Problem 88 - Introductory and Intermediate Algebra for College Students 4th

Question

A bowler's handicap, \(H,\) is often found using the following formula: $$H=0.8(200-A)$$ A bowler's final score for a game is the score for that game increased by the handicap. a. If your average bowling score is \(165,\) what is your handicap? b. What would your final score be if you bowled 140 in a game?

Step-by-Step Solution

Answer

a. The bowler's handicap is 28. b. The bowler's final score for the game is 168.

Step 1: Calculate the Handicap

First we need to find the bowler's handicap. The handicap is calculated using the formula \(H=0.8(200-A)\), where \(A\) is the average bowling score. In this case, \(A = 165\). So, we substitute \(165\) into the formula to get \(H=0.8(200 - 165)\).

Step 2: Simplify

We then simplify to get the handicap. \(H = 0.8(35) = 28\). So the bowler's handicap is \(28\).

Step 3: Calculate the Final Score

Then, we find the final score for a game. This is done by increasing the game score by the handicap. Given that the bowler's game score was \(140\), we add the handicap to this score to get \(140 + 28\).

Step 4: Simplify

Finally, we simplify to get the final score. \( 140 + 28 = 168\). So the bowler's final score for the game is \(168\).

Key Concepts

Algebraic Expressions

Algebraic expressions are mathematical phrases that can contain ordinary numbers, variables (like \(x\) or \(A\)), and operators (such as \( + \) , \( - \), \( * \), and \( / \)). Crucial to algebra, these expressions represent quantities in an abstract way, allowing us to perform general calculations without having to specify the values of the variables involved until we need to.

In the context of the bowler's handicap formula, \( H=0.8(200-A) \), the expression encapsulates a calculation that depends on the average score \(A\) of the bowler. By plugging in different values for \(A\), one can quickly calculate \(H\), the handicap, for any bowler. This is an example of how algebraic expressions are utilized to create formulas that can be used in various situations with different inputs.

Solving Linear Equations

Solving linear equations is one of the fundamental skills in algebra. A linear equation can be recognized by the fact that it has no exponents higher than one and can be drawn as a straight line on a graph. The process of solving a linear equation involves finding the value of the variable that makes the equation true.

In the bowler's handicap formula, we see a practical application of this process. We have already determined the average score \(A\), so to solve for \(H\), we follow a simplified version of the steps to solve linear equations: substitute the known value, multiply through any coefficients, and perform any required addition or subtraction. In the exercise, this provided a quick and clear method to find the bowler's handicap simply and efficiently, demonstrating the importance of knowing how to solve linear equations.

Mathematical Applications

Mathematical applications are all about using mathematical methods and formulas to solve real-world problems and quantifiable phenomena. Math isn't just about abstract ideas; it's widely used in fields like finance, engineering, sports and more.

The bowler's handicap formula is a perfect example of how math can be applied in sports to ensure fairness and competition. The handicap levels the playing field by giving less skilled bowlers a better chance of competing against more skilled ones. It is an application of both algebraic expressions and solving linear equations in a context directly related to people's hobbies and entertainments. By understanding the formula \( H=0.8(200-A) \), participants and organizers can compute handicaps quickly and ensure that the rules of the game are upheld, making the sport enjoyable for everyone, regardless of skill level.