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Question
the table shows the number of hats a company can sell at different prices. use quadratic regressi
price (dollars), x 3 4 5 6 7
hats sold (thousands), y 105 120 125 120 105
they will sell hats.
Step1: Assume the quadratic regression equation is $y = ax^{2}+bx + c$.
We have 5 data - points $(x_1,y_1)=(3,105)$, $(x_2,y_2)=(4,120)$, $(x_3,y_3)=(5,125)$, $(x_4,y_4)=(6,120)$, $(x_5,y_5)=(7,105)$. Substitute them into the equation:
For $(3,105)$: $105=a\times3^{2}+b\times3 + c=9a + 3b + c$.
For $(4,120)$: $120=a\times4^{2}+b\times4 + c=16a+4b + c$.
For $(5,125)$: $125=a\times5^{2}+b\times5 + c=25a + 5b + c$.
Step2: Subtract the first - equation from the second and the second from the third to eliminate $c$.
$(16a + 4b + c)-(9a + 3b + c)=120 - 105$, which simplifies to $7a + b=15$.
$(25a + 5b + c)-(16a + 4b + c)=125 - 120$, which simplifies to $9a + b=5$.
Step3: Subtract the new - formed first equation from the second new - formed equation to solve for $a$.
$(9a + b)-(7a + b)=5 - 15$,
$9a + b-7a - b=-10$,
$2a=-10$, so $a=-5$.
Step4: Substitute $a = - 5$ into $7a + b=15$ to solve for $b$.
$7\times(-5)+b=15$,
$-35 + b=15$,
$b = 50$.
Step5: Substitute $a=-5$ and $b = 50$ into $9a + 3b + c=105$ to solve for $c$.
$9\times(-5)+3\times50 + c=105$,
$-45+150 + c=105$,
$105 + c=105$, so $c = 0$.
The quadratic regression equation is $y=-5x^{2}+50x$.
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The quadratic regression equation for the relationship between the price $x$ (in dollars) and the number of hats sold $y$ (in thousands) is $y=-5x^{2}+50x$. If you want to find the number of hats sold for a particular price, you can substitute the value of $x$ into this equation. For example, if you want to find the maximum of this quadratic function, since for a quadratic function $y = ax^{2}+bx + c$ with $a\lt0$, the $x$ - coordinate of the vertex is $x=-\frac{b}{2a}$. Here, $a=-5$ and $b = 50$, so $x=-\frac{50}{2\times(-5)} = 5$. Substituting $x = 5$ into $y=-5x^{2}+50x$, we get $y=-5\times5^{2}+50\times5=-125 + 250=125$ (in thousands), so the company will sell 125000 hats at the optimal price.