Answer:
within ±1.96 standard deviations of the sample mean
Step-by-step explanation:
A 95% confidence interval is found using the formula C = 1 - α, and some other stuff, but let's focus on that for now. Using the formula:
.95 = 1 - α
α = .05
If α = .05, that means a 2-sided confidence interval would be found using the sample mean and the Z-score Z(subscript α/2), or Z.₀₂₅ because α AKA .05 divided by 2 = .025. From there, you take this either to your calculator or a Z-table (or perhaps you have a chart that lists the common CI values), and see that for the area to be .025 beneath a standard normal curve, your Z value is ±1.96 ("plus or minus" because we're considering a 2-sided confidence interval).
No, you cannot use the Associative properties with either subtraction and division, because it will not end to the same result. 14-8 is not the same as 8-14, and 24/4 is not the same as 4/24, but for example 3+2 is the same as 2+3.
Answer:
2 1/10 miles.
Step-by-step explanation:
She has already jogged 2 1/2 + 3 2/5 miles. This is 2 5/10 + 3 4/10 or 5 9/10. 8 miles is also equal to 7 10/10 miles. So, to find the remaining amount of miles, we can do 7 10/10 - 5 9/10 = 2 1/10.
First, we need to find out how many siblings the students have in total and how many siblings the teachers have in total.
The students: (1 x 4) = 4 + (2 x 7) = 18 + (3 x 5) = 33 + (4 x 2) = 41. The students have a total of 41 siblings.
The teachers: (1 x 3) = 3 + (2 x 2) = 7 + (3 x 4) = 19 + (4 x 5) = 39 + (5 x 3) = 54 + (6 x 1) = 60 + (8 x 1) = 68. The teachers have a total of 68 siblings.
Now that we have this information, we know that the answer is option 1, the students tend to have fewer siblings than the teachers. This is because the students had a total of 41 siblings and the teachers had a total of 68 siblings, and 41 is less than 68 so the students have fewer siblings than the teachers.
I really hope I could help! I put a lot of work into this answer! :D
