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Novosadov [1.4K]
3 years ago
6

If 3a-2b= 8 and a+3b=7 what is the value of 4a+b?

Mathematics
1 answer:
Fittoniya [83]3 years ago
6 0
Step 1: Find the value of one variable in terms of the other.

a+3b=7\\a=7-3b

Step 2: Substitute the value you just found for this variable in the other equation.

3a-2b=8\\3(7-3b)-2b=8\\21-9b-2b=8\\21-11b=8\\21=8+11b\\13=11b\\\frac{13}{11}=b\\ b=1\frac{2}{11}

Step 3: Use your new value for the second variable to find the first.

a+3b=7\\a+3(1\frac{2}{11})=7\\a+3\frac{6}{11}=7\\\\a=3\frac{5}{11}

Now that we know the values for a and b we can find the value of 4a+b.

4a+b\\4(3\frac{5}{11})+1\frac{2}{11}\\12+\frac{20}{11}+1\frac{2}{11}\\\\12+1\frac{9}{11}+1\frac{2}{11}\\\\\boxed{15}
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On Saturday, many tickets for a hockey game were sold. Children and adult tickets were sold in the ratio of 4:7. There are 88 ti
Illusion [34]

Answer:

49 tickets

Step-by-step explanation:

THis is pretty straight forward

So we have the ratio 4:7

We know there were 88 tickets in total but we don’t know the exact amount of adult and kids

So we can Add up the ratio

4+7=11

Now we can divide 88 by 11

you’ll get 7

So by doing this we know there are 7 groups of the ratio for child to adults 4:7

So we just multiply the 7 to each

4*7

7*7

you’ll get 28:49

Now we know there are 28 kid tickets and 49 adult tickets

We know this is correct becuase if you add up the ratio, it’ll be 88

Which is the same as the amount fo people at the hockey game

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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.

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