Problem 79 - Algebra 1

Question

Use the following information. The power generated by a windmill can be modeled by the equation \(w=0.015 s^{3},\) where \(w\) is the power measured in watts and \(s\) is the wind speed in miles per hour. Part A of a test has 10 true-false questions. Part B has 10 multiple-choice questions. Each of the multiple-choice questions has 4 possible answers. There are \(2^{10}\) ways to answer the 10 questions in Part A. There are \(4^{10}\) ways to answer the 10 questions in Part B. a. How many ways are there to answer all 20 questions? b. If you guess the answer to each question, what is the probability that you will get them all right?

Step-by-Step Solution

Answer

a. There are \(2^{10} \times 4^{10}\) ways to answer all the 20 questions. b. The probability of guessing all answers right is \((1/2)^{10} \times (1/4)^{10}\) which is very small and nearly 0.

Step 1: Determining Number of Ways to Answer All

According to the exercise, there are \(2^{10}\) ways to answer the 10 true-false questions in Part A and \(4^{10}\) ways to answer the 10 multiple-choice questions in Part B. To calculate the total number of ways to answer all the 20 questions, these two paths of options should be multiplied together because each decision on each part is independent. Hence, the total number of ways is \(2^{10} \times 4^{10}\).

Step 2: Calculate Probability of Guessing Correctly

Next, calculate the probability of guessing all answers correctly, which is the number of favorable outcomes over the total number of outcomes. In case of multiple choice questions with 4 options, the chance of getting one question right is \(1/4 = 0.25\) and for true/false the chance is \(1/2 = 0.5\). For 10 such questions in both categories, the probability of guessing all answers correctly is given by multiplying the individual probability 10 times each. Considering there are 10 true/false questions, the success outcome will be \((1/2)^{10}\). Similarly, for 10 multiple choice questions, the success outcome will be \((1/4)^{10}\). Therefore, the total probability of guessing all answers right is the product of these two: \((1/2)^{10} \times (1/4)^{10}\).

Key Concepts

Permutations and Combinations

Understanding permutations and combinations is essential in solving problems related to probability and counting. Permutations refer to the number of ways to arrange a set of items where the order matters. For instance, the sequence ABC is different from CBA even though they contain the same letters.

On the other hand, combinations are concerned with the number of ways to select items from a set where the order does not matter. For example, choosing two fruits from an assortment of an apple, banana, and orange can be done in three ways: apple-banana, apple-orange, and banana-orange.

In the exercise provided, we look at the concept of combinations as there are several ways to answer a test with multiple-choice questions. Each question on its own is a combination of four possible answers, where the order of these choices doesn't matter.

Exponential Expressions

Exponential expressions are mathematical notations that involve a base number raised to a power, which represents how many times to multiply the base number by itself. They are written with a superscript, such as in the expression \(2^{10}\), which means that 2 is multiplied by itself 10 times.

These expressions can become large very quickly, which is evident in the exercise's use of \(2^{10}\) and \(4^{10}\) as the numbers of ways to answer true-false and multiple-choice questions, respectively. When you multiply these exponential expressions, you are effectively calculating the total number of possible outcomes of two independent events, illustrating a key property of exponents: \((a^m)\times(a^n) = a^{m+n}\). This property is used to determine the number of ways to answer all 20 questions in the exercise.

Wind Power Algebra Equation

Algebraic equations in fields such as environmental science can represent real-world phenomena like wind power generation. The given equation \(w=0.015s^{3}\) is an example, with \(w\) representing the power output in watts and \(s\) representing the wind speed in miles per hour. Here, the cube of wind speed is used, reflecting the nature of power generation which is not linear but rather exponential.

This cubic relationship between wind speed and power in the exercise indicates that a small increase in wind speed can lead to a significant increase in power generated. The use of algebra in this equation helps in understanding the dynamics of wind energy, providing a clear picture of the potential energy output based on varying speeds, and demonstrating how algebra is a powerful tool in environmental science and engineering.