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3 years ago

7

A bus travel 210 miles in three. Five hours assumed the distance travel is directly proportional to the time traveled right and

solve direct variation equation to find how far the bus will travel in six hours

1 answer:

yarga [219]3 years ago

7 0

God loves all of you, remember that oK. Thts really all u need in life <333 and 210m is

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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:

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

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3 years ago

The value x = 3 is a solution to each of the following except which?

ololo11 [35]

Answer:

$x \: is \: not \: a \: solution \: in \to \: 5x + 4 < 19. \\$

Step-by-step explanation:

$\: if \: x = 3 \: then \\ 4x-1=2x +5 \: is \: same \: as \to \: \\ 4(3) - 1 = 2(3) + 5\\ 12 - 1 = 6 + 5 \\ 11 =11 \to \: \\ \boxed{ x \: is \:a \: solution \: here} \\ \\ if \: x = 3 \\ then \to \\ 5x + 4 < 19\: is \: same \:as \to \\ 5(3) + 4 < 19 \\ 15 + 4 < 19 \\ 19 \: is \: equal \: to \: 19 : \\ 19 \: can \: not \: be \: less \: than \: 19 \\ \boxed{ henc e\: x \: is \: not \: a \: solution \: here}$

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3 years ago

What is the answer to number 23 ?

Vera_Pavlovna [14]

Answer:

C

Step-by-step explanation:

Use πr²

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3 years ago

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The equation that describes the relationship is y =

Rainbow [258]

I just need point tbh so thanks !!

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3 years ago

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Estimate the unit rate.<br> 283 students in 11 classes

Archy [21]

Answer:

So estimate.

Really depends on what needs to be estimated, the answer or the overall equation.
So doing the real one first.
283/11 = 25 8/11

You can’t have “part” of a student so rounding this you can go up since 8 is more then half of 11

So 26 students

If you were to round the overall equation then you’d get
280 / 10 which is 28 students per class

So two options.

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2 years ago