Chapter 3.5, Problem 73AYU

(a)

To determine

The probability that the player who is serving will win the game if the probability of player winning a point on serve is $0.64$, given that the model $P(x)=\frac{x^4\left(-8x^3+28x^2-34x+15\right)}{2x^2-2x+1}$ represents the probability $P$ of the player winning a game in which player is serving the game and $x$ is the probability of winning a point on serve.

(b)

To determine

The value $P(0.62)$ and write its interpretation given that the model $P(x)=\frac{x^4\left(-8x^3+28x^2-34x+15\right)}{2x^2-2x+1}$ represents the probability $P$ of the player winning a game in which player is serving the game and $x$ is the probability of winning a point on serve.

(c)

To determine

The value of $x$ that gives $P(x)=0.9$ given that the model $P(x)=\frac{x^4\left(-8x^3+28x^2-34x+15\right)}{2x^2-2x+1}$ represents the probability $P$ of the player winning a game in which player is serving the game and $x$ is the probability of winning a point on serve.

(d)

To determine

To graph: The function $P(x)=\frac{x^4\left(-8x^3+28x^2-34x+15\right)}{2x^2-2x+1}$ for $0\le x\le 1$ and describes what happens to $P$ as $x$ approaches to 1.