Help me pleaseeee :) for brainliest answer

Find the area of the regular polygon. Round to the nearest tenth.

JulsSmile [24]

Answer:

$110.9\ ft^{2}$

Step-by-step explanation:

we know that

If the triangle of the figure is a regular polygon

then

Is an equilateral triangle

Remember that

The equilateral triangle has three equal sides and three equal interior angles

The measure of each interior angle is 60 degrees

To find the area of the equilateral triangle apply the law of sines

$A=\frac{1}{2} b^{2}sin(60\°)$

we have that

$b=16\ ft$

substitute

$A=\frac{1}{2} (16^{2})sin(60\°)=110.9\ ft^{2}$

8 0

3 years ago

Help find the value of x

Nookie1986 [14]

Answer:

x = 82

Step-by-step explanation:

The bottom angle is the same as the angle next to x + 6

A straight angle is 180 degrees.

You add the given angles: (x + 6) + (x + 10) = 2x + 16

180 = 2x + 16

- 16 -16

164 = 2x

x = 82

8 0

2 years ago

HELP PLEASE!!!!!! Anaya bought a sweater that was on sale. The sweater originally cost $34.50 but was purchased for $27.60. What

Elan Coil [88]

Answer: 80%. Hope this helps, please consider making me Brainliest.

Step-by-step explanation:

To find the percentage, divide the sale cost by the original cost:

27.60/34.50

Let's multiply both sides by 10 to make the operation easier:

27.6/34.5 ( I eliminated the zero's because they kind of have no use) -->

276/345, now solve:

276/345 = 0.8

0.8 = 80%

The percentage is 80%.

5 0

1 year ago

Please need some help on this

vredina [299]

11111111111111117373646328299292922

8 0

3 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3

In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that $\mu = 8.3, \sigma = 3.3$ .

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

$s = \frac{3.3}{\sqrt{37}} = 0.5425$

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

This probability is the pvalue of Z when $X = 8.65$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8.65 - 8.3}{0.5425}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is subtracted by the pvalue of Z when $X = 7.43$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7.43 - 8.3}{0.5425}$

$Z = -1.60$

$Z = -1.60$ has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is the pvalue of Z when $X = 8.65$ subtracted by the pvalue of Z when $X = 7.43$ .

So:

From a), we have that for $X = 8.65$ , we have $Z = 0.65$ , that has a pvalue of 0.7422.

From b), we have that for $X = 7.43$ , we have $Z = -1.60$ , that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0

3 years ago