suppose that a company introduces a new computer game in a city using television advertisements. surveys…

suppose that a company introduces a new computer game in a city using television advertisements. surveys show that p% of the target audience buy the game after x ads are broadcast, satisfying the equation below. complete parts (a) through (d) p(x)=\frac{100}{1 + 51e^{-0.14x}} a) what percentage buy the game without seeing a tv ad (x = 0)? 1.9% (type an integer or a decimal rounded to the nearest tenth as needed.) b) what percentage buy the game after the ad is run 55 times? % (type an integer or a decimal rounded to the nearest tenth as needed.)

suppose that a company introduces a new computer game in a city using television advertisements. surveys show that p% of the target audience buy the game after x ads are broadcast, satisfying the equation below. complete parts (a) through (d) p(x)=\frac{100}{1 + 51e^{-0.14x}} a) what percentage buy the game without seeing a tv ad (x = 0)? 1.9% (type an integer or a decimal rounded to the nearest tenth as needed.) b) what percentage buy the game after the ad is run 55 times? % (type an integer or a decimal rounded to the nearest tenth as needed.)

Answer

Explanation:

Step1: Substitute x = 0 into the formula

When (x = 0), we have (P(0)=\frac{100}{1 + 51e^{-0.14\times0}}). Since (e^{-0.14\times0}=e^{0}=1), then (P(0)=\frac{100}{1 + 51\times1}=\frac{100}{52}\approx1.9).

Step2: Substitute x = 55 into the formula

When (x = 55), we have (P(55)=\frac{100}{1+51e^{-0.14\times55}}). First, calculate the exponent: (-0.14\times55=-7.7). Then (e^{-7.7}\approx0.00045). Next, (51e^{-7.7}=51\times0.00045 = 0.02295). Finally, (P(55)=\frac{100}{1 + 0.02295}\approx97.8).

Answer:

a) 1.9 b) 97.8