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garri49 [273]
3 years ago
11

Drag expressions into order to show how to find 4 × 3/5.

Mathematics
1 answer:
san4es73 [151]3 years ago
5 0

Answer/Explanation:

4 * \frac{3}{5}\\\\= \frac{4}{1} *\frac{3}{5}\\\\= \frac{4*3}{1*5}\\\\= \frac{12}{5}

Without the expressions given, answer by referencing these steps.

4 = 4/1

When multiplying fractions, multiply both numerators and both denominators to get the answer.

In the answer, you may simplify if there are common factors in both the top and bottom number. In this case, 12 and 5 do not have a common factor and is already in the most simplified form.

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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)}

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
A square dartboard contains a red square and a blue square that overlap. A dart hits a random point on the board.
FinnZ [79.3K]
They want to know the probability of landing in the blue and red section at the same time. In other words, they want to know the probability of landing in the purple section.

We'll need the area of the purple square. This square is 1.5 inches by 1.5 inches. This is because 4 - 2.5 = 1.5

So the purple square has an area of 1.5*1.5 = 2.25 square inches

Divide this over the total area of the largest square (which is 9x9) to get 2.25/81 = 0.02777... where the 7's go on forever

Round that to two decimal places. The final answer is 0.03

Side note: 2.25/81 is equivalent to the reduced fraction 1/36 (express 2.25/81 as 225/8100 and then divide both parts by the GCF 225)
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