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91Ó°ÊÓ

Inflation Assuming that the annual rate of inflation averages 4\(\%\) over the next 10 years, the approximate costs \(C\) of goods or services during any year in that decade will be modeled by \(C(t)=P(1.04)^{t},\) where \(t\) is the time in years and \(P\) is the present cost. The price of an oil change for your car is presently \(\$ 23.95 .\) Estimate the price 10 years from now.

Short Answer

Expert verified
The estimated cost of an oil change 10 years from now would be approximately $35.24.

Step by step solution

01

Identify given information

Identify the given information in the problem. An oil change currently costs $23.95 and the problem asks to estimate the price 10 years from now. Also, remember that the inflation rate given is 4% or 0.04.
02

Substitute into the formula

Substitute the known values into the given formula \( C(t) = P(1.04)^t \). Here, \( P = 23.95 \) and \( t = 10 \). So, \( C(t) = 23.95(1.04)^{10} \) .
03

Calculate the future price

Do the calculations. Raise 1.04 to the power of 10 and then multiply the result by 23.95 to get the future cost.

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Key Concepts

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

Understanding Inflation Rate
Inflation is like the rising tide of prices across the board. It refers to the general increase in prices of goods and services in an economy over a period of time. When inflation occurs, each unit of currency buys fewer goods and services than it could before.

The annual rate of inflation is often expressed as a percentage. When we say we have a 4% inflation rate, it means that the cost of essential goods and services increase by 4% compared to the previous year. This slow and steady erosion of purchasing power impacts both consumers and businesses.
  • A higher inflation rate means faster depreciation in the purchasing power of money.
  • Low inflation rates are generally desirable because they indicate a stable economy.
  • Inflation assumptions are often used in financial planning to predict future costs.
Understanding and predicting inflation is crucial for making informed financial decisions. It allows individuals and businesses to prepare for the future by saving or investing accordingly.
Exponential Growth and Its Implications
Exponential growth occurs when a quantity increases at a consistent rate over time. This concept is key to understanding how inflation affects the prices of goods and services.

In mathematics, exponential growth can be modeled using the formula \(C(t) = P(1 + r)^t\), where \(C(t)\) is the future cost, \(P\) is the present cost, \(r\) is the growth rate, and \(t\) is the time period. In the context of inflation:
  • The growth rate \(r\) is equivalent to the inflation rate expressed as a decimal, which is 0.04 for a 4% inflation rate.
  • \(P\) symbolizes the current price of an item or service.
  • \(t\) represents the number of years over which the price is expected to grow.
Exponential growth helps in visualizing the compounding effect of inflation over time. Rather than increasing linearly, the price grows faster because each year's growth is multiplied by the base from the previous year.
Estimating Future Costs
Estimating future costs involves predicting the price of an item at a future time, considering the expected rate of growth, like inflation. It's an essential skill for budgeting and financial planning.

To estimate future costs accurately:
  • Identify the present cost \(P\) of a good or service.
  • Determine the expected inflation rate \(r\).
  • Use the exponential growth formula \(C(t) = P(1+r)^t\).
Let's use these steps to project the price of an oil change in 10 years when inflation is at 4%:
  • Current price, \(P = \$23.95\).
  • Inflation rate, \(r = 0.04\).
  • Years into the future, \(t = 10\).
Substituting these into \(C(t) = 23.95(1.04)^{10}\), you can calculate the future price. The exponent \(10\) signifies how the exponential formula applies the inflation rate across each year cumulatively, leading to a much higher future price than linear growth would suggest. Knowing how to estimate future costs can help avoid financial shortfalls due to underestimated expenses.

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Most popular questions from this chapter

Comparing Models A cup of water at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C} .\) The temperature of the water is measured every 5 minutes during a half-hour period. The results are recorded as ordered pairs of the form \((t, T),\) where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). $$ \begin{array}{l}{\left(0,78.0^{\circ}\right),\left(5,66.0^{\circ}\right),\left(10,57.5^{\circ}\right),\left(15,51.2^{\circ}\right)} \\\ {\left(20,46.3^{\circ}\right),\left(25,42.4^{\circ}\right),\left(30,39.6^{\circ}\right)}\end{array} $$ (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \((t, T)\) and \((t, T-21)\) . (b) An exponential model for the data \((t, T-21)\) is given by \(T-21=54.4(0.964)^{t} .\) Solve for \(T\) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use the graphing utility to plot the points \((t, \ln (T-21))\) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This rasulting line has the form \(\ln (T-21)=a t+b\) . Solve for \(T,\) and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \(y\) -coordinates of the revised data points to generate the points $$ \left(t, \frac{1}{T-21}\right) $$ Use the graphing utility to graph these points and observe that they appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. The resulting line has the form $$ \frac{1}{T-21}=a t+b $$ Solve for \(T,\) and use the graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

Graphical Analysis Use a graphing utility to graph each pair of functions in the same viewing window. Describe any similarities and differences in the graphs. (a) \(y_{1}=2^{x}, y_{2}=x^{2}\) (b) \(y_{1}=3^{x}, y_{2}=x^{3}\)

Expanding a Logarithmic Expression In Exercises \(37-58\) , use properties of logarithms to expand the expression as a sum, difference, and or constant multiple of logarithms. (Assume all variables are positive.) $$\log _{6} \frac{1}{z^{3}}$$

Expanding a Logarithmic Expression In Exercises \(37-58,\) use the properties of logarithms to expand the expression as a sum, difference, and or constant multiple of logarithms. (Assume all variables are positive.) $$\ln \sqrt{\frac{x^{2}}{y^{3}}}$$

In Exercises \(85-88,\) use the following information. The relationship between the number of decibels \(\beta\) and the intensity of a sound I in watts per square meter is given by $$ \boldsymbol{\beta}=10 \log \left(\frac{I}{10^{-12}}\right) $$ Find the difference in loudness between a vacuum cleaner with an intensity of \(10^{-4}\) watt per square meter and rustling leaves with an intensity of \(10^{-11}\) watt per square meter.

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