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Chapter 5: Problem 6

In the United States during the decade of the \(1990 \mathrm{~s}\), live births to unmarried mothers, \(B\), grew according to the exponential model \(B=1.165 \cdot 10^{6}(1.013)^{t},\) where \(t\) is the number of years after \(1990 .\) a. What does the model give as the number of live births to unwed mothers in \(1990 ?\) b. What was the growth factor? c. What does the model predict for the number of live births to unwed mothers in \(1995 ?\) In \(2000 ?\)

Short Answer

Expert verified
Number in 1990: 1,165,000. Growth factor: 1.013. Births in 1995: 1,243,807 and in 2000: 1,322,750.

Step by step solution

01

Identify the given model

The given model for live births to unmarried mothers is \(B = 1.165 \cdot 10^{6} (1.013)^{t}\), where \(t\) is the number of years after 1990.
02

Number of live births in 1990

To find the number of live births to unwed mothers in 1990, set \(t = 0\) in the model. The equation becomes \(B = 1.165 \cdot 10^{6} (1.013)^{0}\). Since any number to the power of zero is 1, the equation simplifies to \(B = 1.165 \cdot 10^{6}\). So, the number of live births in 1990 is \(1,165,000\).
03

Determine the growth factor

The growth factor is the term within the exponent, which in this case is \(1.013\). This factor shows how births are growing each year.
04

Calculate births in 1995

To find the number of live births in 1995, set \(t = 5\) in the model. The equation becomes \(B = 1.165 \cdot 10^{6} (1.013)^{5}\). Calculate \((1.013)^5\) first, which is approximately \(1.0678\). So, \(B = 1.165 \cdot 10^{6} \times 1.0678 = 1,243,807\).
05

Calculate births in 2000

To find the number of live births in 2000, set \(t = 10\) in the model. The equation becomes \(B = 1.165 \cdot 10^{6} (1.013)^{10}\). Calculate \((1.013)^{10}\) first, which is approximately \(1.1350\). So, \(B = 1.165 \cdot 10^{6} \times 1.1350 = 1,322,750\).

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Key Concepts

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

Unmarried Mothers
The term unmarried mothers refers to women who give birth to children without being legally married at the time of the birth. It's a demographic often analyzed in social science and public health studies due to the potential socioeconomic implications.

In the context of our exercise, we are focusing on the number of live births to these mothers in the United States during the 1990s. By understanding these numbers, policymakers and researchers can better tailor support systems and resources.

The exponential growth model tells us how this number increased over time, starting from a base value in 1990.
Live Births
In demographics, a live birth is when a baby shows any sign of life upon delivery, regardless of the length of pregnancy. This can include breathing, heartbeat, umbilical cord pulsation, or voluntary muscle movement.

For our calculation, we use a model that tracks live births specifically to unmarried mothers over time. The model is expressed as:

\( B = 1.165 \times 10^6 (1.013)^t \)
Here, 'B' represents the number of live births, while 't' represents the number of years after 1990.

By setting different values of 't' (e.g., 0 for 1990, 5 for 1995, 10 for 2000), we can predict the number of live births for those specific years.
Growth Factor
The growth factor in our model is crucial. It indicates the rate at which the number of live births to unmarried mothers increases annually. In our model, this factor is represented by the base of the exponent, which is 1.013.

To break it down:
  • If the growth factor is greater than 1, the number of live births is increasing.
  • If it's exactly 1, the numbers are stable鈥攚ith no growth or decline.
  • If it's less than 1, the number of live births is decreasing.

In our model, 1.013 signifies a yearly growth rate of 1.3%. Hence, the number of live births to unmarried mothers increased by 1.3% each year during the 1990s.
College Algebra
College algebra provides the tools we need to understand and manipulate mathematical models like our exponential growth equation. Let's revisit the given model:
\( B = 1.165 \times 10^6 (1.013)^t \)

In this context:
  • \( B \) is the number of live births to unmarried mothers.
  • \( 1.165 \times 10^6 \) is the initial value in 1990.
  • \( (1.013)^t \) represents the exponential growth where 't' is the number of years after 1990.

Understanding this formula allows us to perform predictions and gain insights into trends. Solving problems using similar models in college algebra helps solidify concepts, making abstract ideas more concrete.

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Most popular questions from this chapter

Lead- 206 is not radioactive, so it does not spontaneously decay into lighter elements. Radioactive elements heavier than lead undergo a series of decays, each time changing from a heavier element into a lighter or more stable one. Eventually, the element decays into lead- 206 and the process stops. So, over billions of years, the amount of lead in the universe has increased because of the decay of numerous radioactive elements produced by supernova explosions. Radioactive uranium- 238 decays sequentially into thirteen other lighter elements until it stabilizes at lead-206. The half-lives of the fifteen different elements in this decay chain vary from 0.000164 seconds (from polonium- 214 to lead- 210 ) all the way up to 4.47 billion years (from uranium- 238 to thorium- 234 ). a. Find the decay rate per billion years for uranium- 238 to decay into thorium- 234 . b. Find the decay rate per second for polonium-214 to decay into lead-2.10.

Write the equation of each exponential growth function in the form \(y=C a^{x}\) where: a. The initial population is 350 and the growth factor is \(\frac{4}{3}\). b. The initial population is \(5 \cdot 10^{9}\) and the growth factor is \(1.25 .\) c. The initial population is 150 and the population triples during each time period. d. The initial population of 2 quadruples every time period.

Each of two towns had a population of 12,000 in \(1990 .\) By 2000 the population of town A had increased by \(12 \%\) while the population of town B had decreased by \(12 \%\). Assume these growth and decay rates continued. a. Write two exponential population models \(A(T)\) and \(B(T)\) for towns A and \(\mathrm{B}\), respectively, where \(T\) is the number of decades since 1990 . b. Write two new exponential models \(a(t)\) and \(b(t)\) for towns A and \(\mathrm{B}\), where \(t\) is the number of years since 1990 . c. Now find \(A(2), B(2), a(20)\), and \(b(20)\) and explain what you have found.

Rewrite each expression so that no fraction appears in the exponent and each expression is in the form \(a^{x}\). a. \(3^{w / 4}\) b. \(2^{x / 3}\) c. \(\left(\frac{1}{2}\right)^{x_{4}}\) d. \(\left(\frac{1}{4}\right)^{x / 2}\)

(Graphing program recommended.) You have a chance to invest money in a risky investment at \(6 \%\) interest compounded annually. Or you can invest your money in a safe investment at \(3 \%\) interest compounded annually. a. Write an equation that describes the value of your investment after \(n\) years if you invest \(\$ 100\) at \(6 \%\) compounded annually. Plot the function. Estimate how long it would take to double your money. b. Write an equation that describes the value of your investment after \(n\) years if you invest \(\$ 200\) at \(3 \%\) compounded annually. Plot the function on the same grid as in part (a). Estimate the time needed to double your investment. c. Looking at your graph, indicate whether the amount in the first investment in part (a) will ever exceed the amount in the second account in part (b). If so, approximately when?
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