A: We want to select one person who works out three days a week. Our probability is simply 12/80 ~~ 15%
B: At least 4 days a week means 4 days, 5 days, or 6 days.
Hence, we have: (9 + 20 + 8)/80 = 46%
C: At least once a week.
Let's find out how many DO NOT work out and subtract that from our sample space. This means they must work out at least once a week.
(80 - 15)/80 = 81%
The correct answer is: AB = 3.11
Explanation:
Since
--- (1)
= 50°
base = 2
And hypotenuse = AB
Plug in the values in (1):
(1) => cos(50°) = 2/AB
=> AB = 2/0.643
=> AB = 3.11
Answer:
D: SAS
Step-by-step explanation:
Answer:
The correct answer is option (C)-0.245 = 2.160(0.205)
Step-by-step explanation:
Solution
Given that:
The slope = - 0.245
The size sample = n = 15
The standard error = 0.205
The confidence level = 95
The Significance level= α = (100- 95)% = 0.05
Now,
The freedom of degree = n-2 = 15 -2= 13
Thus,
the critical value = t* = 2.16
By applying Excel = [TINV (0.05, 13)]
The Margin of error is = t* (standard error)
=2.16 *0.205
= 0.4428
