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:
15%
Step-by-step explanation:
We have the followign simple interest formula
FV (aka AV)=PV(1+it)
which means that we have
(800+360)=800(1+3i)
1.45=1+3i
.45=3i
.15=i
15%
In the given statement above, in this case, the answer would be TRUE. It is true that the inequality x + 2y ≥ 3 is satisfied by point (1, 1). In order to prove this, we just have to plug in the values. 1 + 2(1) <span> ≥ 3
So the result is 1 + 2 </span> ≥ 3. 3 <span> ≥ 3, which makes it true, because it states that it is "more than or equal to", therefore, our answer is true. Hope this answer helps.</span>
9 because 18÷2=9
you can use models to help too
get 18 things and put them in 2 groups
Answer:
1/16 or 0.0625
Step-by-step explanation:
