Her grandma will be there 40 minutes early. If you add the hours first, then it makes it 6:35. Add the 50 minutes, 7:25 is the arrival time.
Answer:
1
Step-by-step explanation:
Given :
Mean, μ = 4
Standard deviation, s = 0.8
Sample size, n = 30
The distribution is independent.
Z = (x - μ) / s /sqrt(n)
Probability that downtime period is between 1 and 5
P(1≤ x ≤ 5) :
[(x - μ) / (s /sqrt(n))] ≤ Z ≤ [(x - μ) / (s /sqrt(n))]
[(1 - 4) / (0.8 /sqrt(30))] ≤ Z ≤ [(5 - 4) / (0.8 /sqrt(30)]
[-3 / 0.1460593] ≤ Z ≤] 1 / 0.1460593]
P(-20.539602 ≤ Z ≤ 6.8465342)
P(Z ≤ 6.8465342) - P(Z ≤ - 20.5396)
P(Z ≤ 6.8465342) = 1 (Z probability calculator)
P(Z ≤ - 20.5396) = 0 (Z probability calculator)
1 - 0 = 1
true
Producers --> Consumers --> Primary consumers --> Secondary consumers -->
Tertiary consumers --> Quaternary consumers --> Decomposers --> Producers
and the cycle keeps repeating
<em>Hope it helps... </em>
Its the second one yeah im pretty sure
