The problem has three parts: (a) Find the linear regression equation for the given data set and estimate the value of $x$ when $y=9$. (b) Determine the number of ways to select a coach, a trainer, and an assistant coach from 10 officials. (c) Given the profit function $P(x) = 2x^2 + 40x + 2000$ and the cost function $C(x) = x^2 - 30x + 500$, evaluate the marginal revenue at $x=2$.

1. Problem Description

The problem has three parts:

(a) Find the linear regression equation for the given data set and estimate the value of $x$ when $y=9$.

(b) Determine the number of ways to select a coach, a trainer, and an assistant coach from 10 officials.

(c) Given the profit function $P(x) = 2x^2 + 40x + 2000$ and the cost function $C(x) = x^2 - 30x + 500$, evaluate the marginal revenue at $x=2$.

2. Solution Steps

(a) Linear Regression and Estimation:

The data points are $(2, 3)$, $(4, 7)$, $(6, 5)$, $(8, 10)$.

First, calculate the means of $x$ and $y$:

$\bar{x} = \frac{2+4+6+8}{4} = \frac{20}{4} = 5$

$\bar{y} = \frac{3+7+5+10}{4} = \frac{25}{4} = 6.25$

Next, calculate the slope $m$ of the linear regression equation:

$m = \frac{\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i - \bar{x})^2}$

$m = \frac{(2-5)(3-6.25) + (4-5)(7-6.25) + (6-5)(5-6.25) + (8-5)(10-6.25)}{(2-5)^2 + (4-5)^2 + (6-5)^2 + (8-5)^2}$

$m = \frac{(-3)(-3.25) + (-1)(0.75) + (1)(-1.25) + (3)(3.75)}{(-3)^2 + (-1)^2 + (1)^2 + (3)^2}$

$m = \frac{9.75 - 0.75 - 1.25 + 11.25}{9 + 1 + 1 + 9}$

$m = \frac{19}{20} = 0.95$

Now, find the y-intercept $b$:

$b = \bar{y} - m\bar{x}$

$b = 6.25 - 0.95(5) = 6.25 - 4.75 = 1.5$

The linear regression equation is $y = 0.95x + 1.5$.

To estimate the value of $x$ when $y = 9$, we solve for $x$:

$9 = 0.95x + 1.5$

$0.95x = 9 - 1.5$

$0.95x = 7.5$

$x = \frac{7.5}{0.95} \approx 7.89$

(b) Combinations:

We need to select 3 officials (coach, trainer, and assistant coach) from 10 officials. Since the order matters (each position is distinct), we use permutations.

The number of permutations of $n$ objects taken $r$ at a time is given by:

$P(n, r) = \frac{n!}{(n-r)!}$

In this case, $n = 10$ and $r = 3$.

$P(10, 3) = \frac{10!}{(10-3)!} = \frac{10!}{7!} = 10 \times 9 \times 8 = 720$

(c) Marginal Revenue:

Given the profit function $P(x) = 2x^2 + 40x + 2000$ and cost function $C(x) = x^2 - 30x + 500$.

The revenue function $R(x)$ is related to the profit and cost functions by $P(x) = R(x) - C(x)$, thus $R(x) = P(x) + C(x)$.

$R(x) = (2x^2 + 40x + 2000) + (x^2 - 30x + 500) = 3x^2 + 10x + 2500$

The marginal revenue is the derivative of the revenue function, $R'(x)$.

$R'(x) = \frac{d}{dx}(3x^2 + 10x + 2500) = 6x + 10$

Now, evaluate the marginal revenue at $x = 2$:

$R'(2) = 6(2) + 10 = 12 + 10 = 22$

3. Final Answer

(a) The linear regression equation is $y = 0.95x + 1.5$. The estimated value of $x$ when $y=9$ is approximately $7.89$.

(b) There are 720 ways to select a coach, a trainer, and an assistant coach.

(c) The marginal revenue at $x=2$ is

2

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