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Luden [163]
3 years ago
11

Plz help me I really need it

Mathematics
2 answers:
Furkat [3]3 years ago
6 0

Answer:

Step-by-step explanation: 2= 41/7     3=  37/9     4=  62/7

Zigmanuir [339]3 years ago
4 0

Step-by-step explanation:

There is a wounderful little trick for these. Multiply the whole number (the biggest one) by the denominator (the lower small number), then add the numerator (the small top number). That answer is your numerator for your improper fraction.

Then, using the same denominator as before, write out your fraction as an improper one.

Have a great day, and feel free to comment with questions!

:)

You might be interested in
12. (2v^3 - v + 8) + (-v^3 + v - 3)<br><br> 14. (4h^3 + 3h + 1) - (-5h^3 + 6h - 2)
Lana71 [14]
12. On addition, you can just combine like terms.
2v^3+(-v^3)=v^3
-v+v cancels each other out
8+(-3)=5
So you have v^3+5

14 On subtraction, you have to remember to distribute the negative sign so after you do that you have:
4h^3+3h+1+5h^3-6h+2
Then you can combine like terms
4h^3+5h^3=9h^3
3h-6h=-3h
1+2=3
So you end up with:
9h^3-3h+3

Hope that helps and feel free to ask any questions.
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
Numbers to the nearest ten <br> 48393<br> 1549<br> 37,519<br> 65,810<br> 2175
katen-ka-za [31]

Answer:

48390,"37,520","65,810",2170 or 2180 (both will work)

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3 years ago
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Doss [256]
The type of stress that causes rocks to fold is Compression.
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