A) A ratio system
B) The 4 lb peanuts and the 1 lb mixture because the 4lb added to the 1lb of mixture give the correct percentages.
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
LCM 14 & 24 = 168
then use the highest exponents for the variables
answer = 168x^6y^6z^8
Step-by-step explanation:
6k + 12 + 9=9
6k + 12 = 9 - 9
6k + 12 = 0
12 = -6k
12/-6 = -6k/-6
2/-1 = k
k = -2