Problem 52 - 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. $$ p=0.75, n=1000 $$

Step-by-Step Solution

Answer

The evaluated expression is approximately 0.027.

Step 1: Substitute the values into the formula

The given expression is \(2\sqrt{\frac{p(1-p)}{n}}\). We need to substitute \(p = 0.75\) and \(n = 1000\) into the expression. This results in \(2\sqrt{\frac{0.75(1-0.75)}{1000}}\).

Step 2: Simplify the expression inside the square root

First, calculate \(1 - p\), which is \(1 - 0.75 = 0.25\). Then, multiply \(p\) and \(1 - p\): \(0.75 \times 0.25 = 0.1875\). Now, substitute this back into the expression: \(2 \sqrt{\frac{0.1875}{1000}}\).

Step 3: Calculate the fraction

Divide the product \(0.1875\) by \(1000\) to get \(\frac{0.1875}{1000} = 0.0001875\).

Step 4: Calculate the square root

Find the square root of \(0.0001875\), which is approximately \(0.013693063\).

Step 5: Multiply by 2

Finally, multiply the result of the square root by 2: \(2 \times 0.013693063 = 0.027386126\).

Step 6: Round the result

Round \(0.027386126\) to the nearest thousandth, which gives \(0.027\). Therefore, the evaluated expression is approximately \(0.027\).

Key Concepts

algebra 2 problem solving

Problem-solving in Algebra 2 often involves working with expressions and formulas. It's about using known values to find unknown quantities, much like solving a puzzle.
For instance, in the given problem, the expression \(2\sqrt{\frac{p(1-p)}{n}}\) requires substitutions of given values for \(p\) and \(n\). These types of expressions are common in algebra when dealing with statistical problems and can involve probabilities and sample sizes.

• Start by identifying the expression to be evaluated.
• Substitute the given values for each variable.
• Carry out basic arithmetic operations such as multiplication and division inside the expression.
• Simplify complex expressions step by step.

Breaking down these problems into manageable steps is key to successful algebra problem-solving.

mathematical expressions

Mathematical expressions are combinations of numbers, variables, and operators, like addition, subtraction, multiplication, division, and roots. They can seem daunting at first glance, but by tackling them one step at a time, they can become manageable.

• In the problem given, the expression \(2\sqrt{\frac{p(1-p)}{n}}\) involves several operations: subtraction, multiplication, division, and taking a square root.
• Begin by simplifying the operations inside the parentheses \((1-p)\), then proceed to multiplication \(p(1-p)\).
• Next, divide this product by \(n\).
• Take the square root of the result.

The last operation is to multiply by 2. Each step simplifies the expression, making the transitions from step to step clearer and easier.

rounding to thousandths

Rounding numbers can make complex calculations easier to understand and apply. In mathematics, rounding to a certain decimal place, like the nearest thousandth, helps in achieving a manageable level of precision.
When rounding,

• Identify which decimal place you need to round to — here, it's the third decimal place (thousandths).
• Look at the digit in the fourth place (ten-thousandths) to decide if you round up or keep the current value.
• If the fourth decimal is 5 or more, increase the third decimal by one.
• If it's less than 5, leave the third decimal as is.

In the given problem, we took \(0.027386126\) and rounded it to \(0.027\). This precision is often sufficient in many practical scenarios.