Problem 49 - Algebra 2

Question

Evaluate 2\(\sqrt{\frac{p(1-p)}{n}}\) for the given values of \(p\) and \(n\) Round to the nearest thousandth if necessary. $$ p=0.5, n=100 $$

Step-by-Step Solution

Answer

The evaluated expression is 0.1.

Step 1: Substitute Given Values

Substitute the given values of \( p = 0.5 \) and \( n = 100 \) into the expression \( 2\sqrt{\frac{p(1-p)}{n}} \). This results in the expression \( 2\sqrt{\frac{0.5(1-0.5)}{100}} \).

Step 2: Simplify the Numerator

Calculate the multiplication within the numerator: \( 0.5 \times (1 - 0.5) = 0.5 \times 0.5 = 0.25 \). Thus, the expression becomes \( 2\sqrt{\frac{0.25}{100}} \).

Step 3: Divide the Numerator by the Denominator

Divide the calculated numerator by the denominator: \( \frac{0.25}{100} = 0.0025 \). Now the expression is \( 2\sqrt{0.0025} \).

Step 4: Calculate the Square Root

Evaluate the square root of 0.0025, which is \( \sqrt{0.0025} = 0.05 \). Update the expression to \( 2 \times 0.05 \).

Step 5: Multiply by 2

Multiply 0.05 by 2 to find the final value: \( 2 \times 0.05 = 0.1 \).

Key Concepts

Square Root

The square root is an essential mathematical operation that helps you determine what number, when multiplied by itself, gives the original number. In this problem, we need to calculate the square root of 0.0025.

• The square root of a number 0.0025 is written as \( \sqrt{0.0025} \).
• By inspection or using a calculator, \( \sqrt{0.0025} = 0.05 \).

This operation simplifies the task of solving more complex problems by reducing numbers to simpler forms. Remember that square roots can be positive or negative, but in most cases, we consider the positive value.

Substitution

Substitution is a method used in mathematics to replace variables with their given values to simplify expressions or solve equations. In the exercise, we start by substituting the values of \( p = 0.5 \) and \( n = 100 \) into the expression.

• Begin with the original expression: \( 2\sqrt{\frac{p(1-p)}{n}} \).
• Replace \( p \) with 0.5 and \( n \) with 100 to get \( 2\sqrt{\frac{0.5(1-0.5)}{100}} \).

This makes complex problems more manageable by transforming the expression into one that is easier to evaluate.

Simplification

Simplification involves reducing expressions to their simplest form. It requires performing mathematical operations like multiplication and division. After substitution, you simplify the expression to make calculations easier.

• First, calculate the numerator: \( 0.5 \times (1-0.5) = 0.25 \).
• The expression becomes \( 2\sqrt{\frac{0.25}{100}} \).
• Then divide \( 0.25 \) by 100: \( \frac{0.25}{100} = 0.0025 \).

Simplification helps find results efficiently without unnecessary complication.

Numerator and Denominator

In fractions, understanding the numerator and denominator is crucial for accurate calculations. The numerator is the top number, representing part of the whole, while the denominator is the bottom number, indicating the total quantity.

• In the expression, the numerator is \( 0.25 \) after simplification, and the denominator is 100.
• Division of the numerator by the denominator yields \( \frac{0.25}{100} = 0.0025 \).

Accurate handling of these components ensures precise mathematical operations and results.

Rounding Numbers

Rounding numbers is a key concept in mathematics used to make numbers simpler and easier to work with. It involves adjusting a number to a nearby value that is often more useful in the context given.

• In this problem, once we calculate \( 2 \times 0.05 \), the value is exactly 0.1.
• If a number needed rounding to the nearest thousandth, it means you look at the numbers up to three decimal places.

Rounding can help reduce calculation errors and make numerical results more comprehensible for practical purposes.