A square has a side of 400 feet. how many squares with a side length of 2 can compliantly cover this square

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

3 years ago

Please answer ALL of the following if you do so i will give you brainiest :))))) :

chubhunter [2.5K]

2x^2 if x=1/8 is 1/16
4x^2 if x=1/8 is 1/2

8 0

2 years ago

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Tickets for an American Baseball League game for 3 adults and 3 children cost less than $75, while tickets for 2 adults and 4 ch

nikitadnepr [17]

<h3>No, the cost of 1 adult ticket cannot be $ 20 and the cost of 1 children ticket is not $ 8.</h3>

Step-by-step explanation:

Let us assume the cots of ticket for 1 adult = $ x

Let us assume the cots of ticket for 1 children = $ y

So, the cost of ticket for 3 adults = 3 x ( Cost of 1 adult ticket) = 3 x

The cost of ticket for 3 children = 3 x ( Cost of 1 children ticket) = 3 y

Also, given the combined cost of 3 adult and children ticket is less than $75.

⇒ 3 x + 3 y < 75 ... (1)

Similarly, the cost of ticket for 2 adults = 2 x

The cost of ticket for 4 children = 4 y

⇒ 2 x + 4 y < 62 ... (2)

Now, solving for the values of x and y, we get:

3 x + 3 y < 75 or, x + y < 25 ⇒ x = 25 - y ( substitute in 2)

2 x + 4 y < 62 or, x + 2 y < 31

⇒ 25 - y + 2 y < 31

or, y + 25 < 31

or, y < 6

⇒ x = 25 - 6 = 19

or, x < 19

So, the cost of 1 adult ticket is x and is less than $19.

The cost of 1 child ticket is y and is less than $6.

Hence, by above statement, NO the cost of 1 adult ticket can not be $ 20 and the cost of 1 children ticket is not $ 8.

4 0

3 years ago

ANSWER THIS TO GET 23 POINTS AND BRAINLIEST

puteri [66]

Ohhh, interest problems. I haven't done an equation like this in a long time, but I will attempt.

I would say that the answer is the 2nd option. The equation is i = (5200)(0.06)(2.5).

The traditional interest formula is I = (P)(R)(N).

P = the original amount of money given

R = interest rate

N = the amount of time

I hope this answers your question (and that I understood the question correctly!).

-Toremi

6 0

2 years ago

Need help asap will give brainly, please

Setler [38]

Answer:

g

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers