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sveticcg [70]
3 years ago
6

Janice aud Thomas are roller blading around Sampson Pond It takes them 4 minutes to complete a lap. Write and solve an equation

to how many laps they can complete in one hour if they continue roller at this pace. Brainly
Mathematics
1 answer:
grigory [225]3 years ago
8 0

The total number of laps taken by Janice and Thomas in 1 hour is 15

<h3><u>Solution:</u></h3>

We have been given the following data, Janice and Thomas are roller blading around Sampson Pond it takes them 4 minutes to complete a lap.  

We need to write and solve an equation to how many laps they can complete in one hour if they continue roller at this pace.

We know that, 1 lap = 4mins

\text {The total number of laps }=\frac{\text { total time spent roller blading }}{\text { time taken to complete each lap }}

Plugging in values, we get

\text { Total number of laps }=\frac{1 \text { hour }}{4 \text { mins }}

We know that 1 hour = 60 minutes

=\frac{60 \mathrm{mins}}{4 \mathrm{mins}}=15

Hence, the total number of laps taken by Janice and Thomas in 1 hour is 15

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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
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