What is the GCF for 45 of 60 plz right it out
1 answer:
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Answer:
At a .05 level of significance, it can be concluded that the mean of the population is significantly more than 3 minutes.
Step-by-step explanation:
We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.
At the null hypothesis, we test if the mean is of at most 3 minutes, that is:
At the alternative hypothesis, we test if the mean is of more than 3 minutes, that is:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
3 is tested at the null hypothesis:
This means that
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.
This means that
Value of the test statistic:
P-value of the test and decision:
The p-value of the test is the probability of finding a sample mean above 3.1, which is 1 subtracted by the p-value of z = 2.
Looking at the z-table, z = 2 has a p-value of 0.9772.
1 - 0.9772 = 0.0228
The p-value of the test is of 0.0228 < 0.05, meaning that the is significant evidence to conclude that the mean of the population is significantly more than 3 minutes.
Answer:
There should be at most 24 lucky numbers in the third bag.
Step-by-step explanation:
Initially, there are 200 numbers. Two bags with 100 each. There are 31+18 = 49 lucky numbers. So there is a 49/200 = 0.245 probability that a randomly selected number from a random bag is the lucky number.
Now with 300 numbers, we want this probability to be lower than 24.5%. So we should solve the following rule of three:
200 - 49
300 - x
With the third bag, the probability will be the same if 73.5-49 = 24.5 lucky numbers are added. So there should be at most 24 lucky numbers in the third bag.