Matlab
Matlab
6th Edition
ISBN: 9781119299257
Author: Amos Gilat
Publisher: WILEY CONS
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Chapter 4, Problem 15P

A person at point A spots a child in trouble at point B across the river. The person can run at a speed of 8.6 ft/s and can swim at a speed of 3.9 ft/s. In order a to reach he child in the shortest time the person runs to point C and then swims to point B. as shown. Write a MATLAB program that determines the distance x w point C that minimizes the time the person can reach the child. In the program define a vector x with values ranging from 0 to 5.000 with increments of 1. Use this vector to calculate the corresponding values of x. Then use MATLAB’s built-in function mm to find the value of x that corresponds to the shortest time.

Chapter 4, Problem 15P, A person at point A spots a child in trouble at point B across the river. The person can run at a

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For each of the time​ series, construct a line chart of the data and identify the characteristics of the time series​ (that is,​ random, stationary,​ trend, seasonal, or​ cyclical). Year    Month    Rate (%)2009    Mar    8.72009    Apr    9.02009    May    9.42009    Jun    9.52009    Jul    9.52009    Aug    9.62009    Sep    9.82009    Oct    10.02009    Nov    9.92009    Dec    9.92010    Jan    9.82010    Feb    9.82010    Mar    9.92010    Apr    9.92010    May    9.62010    Jun    9.42010    Jul    9.52010    Aug    9.52010    Sep    9.52010    Oct    9.52010    Nov    9.82010    Dec    9.32011    Jan    9.12011    Feb    9.02011    Mar    8.92011    Apr    9.02011    May    9.02011    Jun    9.12011    Jul    9.02011    Aug    9.02011    Sep    9.02011    Oct    8.92011    Nov    8.62011    Dec    8.52012    Jan    8.32012    Feb    8.32012    Mar    8.22012    Apr    8.12012    May    8.22012    Jun    8.22012    Jul    8.22012    Aug    8.12012    Sep    7.82012    Oct…
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For each of the time​ series, construct a line chart of the data and identify the characteristics of the time series​ (that is,​ random, stationary,​ trend, seasonal, or​ cyclical) Date    IBM9/7/2010    $125.959/8/2010    $126.089/9/2010    $126.369/10/2010    $127.999/13/2010    $129.619/14/2010    $128.859/15/2010    $129.439/16/2010    $129.679/17/2010    $130.199/20/2010    $131.79

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