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3 years ago

Could anyone help, please?

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

4 0

First, find the probability of each event:

1) the probability that the spinner will land on a 7.

Since the spinner is split 4 equal sections and there is only 1 sector with 7, we can say the probability of getting a 7 is 1/4 as there is only 1 of 7 out of the total of 4 sections.

and

2) the probability that the spinner will land on B.

Since the spinner is split into 3 equal sections, and there is only 1 sector for B, we can say the probability of getting B is 1/3.

To find the probability of 2 events, we need to multiply the two probabilities.

1/4*1/3 = 1/12

So the answer is 1/12.

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Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o

timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

Step-by-step explanation:

Let the random variable X = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, λt = 8 per hour.

(a)

For t = 1 the average number of aircraft arrival is:

$\lambda t=8\times 1=8$

The probability distribution of a Poisson distribution is:

$P(X=x)=\frac{e^{-8}(8)^{x}}{x!}$

Compute the value of P (X = 6) as follows:

$P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214$

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

$P(X\geq 6)=1-P(X$

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

$P(X\geq 10)=1-P(X$

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For t = 90 minutes = 1.5 hour, the value of λ, the average number of aircraft arrival is:

$\lambda t=8\times 1.5=12$

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

$SD=\sqrt{\lambda t}=\sqrt{12}=3.464$

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For t = 2.5 the value of λ, the average number of aircraft arrival is:

$\lambda t=8\times 2.5=20$

Compute the value of P (X ≥ 20) as follows:

$P(X\geq 20)=1-P(X$

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

$P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108$

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0

3 years ago

Help me with this one please

lesya [120]

For me, it’s easiest when i distribute the negative sign if i need to and then reorder to put the like terms together. and then solve.
(also sorry if it’s a little confusing with all the parentheses, i use them because it helps me organize everything)

21. (4x-9y) + (6x+10) + (8y-4)
= 4x + 6x - 9y + 8y + 10 - 4
= 10x - y + 6
—> D

22. (6x+9y-15) + (2x-9y+8)
= 6x + 2x + 9y - 9y - 15 + 8 (the 9y - 9y = 0, so you can leave it out of the final equation)
= 8x - 7
—> D

23. (9x^2-8x+3) - (5x^2-6x+4)
= 9x^2 - 8x + 3 - 5x^2 -(-6x) - 4
= 9x^2 - 5x^2 - 8x + 6x + 3 - 4 (remember that two - signs next to each other make a + sign)
= 4x^2 - 2x - 1
—> A

24. (9x^3-7x+8) - (5x^2+7x-10)
= 9x^3 - 7x + 8 - 5x^2 - 7x -(-10)
= 9x^3 - 5x^2 - 7x - 7x + 8 + 10
= 9x^3 - 5x^2 - 14x + 18
—> D

25. (6x+14y) - ((7x+5y) + (x-8y))
= (6x+14y) - (7x + x + 5y - 8y)
= (6x+14y) - (8x-3y)
= 6x + 14y - 8x -(-3y)
= 6x - 8x + 14y + 3y
= -2x + 17y
—> B

5 0

2 years ago

Prove the following statements: (a) For every integer x, if x is even, then for every integer y, xy is even. (b) For every integ

Murrr4er [49]

Answer and Step-by-step explanation:

(a) Given that x and y is even, we want to prove that xy is also even.

For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.

(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:

Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.

(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.

3 0

3 years ago

What is 2(2x+1) equivalent to

Nostrana [21]

2(2x+1)= (2 x 2x) + (2 x 1) = 4x + 2

6 0

3 years ago

Read 2 more answers

a company makes these biscuits at a cost of $1.35 per packet. these biscuits are sold for $1.89 per packet. calculate the percen

nydimaria [60]

Answer:

They Make 50 cents profit a pack

Step-by-step explanation:

1.89-1.35

8 0

2 years ago