Step-by-step answer:
Given:
mean, mu = 200 m
standard deviation, sigma = 30 m
sample size, N = 5
Maximum deviation for no damage, D = 100 m
Solution:
Z-score for maximum deviation
= (D-mu)/sigma
= (100-200)/30
= -10/3
From normal distribution tables, the probability of right tail with
Z= - 10/3
is 0.9995709, which represents the probability that the parachute will open at 100m or more.
Thus, by the multiplication rule, the probability that all five parachutes will ALL open at 100m or more is the product of the individual probabilities, i.e.
P(all five safe) = 0.9995709^5 = 0.9978565
So there is an approximately 1-0.9978565 = 0.214% probability that at least one of the five parachutes will open below 100m
Answer:
There are 40 more females than males.
In total there would be 60 females, and 20 males.
Step-by-step explanation:
Step-by-step explanation:
(i) Sum the forces at point M in the x direction.
∑F = ma
-T sin 30° − T sin 30° + 5 N = 0
2T sin 30° = 5 N
T = 5 N
(ii) Sum the forces on the ring in the x direction.
∑F = ma
T sin 30° − N = 0
N = 2.5 N
Sum the forces on the ring in the y direction.
∑F = ma
T cos 30° − mg − Nμ = 0
Nμ = T cos 30° − mg
2.5 μ = 5 cos 30° − 2
μ = 0.932
(iii) Sum the forces on the ring in the y direction.
∑F = ma
T cos 30° − m₁g − m₂g + Nμ = 0
m₂g = T cos 30° − m₁g + Nμ
m (10) = 5 cos 30° − 2 + (2.5)(0.932)
m = 0.466 kg
Brianna's thinking is incorrect.
Expessions A, and C. are equivalent.
For C, you add 5x and x together (because they have the same variable) to get 6x - 4.
For A, you subtract 9x and 3x (because they have the same variable) to get 6x- 4.
Answer:
5:2
Step-by-step explanation:
35÷7= 5
14÷7=2
You get the ratio by dividing by a common multiple
