Use a table of values to find the solution to the equation f(x)=h(x) where

Mathematics

1 answer:

GuDViN [60]3 years ago

6 0

9514 1404 393

Answer:

x = 2 1/3

Step-by-step explanation:

We can examine the equations to see where the solution lies.

f(x) = (2/3) -x

This has an x-intercept where y=0, at x=2/3. It has a y-intercept where x=0, at y=2/3. Its slope is -1.

h(x) = 3 -2x

This has an x-intercept where y=0, at x=3/2. It has a y-intercept where x=0, at y=3. Its slope is -2.

In the first quadrant, the graph of h(x) is farther from the origin and steeper than the graph of f(x). The lines must cross in the 4th quadrant at some value of x that is greater than 3/2. The fraction in the definition of f(x) suggests that the solution will be a multiple of 1/3.

The attached table shows a couple of guesses at values of x that would make f(x) = h(x). We find that x = 7/3 is the solution we're looking for.

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Additional comment

Repetitive function evaluations are done conveniently and with fewer errors by a calculator or spreadsheet that can work with tables of values. Here, we have used a graphing calculator. These tools are readily available for free on almost any phone, tablet, or desktop computer platform.

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Customers arrive at a service facility according to a Poisson process of rate λ customers/hour. Let X(t) be the number of custom

mash [69]

Answer:

Step-by-step explanation:

Given that:

X(t) = be the number of customers that have arrived up to time t.

$W_1,W_2$ ... = the successive arrival times of the customers.

(a)

Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;

$E(W_!|X(t)=2) = \int\limits^t_0 {X} ( \dfrac{d}{dx}P(X(s) \geq 1 |X(t) =2))$

$= 1- P (X(s) \leq 0|X(t) = 2) \\ \\ = 1 - \dfrac{P(X(s) \leq 0 , X(t) =2) }{P(X(t) =2)}$

$= 1 - \dfrac{P(X(s) \leq 0 , 1 \leq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

$= 1 - \dfrac{P(X(s) \leq 0 ,P((3 \eq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

Now $P(X(s) \leq 0) = P(X(s) = 0)$

(b) We can Determine the conditional mean E[W3|X(t)=5] as follows;

$E(W_1|X(t) =2 ) = \int\limits^t_0 X (\dfrac{d}{dx}P(X(s) \geq 3 |X(t) =5 )) \\ \\ = 1- P (X(s) \leq 2 | X (t) = 5 ) \\ \\ = 1 - \dfrac{P (X(s) \leq 2, X(t) = 5 }{P(X(t) = 5)} \\ \\ = 1 - \dfrac{P (X(s) \LEQ 2, 3 (t) - X(s) \leq 5 )}{P(X(t) = 2)}$