1. 4.625
2. 0.708
3. 8.75
4. -3.16
Hope it helps~
In Q.2 is it 22/3-11/8 or 2/3-11/8?
In Q.4 is it 41/3-21/6 or 1/3-21/6?
Yes. By all means yes. Its very very possible. In fact it can go in there ten times over.
Answer:
1607
Step-by-step explanation:
The mean (or average) of a data set is the sum of all of the data divided by the number of data in the set. In this case, that would be:
(1606 + 1607 + 1606 + 1607 + 1607 + 1609) / 6
= 9642 / 6
= 1607
Answer:
For number 13, I think it is rational, integer, and whole number
You should use a T distribution to find the critical T value based on the level of confidence. The confidence level is often given to you directly. If not, then look for the significance level alpha and compute C = 1-alpha to get the confidence level. For instance, alpha = 0.05 means C = 1-0.05 = 0.95 = 95% confidence
Use either a table or a calculator to find the critical T value. When you find the critical value, assign it to the variable t.
Next, you'll compute the differences of each pair of values. Form a new column to keep everything organized. Sum everything in this new column to get the sum of the differences, which then you'll divide that by the sample size n to get the mean of the differences. Call this dbar (combination of d and xbar)
After that, you'll need the standard deviation of the differences. I recommend using a calculator to quickly find this. A spreadsheet program is also handy as well. Let sd be the standard deviation of the differences
The confidence interval is in the form (L, U)
L = lower bound
L = dbar - t*sd/sqrt(n)
U = upper bound
U = dbar + t*sd/sqrt(n)
