Problem 80 - Beginning and Intermediate Algebra

Question

Dawn buys a 27 -oz box of cereal. The possible error in this amount, however, is \(\pm 0.5\) oz. Let \(c\) represent the range of values for the amount of cereal in the box. Write an absolute value inequality to represent the range for the number of ounces of cereal in the box, then solve the inequality and explain the meaning of the answer.

Step-by-Step Solution

Answer

The actual weight of the cereal in the box can range between 26.5 ounces and 27.5 ounces. This means that the actual amount of cereal in the box is within 0.5 oz of the stated weight of 27 oz.

Step 1: Define the variable for the actual weight

Let's denote the actual weight of the cereal in the box as \(w\) in ounces.

Step 2: Write the absolute value inequality for the range

We are given that the possible error in the weight of cereal is \(\pm 0.5\) oz. So, the difference between the actual weight \(w\) and the stated weight 27 oz must be within this error range. We can write the absolute value inequality as: \[|w - 27| \le 0.5\]

Step 3: Solve the absolute value inequality

To solve the absolute value inequality, we need to separate it into two inequalities: 1. \(w - 27 \le 0.5\) 2. \(-(w - 27) \le 0.5\) Now, solve each inequality: 1. For the first inequality, add 27 to both sides: \(w \le 27.5\) 2. For the second inequality, distribute the negative sign: \(-w + 27 \le 0.5\) Now, add \(w\) and subtract 27 from both sides: \(-0.5 \le w - 27\) Add 27 to both sides: \[26.5 \le w\] Combine the two inequalities to find the range of values for the actual weight \(w\): \[26.5 \le w \le 27.5\]

Step 4: Interpret the solution

The solution tells us that the actual weight of the cereal in the box can range between 26.5 ounces and 27.5 ounces, which means that the actual amount of cereal in the box is within 0.5 oz of the stated weight of 27 oz.

Key Concepts

Understanding Error Range

When calculating measurements, especially in a real-world scenario like buying cereal, there is often an **error range**. This refers to the possible difference from the stated or measured value. In this example, Dawn's cereal has a possible error of \(\pm 0.5\) ounces from its stated weight of 27 ounces. This means:

• The actual weight could be a bit more or a bit less.
• The error range gives us the boundaries for this fluctuation.

By capturing this uncertainty using absolute value, we express that the actual weight of the cereal \((w)\) may not be exactly 27 ounces. Instead, it might vary, allowing for a little extra or a little less, maintaining the precision of our measurement.

Algebra Problem Solving with Inequalities

Algebra helps us solve real-life problems by setting up and solving equations or inequalities. To solve an **absolute value inequality** like \(|w - 27| \le 0.5\), we take the following steps:

• Understand that absolute value \(|x|\) measures the distance from zero on a number line, so \(|w - 27|\) measures how far \(w\) is from 27.
• This is split into two separate inequalities: \(w - 27 \le 0.5\) and \(-(w - 27) \le 0.5\).
• Solving these gives us the range of possible values for \(w\), indicating how much the actual weight might differ from the stated value of 27 ounces.

Integrating algebraic techniques into problem-solving involves isolating variables and making logical conclusions, ensuring our results are both accurate and interpretative.

Interpreting Inequalities

Once we establish the inequalities \(26.5 \le w \le 27.5\), we must **interpret** the result:

• This tells us that the weight of cereal could be anywhere between 26.5 to 27.5 ounces.
• The double inequality indicates a range, showcasing that the potential variation from the expected value is well-defined.

Understanding and interpreting inequalities helps in recognizing their real-world implications. It showcases not only the boundaries within which a solution lies but also provides a clear picture of how precise or uncertain a measurement can be.