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Gnom [1K]
3 years ago
11

4 times the quotient of 3 and 4

Mathematics
1 answer:
MaRussiya [10]3 years ago
4 0

Answer:

3

Step-by-step explanation:

You would start by dividing

3 by 4. Then multiply 3/4 and 4. The 4s would cancel out leaving you with 3.  3/ 4×4=3

hope this helps

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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
Let C = C1 + C2 where C1 is the quarter circle x^2+y^2=4, z=0,from (0,2,0) to (2,0,0), and where C2 is the line segment from (2,
trapecia [35]
Not much can be done without knowing what \mathbf F(x,y,z) is, but at the least we can set up the integral.

First parameterize the pieces of the contour:

C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)
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where 0\le t_1\le\dfrac\pi2 and 0\le t_2\le1. You have

\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1
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and so the work is given by the integral

\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r
=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1
{}\displaystyle\,\,\,\,\,\,\,\,+\int_0^1\mathbf F(2+t_2,3t_2,3t_2)\cdot(1,3,3)\,\mathrm dt_2
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3 years ago
Plssss help in need of answers soon
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My guess is that you're doing the Law of Cosines? You have everything you need for that except the angle theta, which is the thing you need to find.  It's set up like this: (8)^2 = (10)^2 + (5)^2 -[2(10)(5)cos A]  I used A instead of theta. Doing that math, you have: 64 = 100 + 25 -[ 100 cos A]; 64 = 125 - 100 cos A;
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On a coordinate plane, solid circles appear at the following points: (negative 3, negative 3), (negative 3, 2), (1, 1), (2, nega
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ITS (3-2.) Thank me later

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600 feet of tape because 4 +4 for two sides of the window is 8 and 3+ 3 is 6, 14 plus the foot of overlap is 15 x 40 is 600.
3 0
3 years ago
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