Math  /  Data & Statistics

QuestionShown to the right is a certain population, in billions, for seven selected years from 1950 through 2007. Using a graphing utility's logistic regression option, we obtain the logistic growth model shown below for population, f(x)f(x), in billions, xx years after 1949. How well does the function model the data for 2007? f(x)=10.231+3.55e0.030xf(x)=\frac{10.23}{1+3.55 e^{-0.030 x}} \begin{tabular}{|c|c|} \hline x, Number of Years after 1949 & y, Population (billions) \\ \hline 1(1950)1(1950) & 2.4 \\ \hline 11(1960)11(1960) & 2.8 \\ \hline 21(1970)21(1970) & 3.5 \\ \hline 31(1980)31(1980) & 4.3 \\ \hline 41(1990)41(1990) & 5.1 \\ \hline 51(2000)51(2000) & 5.9 \\ \hline 58(2007)58(2007) & 6.3 \\ \hline \end{tabular}
For 2007, the function \square the population to one decimal place. slightly overestimates slightly underestimates accurately predicts

Studdy Solution

STEP 1

What is this asking? We're checking how well a logistic growth model predicts the world population in 2007, comparing the model's prediction to the actual population given in the table. Watch out! Make sure to correctly plug in the right value for xx into the formula, representing the number of years after 1949.

STEP 2

1. Calculate xx
2. Calculate the predicted population
3. Compare with the actual population

STEP 3

We're told that xx represents the number of years after 1949.
Since we're looking at the year 2007, we need to find the difference between 2007 and 1949.

STEP 4

20071949=582007 - 1949 = \textbf{58} So, our **xx value is 58**.

STEP 5

Now, let's **plug in our xx value** into the logistic growth model: f(x)=10.231+3.55e0.030xf(x) = \frac{10.23}{1 + 3.55 e^{-0.030 x}} Substituting x=58x = \textbf{58}, we get: f(58)=10.231+3.55e0.03058f(\textbf{58}) = \frac{10.23}{1 + 3.55 e^{-0.030 \cdot \textbf{58}}}

STEP 6

Let's **evaluate the exponent** first: 0.03058=-1.74-0.030 \cdot \textbf{58} = \textbf{-1.74} So, the equation becomes: f(58)=10.231+3.55e-1.74f(\textbf{58}) = \frac{10.23}{1 + 3.55 e^{\textbf{-1.74}}}

STEP 7

Now, let's **calculate** e-1.74e^{\textbf{-1.74}}: e-1.740.1755e^{\textbf{-1.74}} \approx \textbf{0.1755} Substituting this value back into our equation: f(58)=10.231+3.550.1755f(\textbf{58}) = \frac{10.23}{1 + 3.55 \cdot \textbf{0.1755}}

STEP 8

Next, let's **multiply** 3.553.55 by 0.1755\textbf{0.1755}: 3.550.17550.62303.55 \cdot \textbf{0.1755} \approx \textbf{0.6230} Our equation now looks like this: f(58)=10.231+0.6230f(\textbf{58}) = \frac{10.23}{1 + \textbf{0.6230}}

STEP 9

Let's **add** 1 to 0.6230\textbf{0.6230}: 1+0.6230=1.62301 + \textbf{0.6230} = \textbf{1.6230} Now, we have: f(58)=10.231.6230f(\textbf{58}) = \frac{10.23}{\textbf{1.6230}}

STEP 10

Finally, let's **divide** 10.2310.23 by 1.6230\textbf{1.6230} to get our **predicted population**: f(58)6.30f(\textbf{58}) \approx \textbf{6.30}

STEP 11

The table tells us that the actual population in 2007 was **6.3 billion**.
Our model predicted **6.30 billion**.

STEP 12

The model's prediction of 6.30\textbf{6.30} billion is very close to the actual value of 6.3\textbf{6.3} billion.

STEP 13

The function accurately predicts the population to one decimal place.

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