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Internet access: The following table gives the number \(I=I(t)\), in millions, of Americans with Internet access in year \(t\). $$ \begin{array}{|l|c|c|c|} \hline t=\text { Year } & 1997 & 2000 & 2005 \\ \hline \begin{array}{l} t=\text { Millions } \\ \text { of Americans } \end{array} & 46 & 113 & 172 \\ \hline \end{array} $$ a. Find \(I(2000)\) and explain what it means. b. Find the average rate of change per year during the period from 2000 to 2005 . c. Estimate the value of \(I(2002)\). Explain how you got your answer.

Short Answer

Expert verified
a. 113 million; b. 11.8 million/year; c. Approximately 136.6 million.

Step by step solution

01

Determine I(2000)

From the table, we can directly find that the number of Americans with internet access in the year 2000 is 113 million. This value, \( I(2000) = 113 \), means that in the year 2000, there were 113 million people in America who had access to the internet.
02

Calculate the Average Rate of Change from 2000 to 2005

The average rate of change of a function between two points is given by the formula: \( \text{Average Rate of Change} = \frac{I(t_2) - I(t_1)}{t_2 - t_1} \). Here, \( t_1 = 2000 \) and \( t_2 = 2005 \) with \( I(2000) = 113 \) and \( I(2005) = 172 \). Therefore, the average rate of change is: \[ \frac{172 - 113}{2005 - 2000} = \frac{59}{5} = 11.8 \text{ million per year.} \]
03

Estimate I(2002) using Linear Interpolation

To estimate \( I(2002) \), perform linear interpolation between the years 2000 and 2005. This assumes no abrupt changes and a linear increase. The formula for interpolation is: \[ I(2002) = I(2000) + \frac{2}{5} \times (I(2005) - I(2000)) \]. Substituting the values: \[ \begin{align*} I(2002) & = 113 + \frac{2}{5} \times (172 - 113) \ & = 113 + \frac{2}{5} \times 59 \ & = 113 + 23.6 = 136.6. \end{align*} \] Thus, \( I(2002) \) is estimated to be approximately 136.6 million.

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Key Concepts

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

Linear Interpolation
Linear interpolation is a method used to estimate values that lie between two known data points. It is particularly useful when you expect that changes between data points are linear or relatively even. Essentially, it helps us predict a value by assuming that the change between known values is consistent. Let's break down the process:- **Identify Known Points**: In the context of internet usage, you know that 113 million Americans had internet access in 2000 and 172 million in 2005.- **Choose a Target Year**: Suppose you wish to estimate the number of internet users in 2002.- **Apply the Linear Interpolation Formula**: \( I(2002) = I(2000) + \frac{2}{5} \times (I(2005) - I(2000)) \) This formula takes the increase in users between the years you have data for (from 113 to 172 million), and spreads it equally over the five years between 2000 and 2005. For 2002, two-fifths of this increase are added to the 2000 figure, leading to an estimated 136.6 million users. By assuming linear growth, you simplify many real-world applications where actual data follows this trend closely.
Internet Usage Statistics
Internet usage statistics help us understand how technology adoption changes over time. By examining the number of internet users over different years, you can gauge technological growth and market potential, which is vital for businesses, policymakers, and researchers. ### Why Are These Statistics Important? - **Trend Analysis**: Statistics can reveal trends, such as accelerating growth rates during various periods. - **Market Predictions**: Businesses use these trends to forecast demand and guide product development. - **Policy Making**: Governments may set regulations or funding initiatives based on connectivity growth. ### An Example in Action From the example table: - In 1997: 46 million had access to the internet. - In 2000: That number jumped to 113 million. - By 2005: It further increased to 172 million. Each of these numbers gives insights into technological dissemination and infrastructure development. Such data has played a key role in shaping policies around digital inclusion, access inequality, and educational outreach.
Function Modeling
Function modeling involves representing real-world situations through mathematical functions, facilitating analysis and insights. Here, by modeling internet usage with a function, you can examine how it changes over time and predict future behaviors. ### Purpose of Function Modeling in This Context- **Simplifies Complexity**: Condenses real-world complexities to simple, understandable mathematical terms.- **Enables Predictions**: By understanding how variables interact in a function, predictions become more accurate.- **Informs Decisions**: Helps decision-makers in developing strategies by predicting potential future scenarios.### Applying Function Modeling to Internet UsageThe data table is a form of function modeling, with year \(t\) providing the input and the number of users \(I(t)\) serving as the output. By estimating the value of \(I(t)\) for years not directly listed, through methods like linear interpolation, you creatively extend the usefulness of your data.Thus, not only do function models act as essential tools for prediction, they also help in making informed decisions and planning for future developments, all based on a firm understanding of existing trends and movements.

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Most popular questions from this chapter

Stock turnover rate: In a retail store the stock turnover rate of an item is the number of times that the average inventory of the item needs to be replaced as a result of sales in a given time period. It is an important measure of sales demand and merchandising efficiency. Suppose a retail clothing store maintains an average inventory of 50 shirts of a particular brand. a. Suppose that the clothing store sells 350 shirts of that brand each year. How many orders of 50 shirts will be needed to replace the items sold? b. What is the annual stock turnover rate for that brand of shirt if the store sells 350 shirts each year? c. What would be the annual stock turnover rate if 500 shirts were sold? d. Write a formula expressing the annual stock turnover rate as a function of the number of shirts sold. Identify the function and the variable, and state the units.

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United States population growth: In 1960 the population of the United States was about \(180 \mathrm{mil-}\) lion. Since that time the population has increased by approximately \(1.2 \%\) each year. This is a verbal description of the function \(N=N(t)\), where \(N\) is the population, in millions, and \(t\) is the number of years since 1960 . a. Express in functional notation the population of the United States in 1963. Calculate its value. b. Use the verbal description of \(N\) to make a table of values that shows U.S. population in millions from 1960 through \(1965 .\) c. Make a graph of U.S. population versus time. Be sure to label your graph appropriately. d. Verify that the formula \(180 \times 1.012^{t}\) million people, where \(t\) is the number of years since 1960 , gives the same values as those you found in the table in part b. (Note: Because \(t\) is the number of years since 1960, you would use \(t=2\) to get the population in 1962.) e. Assuming that the population has been growing at the same percentage rate since 1960 , what value does the formula above give for the population in 2000? (Note: The actual population in 2000 was about 281 million.)

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Arterial blood flow: Medical evidence shows that a small change in the radius of an artery can indicate a large change in blood flow. For example, if one artery has a radius only \(5 \%\) larger than another, the blood flow rate is \(1.22\) times as large. Further information is given in the table below. $$ \begin{array}{|c|c|} \hline \text { Increase in radius } & \begin{array}{c} \text { Times greater blood } \\ \text { flow rate } \end{array} \\ \hline 5 \% & 1.22 \\ \hline 10 \% & 1.46 \\ \hline 15 \% & 1.75 \\ \hline 20 \% & 2.07 \\ \hline \end{array} $$ a. Use the average rate of change to estimate how many times greater the blood flow rate is in an artery that has a radius \(12 \%\) larger than another. b. Explain why if the radius is increased by \(12 \%\) and then we increase the radius of the new artery by \(12 \%\) again, the total increase in the radius is \(25.44 \%\). c. Use parts a and \(b\) to answer the following question: How many times greater is the blood flow rate in an artery that is \(25.44 \%\) larger in radius than another? d. Answer the question in part c using the average rate of change.

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