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Drupady [299]
3 years ago
13

How do u write 5.6 as a fraction in simplest form

Mathematics
2 answers:
krok68 [10]3 years ago
5 0
The first answer to your question is 5 6/10 because you pronounce the decimal as six tenths. There is a simpler form of 5 6/10, it is 5 3/5. This is because you divide six by 2 and get 3 and divide 10 by two and get 5. 
ikadub [295]3 years ago
4 0
The answer to your question is 5 6/10

Your 5 is a whole number and your 6 is past the decimal point and is in the tenth column therefore that's why you get your 6/10.

Another Example: What is 2.5 as a fraction?
Answer:
2 5/10
You might be interested in
How many times does 28 go in to 45
taurus [48]

Answer:

<em>About 1.61 times </em>

Step-by-step explanation:

45/28≈1.61 times

So the answer is 1.61 Times

<em>Hope I Helped</em>

4 0
3 years ago
Py+7=6y+q solve for y
andreyandreev [35.5K]

We know that we are solving for y.


This is a step by step procedure to get the value of y.


First: Move all terms to the left side and set equal to zero.


Second: Then set each factor equal to zero.

 

The application is:


Given: py+7=6y+q 


-6y -7 -6y -7 = 0


(p-6)y = q-7



divide both sides by p-6
y=(q-7)/(p-6)


Answer is y = (q – 7) / (p – 6) 

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
Help solve 10×-6y=7​
spayn [35]

Answer:

y = -7/60

There you go

7 0
3 years ago
Ed invests $912.08 in stocks and $487.78 in bonds. Then, Ed moves $378.29 from stocks to bonds. How much money does Ed now have
allochka39001 [22]

Answer:

I believe it would be $866.07

Step-by-step explanation:

All you need to do is 487.78+378.29 right?

I could be wrong-

4 0
3 years ago
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