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
The percentage value of the decimals are 475%, 99% and 1040% respectively.
Data;
<h3>Conversion of Decimals to Percentage</h3>
a) 4.75
To convert a decimal value to percentage, all we simply need to do is multiply that number by 100.

b) 0.99
To convert this decimal value to percentage, we just need to multiply this number by 100.

c) 10.4
To convert this decimal value to percentage, we just need to multiply this number by 100.

Learn more on conversion of decimal value to percentage here;
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Answer:
The last one, 8 / 0
Step-by-step explanation:
The last expression is undefined because in math it is forbidden to divide by zero.
Hope this helps)))