{"id":2126,"date":"2023-04-26T14:28:33","date_gmt":"2023-04-26T14:28:33","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=2126"},"modified":"2025-08-28T19:28:18","modified_gmt":"2025-08-28T19:28:18","slug":"modeling-linear-growth-learn-it-1","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/modeling-linear-growth-learn-it-1\/","title":{"raw":"Modeling Linear Growth: Learn It 1","rendered":"Modeling Linear Growth: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Create a linear model that describes a real-world situation<\/li>\r\n\t<li>Use linear regression to analyze a data set and find the best-fit line<\/li>\r\n\t<li>Calculate and use the coefficient of determination to determine how well a linear model fits the data<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Modeling Linear Growth<\/h2>\r\n<p>Situations involving growth (or decay) in which the output quantity changes by a constant amount per unit of input are said to exhibit linear growth. That is, from one input to the next, there is a common difference between the output values.<\/p>\r\n\r\n[caption id=\"attachment_3372\" align=\"alignright\" width=\"300\"]<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4685\/2019\/11\/01024539\/Morgan_silver_dollar.jpg\"><img class=\"wp-image-3372 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4685\/2019\/11\/01024539\/Morgan_silver_dollar-300x288.jpg\" alt=\"A Morgan silver dollar\" width=\"300\" height=\"288\" \/><\/a> <center>Figure 1. A Morgan silver dollar<\/center>[\/caption]\r\n\r\n<p>For example, let's say you inherit a collection of [latex]47[\/latex] silver dollars minted between 1878 and 1935. After doing some research, you learn that they are worth between $14 and $30 each. Rather than just cashing them in, you decide to grow the collection by purchasing an additional silver dollar each month. You can write a mathematical model to quickly predict how many silver dollars you'll have in the collection at any time in the future assuming you increase your collection by [latex]12[\/latex] dollars per year.\u00a0 We can write a mathematical model to describe this situation from the information given about the starting amount and the constant quantity of change.<\/p>\r\n<p>Let the input variable (the independent variable) [latex]x[\/latex] represent the number of years you've owned the collection of silver dollars. The output variable (the dependent variable) [latex]y[\/latex] will represent the total number of dollars in the collection in any year [latex]x[\/latex].<\/p>\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 16.6667%; text-align: center;\">The number of dollars in the collection<\/td>\r\n<td style=\"width: 16.6667%; text-align: center;\">is obtained from<\/td>\r\n<td style=\"width: 16.6667%; text-align: center;\">the starting number, 47<\/td>\r\n<td style=\"width: 16.6667%; text-align: center;\">together with<\/td>\r\n<td style=\"width: 16.6667%; text-align: center;\">12 more for each year you've owned it<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 16.6667%; text-align: center;\">[latex]y[\/latex]<\/td>\r\n<td style=\"width: 16.6667%; text-align: center;\">[latex]=[\/latex]<\/td>\r\n<td style=\"width: 16.6667%; text-align: center;\">[latex]47[\/latex]<\/td>\r\n<td style=\"width: 16.6667%; text-align: center;\">[latex]+[\/latex]<\/td>\r\n<td style=\"width: 16.6667%; text-align: center;\">[latex]12\\ast x[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center;\">[latex]y=47+12x[\/latex]<\/p>\r\n<p>Equivalently, in the slope-intercept form of a linear equation,<\/p>\r\n<p style=\"text-align: center;\">[latex]y=mx+b[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]y=12x+47[\/latex]<\/p>\r\n<section class=\"textbox recall\"><strong>equations of lines in slope-intercept form<\/strong>\r\n<ul>\r\n\t<li>[latex] \\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex] and [latex] \\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}[\/latex] where [latex]m=\\text{slope}[\/latex]\u00a0and [latex] \\displaystyle ({{x}_{1}},{{y}_{1}})[\/latex] and [latex] \\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex] are two points on the line.<\/li>\r\n\t<li>The slope-intercept form of the equation of line is [latex]y=mx+b[\/latex] where [latex]m[\/latex] represents the slope of the line, [latex]b[\/latex] represents the [latex]y[\/latex]-intercept, and [latex]x[\/latex] and [latex]y[\/latex] may be substituted by the coordinates of any point contained on the graph of the equation to satisfy it.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox example\">A car is taken to a mechanic for repair. \u00a0The mechanic charges [latex]140[\/latex] dollars for parts, plus [latex]33 [\/latex] dollars per hour for labor.\u00a0Write a linear equation, in slope-intercept form, that models this situation.[reveal-answer q=\"920714\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"920714\"]We are interested in the total cost of the car repair, which will vary depending upon the number of hours of the mechanic needs to work on it. So, the output variable, the total cost, is dependent on the input variable, the total number of hours.\r\n\r\n<p>Let [latex]y[\/latex] represent the total bill for the repair.<\/p>\r\n<p>Let [latex]x[\/latex] represent the number of hours of labor<\/p>\r\n<p>The initial cost for the repair is [latex]$140[\/latex], which is increased by\u00a0[latex]$33[\/latex] per hour worked.<\/p>\r\n<p>In the slope-intercept form of a linear equation [latex]y=mx+b[\/latex], we have:<\/p>\r\n<p style=\"text-align: center;\">[latex]y=33x+140[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Create a linear model that describes a real-world situation<\/li>\n<li>Use linear regression to analyze a data set and find the best-fit line<\/li>\n<li>Calculate and use the coefficient of determination to determine how well a linear model fits the data<\/li>\n<\/ul>\n<\/section>\n<h2>Modeling Linear Growth<\/h2>\n<p>Situations involving growth (or decay) in which the output quantity changes by a constant amount per unit of input are said to exhibit linear growth. That is, from one input to the next, there is a common difference between the output values.<\/p>\n<figure id=\"attachment_3372\" aria-describedby=\"caption-attachment-3372\" style=\"width: 300px\" class=\"wp-caption alignright\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4685\/2019\/11\/01024539\/Morgan_silver_dollar.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3372 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4685\/2019\/11\/01024539\/Morgan_silver_dollar-300x288.jpg\" alt=\"A Morgan silver dollar\" width=\"300\" height=\"288\" \/><\/a><figcaption id=\"caption-attachment-3372\" class=\"wp-caption-text\">\n<div style=\"text-align: center;\">Figure 1. A Morgan silver dollar<\/div>\n<\/figcaption><\/figure>\n<p>For example, let&#8217;s say you inherit a collection of [latex]47[\/latex] silver dollars minted between 1878 and 1935. After doing some research, you learn that they are worth between $14 and $30 each. Rather than just cashing them in, you decide to grow the collection by purchasing an additional silver dollar each month. You can write a mathematical model to quickly predict how many silver dollars you&#8217;ll have in the collection at any time in the future assuming you increase your collection by [latex]12[\/latex] dollars per year.\u00a0 We can write a mathematical model to describe this situation from the information given about the starting amount and the constant quantity of change.<\/p>\n<p>Let the input variable (the independent variable) [latex]x[\/latex] represent the number of years you&#8217;ve owned the collection of silver dollars. The output variable (the dependent variable) [latex]y[\/latex] will represent the total number of dollars in the collection in any year [latex]x[\/latex].<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 16.6667%; text-align: center;\">The number of dollars in the collection<\/td>\n<td style=\"width: 16.6667%; text-align: center;\">is obtained from<\/td>\n<td style=\"width: 16.6667%; text-align: center;\">the starting number, 47<\/td>\n<td style=\"width: 16.6667%; text-align: center;\">together with<\/td>\n<td style=\"width: 16.6667%; text-align: center;\">12 more for each year you&#8217;ve owned it<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 16.6667%; text-align: center;\">[latex]y[\/latex]<\/td>\n<td style=\"width: 16.6667%; text-align: center;\">[latex]=[\/latex]<\/td>\n<td style=\"width: 16.6667%; text-align: center;\">[latex]47[\/latex]<\/td>\n<td style=\"width: 16.6667%; text-align: center;\">[latex]+[\/latex]<\/td>\n<td style=\"width: 16.6667%; text-align: center;\">[latex]12\\ast x[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">[latex]y=47+12x[\/latex]<\/p>\n<p>Equivalently, in the slope-intercept form of a linear equation,<\/p>\n<p style=\"text-align: center;\">[latex]y=mx+b[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]y=12x+47[\/latex]<\/p>\n<section class=\"textbox recall\"><strong>equations of lines in slope-intercept form<\/strong><\/p>\n<ul>\n<li>[latex]\\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex] and [latex]\\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}[\/latex] where [latex]m=\\text{slope}[\/latex]\u00a0and [latex]\\displaystyle ({{x}_{1}},{{y}_{1}})[\/latex] and [latex]\\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex] are two points on the line.<\/li>\n<li>The slope-intercept form of the equation of line is [latex]y=mx+b[\/latex] where [latex]m[\/latex] represents the slope of the line, [latex]b[\/latex] represents the [latex]y[\/latex]-intercept, and [latex]x[\/latex] and [latex]y[\/latex] may be substituted by the coordinates of any point contained on the graph of the equation to satisfy it.<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\">A car is taken to a mechanic for repair. \u00a0The mechanic charges [latex]140[\/latex] dollars for parts, plus [latex]33[\/latex] dollars per hour for labor.\u00a0Write a linear equation, in slope-intercept form, that models this situation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q920714\">Show Solution<\/button><\/p>\n<div id=\"q920714\" class=\"hidden-answer\" style=\"display: none\">We are interested in the total cost of the car repair, which will vary depending upon the number of hours of the mechanic needs to work on it. So, the output variable, the total cost, is dependent on the input variable, the total number of hours.<\/p>\n<p>Let [latex]y[\/latex] represent the total bill for the repair.<\/p>\n<p>Let [latex]x[\/latex] represent the number of hours of labor<\/p>\n<p>The initial cost for the repair is [latex]$140[\/latex], which is increased by\u00a0[latex]$33[\/latex] per hour worked.<\/p>\n<p>In the slope-intercept form of a linear equation [latex]y=mx+b[\/latex], we have:<\/p>\n<p style=\"text-align: center;\">[latex]y=33x+140[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":18,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":87,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2126"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":13,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2126\/revisions"}],"predecessor-version":[{"id":15876,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2126\/revisions\/15876"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/parts\/87"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2126\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=2126"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=2126"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=2126"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=2126"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}