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Licemer1 [7]
3 years ago
10

Simplify x + (7+14x)​

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
8 0

Answer:

15x + 7

Step-by-step explanation:

x + (7 + 14x)

x + 7 + 14x

15x + 7

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From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. Wha
Ede4ka [16]

Answer:

D. 2/7

Step-by-step explanation:

The probability is calculate as a division between the number of ways in which Andrew will be selected and karen will not and the total ways in which they can select 4 people from a group of 8 volunteers.

The number of ways in which they can select 4 people from a group of 8 volunteers can be calculated using:

nCk=\frac{n!}{k!(n-k)!}

Where nCk give as the number of ways in which we can form groups of k elements from a group of n elements.

So, if we replace n by 8 and k by 4, we get:

8C4=\frac{8!}{4!(8-4)!}=70

Then, there are 70 ways to select 4 people from a group of 8 volunteers.

At the same way, the number of ways in which Andrew will be selected and karen will not can be calculate replacing n by 6 and k by 3 in the same equation as:

6C3=\frac{6!}{3!(6-3)!}=20

We replace n by 6 because from the 8 volunteers, Andrew is going to be selected and Karen will not, so, the people that affect the result are just the 6 remaining volunteers. Also, k is equal to 3 because Andrew is already selected and we just need to select 3 more people.

Finally, the probability is:

P=\frac{20}{70} = \frac{2}{7}

4 0
3 years ago
Can you help me pls 2
uysha [10]
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5 0
3 years ago
Pls help; tysmmmmmmn <3
Alexxx [7]
8 is the answer, yw have a nice day:)
3 0
3 years ago
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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
X = ? ? ? ? ? ? ? ? ?
natita [175]

Answer:

7

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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