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s2008m [1.1K]
3 years ago
14

Can someone help me with this I’ll mark as brainliest.

Mathematics
2 answers:
nikdorinn [45]3 years ago
7 0
1 = 10
2 = 12
3 = 14
4 = 16
5 = 18
6 = 20


Δweight
------------- =
Δ cost

2-1
------- =
12-10

1
—
2
vodka [1.7K]3 years ago
3 0

Answer:

The W/C is divided by the numbers on top to the number on the bottom

may i have brainlist??

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Answer: 12+20 = 12 + 20 = 4(3+5)

Step-by-step explanation:

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Simplify open parentheses x to the 2 third power close parentheses to the 4 fifths power.
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X to the 8/15 power. When you have a power raised to another power, you multiply the exponents.
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The base of a 15 foot long guy wire is located 5 ft from the telephone pole it is anchoring. how high up the pole does the guy w
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6 0
3 years ago
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
Correct answer gets brainliest and 5 stars
Serggg [28]

Answer:

A is false

Step-by-step explanation:

notice how in both diagrams, there are 4 triangles with lengths a, b, c, but in each diagram the triangles are just rotated and positioned differently. if there are the same amount of triangles in both diagrams, and the have equal lengths then the area of the remaining figure, should both match up.

basically the area of the big square in Step 2, is equal to both of the shaded squares, combined area, in Step 1.

3 0
3 years ago
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