Using UN data from $1975$ and projections to $2050$ , the number of millions of people age $15$ to $59$ in a country can be modeled by $y=-0.224x^{2}+23x+371.7$ , with $x$ equal to the number of years after $1970$ and $y$ equal to the number of millions of people in this labor pool. Use technology with the model to find when this population is expected to be equal to $920.3$ million.

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Q. Using UN data from

1975

and projections to

2050

, the number of millions of people age

15

59

in a country can be modeled by

y=-0.224x^{2}+23x+371.7

, with

x

equal to the number of years after

1970

and

y

equal to the number of millions of people in this labor pool. Use technology with the model to find when this population is expected to be equal to

920.3

million.

Given Quadratic Model: We are given the quadratic model $y = -0.224x^2 + 23x + 371.7$ , where $y$ represents the number of millions of people aged $15$ to $59$ and $x$ represents the number of years after $1970$ . We want to find the value of $x$ when $y$ is equal to $920.3$ million.
Set Equation and Solve: Set the equation equal to $920.3$ million to solve for $x$ : $\newline$ $-0.224x^2 + 23x + 371.7 = 920.3$
Simplify and Calculate Discriminant: Subtract $920.3$ from both sides to set the equation to zero: $\newline$ $-0.224x^2 + 23x + 371.7 - 920.3 = 0$
Use Quadratic Formula: Simplify the equation: $-0.224x^2 + 23x - 548.6 = 0$
Calculate Solutions: Use a quadratic formula or technology (such as a graphing calculator or computer software) to solve for $x$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -0.224$ , $b = 23$ , and $c = -548.6$ .
Solve for $x_1$ and $x_2$ : Calculate the discriminant $(b^2 - 4ac)$ :
Discriminant = $23^2 - 4(-0.224)(-548.6)$
Discriminant = $529 - 4(0.224)(548.6)$
Discriminant = $529 - 491.3664$
Discriminant = $37.6336$
Calculate Numerical Values: Since the discriminant is positive, there are two real solutions. Calculate the solutions using the quadratic formula: $\newline$ $x = \frac{-23 \pm \sqrt{37.6336}}{2 \times -0.224}$
Find Actual Years: Calculate the two possible values for $x$ : $x_1 = \frac{{-23 + \sqrt{37.6336}}}{{-0.448}}$ $x_2 = \frac{{-23 - \sqrt{37.6336}}}{{-0.448}}$
Final Prediction: Solve for $x_1$ and $x_2$ :
$x_1 = \frac{{-23 + 6.13524}}{{-0.448}}$
$x_2 = \frac{{-23 - 6.13524}}{{-0.448}}$
Final Prediction: Solve for $x_1$ and $x_2$ :
$x_1 = \frac{-23 + 6.13524}{-0.448}$
$x_2 = \frac{-23 - 6.13524}{-0.448}$ Calculate the numerical values for $x_1$ and $x_2$ :
$x_1 = \frac{-16.86476}{-0.448} \approx 37.67$
$x_2 = \frac{-29.13524}{-0.448} \approx 65.03$
Final Prediction: Solve for $x_1$ and $x_2$ :
$x_1 = \frac{{-23 + 6.13524}}{{-0.448}}$
$x_2 = \frac{{-23 - 6.13524}}{{-0.448}}$ Calculate the numerical values for $x_1$ and $x_2$ :
$x_1 = \frac{{-16.86476}}{{-0.448}} \approx 37.67$
$x_2 = \frac{{-29.13524}}{{-0.448}} \approx 65.03$ Since $x$ represents the number of years after $1970$ , we need to add $1970$ to each solution to find the actual years:
Year $x_2$ $1$
Year $x_2$ $2$
Final Prediction: Solve for $x_1$ and $x_2$ :
$x_1 = \frac{-23 + 6.13524}{-0.448}$
$x_2 = \frac{-23 - 6.13524}{-0.448}$ Calculate the numerical values for $x_1$ and $x_2$ :
$x_1 = \frac{-16.86476}{-0.448} \approx 37.67$
$x_2 = \frac{-29.13524}{-0.448} \approx 65.03$ Since $x$ represents the number of years after $1970$ , we need to add $1970$ to each solution to find the actual years:
Year $x_2$ $1$
Year $x_2$ $2$ The model predicts that the population will be $x_2$ $3$ million in approximately the years $x_2$ $4$ and $x_2$ $5$ . However, since we are looking for a future projection and the year $x_2$ $4$ has already passed, we consider the year $x_2$ $5$ as the answer.