This is something you'll need a T table for, or a calculator that can compute critical T values. Either way, we have n = 10 as our sample size, so df = n-1 = 10-1 = 9 is the degrees of freedom.
If you use a table, look at the row that starts with df = 9. Then look at the column that is labeled "95% confidence"
I show an example below of what I mean.
In that diagram, the row and column mentioned intersect at 2.262 (which is approximate). This value then rounds to 2.26
<h3>
Answer: 2.26</h3>
Answer: The average number of hours she danced per day is 1.9 hours (rounded to the nearest tenth)
Step-by-step explanation: We start by calculating how many hours she danced all together which can be derived as follows;
Summation = 3 +2 +2 + 1 + 1.5 + 2 = 11.5
The number of days she danced which is the observed data is 6 days (she did not dance at all on Wednesday).
The average (or mean) hours she danced each day can be calculated as
Average = ∑x ÷ x
Where ∑x is the summation of all data and x is number of observed data
Average = (3+2+2+1+1.5+2) ÷ 6
Average = 11.5 ÷ 6
Average = 1.9166
Approximately, average hours danced is 1.9 hours (to the nearest tenth)
Answer:
8r^4
Step-by-step explanation:

The expression 5h+8 represents Harriet's gross pay each day.
Step-by-step explanation:
Given,
Gross pay of Harriet = 35h + 56
This expression represent Harriet's pay for 7 day.
Harriet's gross pay per day = 
Harriet's gross pay per day = 
Harriet's gross pay per day = 
Harriet's gross pay per day = 5h+8
The expression 5h+8 represents Harriet's gross pay each day.
Keywords: division, addition
Learn more about division at:
#LearnwithBrainly
Answer:
The proportion of student heights that are between 94.5 and 115.5 is 86.64%
Step-by-step explanation:
We have a mean
and a standard deviation
. For a value x we compute the z-score as
, so, for x = 94.5 the z-score is (94.5-105)/7 = -1.5, and for x = 115.5 the z-score is (115.5-105)/7 = 1.5. We are looking for P(-1.5 < z < 1.5) = P(z < 1.5) - P(z < -1.5) = 0.9332 - 0.0668 = 0.8664. Therefore, the proportion of student heights that are between 94.5 and 115.5 is 86.64%