abercombie and fitch stock had a price given as p = 0.2t² - 5.6t + 50.2, where t is the time in months from…

Explanation:

Step1: Set up the equation

Set $P = 30$ in the equation $P=0.2t^{2}-5.6t + 50.2$. So we have $0.2t^{2}-5.6t + 50.2=30$.

Step2: Rearrange the equation

Subtract 30 from both sides to get a quadratic - equation in standard form: $0.2t^{2}-5.6t+50.2 - 30 = 0$, which simplifies to $0.2t^{2}-5.6t + 20.2 = 0$. Multiply through by 10 to clear the decimals: $2t^{2}-56t + 202 = 0$. Divide by 2: $t^{2}-28t + 101 = 0$.

Step3: Use the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For $t^{2}-28t + 101 = 0$, $a = 1$, $b=-28$, and $c = 101$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-28)^{2}-4\times1\times101=784 - 404 = 380$. Then $t=\frac{28\pm\sqrt{380}}{2}=\frac{28\pm2\sqrt{95}}{2}=14\pm\sqrt{95}$.

Step4: Calculate the values of t

$\sqrt{95}\approx9.75$. So $t_1=14 - 9.75 = 4.25\approx4$ and $t_2=14 + 9.75 = 23.75\approx24$. Since $t = 1$ is January 1999, $t = 4$ corresponds to April 1999 and $t = 24$ corresponds to December 2000.

Answer:

C. April 1999 and December 2000