A square-based tent covers an area of 81 square feet. What is the length of one side of the tent?
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Answer: The greatest number of rows Li Na can plant is 9.
Step-by-step explanation:
Given: Li Na is going to plant 63 tomato plants and 81 rhubarb plants.
Li Na would like to plant the plants in rows where each row has the same number of tomato plants and each row has the same number of rhubarb plants.
To find the greatest number of rows Li Na can plant, we need to find the GCF of 63 and 81.
Since ,
Clearly, GCF(63,81)=9
Therefore, the greatest number of rows Li Na can plant is 9.
Answer:
a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes
c) 0.2971 = 29.71% 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:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean of 8.9 minutes and a standard deviation of 2.9 minutes.
This means that
Sample of 37:
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?
320/37 = 8.64865
Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So
By the Central Limit Theorem
has a p-value of 0.2981
0.2981 = 29.81% 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?
275/37 = 7.4324
Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So
has a p-value of 0.001
1 - 0.001 = 0.999
0.999 = 99.9% 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?
Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So
0.2981 - 0.0010 = 0.2971
0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes