Math  /  Data & Statistics

QuestionUse Regression to Find and Use an Exponential Model to answer the questions: The numbers of polio cases in the world are shown in the table for various years. \begin{tabular}{|c|c|} \hline Year & Number of Polio Cases (thousands) \\ \hline 1988 & 350 \\ \hline 1992 & 138 \\ \hline 1996 & 33 \\ \hline 2000 & 5 \\ \hline 2005 & 3.2 \\ \hline 2007 & 1.3 \\ \hline \end{tabular}
Let f(t)f(t) be the number of polio cases in the world tt years since 1980 . The exponential regression equation for the data above is: f(t)=3768.57(0.74)tf(t)=3768.57(0.74)^{t}
Predict the number of polio cases in 2011. \square Hint
Predict in which year there will be 1 case of polio. \square Find the approximate half-life of the number of polio cases. Round your answer to one decimal place. \square years Question Help: Post to forum

Studdy Solution

STEP 1

What is this asking? We're going to use a given exponential equation, which models the decline of polio cases over time, to predict the number of cases in 2011, the year when there will be only one case, and the half-life of the disease based on the model. Watch out! Make sure to correctly substitute the *t* value, representing the years since 1980, not the actual year.
Also, remember the difference between predicting the number of cases and predicting the year!

STEP 2

1. Predict the number of polio cases in 2011.
2. Predict the year when there will be one case of polio.
3. Find the approximate half-life of the number of polio cases.

STEP 3

Since our function f(t)f(t) uses *t* as the years since 1980, we need to figure out what *t* is for the year 2011.
We can do this by subtracting 1980 from 2011.
So, t=20111980=31t = 2011 - 1980 = \textbf{31}.

STEP 4

Now, we **plug** our *t*-value into the given exponential equation: f(t)=3768.57(0.74)tf(t) = 3768.57 \cdot (0.74)^{t} f(31)=3768.57(0.74)31f(\textbf{31}) = 3768.57 \cdot (0.74)^{\textbf{31}}

STEP 5

Let's **calculate** the result: f(31)=3768.57(0.74)313768.570.00006120.230f(31) = 3768.57 \cdot (0.74)^{31} \approx 3768.57 \cdot 0.0000612 \approx \textbf{0.230} So, our model predicts approximately 0.230\textbf{0.230} thousand, or 230\textbf{230} cases of polio in 2011.

STEP 6

We want to find the year when there will be one thousand cases, so we **set** f(t)=1f(t) = 1 and **solve** for *t*: 1=3768.57(0.74)t1 = 3768.57 \cdot (0.74)^{t}

STEP 7

To **isolate** the exponential term, we'll **divide** both sides by 3768.57: 13768.57=(0.74)t\frac{1}{3768.57} = (0.74)^{t} 0.000265(0.74)t0.000265 \approx (0.74)^{t}

STEP 8

Now we can use logarithms to **solve** for *t*.
Taking the logarithm base 0.74 of both sides: log0.74(0.000265)t\log_{0.74}(0.000265) \approx t Using the change of base formula: tln(0.000265)ln(0.74)t \approx \frac{\ln(0.000265)}{\ln(0.74)} t8.2350.30127.4t \approx \frac{-8.235}{-0.301} \approx \textbf{27.4}

STEP 9

Since *t* represents years since 1980, we **add** 27.4 to 1980 to get the year: 1980+27.4=2007.41980 + \textbf{27.4} = \textbf{2007.4} So, our model predicts that there will be one thousand cases of polio around mid-2007.

STEP 10

The half-life is the time it takes for the number of cases to decrease by half.
We can find this by setting f(t)f(t) equal to half of the **initial value**.
The initial value is the number of cases when t=0t = 0 (in 1980), which is f(0)=3768.57(0.74)0=3768.571=3768.57f(0) = 3768.57 \cdot (0.74)^0 = 3768.57 \cdot 1 = 3768.57.
Half of this is 3768.572=1884.285\frac{3768.57}{2} = 1884.285.

STEP 11

We **set** f(t)=1884.285f(t) = 1884.285 and **solve** for *t*: 1884.285=3768.57(0.74)t1884.285 = 3768.57 \cdot (0.74)^{t}

STEP 12

**Divide** both sides by 3768.57: 0.5=(0.74)t0.5 = (0.74)^{t} Now, we can use logarithms.
Taking the logarithm base 0.74 of both sides: log0.74(0.5)=t\log_{0.74}(0.5) = t Using the change of base formula: t=ln(0.5)ln(0.74)t = \frac{\ln(0.5)}{\ln(0.74)} t0.6930.3012.3t \approx \frac{-0.693}{-0.301} \approx \textbf{2.3}

STEP 13

The half-life is approximately 2.3\textbf{2.3} years.
This means that every 2.3 years, the number of polio cases is reduced by half, according to our model.

STEP 14

In 2011, there will be approximately 230 cases of polio.
There will be approximately 1000 cases of polio mid-2007.
The approximate half-life of the number of polio cases is 2.3 years.

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