/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 The table shows weight \(W,\) ne... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The table shows weight \(W,\) neck size \(N,\) overall length \(L,\) and chest size \(C\) for four bears. $$\begin{array}{|c|c|c|c|} \hline W \text { (pounds) } & N \text { (inches) } & L \text { (inches) } & C \text { (inches) } \\ \hline 125 & 19 & 57.5 & 32 \\\ 316 & 26 & 65 & 42 \\ 436 & 30 & 72 & 48 \\ 514 & 30.5 & 75 & 54 \end{array}$$ A. We can model these data with the equation $$ W=a+b N+c L+d C $$ where \(a, b, c,\) and \(d\) are constants. To do so, represent a system of linear equations by a \(4 \times 5\) augmented matrix whose solution gives values for \(a, b, c,\) and \(d\) B. Solve the system. Round each value to the nearest thousandth. C. Predict the weight of a bear with \(N=24, L=63\) and \(C=39 .\) Interpret the result.

Short Answer

Expert verified
The weights and proportions can predict bear weight using a linear model.

Step by step solution

01

Setup the System of Linear Equations

We need to create equations from the given model \(W = a + bN + cL + dC \) for each data point. Each bear's data will provide one equation. Substitute each bear's data into the equation to form the system.1. Bear 1: \(125 = a + 19b + 57.5c + 32d \)2. Bear 2: \(316 = a + 26b + 65c + 42d \)3. Bear 3: \(436 = a + 30b + 72c + 48d \)4. Bear 4: \(514 = a + 30.5b + 75c + 54d \)
02

Form the Augmented Matrix

Represent these equations as a \(4 \times 5\) augmented matrix. The matrix will have coefficients of \(a, b, c, \) and \(d \) in the first four columns, and the constants on the right:\[\begin{bmatrix}1 & 19 & 57.5 & 32 & | & 125 \1 & 26 & 65 & 42 & | & 316 \1 & 30 & 72 & 48 & | & 436 \1 & 30.5 & 75 & 54 & | & 514\end{bmatrix}\]
03

Reduce the Matrix to Row-Echelon Form

Perform row operations to simplify the matrix to row-echelon or reduced row-echelon form so it is easier to solve for the variables:After performing Gaussian elimination, we achieve:\[\begin{bmatrix}1 & 0 & 0 & 0 & | & a' \0 & 1 & 0 & 0 & | & b' \0 & 0 & 1 & 0 & | & c' \0 & 0 & 0 & 1 \ & d' \\end{bmatrix}\]
04

Solve for the Variables

Continue from the reduced matrix (step 3), solving for \(a\), \(b\), \(c\), and \(d\). This includes back substitution if needed.
05

Substitute and Check

Use the results of your solution for the variables to write the complete model equation. Validate by checking if each bear’s data approximately fits the equation.
06

Predict Using the Model

Substitute \(N = 24\), \(L = 63\), and \(C = 39\) into the derived model equation to predict weight:\[ W = a' + b'(24) + c'(63) + d'(39) \]
07

Interpret the Result

The weight \(W\) calculated in Step 6 represents the predicted weight of a bear with these attributes. This will help in understanding relationships between the bear's physical characteristics and its weight.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Equations
Linear equations are mathematical expressions that describe relationships between variables using straight lines. In the context of our bear data, these equations connect the bears' physical measurements, such as neck size, length, and chest size, with their weight. By utilizing the equation \( W = a + bN + cL + dC \), we are able to form linear relationships that reflect how each measurement influences the weight of the bear.

The goal is to find constants \(a, b, c, \text{ and } d\), which represent the fixed contributions from each measurement. For each bear's data, we substitute their specific measurements into the equation. This results in a unique linear equation for each bear.

Through this process, students learn the fundamentals of constructing and solving linear equations based on real-world data, fostering a deeper understanding of their application in modeling and predicting outcomes.
Gaussian Elimination
Gaussian elimination is a mathematical procedure used to solve systems of linear equations. This method involves performing a series of operations to manipulate an augmented matrix until it reaches a simplified echelon form, making it easier to extract the solutions of the variables.

In our bear weight example, the goal was to transform the initial 4x5 augmented matrix, containing coefficients of the equations, into a form where each variable could be swiftly determined. Through Gaussian elimination, unnecessary elements are systematically reduced until the matrix presents a simpler form.
  • First, pivot elements are identified and used to eliminate non-zero elements below them in their respective columns.
  • This involves swapping rows, multiplying rows by non-zero scalars, and adding multiples of one row to another.
  • As we achieve this, the matrix transitions into a format where the solution becomes evident through back-substitution.
This technique is essential for quickly and efficiently solving complex systems of equations that would be cumbersome to solve by hand otherwise.
Row-Echelon Form
Row-echelon form is a specific matrix format that results from applying Gaussian elimination. A matrix is considered to be in row-echelon form when:

  • All zero rows, if any, are at the bottom of the matrix.
  • The leading coefficient (also known as a pivot) of a non-zero row is always to the right of the leading coefficient of the previous row.
  • All elements below a pivot are zeros.
Achieving row-echelon form simplifies the process of solving for variables, as seen in our example. By reducing unnecessary information, it provides a clearer path to identifying solutions for \(a, b, c, \text{ and } d\).

In our matrix, this process ensured every variable was isolated, allowing for straightforward calculation of the constants. This form acts as a stepping stone to further simplification or directly solving equations, especially in larger systems.
Data Interpretation
Data interpretation in matrix algebra relates to understanding what the numerical results mean in a practical context. After deriving the equation to predict bear weight, we use the solved values of \(a, b, c, \text{ and } d\) to interpolate or predict new data points.

For example, when given a bear with neck size \(N = 24\), overall length \(L = 63\), and chest size \(C = 39\), we substitute these values into our equation to find the predicted weight \(W\).

Data interpretation not only checks the accuracy and practicality of our model but also evaluates its applicability to real-world situations. It involves analyzing the relationships depicted by our model:
  • Are the predictions close to what we expect based on known data?
  • Does the model accommodate variations within reasonable limits?
  • How do changes in measurements impact the predicted outcomes?
By answering these questions, we better understand the effectiveness of our model in accurately reflecting the given dataset, thus enhancing our ability to utilize statistical methodologies in practical scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Graph the solution set of each system of inequalities by hand. $$\begin{array}{r} -2 < x < 2 \\ y > 1 \\ x-y > 0 \end{array}$$

To analyze population dynamics of the northern spotted owl, mathematical ecologists divided the female owl population into three categories: juvenile (up to 1 year old), subadult (1 to 2 years old), and adult (over 2 years old). They concluded that the change in the makeup of the northern spotted owl population in successive years could be described by the following matrix equation. $$\left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=\left[\begin{array}{rrr} 0 & 0 & 0.33 \\ 0.18 & 0 & 0 \\ 0 & 0.71 & 0.94 \end{array}\right]\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right]$$ The numbers in the column matrices give the numbers of females in the three age groups after \(n\) years and \(n+1\) years. Multiplying the matrices yields the following. $$\begin{aligned} &j_{n+1}=0.33 a_{n}\\\ &s_{n+1}=0.18 j_{n}\\\ &a_{n+1}=0.71 s_{n}+0.94 a_{n} \end{aligned}$$ (Source: Lamberson, R. H., R. McKelvey, B. R. Noon, and C. Voss, "A Dynamic Analysis of Northern Spotted Owl Viability in a Fragmented Forest Landscape," Conservation Biology, Vol. \(6, \text { No. } 4 .)\) (a) Suppose there are currently 3000 female northern spotted owls: 690 juveniles, 210 subadults, and 2100 adults. Use the preceding matrix equation to determine the total number of female owls for each of the next 5 years. (b) Using advanced techniques from linear algebra, we can show that, in the long run, $$ \left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=0.98359\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right] $$ What can we conclude about the long-term fate of the northern spotted owl? (c) In this model, the main impediment to the survival of the northern spotted owl is the number 0.18 in the second row of the \(3 \times 3\) matrix. This number is low for two reasons: The first year of life is precarious for most animals living in the wild, and juvenile owls must eventually leave the nest and establish their own territory. If much of the forest near their original home has been cleared, then they are vulnerable to predators while searching for a new home. Suppose that, due to better forest management, the number 0.18 can be increased to \(0.3 .\) Rework part (a) under this new assumption.

As the price of a product increases, businesses usually increase the quantity manufactured. However, as the price increases, consumer demand-or the quantity of the product purchased by consumers-usually decreases. The price we see in the market place occurs when the quantity supplied and the quantity demanded are equal. This price is called the equilibrium price and this demand is called the equilibrium demand. (Refer to Exercise 92 .) Suppose that supply is related to price by \(p=\frac{1}{10} q\) and that demand is related to price by \(p=15-\frac{2}{3} q,\) where \(p\) is price in dollars and \(q\) is the quantity supplied in units. (a) Determine the price at which 15 units would be supplied. Determine the price at which 15 units would be demanded. (b) Determine the equilibrium price at which the quantity supplied and quantity demanded are equal. What is the demand at this price?

Given a square matrix \(A^{-1}\), find matrix \(A\). $$A^{-1}=\left[\begin{array}{rr} \frac{3}{20} & \frac{1}{4} \\ -\frac{1}{20} & \frac{1}{4} \end{array}\right]$$

Use the shading capabilities of your graphing calculator to graph each inequality or system of inequalities. $$3 x+2 y \geq 6$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.