<span>They need to drink 3620 ml combined to meet the recommended amounts for the day.</span>
Answer:
The 90% confidence interval for the mean time required by all college graduates is between 5.36 years and 5.44 years.
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 
.
So it is z with a pvalue of 
, so 
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.
The lower end of the interval is the sample mean subtracted by M. So it is 5.4 - 0.04 = 5.36 years.
The upper end of the interval is the sample mean added to M. So it is 5.4 + 0.04 = 5.44 years.
The 90% confidence interval for the mean time required by all college graduates is between 5.36 years and 5.44 years.
Well you are given the equation so let's plug in for kaylib and see how many miles she can see
distance = sqrt [(3 * height) / 2]
d = sqrt [(3 *48) / 2]
d = sqrt (144 / 2)
d = sqrt (72)
d = sqrt (3 * 3 * 2 * 2 * 2)
d = 6 * sqrt (2)
You you did not list Addisons height but I will say she is at x feet above sea level. we plug in x for height:
d = sqrt [(3x) / 2]
It it says how much farther for Addison which means she can see farther. to find difference we just subtract kaylibs distance from Addison. so:
sqrt [(3x) / 2] - 6 * sqrt (2)
plug in your x and use a calculator to get a decimal approximation
Answer:
any score that lies between 88.8 and 97.2 is within one std. dev. of the mean
Step-by-step explanation:
One std. dev. above the mean would be 93 + 4.2, or 97.2. One std. dev. below the mean would be 93 - 4.2, or 88.8.
So: any score that lies between 88.8 and 97.2 is within one std. dev. of the mean.
2 for the whole number and 2/1 for the fraction
