TanA/secA-1-sinA/1+cosA =2 cotA
1 answer:
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Answer:
0.0668 = 6.68% probability that the worker earns more than $8.00
Step-by-step explanation:
When the distribution is normal, we use 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.
The average hourly wage of workers at a fast food restaurant is $7.25/hr with a standard deviation of $0.50.
This means that
If a worker at this fast food restaurant is selected at random, what is the probability that the worker earns more than $8.00?
This is 1 subtracted by the pvalue of Z when X = 8. So
has a pvalue of 0.9332
1 - 0.9332 = 0.0668
0.0668 = 6.68% probability that the worker earns more than $8.00