Answer:
The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?
90th percentile
The 90th percentile is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. So




The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds
Answer:
The solution is g = 4
Step-by-step explanation:
* 6(-2g - 1) = -(13g + 2)
- We need to solve it to find the value of g
- Let us simplify each side and then solve the equation
∵ 6(-2g - 1) = (6)(-2g) - (6)(1)
∴ 6(-2g - 1) = -12g - 6
∴ The simplify of 6(-2g - 1) is -12g - 6 ⇒ (1)
∵ -(13g + 2) = -13g - 2
∴ The simplify of -(13g + 2) is -13g - 2 ⇒ (2)
- Equate (1) and (2)
∴ -12g - 6 = -13g - 2
- Add 13g to both sides
∴ g - 6 = - 2
- Add 6 to both sides
∴ g = 4
* The solution is g = 4
Answer:
a. Second Option
b. Second Option
c. Third Option
d.ASA
Step-by-step explanation:
Answer:
The answer is 4 hours and 49 minutes
Step-by-step explanation:
Answer:
100
Step-by-step explanation:
because 1 x 100 = 100