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siniylev [52]
3 years ago
11

Find the distance between (3,-2) and (2,-4)

Mathematics
1 answer:
Andreas93 [3]3 years ago
8 0

Answer:

7

Step-by-step explanation:

look at a graph. Hope this helps.

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
schepotkina [342]
I need you to be more specific
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3 years ago
Read 2 more answers
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.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

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

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 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 <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, 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 <em>t</em> = 2.5 the value of <em>λ</em>, 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
Karen is buying pens and pencils for the new school year. She wants to have no more than 25 writing utensils in all. She also wa
bezimeni [28]
For this case, the first thing we must do is define variables.
 We have then:
 x: number of pens
 y: number of pencils
 We now write the system of inequations:
 x + y  \leq  25&#10;&#10;y  \geq  (x - 3) ^ 2&#10;
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 Note: see attached image.

3 0
4 years ago
The length of the rectangle is 9 inches. The area is 72 square inches. What is the width of the rectangle?
mote1985 [20]

Answer:

8

Step-by-step explanation:

Formula for area = Length times width

72 divided by 9 = 8

Double check: 9(length) x 8 (width) = 72 (area).

Width = 8 inches

Hope this helped! If it did, please give brainliest! :)

4 0
3 years ago
Please break this up:<br> a^2-b^2-4b-4
hoa [83]

(a + b) (a-b ) - 4b - 4

<u>Explanation:</u>

Given:

a² - b² - 4b - 4

(a + b) (a-b ) - 4b - 4

This equation cannot be solved further.

Thus, the equation is (a + b) (a-b ) - 4b - 4

6 0
3 years ago
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