Ask question

Ask question

All categories

4 years ago

13

An Electrician charges in initial fee of $50 and $190 after four hours of work. Write a linear model that represents the total c

ost as a function of the number work

1 answer:

stiv31 [10]4 years ago

4 0

The equation for the line to graph would be y = 47.5x + 50

You might be interested in

Based on article 39, infer how some people had been abused by King John´s rule

Veseljchak [2.6K]

Answer:King John wanted to wage a war with France to win back his ancestors land, and taxed the barons and people heavily.

Step-by-step explanation:

5 0

3 years ago

Free pointzzzzzzzzzzzzzzzzzzzzzzzzzzz

DiKsa [7]

WOWW LOL THNX THO teeeheee

4 0

3 years ago

Read 2 more answers

Jello someone help is this correct answer <br> No links or files <br> Correct answer please

Dvinal [7]

Answer:

c is the right answer

Step-by-step explanation:

5 0

3 years ago

Hello! I need help with this problem thanks!!

Helen [10]

Answer:

10 tomato plants and

15 rose plants

Step-by-step explanation:

Let t be the no. of tomato plants,

And r the no. of rose plants

t + r = 25

t + 2r = 40

t = 25 - r

25 - r + 2r = 40

r = 15

t = 25 - 15

t = 10

8 0

3 years ago

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;

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

Now; $P (X(s) \leq 2 ) = P(X(s) = 0 ) + P(X(s) = 1) + P(X(s) = 2)$

(c) Determine the conditional probability density function for W2, given that X(t)=5.

So ; the conditional probability density function of $W_2$ given that X(t)=5 is:

$f_{W_2|X(t)=5}}= (W_2|X(t) = 5) \\ \\ =\dfrac{d}{ds}P(W_2 \leq s | X(t) =5 ) \\ \\ = \dfrac{d}{ds}P(X(s) \geq 2 | X(t) = 5)$

7 0

4 years ago