Answer:
The 95% confidence interval for the average number of units that students in their college are enrolled in is between 11.7 and 12.5 units.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 45 - 1 = 44
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 12.1 - 0.4 = 11.7 units
The upper end of the interval is the sample mean added to M. So it is 12.1 + 0.4 = 12.5 units
The 95% confidence interval for the average number of units that students in their college are enrolled in is between 11.7 and 12.5 units.
Answer:
Step-by-step explanation:
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Answer:
11x 5 =55-22=33
Step by step explanation
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The probability any one system works is 0.99
So the probability of any one system failing is 1-0.99 = 0.01, so basically a 1% chance of failure for any one system
Multiply out the value 0.01 with itself four times
0.01*0.01*0.01*0.01 = 0.000 000 01
I'm using spaces to make the number more readable
So the probability of all four systems failing is 0.00000001
Subtract this value from 1 to get
1 - 0.00000001 = 0.99999999
The answer is 0.99999999 which is what we'd expect. The probability of at least one of the systems working is very very close to 1 (aka 100%)
900 x 8= 7200 is your answer