I think the answer is letter B
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:
GCF=4
Step-by-step explanation:
16/4=4
24/4=6
36/4=9
For two shots Pedro has four outcomes:
<u>1 shot | 2 shot</u>
score | score
score | not score
not score | score
not score | not score
The probability Pedro's shot will score in a lacrosse game is 0.30 and the probability his shot will not score in a lacrosse game is 1-0.30=0.70. So you can count probabilities for all cases:
1. 0.3·0.3=0.09;
2. 0.3·0.7=0.21;
3. 0.7·0.3=0.21;
4. 0.7·0.7=0.49.
In total 0.09+0.21+0.21+0.49=1. The first outcomes is that what you need.
Answer: 0.09.
For this case we must simplify the following expression
We know that, by definition:
So, rewriting the expression we have:
We add similar terms taking into account that:
- Equal signs are added and the same sign is placed.
- Different signs are subtracted and the major's sign is placed:
Answer:
