Help me solve please
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Answer:
70 students need to be surveyed.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
That is z with a pvalue of , so Z = 1.96.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
Assuming that the standard deviation of the above sample is a reasonable estimate of the standard deviation of sleep time at your university.
This means that
How many students do you need to survey to estimate the mean sleep time of students at your university with 95% confidence and a margin of error of 0.5 hours?
n students need to be surveyed, and n is found when M = 0.5. So
Multiplying by 2 on both sides:
Rounding up
70 students need to be surveyed.
2c^5 + 44c^4 + 242c^3
Take out common factors.
= 2c^3(c^2 + 22c + 121)
( Refer for attachment for next step )
= 2c^3(c + 11)(c + 11)
= 2c^3(c + 11)^2
Take out common factors.
= 2c^3(c^2 + 22c + 121)
( Refer for attachment for next step )
= 2c^3(c + 11)(c + 11)
= 2c^3(c + 11)^2