Questions: Analyze the trend of nuclear power plant capacity over time: Year Percent 1975 56 1980 59 1985 58 1990 70 1995 76 2000 88 2004 89 Let f(t) be the capacity (in percent) at which U.S. nuclear power plants are working at t years since 1970. A model of the situation is f(t)=0.027 t^2+0.216 t+53.296.

Analyze the trend of nuclear power plant capacity over time:

Year Percent
1975 56
1980 59
1985 58
1990 70
1995 76
2000 88
2004 89

Let f(t) be the capacity (in percent) at which U.S. nuclear power plants are working at t years since 1970. A model of the situation is f(t)=0.027 t^2+0.216 t+53.296.

Transcript text: Analyze the trend of nuclear power plant capacity over time: \begin{tabular}{|c|c|} \hline Year & Percent \\ \hline 1975 & 56 \\ \hline 1980 & 59 \\ \hline 1985 & 58 \\ \hline 1990 & 70 \\ \hline 1995 & 76 \\ \hline 2000 & 88 \\ \hline 2004 & 89 \\ \hline \end{tabular} Let $f(t)$ be the capacity (in percent) at which U.S. nuclear power plants are working at t years since 1970. A model of the situation is $f(t)=0.027 t^{2}+0.216 t+53.296$.

Solution

Solution Steps

To analyze the trend of nuclear power plant capacity over time, we can plot the given data points and the model function $ f(t) = 0.027t^2 + 0.216t + 53.296 $. This will allow us to visually compare the actual data with the model's predictions. We can use Python's plotting libraries to create these visualizations.

Step 1: Understanding the Problem

We are given a table showing the percentage capacity of U.S. nuclear power plants over several years. We also have a quadratic model $ f(t) = 0.027 t^2 + 0.216 t + 53.296 $ that represents the capacity as a function of $ t $, where $ t $ is the number of years since 1970. Our task is to analyze this trend using the given model.

Step 2: Calculate the Capacity for Each Year

We will use the model $ f(t) = 0.027 t^2 + 0.216 t + 53.296 $ to calculate the capacity for each year provided in the table.

For 1975 ($ t = 5 $):

\[ f(5) = 0.027 \times 5^2 + 0.216 \times 5 + 53.296 \] \[ = 0.027 \times 25 + 1.08 + 53.296 \] \[ = 0.675 + 1.08 + 53.296 \] \[ = 55.051 \]

For 1980 ($ t = 10 $):

\[ f(10) = 0.027 \times 10^2 + 0.216 \times 10 + 53.296 \] \[ = 0.027 \times 100 + 2.16 + 53.296 \] \[ = 2.7 + 2.16 + 53.296 \] \[ = 58.156 \]

For 1985 ($ t = 15 $):

\[ f(15) = 0.027 \times 15^2 + 0.216 \times 15 + 53.296 \] \[ = 0.027 \times 225 + 3.24 + 53.296 \] \[ = 6.075 + 3.24 + 53.296 \] \[ = 62.611 \]

Step 3: Compare Model Predictions with Actual Data

Now, let's compare the calculated values with the actual data from the table: