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Delvig [45]
3 years ago
15

How many inches are in 508 centimeters? 1 in. = 2.54 cm Enter your answer in the box. [ ] in.

Mathematics
2 answers:
lora16 [44]3 years ago
5 0
Set a proportion
1/2.54=x/508..___Cross multiply.
2.54x=508
2.54/2.54x=508/2.54
x=200
Therefore there are 200 inches in 508 centimeters
leva [86]3 years ago
4 0

Answer: 200 i reviewd the test.

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Please select the word from the list that best fits the definition
sp2606 [1]

Answer:

Epics

Step-by-step explanation:

Two dictionary definitions are

1. a long poem, typically one derived from ancient oral tradition, narrating the deeds and adventures of heroic or legendary figures or the history of a nation.

2. a long film, book, or other work portraying heroic deeds and adventures or covering an extended period of time.

7 0
3 years ago
Read 2 more answers
Mark is in training for a bicycle race. He needs to ride 50 kilometers a week as part of his training. He rode 18.23 kilometers
marishachu [46]

Answer:

17.83 km

Step-by-step explanation:

7 0
3 years ago
Quadrilateral ABCD is inscribed. The measure of ∠A=67°. What is the measure of ∠C?
WARRIOR [948]

Answer:

angle C =113 degree

Step-by-step explanation:

angle A + angle C =180 degree

angle C = 180 degree - angle A  

angle C = 180 degree - 67 degree

=113 degree

7 0
3 years ago
What is the area of the triangle?<br> 9 4 12
LUCKY_DIMON [66]

Answer:

  13.636

Step-by-step explanation:

The area can be found using Heron's formula:

  A = √(s(s-a)(s-b)(s-c))

where s=(a+b+c)/2.

For the given triangle, ...

  a=9, b=4, c=12, s=(9+4+12)/2 = 12.5

  A = √(12.5(3.5)(8.5)(0.5)) = √185.9375 ≈ 13.636

The area of the triangle is about 13.636 square units.

7 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
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