Again pls help me :((

A mosaic at the art museum is divided into 6 sections that each contain about the same number of tiles. There are 2,889 tiles to

AlexFokin [52]

Answer

Find out the how many tiles are in each section .

To prove

Let us assume that the number of tiles are in section are be x.

As given

A mosaic at the art museum is divided into 6 sections that each contain about the same number of tiles.

There are 2,889 tiles total in the mosiac .

Than the equation becomes

6 × x = 2889

$x = \frac{2889}{6}$

x = 481.5

Therefore the total number of tiles in each sections are 481.5 .

7 0

3 years ago

On a camping trip with friends, you read about mosquito activity after dark. The number of mosquitos in your campground can be m

Dmitrij [34]

Answer:

The answer is 54

Step-by-step explanation:

If you follow PEMDAS, and x = 3, then you get 54. As an example, x4 is 12. You change X to 3, then multiply (3)4=12. If you follow the rest of the equation, you get 54.

6 0

3 years ago

Read 2 more answers

A person on a moving sidewalk travels 6 feet in 3 seconds. The moving sidewalk has a length of 80 feet. How long will it take to

tekilochka [14]

Answer:

40.5 seconds

Step-by-step explanation:

5 0

2 years ago

What is the area of a rectangle with a length of 88 inches and width of 56 inches?

Alecsey [184]

Answer:

C. 4928 in2

Step-by-step explanation:

88 X 56 = 4928 in2

4 0

3 years ago

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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.

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

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