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timofeeve [1]
3 years ago
12

A line passes throuh the ordered pair (7,2) and (5,4). What is the slope? PLEASE HELP DUE TODAY

Mathematics
2 answers:
Akimi4 [234]3 years ago
5 0
The formula for slope is the change in y over the change in x
m=slope
m=4-2/5-7
m=2/-2
m=-1
it has a slope of -1
Solnce55 [7]3 years ago
5 0
The answer is -1, for every one it goes over, it goes down one
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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
4 years ago
Please answer both questions thanks​
Otrada [13]

Answer:

5. tan(25) = 12/x

x · tan(25) = 12

x = 12/tan(25)

= 25.734

≈ 25.73

6. tan(52) = x/3.4

x = 3.4tan(52)

= 4.352

≈ 4.4

6 0
3 years ago
Suppose a football is kicked with an initial velocity of 82 ft/sec., at an angle of
satela [25.4K]

Answer:

The position P is:

P = 87\^x + 75\^y ft     <u><em> Remember that the position is a vector. Observe the attached image</em></u>

Step-by-step explanation:

The equation that describes the height as a function of time of an object that moves in a parabolic trajectory with an initial velocity s_0 is:

y(t) = y_0 + s_0t -16t ^ 2

Where y_0 is the initial height = 0 for this case

We know that the initial velocity is:

82 ft/sec at an angle of 58 ° with respect to the ground.

So:

s_0 = 82sin(58\°) ft/sec

s_0 = 69.54 ft/sec

Thus

y(t) = 69.54t -16t ^ 2

The height after 2 sec is:

y(2) = 69.54 (2) -16 (2) ^ 2

y(2) = 75\ ft

Then the equation that describes the horizontal position of the ball is

X(t) = X_0 + s_0t

Where

X_ 0 = 0 for this case

s_0 = 82cos(58\°) ft / sec

s_0 = 43.45 ft/sec

So

X(t) = 43.45t

After 2 seconds the horizontal distance reached by the ball is:

X (2) = 43.45(2)\\\\X (2) = 87\ ft

Finally the vector position P is:

P = 87\^x + 75\^y ft

6 0
3 years ago
Plz help me :) dropdowns
ollegr [7]

Answer:

1

Step-by-step explanation:

3 0
3 years ago
Please help me answer this ASAP
Cerrena [4.2K]
I thick its C for the answer
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