Question

question 13
the numbers of polio cases in the world are shown in the table for various years.
year number of polio cases (thousands)
1988 350
1992 138
1996 31
2000 6
2005 3.2
2007 1.3
let f(t) be the number of polio cases in the world t years since 1980.
use a graphing calculator to draw a scattergram of the data. is it better to model the data by using a linear or exponential model? select an answer
find an equation of f. hint
f(t)= round the coefficients to 2 decimal places.
the number of polio cases is select an answer by select an answer per year.
predict the number of polio cases in 2015.
hint
predict in which year there will be 1 case of polio.
find the approximate half - life of the number of polio cases. hint
question help: video post to forum

Explanation:

Step1: Determine the model type

By observing the data, the number of polio - cases is decreasing rapidly. The ratio of the number of cases in consecutive years is relatively constant. For example, from 1988 to 1992 ($t = 8$ to $t = 12$), $\frac{138}{350}\approx0.394$, from 1992 to 1996 ($t = 12$ to $t = 16$), $\frac{31}{138}\approx0.225$. An exponential model $f(t)=ab^{t}$ is more appropriate. Let $t$ be the number of years since 1980. For 1988 ($t = 8$), $f(8)=ab^{8}=350$, for 1992 ($t = 12$), $f(12)=ab^{12}=138$.

Step2: Solve for $a$ and $b$

Divide $ab^{12}$ by $ab^{8}$: $\frac{ab^{12}}{ab^{8}}=b^{4}=\frac{138}{350}\approx0.394$. Then $b=\sqrt[4]{0.394}\approx0.82$. Substitute $b\approx0.82$ into $ab^{8}=350$, we get $a\times(0.82)^{8}=350$. So $a=\frac{350}{(0.82)^{8}}\approx\frac{350}{0.1677}\approx2087.00$. The function is $f(t)=2087.00\times(0.82)^{t}$.

Step3: Analyze the rate of change

The number of polio cases is decreasing exponentially. The decay factor is $b = 0.82$, which means the number of cases is decreasing by $(1 - 0.82)=0.18$ or 18% per year.

Step4: Predict the number of cases in 2015

For 2015, $t=2015 - 1980 = 35$. Then $f(35)=2087.00\times(0.82)^{35}\approx2087.00\times0.00019\approx0.40$ (in thousands), so $f(35)\approx0.40\times1000 = 400$ cases.

Step5: Predict the year with 1 case

Set $f(t)=1$ (in thousands), so $1 = 2087.00\times(0.82)^{t}$. Then $(0.82)^{t}=\frac{1}{2087.00}\approx0.00048$. Take the natural - logarithm of both sides: $\ln(0.82^{t})=\ln(0.00048)$. Using the property of logarithms $\ln(a^{b})=b\ln(a)$, we have $t\ln(0.82)=\ln(0.00048)$. So $t=\frac{\ln(0.00048)}{\ln(0.82)}\approx\frac{-7.64}{-0.198}\approx38.6$. Since $t$ is the number of years since 1980, the year is approximately $1980 + 39=2019$.

Step6: Find the half - life

Set $f(t + h)=\frac{1}{2}f(t)$, where $h$ is the half - life. If $f(t)=ab^{t}$ and $f(t + h)=ab^{t + h}$, then $ab^{t + h}=\frac{1}{2}ab^{t}$. Canceling out $a$ and $b^{t}$, we get $b^{h}=\frac{1}{2}$. Take the natural - logarithm of both sides: $\ln(b^{h})=\ln(\frac{1}{2})$. Using the property $\ln(a^{b})=b\ln(a)$, we have $h\ln(b)=\ln(\frac{1}{2})$. Since $b\approx0.82$, $h=\frac{\ln(0.5)}{\ln(0.82)}\approx\frac{-0.693}{-0.198}\approx3.5$.

Answer:

1. Exponential
2. $f(t)=2087.00\times(0.82)^{t}$
3. Decreasing, 18%
4. 400
5. 2019
6. 3.5