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- 47 → d
using the ' product rule '
(uv) = uv' + vu' = (3 × - 4) + (7 × - 5 ) = - 12 - 35 = - 47
Answer:
The minimum score required for an A grade is 88.
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:
Find the minimum score required for an A grade.
Top 12%, which is at least the 100-12 = 88th percentile, which is the value of X when Z has a pvalue of 0.88. So it is X when Z = 1.175.
Rounding to the nearest whole number
The minimum score required for an A grade is 88.