Which of the following graphs shows the solution set to 2x < 6 and 3x + 2 > -4?

Mathematics

1 answer:

Ilya [14]2 years ago

5 0

Answer:

The third picture

Step-by-step explanation:

Solve for x in both equations

2x<6

Divide both sides by 2:

x<3

3x+2>-4

Subtract 2 from both sides:

3x>-6

Divide both sides by 3:

x>-2

There is this trick you can use when x is on the left side of the equation to find out which way to shade in you graph. Keep in mind this is only for the left side, it will not work if your variable is on the right.

When the symbol is facing left < then shade left, imagine it is pointing which way to shade. x<3 is represented by the 3 picture on the left. When the symbol is facing right > then shade right, again it is pointing which way to shade. x>-2 is represented by the 3 picture on the right.

The circles are not filled in because the symbol is < and > rather than $\leq and \geq$ . When it is greater than or equal to or less than and equal to (represented by the line under the symbol), then the circle is shaded in.

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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;

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