Chapter 3, Problem 122CR

(a)

To determine

To find: The linear model, an exponential model and power model of the data.

Answer to Problem 122CR

Linear model: $N=35.685t+2773.3$

Exponential model: $N=2795.3{\left(1.0113\right)}^t$

Power model: $N=2379.9t^{0.1213}$

Explanation of Solution

Given:Each ordered pair $\left(t,N\right)$ represents the year t and the number N ( in thousands) of female participants in high school athelic programs during 12 school years, with $t=6$ corresponding to the 2005 − 2006 school year.

  $\begin{array}{l}\left(6,2953\right),\left(7,3022\right),\left(8,3057\right),\left(9,3114\right),\left(10,3173\right),\left(11,3174\right),\\ \left(12,3208\right),\left(13,3223\right),\left(14,3268\right),\left(15,3288\right),\left(16,3324\right),\left(17,3400\right)\end{array}$

Using graphing regression calculator to find the model.

Linear model:

Equation, $N=35.685t+2773.3$

Determination coefficient, $R^2=0.9734$

Exponential model:

Equation, $N=2795.3{\left(1.0113\right)}^t$

Determination coefficient, $R^2=0.9705$

Power model:

Equation, $N=2379.9t^{0.1213}$

Determination coefficient, $R^2=0.9797$

(b)

To determine

To graph: The linear model, an exponential model and power model of the data.

Explanation of Solution

Given: Each ordered pair $\left(t,N\right)$ represents the year t and the number N ( in thousands) of female participants in high school athelic programs during 12 school years, with $t=6$ corresponding to the 2005 − 2006 school year.

  $\begin{array}{l}\left(6,2953\right),\left(7,3022\right),\left(8,3057\right),\left(9,3114\right),\left(10,3173\right),\left(11,3174\right),\\ \left(12,3208\right),\left(13,3223\right),\left(14,3268\right),\left(15,3288\right),\left(16,3324\right),\left(17,3400\right)\end{array}$

Using graphing utility to graph linear model.

  

Graph exponential model.

  

Graph power model,

  

(c)

To determine

To describe: The best fit model of the data.

Answer to Problem 122CR

Power model: $N=2379.9t^{0.1213}$

Explanation of Solution

Given: Each ordered pair $\left(t,N\right)$ represents the year t and the number N ( in thousands) of female participants in high school athelic programs during 12 school years, with $t=6$ corresponding to the 2005 − 2006 school year.

  $\begin{array}{l}\left(6,2953\right),\left(7,3022\right),\left(8,3057\right),\left(9,3114\right),\left(10,3173\right),\left(11,3174\right),\\ \left(12,3208\right),\left(13,3223\right),\left(14,3268\right),\left(15,3288\right),\left(16,3324\right),\left(17,3400\right)\end{array}$

Using graphing regression calculator to find the model.

Linear model:

Equation, $N=35.685t+2773.3$

Determination coefficient, $R^2=0.9734$

Exponential model:

Equation, $N=2795.3{\left(1.0113\right)}^t$

Determination coefficient, $R^2=0.9705$

Power model:

Equation, $N=2379.9t^{0.1213}$

Determination coefficient, $R^2=0.9797$

The coefficient of determination of power model is more nearest to 1 as compare to linear and exponential mode.

Therefore, power model is best fit for given data.

(d)

To determine

To describe: The best fit model of the data.

Answer to Problem 122CR

2072-2073

Explanation of Solution

Given: Each ordered pair $\left(t,N\right)$ represents the year t and the number N ( in thousands) of female participants in high school athelic programs during 12 school years, with $t=6$ corresponding to the 2005 − 2006 school year.

  $\begin{array}{l}\left(6,2953\right),\left(7,3022\right),\left(8,3057\right),\left(9,3114\right),\left(10,3173\right),\left(11,3174\right),\\ \left(12,3208\right),\left(13,3223\right),\left(14,3268\right),\left(15,3288\right),\left(16,3324\right),\left(17,3400\right)\end{array}$

The best fit model Equation, $N=2379.9t^{0.1213}$

In which year number of female will be 4 million.

As N represents number of female in thousands.

So, 4 million = 4000 thousands

So put $N=4000$ in model

  $\begin{array}{c}4000=2379.9t^{0.1213}\\ t^{0.1213}=1.6807\\ t=72.28\\ t\approx 72\end{array}$

  $t=6$ corresponding to the 2005-2006

So, $t=72$ would be 2072-2073