PLEASE HELP I REALLY NEED IT!! ILL MARK BRAINLIEST!! The question is in the picture below!!!
2 answers:
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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.
Answer:
Part A : 2y( x³ + 9x - 5x² - 45 ), Part B : 2y( x - 5 )( x² + 9 )
Step-by-step explanation:
Part A : Let's break every term down here to their " prime factors ", and see what is common among them,
2x³y + 18xy − 10x²y − 90y -
2x³y = 2 x³
y,
18xy = 2 3
3
x
y,
− 10x²y = 2 - 5
x²
y, - so as you can see for this example I purposely broke down - 10 into 2 and - 5. I could have placed the negative on the 2, but as that value was must likely common among all the terms, I decided to place it on the 5. The same goes for " − 90y. " I placed the negative there on the 5 once more.
− 90y = 2 - 5
3
3
y
The terms common among each term are 2 and y. Therefore, the GCF ( greatest common factor ) is 2x. Let's now factor the expression using this value.
2y( x³ + 9x - 5x² - 45 )
Part B : Let's simply factor this entire expression. Of course starting with the " factored " expression : 2y( x³ + 9x - 5x² - 45 ),
- Factor out "
" by grouping,
- Factor 9 from 9x - 45 and x² from x³ - 5x²,
- Factor out common term x - 5,
- And our solution is thus 2y( x - 5 )( x² + 9 )