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2 years ago

Please i need help with this question

Mathematics

1 answer:

topjm [15]2 years ago

4 0

Answer:

Ounces: 10, 15

Distance: 4, 6

Step-by-step explanation:

The ratio is constantly 5:2, which is equal to 10:4 and 15:6

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Please help on this one?

ki77a [65]

m = $\frac{1}{2}$ , (x₁ , y₁) = (-2, -4)

y - y₁ = m(x - x₁)

y -(-4) = $\frac{1}{2}$ (x - (-2))

y + 4 = $\frac{1}{2}$ (x +2)

y + 4 = $\frac{1}{2}x$ + 1

y = $\frac{1}{2}x$ - 3

Answer: A

5 0

3 years ago

PLEASE HELP DUE IN 15 MINS!!!!!!!!!!!!

Alborosie

$∠3 =∠4$

$∠S +∠T=90°$

$∠X+∠Y=180°$

$∠2+∠4=90°$

$∠J+∠K = 180°$

4 0

2 years ago

You owe your mom 10$. You pay her back 7$. Write an expression that can be used to represent this situation.

mezya [45]

Answer:

10$-7$=3$

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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; $P (X(s) \leq 2 ) = P(X(s) = 0 ) + P(X(s) = 1) + P(X(s) = 2)$

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

Plz hurry more questions on the way

Scrat [10]

The answer is A, I’m pretty sure

7 0

2 years ago

Read 2 more answers