750% was gained
.............
Answer:
The 95% confidence interval for the mean time spent studying for the intro statistics final exam by all students is between 6.05 hours and 9.84 hours.
Step-by-step explanation:
We have the standard deviation for the sample, which meas that the t-distribution is used 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 = 18 - 1 = 17
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.11
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 7.74 - 1.69 = 6.05 hours
The upper end of the interval is the sample mean added to M. So it is 7.74 + 1.69 = 9.84 hours.
The 95% confidence interval for the mean time spent studying for the intro statistics final exam by all students is between 6.05 hours and 9.84 hours.
I'm not sure if this is correct,
but, I got:
130cm x 100cm ?
13,000cm ?
750 divided by 600 x 100= 125%
1 oz= 29.5735296 ml
Convert ounces to milliliters by multiplying milliliters in 1 ounce by 0.4 ounces.
=0.4 oz * 29.5735296 ml/oz
=11.829 ml
Divide 60 ml by the portion size (11.829 ml)
=60 ÷ 11.829
= 5.07 times
ANSWER: 5.07 times
Hope this helps! :)