Question

Between 2005 -2009, the percent (in decimal form)of network commercials that lasted exactly 18 seconds could be modeled by the equation y=0.38x2-1.8\0.56x2+100,where x is the number of years since 2005 what percent of network commercials lasted exactly 18 seconds in 2008?Round you answer to the nearest percent

Answer 1

Answer:

Approximately 1.54% of the commercials lasted exactly 18 seconds in 2008.

Step-by-step explanation:

We are given the following in the question:

$y=\dfrac{0.38x^2-1.8}{0.56x^2+100}$

y is the percent of network commercials that lasted exactly 18 seconds and x is the number of years since 2005.

We have to estimate the percentage of advertisement that  lasted exactly 18 seconds in 2008.

Thus, we put x = 3 in the equation:

$y(3)=\dfrac{0.38(3)^2-1.8}{0.56(3)^2+100} = 0.0154 = 1.54\%$

Thus, approximately 1.54% of the commercials lasted exactly 18 seconds in 2008.