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2 years ago

Find 35% more than 90

1 answer:

4 0

90 x 135% = 90 x 1.35 = 121.5 (notice that instead of finding 35% and adding it on, I included the 90 by mutliplying by 1.xx) - in this case xx - 35 since it was a 35% increase.

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I need help plssssssssssssssss

liubo4ka [24]

Answer:

wha do you need help on i can help if you are in 7th grade or lower so ya

Step-by-step explanation:

5 0

3 years ago

List these from least to greatest π2, 12, √80

Evgesh-ka [11]

Answer:

π2,√80,12

Step-by-step explanation:

π2= 6.28

√80= about 8

12=12

Hope I could help!

5 0

2 years ago

Below are several points on the graph of a polynomial. Which of the following statements are true about the graph of this polyno

FinnZ [79.3K]

Notice that we need to look for the intervals of x and also for the intervals of y.<span> We can observe that t</span>here is a minimum between x = 0 and x = 1 because the graph is descending and then ascending and also the graph has a maximum value between y = -3 and y = -1.

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2 years ago

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

Please answer this Yall

lora16 [44]

I believe the answer would be b^4.

Explanation:
Use Quotient Rule: x^a/x^b=x^a-b
b^10-6

Then Simplify 10-6 to 4
b^4

5 0

2 years ago