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Elanso [62]
3 years ago
11

What is 20,386.145 rounded to nearest tenth

Mathematics
1 answer:
const2013 [10]3 years ago
8 0
You would first start by finding the tenths place. When you find it, The number behind it., if it is 5+ you make the tenths place higher. If it is lower than 5, then you make the number lower.
You might be interested in
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
Aaron has a triathlon. Sunday he bikes 12 5/8 miles and swims 5 2/3 miles. Monday, he runs 6 3/8 miles. How many total miles Aar
Alla [95]
Let’s put these into one fraction each:

Biking:
= (12*5)/8 miles
= 60/8 miles
This can be reduced to,
= 15/2 miles

Swimming:
= (5*2)/3 miles
= 10/3 miles

Running:
= (6*3)/8 miles
= 18/8 miles
This can be reduced to,
= 9/4

Now we add all the distances up:
= (15/2) + (10/3) + (9/4)

Now we just can add the numerators and the denominators. We must find the lowest common factor (LCF) for our 3 denominators (2,3,4). Our LCF turns out to be 12. *Remember what we do to the denominator we must do to the numerator. So:
= (15/2) + (10/3) + (9/4)
= [(15*6)/12] + [(10*4)/12] + [(9*3)/12]
= (90/12) + (40/12) + (27/12)
= (157/12)
Therefore Aaron Ran 157/12 miles or 13.08 miles.

Hope this helps!
4 0
3 years ago
Write 4.14 in a fractional notation ?
tatiyna
If it's in a fraction you want it it's 207/50 irregular
And 4 7/50 as a mixed fraction
3 0
3 years ago
Exactly how many plants contain points J, K and N
Evgesh-ka [11]
The contain points j,k and N is = X plants
7 0
3 years ago
If Fran were to paint her living room alone, it would take 6 hours. Her sister Denise
GalinKa [24]

Answer:

It will take Fran and Denise 3\frac{3}{7} hours to paint the living room.

Step-by-step explanation:

Given that:

Time taken by Fran to paint living house = 6 hours

Time taken by Denise = 8 hours

Let,

x be the time taken by both of them.

\frac{1}{Fran's\ share}+\frac{1}{Denise's\ share}=\frac{1}{x}\\\\\frac{1}{6}+\frac{1}{8}=\frac{1}{x}\\\\\frac{4+3}{24}=\frac{1}{x}\\\\\frac{7}{24}=\frac{1}{x}\\\\x = \frac{24}{7}

x = 3\frac{3}{7} hours

Hence,

It will take Fran and Denise 3\frac{3}{7} hours to paint the living room.

4 0
3 years ago
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