Question

Let f(t) be the number of polio cases in thousands 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 4 decimal places.

The number of polio cases is Select an answer by Select an answer per year.

Predict the number of polio cases in 2017.

Hint

Predict in which year there will be 1 case of polio. Hint

Find the approximate half-life of the number of polio cases. Hint

year


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Answer 1

The completed statements can be presented as follows;

f(t) =
`\underline{2* 10^(265)\cdot e^((-0.304\cdot t))}`

The number of polio cases decreases by 30.4% per year

Predicted number of polio cases in 2017, are about 101 polio cases

Predicted year in which there will be 1 polio case in the year 2032

Half-life of the number of polio cases is; 4 years

The steps by which the above values are found are presented as follows;

The scatter diagram obtained using the data in the table indicates that the square of the R-squared value for the linear model is R² = 0.7195, and the R-squared for the exponential model is; R² = 0.9894, which indicates that it is better to model the data using an exponential model

The equation for f, obtained using the exponential model can be presented as follows;

f(t) = 2×10²⁶⁵×
`e^((-0.304\cdot t))`

The exponential model indicates that the number of polio cases decreases by 30.4% per year

The number of polio cases in 2017, can be found by plugging in 2017 in the the equation, f(t) = 2×10²⁶⁵×
`e^((-0.304\cdot t))`, as follows;

f(2017) = 2×10²⁶⁵×
`e^((-0.304* 2017))`

2×10²⁶⁵×
`e^((-0.304* 2017))` ≈ 0.1013

Therefore, the number of polio cases in 2017 is about; 1000 × 0.1013 = 101.3

The year in which there will be 1 case of polio, can be found using the following equation

f(t) = 2×10²⁶⁵×
`e^((-0.304\cdot t))`

f(t) = 1

2×10²⁶⁵×
`e^((-0.304\cdot t))` = 1

`e^((-0.304\cdot t))=(1)/(2* 10^(265))`

`\ln e^((-0.304\cdot t))=\ln ((1)/(2* 10^(265)))`

-0.304·t =
`\ln ((1)/(2* 10^(265)))`

`t=(\ln ((1)/(2* 10^(265))))/(-0.304)`

`(\ln ((1)/(2* 10^(265))))/(-0.304)\approx 2032`

The year in which there will be 1 polio case is about 2032

The approximate half life can be found as follows;

The number of polio cases in 1988 is 350,000

When the number is half the above amount or 350,000/2 = 175,000, we get;

2 × 10²⁶⁵ ×
`e^((-0.304\cdot t))` = 175

ln
`e^((-0.304\cdot t))` =
`\ln ((175)/(2* 10^(265)))`

-0.304·t =
`\ln ((175)/(2 * 10^(265)))`

`t = (\ln ((175)/(2* 10^(265))) )/(-0.304)`

t ≈ 1992

Therefore, the half life is about 1992 - 1988 = 4 years