizes and densities.
3. Robustness to outliers.

**DBSCAN**

DBSCAN is a density-based algorithm that clusters data points into dense regions separated by low-density regions. It is robust to noise and can handle clusters of varying sizes and densities.

**Cluster Validity**

Cluster validity refers to the evaluation of the goodness of a clustering structure. There are several measures of cluster validity, including:

1. External index: measures the extent to which cluster labels match externally supplied class labels.
2. Internal index: measures the goodness of a clustering structure without respect to external information.
3. Relative index: compares two different clusterings or clusters.

**Measures of Cluster Validity**

There are several measures of cluster validity, including:

1. Sum of Squared Error (SSE): measures the distance between each point and its nearest cluster centroid.
2. Entropy: measures the amount of uncertainty or randomness in a clustering structure.
3. Silhouette coefficient: combines ideas of cohesion and separation for individual points, clusters, and clusterings.

**Framework for Cluster Validity**

A framework for cluster validity is necessary to interpret any measure of cluster validity. This framework can be statistical, providing a basis for comparing the values of an index that result from random data or clusterings to those of a clustering result.


# Fortune Hunter/Education

## Classroom Application
The game is designed for 4th-grade students to assist with math topics and provide excellent reinforcement material.

## Standards Addressed
The game covers elements of the 4th-grade math curriculum in Boston, MA, USA, including:

* NUMBER SENSE AND OPERATIONS:
	+ 4.N.3: Understanding fractions as parts of unit wholes, collections, and number lines.
	+ 4.N.4: Relating common fractions and mixed numbers, finding equivalent fractions, and ordering fractions.
	+ 4.N.5: Identifying equivalent forms of decimals and fractions less than one whole.
	+ 4.N.6: Understanding the base ten number system, reading, naming, and writing decimals between 0 and 1 up to the hundredths.
	+ 4.N.9: Using commutative, associative, and identity properties of operations on whole numbers.
	+ 4.N.11: Knowing multiplication facts through 12 x 12 and related division facts.
	+ 4.N.12: Adding, subtracting, and multiplying accurately and efficiently.
	+ 4.N.14: Using conventional algorithms for addition, subtraction, and multiplication.
* GEOMETRY:
	+ 4.G.1: Comparing and analyzing attributes of two- and three-dimensional geometric shapes.
	+ 4.G.3: Recognizing similar figures.
	+ 4.G.4: Identifying angles as acute, right, or obtuse.
	+ 4.G.7: Describing and applying techniques for determining congruent shapes.
	+ 4.G.9: Predicting and validating results of partitioning, folding, and combining shapes.
* MEASUREMENT:
	+ 4.M.2: Carrying out simple unit conversions within a system of measurement.

## Preparation/Materials
Before playing, students should:
* Read the Instructions/Tutorial.
* Have a pen, pencil, and paper for notes.
* Introduction to topics covered is encouraged but not required.

## Goals
* Provide an introduction to topics through the Introduction/Tutorial.
* Become familiar with topics in a contextual setting.

## Objectives
* Guide students through Intro slides to understand game interaction.
* Defeat enemies using math concept-based attacks.
* Overcome in-game challenges by solving math problems.

## Vocabulary
* Dungeon: The game world covering the current lesson.
* HP: Player's health, which reaches 0 results in loss.
* Power: A bar that diminishes during attacks, affecting damage.
* Damage/Dmg.: HP lost by a character or opponent.
* Hit: A successful attack.
* Critical Hit: An effective attack requiring a simple multiplication problem.
* Miss: An unsuccessful attack with 0 damage.
* Attack: How the player defeats an opponent.
* Basic Attack: A move with a chance of hitting, missing, or critical hit.
* Fraction Attack: An attack requiring fraction problem-solving.
* Geometry Attack: An attack requiring geometry and graphing problem-solving.
* Shop: Where players purchase equipment and items using in-game currency.
* Boss: A powerful enemy requiring specialized problem completion.

## Rationale
The game promotes math education through interactive software, teaching students to apply math skills in a non-workbook setting.

## Supporting Materials
* http://wiki.sugarlabs.org/go/Fortune_Hunter
* http://wiki.sugarlabs.org/go/Math4Team/Resources/Curriculum_Chart


# Adding Problems

#### Number of problems found: 423

- At midmorning: The temperature at midmorning was 31°C. After dropping by 5°C in the evening, the resulting temperature is 31 - 5 = 26°C.

- Granddaughter: In 2014, the sum of Meghan's aunt, daughter, and granddaughter's ages was 100 years. Given their ages are powers of two, we need to find the year the granddaughter was born.

- Below sea level: A submarine was 800 ft below sea level and ascended 250 ft. Its new position is -800 + 250 = -550 ft below sea level.

- Clock mathematics: If it's 7:38 pm, we need to calculate the time after 30,033,996,480 minutes.

- The temperature 10: The temperature in a freezer is -15°C and increases by 3°C, making it -15 + 3 = -12°C.

- Eight chairs: With eight chairs and one more child on each, and twice as many girls as boys, we need to determine the number of girls and boys.

- A trout: A trout was 9 feet below the surface and dove 8 more feet, making its position 9 + 8 = 17 feet below the surface.

- Sum 1-6: The sum of the geometric progression 3, 15, 75,… to six terms can be calculated using the formula for the sum of a geometric series.

- Andy had: Andy had $5.95 and received $3.55 from his father, making a total of $5.95 + $3.55 = $9.50.

- The temperature 9: The temperature at 9 AM was 2 degrees and rose 3 degrees by noon, making it 2 + 3 = 5 degrees.

- Temperature rise and fall: The temperature was 82°F on Friday, changed by -2°F on Saturday, and then by 5°F on Sunday, resulting in 82 - 2 + 5 = 85°F.

- Insert 4: We need to insert three arithmetic means between 3 and 63.

- Sum of odd numbers: The sum of all odd integers from 13 to 781 can be calculated using the formula for the sum of an arithmetic series.

- Rectangular garden: A rectangular garden has a length of 48.7 m and a width 6.3 m shorter, so 48.7 - 6.3 = 42.4 m. The area is 48.7 * 42.4 = 2065.28 m².

- Extending square garden: Mrs. Petrová's garden was a square with a side length of 15 m, enlarged by 64 m², resulting in a new area of 15² + 64 = 225 + 64 = 289 m², making the new side length √289 = 17 m, so each side was extended by 17 - 15 = 2 m.

- At noon: The temperature at noon was 6°C and increased 4°C, making it 6 + 4 = 10°C.

- On the floor: With 2 dogs, 4 horses, 1 giraffe, and 1 duck on the bed, and 3 chickens flying, only the dogs, horses, and giraffe have feet on the floor, totaling 2 * 4 + 4 * 4 + 4 = 8 + 16 + 4 = 28 feet.

- Adding mixed 4: 2 and 1/8th plus 1 and 1/3rd can be calculated by finding a common denominator and adding the fractions.

- Outdoor temperature: The temperature was 8 degrees at midnight and declined 5 degrees each hour for 3 hours, resulting in 8 - 5 * 3 = 8 - 15 = -7 degrees.

- How many 3: The Andersons watched TV for 3/1 hr on Wednesday, 2/3 hr on Thursday, 4/5 hr on Friday, and 3/4 hr on Saturday, which can be added to find the total hours watched.


Fano Varieties

**Legend for Period Sequences:**
- n/a: known, not in Minkowski list
- #: known, number # from Minkowski list
- A#: known, computed with A/NA correlation
- ?*#: unknown, predicted by STD ansatz
- ?: unknown, no prediction

**Rank 1 Fano 3folds:**

| Name | Fano Index | Period Sequence | Construction |
| --- | --- | --- | --- |
|  | 4 | 12 | Projective 3-space |
|  | 3 | 4 | Quadric 3fold |
|  | 2 | 14 | Linear section of |
|  | 2 | 2 | (2,2) complete intersection in |
|  | 2 | 0 | Cubic 3fold |
|  | 2 | 1 | Quartic in |
|  | 2 | n/a | Sextic in |
|  | 1 | n/a | Sextic in |
|  | 1 | 15 | Quartic in |
|  | 1 | 19 | (2,3) complete intersection in |
|  | 1 | 5 | (2,2,2) complete intersection in |
|  | 1 | 9 | Section of by intersection of quadric and linear space |
|  | 1 | 7 | Linear section of spinor variety |
|  | 1 | 18 | Linear section of |
|  | 1 | 20 | Linear section of |
|  | 1 | 10 | Linear section of |
|  | 1 | 17 | Section of triple universal bundle on |

**Rank 2 Fano 3folds:**

| Number | Period Sequence | Description |
| --- | --- | --- |
| 1 | 4 | n/a | Blowup of with centre an elliptic curve |
| 2 | 6 | n/a | Double cover of branched along a divisor of bidegree (2,4) |
| 3 | 8 | n/a | Blowup of with centre an elliptic curve |
| 4 | 10 | 49 | Blowup of with centre an intersection of two cubics |
| 5 | 12 | 34 | Blowup of with centre a plane cubic on it |
| 6 | 12 | 11 | Divisor of bidegree (2,2) in |
| 7 | 14 | 51 | Blowup of a quadric with centre the intersection of two members of |
| 8 | 14 | 26 | Double cover of with branch locus a member of |
| 9 | 16 | 62 | Blowup of in a curve of degree 7 and genus 5 |
| 10 | 16 | 40 | Blowup of with centre an elliptic curve |
| 11 | 18 | 56 | Blowup of with centre a line on it |
| 12 | 20 | 13 | Blowup of in a curve of degree 6 and genus 3 |
| 13 | 20 | 52 | Blowup of a 3-dimensional quadric in a curve of genus 2 and degree 6 |
| 14 | 20 | A 39 | Blowup of with center an elliptic curve |
| 15 | 22 | 35 | Blowup of with center the intersection of a quadric and a cubic |
| 16 | 22 | 59 | Blowup of with center a conic on it |
| 17 | 24 | A 38 | Blowup of a 3-dimensional quadric with center an elliptic curve |
| 18 | 24 | 60 | Double cover of with branch locus a divisor of bidegree (2,2) |
| 19 | 26 | 55 | Blowup of with center a line on it |
| 20 | 26 | A 46 | Blowup of with center a twisted cubic on it |
| 21 | 28 | A 8 | Blowup of with center a twisted quartic on it |
| 22 | 30 | A 50 | Blowup of with center a conic on it |
| 23 | 30 | 29 | Blowup of a quadric with center an intersection of and |
| 24 | 30 | 66 | Divisor of bidegree (1,2) on |
| 25 | 32 | 28 | Blowup of with centre an elliptic curve |
| 26 | 34 | ?*47 | Blowup of with center a line on it |
| 27 | 38 | 61 | Blowup of with center a twisted cubic |
| 28 | 40 | 33 | Blowup of with centre a plane cubic |
| 29 | 40 | 42 | Blowup of a quadric 3fold with centre a conic on it |
| 30 | 46 | 70 | Blowup of with center a conic |
| 31 | 46 | 48 | Blowup of a quadric 3fold with center a line on it |
| 32 | 48 | 6 | Divisor on of bidegree |
| 33 | 54 | 54 | Blowup of with center a line |
| 34 | 54 | 44 |  |
| 35 | 56 | 30 | Blowup of at a point |
| 36 | 62 | 58 | Scroll over |

**Rank 3 Fano 3folds:**

| Number | Period Sequence | Description |
| --- | --- | --- |
| 1 | 12 | 22 | Double cover of with branch locus a divisor of tridegree (2,2,2) |
| 2 | 14 | 97 | Member of on the bundle over |
| 3 | 18 | 31 | Divisor on of tridegree |
| 4 | 18 | 151 | Blowup of with center a smooth fiber of |
| 5 | 20 | 109 | Blowup of with center a curve of bidegree (5,2) |
| 6 | 22 | 146 | Blowup of with center a disjoint union of a line and an elliptic curve |
| 7 | 24 | 36 | Blowup of with center an elliptic curve |
| 8 | 24 | 85 | Member of the linear system on |
| 9 | 26 | 68 | Blowup of the cone over the Veronese surface with center a disjoint union |
| 10 | 26 | 67 | Blowup of with center a disjoint union of two conics on it |
| 11 | 28 | 107 | Blowup of with center a complete intersection of two general members of |
| 12 | 28 | 144 | Blowup of with centre a disjoint union of a twisted cubic and a line |
| 13 | 30 | 16 | Blowup |
| 14 | 32 | 148 | Blowup |
| 15 | 32 | 112 | Blowup of a quadric with center the disjoint union of a line on and a conic on |
| 16 | 34 | 119 | Blowup |
| 17 | 36 | 37 | Smooth divisor on of tridegree |
| 18 | 36 | 160 | Blowup |
| 19 | 38 | 57 | Blowup |
| 20 | 38 | 63 | Blowup |
| 21 | 38 | 98 | Blowup of along a curve of bidegree (2,1) |
| 22 | 40 | 152 | Blowup of along a curve of bidegree (0,2) |
| 23 | 42 | 158 | Blowup |
| 24 | 42 | 86 | Fiber product |
| 25 | 44 | 41 | Blowup |
| 26 | 46 | 113 | Blowup of with center a disjoint union of a point and a line |
| 27 | 48 | 21 |  |
| 28 | 48 | 90 |  |
| 29 | 50 | 163 | Blowup |
| 30 | 50 | 84 | Blowup |
| 31 | 52 | 53 | Blowup |

**Rank 4 Fano 3folds:**

| Number | Period Sequence | Description |
| --- | --- | --- |
| 1 | 24 | 3 | Smooth divisor on of polydegree |
| 2 | 28 | 32 |  |
| 3 | 30 | 122 |  |
| 4 | 32 | 103 |  |
| 5 | 32 | 147 |  |
| 6 | 34 | 65 |  |
| 7 | 36 | 69 |  |
| 8 | 38 | 105 |  |
| 9 | 40 | 102 |  |
| 10 | 42 | 142 |  |
| 11 | 44 | 93 |  |
| 12 | 46 | 150 |  |
| 13 | 26 | 88 | Blowup of curve of tridegree on |

**Rank 5 Fano 3folds:**

| Number | Period Sequence | Description |
| --- | --- | --- |
| 1 | 28 | 114 |  |
| 2 | 36 | 87 | Smooth toric, but not 53 |
| 3 | 36 | 43 |  |

**Fano 3folds with Picard Number:**

|  | Period Sequence | Description |
| --- | --- | --- |
| 6 | 30 | 64 |  |
| 7 | 24 | 71 |  |
| 8 | 18 | 45 |  |
| 9 | 12 | n/a |  |
| 10 | 6 | n/a |  |

**Discrepancy Note:**
There is a discrepancy between tables, requiring replacement of  with  in table 2.

**Period Sequences with N=4:**
49 period sequences have N=4, with 45 corresponding to Fano threefolds. However, Fano variety 4.7 (period sequence 69) has an expected N of 5, contradicting the expected N from Matsuki. The remaining 4 period sequences (23, 24, 25, and 27) are considered parasitic.


Complex numbers, also known as imaginary numbers, have been a topic of interest for many. The concept of complex numbers was first introduced to deal with problems that couldn't be solved using real numbers alone. Leonard Euler's idea of writing i = sqrt(-1) laid the foundation for complex analysis.

The real number line is well-known, and complex numbers are plotted on a number line perpendicular to the real number line, forming a complex plane. The imaginary axis is represented by 'Im', and the real axis is represented by 'Re'. Multiplying a number by 'i' results in a rotation of 90 degrees. For example, multiplying 1 by 'i' gives 'i', then '-1', and then '-i', tracing a circle.

Multiplying by 'i' can be thought of as rotation, while multiplying by a factor of 'i' (e.g., 2i or 3i) results in scaling and rotation. Understanding the complex plane makes it easy to locate any complex number, represented as 'a + ib', where 'a' is on the real axis and 'b' is on the imaginary axis.

Complex functions have numerous applications, particularly in solving differential equations. A simple complex function is defined as the exponential of (ix), which can be simplified using Euler's formula. Visualizing this function in the complex plane reveals a helical structure with cosine on the real part and sine on the imaginary part.

Any complex function F(z) can be written as a linear combination of functions 'u' and 'v', i.e., F(z) = u + iv. Limits, derivatives, and integrals of complex functions are defined similarly to those of real functions.

**A) Limits:** Finding limits of complex numbers is straightforward, as F(z) = u + iv, and limits can be taken separately for 'u' and 'v'.

**B) Differentiation:** Differentiation of complex functions follows the same rules as real functions. The Cauchy-Riemann equations play a crucial role in determining if a function F(z) is complex differentiable. For F(z) to be complex differentiable, it must satisfy the Cauchy-Riemann equations.

**C) Integration:** Only analytic functions in complex planes can be integrated. The residue method is a popular technique for solving integrals of complex functions.

The history of complex numbers dates back to the 14th-15th century, when mathematicians first encountered quadratic equations. The general solution to these equations led to the development of complex numbers. Mathematicians such as Cardano worked on solving cubic equations, which ultimately led to the realization of imaginary numbers. Leonard Euler's work on complex functions and their integration and differentiation laid the foundation for transformation techniques.

The development of complex numbers took approximately two centuries, from the 14th to the 16th century. Euler's contributions to the field of complex analysis were significant, and his work paved the way for future discoveries. For a more in-depth look at the history of complex numbers, the YouTube channel "Veritasium" offers an informative video on the topic.


In Julia, it's generally recommended to use the `complex` constructor instead of the `Complex` type. The `Complex` type can be useful when you want to force a specific parameter, such as `Complex{Float64}(3, -2)`, which returns `3.0 - 2.0im`. In all other cases, using `complex` is preferred.

Note that `float` converts a number to the most appropriate floating-point number, but there is no associated `Float` type. In Julia, type names start with a capital letter, while function names are in lowercase.

Julia does not have a built-in `Irrational` type for numbers like `2pi`. However, the `Tau.jl` module from GitHub provides a definition of the circle constant `tau` (equivalent to `2π`). Irrational numbers cannot be represented numerically by a finite-memory computer, so they are often represented as float numbers. Julia does not perform symbolic algebra.

The convenience of using irrational numbers in Julia is that they are automatically converted to the most appropriate precision when they come into contact with another number. For example, `pi * 1` returns `3.141592653589793`, which is precise up to the 53-bit significand of a `Float64`. In contrast, `pi * big(1)` returns a more precise result: `3.141592653589793238462643383279502884197169399375105820974944592307816406286198`.

When working with irrational numbers, you can use `float(x)` to convert to `Float64`, or `big(x)` for maximum precision. If your function has another argument, consider using a promotion rule. For instance, `big(pi * 1)` returns `3.141592653589793115997963468544185161590576171875`, which loses precision after the 16th significant digit.


I need to determine whether the eigenvalues of a large $2n$ by $2n$ block matrix $M$ are inside the unit circle. The matrix $M$ is defined as:

\[
\begin{pmatrix}
A (\equiv I + 3\alpha J) & B (\equiv -\alpha J) \\
I & 0
\end{pmatrix}
\]

where $J$ is an $n$-by-$n$ complex matrix, $I$ is the identity, and $\alpha \in (0,1]$ is a parameter.

The problem comes from the discretization of a PDE, with $n$ potentially on the order of 5000. I am looking for a numerical method to solve this problem without having to find the eigenvalues of the $2n$ by $2n$ matrix $M$ directly.

By definition, $\lambda$ is an eigenvalue of $M$ if and only if there exist $(p_1, p_2) \in (\mathbb{C}^{n})^2 \setminus \{(0,0)\}$ such that:

\[
\begin{pmatrix}
A & B \\
I & 0
\end{pmatrix}
\begin{pmatrix}
p_1 \\
p_2
\end{pmatrix}
= \lambda
\begin{pmatrix}
p_1 \\
p_2
\end{pmatrix}
\]

This leads to the non-linear eigenvalue problem:

\[
\lambda(A - \lambda I)p_2 = -B p_2
\]

with $p_1 = \lambda p_2$, which in terms of $J$ is:

\[
\lambda(3 \alpha J - (\lambda - 1)I)p_2 = \alpha J p_2
\]

Is there an efficient algorithm to solve this problem? Alternatively, can the eigenvalues of $M$ be determined directly from those of $J$?


# Lesson 23: Estimating a Hemisphere

## 23.1: Introduction to Hemispheres

We will estimate the volume of hemispheres using known shapes.

## 23.2: Hemispheres in Boxes

Mai has a glass hemisphere paperweight with a radius of 3 cm. To design a box for it, we need to determine the dimensions of the inside of the box to make it as small as possible.

- The box's dimensions should be twice the radius of the hemisphere in diameter (6 cm) and equal to the radius in height (3 cm) to fit the hemisphere snugly.
- The volume of the box is calculated as length × width × height, which equals 6 cm × 6 cm × 6 cm = 216 cubic cm.
- A reasonable estimate for the volume of the paperweight is less than the volume of the box.

Tyler has a box with side lengths twice as long as Mai's box, which can hold a different glass paperweight.
- The volume of Tyler's box is 12 cm × 12 cm × 12 cm = 1728 cubic cm.
- A reasonable estimate for the volume of this glass paperweight is less than the volume of Tyler's box.
- The volume of the paperweight in Tyler's box is likely 8 times bigger than Mai's paperweight because the volume of the box increased by a factor of 8 (2^3).

## 23.3: Estimating Hemispheres

A hemisphere with a radius of 5 units fits snugly into a cylinder of the same radius and height.
- The volume of the cylinder is πr^2h = π(5)^2(5) = 125π cubic units ≈ 392.7 cubic units.
- The volume of the hemisphere is estimated to be less than the volume of the cylinder.

A cone fits snugly inside a hemisphere with a shared radius of 5 units.
- The volume of the cone is (1/3)πr^2h = (1/3)π(5)^2(5) = (1/3)125π cubic units ≈ 65.45 cubic units.
- The volume of the hemisphere is estimated to be more than the volume of the cone.

Comparing estimates, the hemisphere's volume is between the cone's and cylinder's volumes.

## Summary

To estimate the volume of a hemisphere, compare it to known shapes. A hemisphere of radius 1 unit fits inside a cylinder with a radius of 1 unit and height of 1 unit, and its volume is less than the cylinder's volume (approximately 3.14 cubic units). A cone of radius 1 unit and height 1 unit fits inside the hemisphere, and its volume is approximately 1.05 cubic units, making it a reasonable under-estimate for the hemisphere's volume.

Averaging the volumes of the cylinder and the cone, we estimate the hemisphere's volume to be about 2.10 cubic units. Since a hemisphere is half of a sphere, a sphere with a radius of 1 unit would have a volume of approximately 4.20 cubic units.


**Math Problems and Solutions**

### System of Equations
Three students are working to find the solution set of the system of equations:
y = 3x + 10
2y = 6x – 4

### Average Rate of Change
Find the average rate of change for f(x) = x^2 + 7x + 10 from x = 0 to x = 2.

### Similar Triangles
Triangle ABC is similar to triangle DEF. The length of AC is 12 cm. The length of BC is 18 cm. The length of DF is 10 cm. What is the length of EF?

### Probability
The following frequency table summarizes the number of children that dads in Dads Club have. Based on this data, what is a reasonable estimate of the probability that the next dad to join Dads Club has fewer than 3 children?

### Similarity Between Figures
What are the elements needed to prove similarity between two figures? Explain.

### Volume of a Rectangular Prism
A rectangular prism has a length of 3 1/2 inches, a width of 5 inches, and a height of 1 1/2 inches. What is the volume of the prism?

### Ordering Lengths
Rewrite the following lengths in increasing order: 537 km, 790 dm, 41 cm, 8 mm, 788 m.

### Net Worth
Laura could sell her car for $18,000, and she owes $5,500 on it. Which statement best describes how this situation affects her net worth?
A. Laura has an $18,000 asset and a $5,500 liability.
B. Laura has a $15,500 asset and an $18,000 liability.
C. Laura's car is considered a liability only because she still owes money on it.
D. Laura's car is considered an asset only because it is worth more than she owes.

### Investment
How much money has to be invested at 2.9% interest compounded continuously to have $34,000 after 18 years?

### Volume of a Cone
A cone with a radius of 2 units has a volume of 29 cubic units. Find the height of the cone.

### Feasible Region
Choose the correct feasible region for the following constraints:
5X + 5Y < 80
2X + 6Y < 72
3X + 2Y < 42
X, Y > 0

### Isosceles Triangle
Triangle XYZ is isosceles. The measure of the vertex angle, Y, is twice the measure of a base angle. What is true about triangle XYZ?
a. Angle Y is a right angle.
b. The measure of angle Z is 45°.
c. The measure of angle X is 36°.
d. The measure of the vertex angle is 72°.
e. The perpendicular bisector of creates two smaller isosceles triangles.

### Taylor Series
Given f(3) = 9 and f(5) = 28, using the first two terms of the Taylor series, the approximate value of f'(3) is.

### Probability of Rain
The probability that it will not rain on Thursday is 67%. What is the probability that it will rain on Thursday?

### Cookie Recipe
A baker is using a cookie recipe that calls for 2 1/4 cups of flour to yield 36 cookies. How much flour will the baker need to make 90 cookies using the same recipe?

### Grouping Students
Fifteen students are going hiking on their spring break. They plan to travel in three vehicles—one seating 7, one seating 5, and one seating 3 students. In how many ways can the students group themselves for their trip?

### Scales of Measurement
Suppose you conducted a survey of your classmates to determine how much money they intended to spend on Halloween candy this year. Which of the following scales of measurement would your measure fall under?
A. ordinal
B. interval
C. ratio

### Inverse Variation
The number of new songs that a rock group needs to stay at the top each year is inversely proportional to the year and that the group has been in the industry.
a. Write an equation of variation.
b. If after 3 years in the industry the group needs 50 new songs, how many songs will you need after 5 years?

### Candy Bags
Hannah was preparing to make 'goody bags' for trick-or-treaters on Halloween. She had large bags of snack pack candies: one bag held 48 pieces of chocolate, a second held 64 pieces lollipops, and the third held 80 pieces of bubblegum. How many bags can she prepare if she does not want any candy left over? How many of each type will go into the bags?

### Separable Equation
Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable.
dy/dx = (xy + y^2) / (3x^2 + y^2) becomes integral (u^2 + A) / (u^3 + Bu^2 - 4u)


# Category: Limits

This category contains results about **Limits**. Definitions specific to this category can be found in Definitions/Limits.

Let $\sequence {x_n}$ be a sequence in $\R$ that converges to a value $l \in \R$. Then $l$ is a **limit of $\sequence {x_n}$ as $n$ tends to infinity**.

## Subcategories

This category has 6 subcategories, including:
- Limits of Complex Functions
- Limits of Sequence of Sets

## Pages in category "Limits"

There are 3 pages in this category, providing information on various aspects of limits.


Grades 2, 3, and 4 can benefit from introducing multiplication using arrays, which helps students model multiplication as repeated addition and understand its properties. This foundation is crucial for learning division, fractions, and more. To achieve automaticity with multiplication and division facts, students need to make connections through a deep understanding of the concept and its properties.

When introducing multiplication, the following steps are essential:
- Multiplication as repeated addition: using arrays and area models to grasp the concept
- Practicing the commutative property of multiplication, which states that the order of numbers does not matter
- Practicing the associative property of multiplication, which states that the grouping of numbers does not matter when multiplying a series of values together, e.g., (5×4)×2 = 5×(4×2)

Manipulatives like building bricks are effective for introducing multiplication and arrays. They allow students to create arrays, represent multiplication expressions, and understand multiplication properties. To support virtual teaching, printable and digital activity cards have been created based on building bricks. These cards help practice multiplication and its properties.

The activity cards offer multiple levels:
- Building arrays based on given expressions
- Solving problems after building arrays for given expressions
- Building different arrays for a number (starting with basic multiplication facts)
- Selecting the correct expressions for already built arrays (multiplication properties)
- Completing expressions for already built arrays (multiplication properties)
- Writing expressions for attached brick combinations (problem-solving)

The printable and digital activity cards can be used as virtual manipulatives, for warm-ups, practice, assessment, centers, homework, and more. They are ideal for initiating discussions and comparing different ways of representing a number and solving a multiplication problem. The activity cards are available in a free version (Google Slides) and a premium version, which includes 46 printable activity cards and 100 digital activity cards.