/*! 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} Q12E [M] Budget Rent A car in Wichit,... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Q12E (page 1)

[M] Budget Rent A car in Wichit, Kanas, has a fleet of about 500 cars, at three locations. A car rented at one location may be returned to any of the three locations. The various fraction of cars returned to the three locations are shown in the matrix below. Suppose that on Monday there are 295 cars at the airport (or rented from there), 55 cars at the east side office, and 150 cars at the west side of office. What will be the approximate distribution of cars on Wednesday?

Cars Rented Form

Returned To

Airport

East

West

0.97

0.05

0.10

Airport

0.00

0.90

0.05

East

0.03

0.05

0.85

West

Short Answer

Expert verified

\(\left[ {\begin{array}{*{20}{c}}{312}\\{58}\\{130}\end{array}} \right]\)

Step by step solution

01

Form the difference equation

The matrix \(M = \left[ {\begin{array}{*{20}{c}}{0.97}&{0.05}&{0.10}\\{0.00}&{0.90}&{0.05}\\{0.03}&{0.05}&{0.85}\end{array}} \right]\) and the matrix \({{\bf{x}}_k}\) represent the car available at the airport on a particular day.

Let \(k = 0\) represent Monday. Then, by the difference equation, \({{\bf{x}}_{k + 1}} = M{{\bf{x}}_k}\).

\({{\bf{x}}_0} = \left[ {\begin{array}{*{20}{c}}{295}\\{55}\\{150}\end{array}} \right]\)

02

Calculate the cars available on Tuesday

For \(k = 0\), by the equation \({{\bf{x}}_{k + 1}} = M{{\bf{x}}_k}\),

\(\begin{array}{c}{{\bf{x}}_1} = \left[ {\begin{array}{*{20}{c}}{0.97}&{0.05}&{0.10}\\{0.00}&{0.90}&{0.05}\\{0.03}&{0.05}&{0.85}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{295}\\{55}\\{150}\end{array}} \right]\\ \approx \left[ {\begin{array}{*{20}{c}}{304}\\{57}\\{139}\end{array}} \right].\end{array}\)

03

Calculate the cars available on Wednesday

For \(k = 1\), by the equation \({{\bf{x}}_{k + 1}} = M{{\bf{x}}_k}\).

\(\begin{array}{c}{{\bf{x}}_2} = \left[ {\begin{array}{*{20}{c}}{0.97}&{0.05}&{0.10}\\{0.00}&{0.90}&{0.05}\\{0.03}&{0.05}&{0.85}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{304}\\{57}\\{139}\end{array}} \right]\\ \approx \left[ {\begin{array}{*{20}{c}}{312}\\{58}\\{130}\end{array}} \right].\end{array}\)

So, the matrix \(\left[ {\begin{array}{*{20}{c}}{312}\\{58}\\{130}\end{array}} \right]\) represents the cars at three locations on Wednesday.

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 91影视!

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

Suppose an experiment leads to the following system of equations:

\(\begin{aligned}{c}{\bf{4}}.{\bf{5}}{x_{\bf{1}}} + {\bf{3}}.{\bf{1}}{x_{\bf{2}}} = {\bf{19}}.{\bf{249}}\\1.6{x_{\bf{1}}} + 1.1{x_{\bf{2}}} = 6.843\end{aligned}\) (3)

  1. Solve system (3), and then solve system (4), below, in which the data on the right have been rounded to two decimal places. In each case, find the exactsolution.

\(\begin{aligned}{c}{\bf{4}}.{\bf{5}}{x_{\bf{1}}} + {\bf{3}}.{\bf{1}}{x_{\bf{2}}} = {\bf{19}}.{\bf{25}}\\1.6{x_{\bf{1}}} + 1.1{x_{\bf{2}}} = 6.8{\bf{4}}\end{aligned}\) (4)

  1. The entries in (4) differ from those in (3) by less than .05%. Find the percentage error when using the solution of (4) as an approximation for the solution of (3).
  1. Use your matrix program to produce the condition number of the coefficient matrix in (3).

In Exercises 32, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

32. \(\left[ {\begin{array}{*{20}{c}}1&2&{ - 5}&0\\0&1&{ - 3}&{ - 2}\\0&{ - 3}&9&5\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&2&{ - 5}&0\\0&1&{ - 3}&{ - 2}\\0&0&0&{ - 1}\end{array}} \right]\)

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation, and suppose \(T\left( u \right) = {\mathop{\rm v}\nolimits} \). Show that \(T\left( { - u} \right) = - {\mathop{\rm v}\nolimits} \).

Explain why a set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) in \({\mathbb{R}^5}\) must be linearly independent when \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly independent and \({{\mathop{\rm v}\nolimits} _4}\) is not in Span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\).

Construct a \(3 \times 3\) matrix\(A\), with nonzero entries, and a vector \(b\) in \({\mathbb{R}^3}\) such that \(b\) is not in the set spanned by the columns of\(A\).
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation
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.