Serena wants to solve the following system of equations in the most efficient way.
2 answers:
Answer:
Serina should have solved for x in the second equation because it has a coefficient of 1.
Step-by-step explanation:
we have
----> equation A
---> equation B
we know that
To solve the system of equations in the most efficient way, solve for x equation B and then substitute the value of x in equation A
so
----> equation C
substitute equation C in equation A
solve for y
Find the value of x
substitute the value of y in the equation C
The solution is (3,4)
therefore
Serina should have solved for x in the second equation because it has a coefficient of 1.
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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 and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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 .
(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:
Total time of 320 minutes for 37 jets, so
This probability is the pvalue of Z when . So
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
This probability is subtracted by the pvalue of Z when
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
Total time of 275 minutes for 37 jets, so
This probability is the pvalue of Z when subtracted by the pvalue of Z when
.
So:
From a), we have that for , we have
, that has a pvalue of 0.7422.
From b), we have that for , we have
, 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.