(39-7/4^2
32/16=2
I think the question is straight forward
Answer:
The percent of the pairs that last longer than six months—that is, 180 days is 95.637%
Step-by-step explanation:
Mean = xbar = 208 days
Standard deviation = σ = 14
The standardized score for 180 days is the value minus the mean then divided by the standard deviation.
z = (x - xbar)/σ = (180 - 208)/14 = - 1.71
To determine the percent of boots that last longer than 180 days, we need this probability, P(x > 180) = P(z > (-1.71))
We'll use data from the normal probability table for these probabilities
P(x > 180) = P(z > (-1.71)) = 1 - P(z ≤ (-1.71)) = 1 - 0.04363 = 0.95637 = 95.637%
Answer:
-4
Step-by-step explanation:
- 4(m + 6) = - 8
(- 4)(m) + (- 4)(6) = - 8
- 4m - 24 = -8
- 4m = 16
m = - 4
-3r + 15 ≥ 4(r - 2)
-3r + 15 ≥ 4(r) - 4(2)
-3r + 15 ≥ 4r - 8
+ 3r + 3r
15 ≥ 7r - 8
+ 8 + 8
23 ≥ 7r
7 7
3²/₇ ≥ r
r ≤ 3²/₇
Solution Set: x ∈ (-∞, 3²/₇]