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100% equals 1 because it represents the fraction $100/100$. When the numerator and denominator of a fraction are the same nonzero number, the value of the fraction is exactly one whole.

Conditions: The percentage value is 100.; The denominator is 100.

100% equals 1 because it represents the fraction $100/100$. When the numerator and denominator of a fraction are the same nonzero number, the value of the fraction is exactly one whole.

Conditions: The percentage value is 100.; The denominator is 100.

The expected number of heads is 50 because the expected value for a counting experiment like this is calculated as the number of trials multiplied by the probability of success on each trial. With 100 tosses and a fair coin (probability of heads = 0.5), the calculation is $100 \cdot 0.5 = 50$.

Conditions: The coin is fair.; There are 100 tosses per trial.; $X$ counts heads.

The expected number of heads is 50 because the expected value for a counting experiment like this is calculated as the number of trials multiplied by the probability of success on each trial. With 100 tosses and a fair coin (probability of heads = 0.5), the calculation is $100 \cdot 0.5 = 50$.

Conditions: The coin is fair.; There are 100 tosses per trial.; $X$ counts heads.

To derive the linear pricing equations, you impose the no-arbitrage condition that the current price of each traded asset must equal the sum of its state-contingent payoffs weighted by the corresponding state prices. This produces one linear equation per traded asset, with the state prices acting as the unknowns.

Conditions: One-period model with a finite set of states.; Prices of all traded assets are known today.; State-contingent payoffs for each asset are known.; The unknowns are the state prices (Arrow-Debreu security prices).

To derive the linear pricing equations, you impose the no-arbitrage condition that the current price of each traded asset must equal the sum of its state-contingent payoffs weighted by the corresponding state prices. This produces one linear equation per traded asset, with the state prices acting as the unknowns.

Conditions: One-period model with a finite set of states.; Prices of all traded assets are known today.; State-contingent payoffs for each asset are known.; The unknowns are the state prices (Arrow-Debreu security prices).

The optimal solution uses machines 2 and 3, while machine 1 remains unused. The production quantities are $X_1=0$, $X_2=550$, and $X_3=450$.

Conditions: The model includes three machines with specific costs and capacities.; The demand is exactly 1000 units.; The objective is to minimize total cost.; The solution is derived from solving the integer linear program.

The optimal solution uses machines 2 and 3, while machine 1 remains unused. The production quantities are $X_1=0$, $X_2=550$, and $X_3=450$.

Conditions: The model includes three machines with specific costs and capacities.; The demand is exactly 1000 units.; The objective is to minimize total cost.; The solution is derived from solving the integer linear program.

0% means zero over one hundred, which is a zero fraction with a value of 0. It is allowed because zero may appear on the top (numerator) of a fraction, but not on the bottom (denominator).

Conditions: The numerator is 0.; The denominator is 100.

0% means zero over one hundred, which is a zero fraction with a value of 0. It is allowed because zero may appear on the top (numerator) of a fraction, but not on the bottom (denominator).

Conditions: The numerator is 0.; The denominator is 100.

To convert a percentage into a fraction, place the numerical value of the percentage over 100. The number before the percent sign becomes the numerator, and the denominator is fixed at 100.

Conditions: The starting value is in percent form.; The denominator must be 100.

To convert a percentage into a fraction, place the numerical value of the percentage over 100. The number before the percent sign becomes the numerator, and the denominator is fixed at 100.

Conditions: The starting value is in percent form.; The denominator must be 100.

An equality constraint is used because the customer order requires exactly 1000 units to be supplied. The problem specifies that the order must be met precisely, meaning neither underproduction (shortfall) nor overproduction (surplus) is permitted.

Conditions: The problem involves fulfilling a specific customer order.; The order quantity is fixed and must be met exactly.; Overproduction or underproduction is not allowed or desired.

An equality constraint is used because the customer order requires exactly 1000 units to be supplied. The problem specifies that the order must be met precisely, meaning neither underproduction (shortfall) nor overproduction (surplus) is permitted.

Conditions: The problem involves fulfilling a specific customer order.; The order quantity is fixed and must be met exactly.; Overproduction or underproduction is not allowed or desired.

To find the left-hand limit as x approaches a value c (denoted $\lim_{x \to c^-} f(x)$), observe the behavior of the function's curve strictly for input values less than c, moving towards c. If the y-values grow without bound in either the positive or negative direction, the finite real limit does not exist.

Conditions: The function has a vertical asymptote or discontinuity at $x=c$.; You are evaluating the limit from the left side (inputs strictly less than c).

To find the left-hand limit as x approaches a value c (denoted $\lim_{x \to c^-} f(x)$), observe the behavior of the function's curve strictly for input values less than c, moving towards c. If the y-values grow without bound in either the positive or negative direction, the finite real limit does not exist.

Conditions: The function has a vertical asymptote or discontinuity at $x=c$.; You are evaluating the limit from the left side (inputs strictly less than c).

The example concludes this because the conditional expectation of the discounted time-3 stock price, given information at time 2, equals the discounted time-2 stock price. Specifically, $\mathbb{E}_2\left[\frac{S_3}{(1+r)^3}\right] = 82.64$ and $\frac{S_2}{(1+r)^2} = 82.64$, satisfying the martingale condition $M_n = \mathbb{E}_n[M_{n+1}]$.

Conditions: The expectation is taken under the risk-neutral probability $p = 2/3$.; Discounting uses the interest rate $r = 0.1$.

The example concludes this because the conditional expectation of the discounted time-3 stock price, given information at time 2, equals the discounted time-2 stock price. Specifically, $\mathbb{E}_2\left[\frac{S_3}{(1+r)^3}\right] = 82.64$ and $\frac{S_2}{(1+r)^2} = 82.64$, satisfying the martingale condition $M_n = \mathbb{E}_n[M_{n+1}]$.

Conditions: The expectation is taken under the risk-neutral probability $p = 2/3$.; Discounting uses the interest rate $r = 0.1$.

To scale an expected value, calculate the expectation for a single unit and then multiply it by the total number of units. This relies on the linearity of expectation, which holds regardless of whether the units are independent, provided the individual expectations are defined.

Conditions: The expectation for a single unit is known or calculable.; The number of units is fixed and finite.

To scale an expected value, calculate the expectation for a single unit and then multiply it by the total number of units. This relies on the linearity of expectation, which holds regardless of whether the units are independent, provided the individual expectations are defined.

Conditions: The expectation for a single unit is known or calculable.; The number of units is fixed and finite.

The sample mean converges to 50 because 50 is the expected value $E(X)$ of the random variable $X$ (number of heads in 100 fair-coin tosses). According to the law of large numbers, as the number of trials $n$ approaches infinity, the sample mean $\overline{X}_n$ approaches the expected value.

Conditions: The coin is fair.; $X$ counts heads after 100 tosses.; $n$ approaches infinity.

The sample mean converges to 50 because 50 is the expected value $E(X)$ of the random variable $X$ (number of heads in 100 fair-coin tosses). According to the law of large numbers, as the number of trials $n$ approaches infinity, the sample mean $\overline{X}_n$ approaches the expected value.

Conditions: The coin is fair.; $X$ counts heads after 100 tosses.; $n$ approaches infinity.

Min-Max normalization is a method that linearly transforms raw data into the interval [0, 1]. The formula is $X' = \frac{X - \min}{\max - \min}$, where $X$ is the original value, and $\min$ and $\max$ are the minimum and maximum values of that feature column, respectively.

Conditions: The dataset has clear min and max values.; max is not equal to min

Min-Max normalization is a method that linearly transforms raw data into the interval [0, 1]. The formula is $X' = \frac{X - \min}{\max - \min}$, where $X$ is the original value, and $\min$ and $\max$ are the minimum and maximum values of that feature column, respectively.

Conditions: The dataset has clear min and max values.; max is not equal to min