A haircut cost $15. I left a 20% tip. How much was the haircut including the tip? PLEASE ANSWER WITH WORK
2 answers:
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Answer:
3 and 28
Step-by-step explanation:
note ( )(x ) =
= and
( )(4) =
=
=
= 3
-------------------------------------------------------
(f + g)(x) = f(x) + g(x)
f(x) + g(x)
= x² + 2x - 3 + x² - 9 = 2x² + 2x - 12 and
(f + g)(4) = 2(4)² + 2(4) - 12 = 2(16) + 8 - 12 = 32 + 8 - 12 = 28
Answer:
52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem:
Normal probability distribution:
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.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct
This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 - 339 = 30054. So
X = 30732
By the Central Limit Theorem
has a pvalue of 0.7642.
X = 30054
has a pvalue of 0.2358
0.7642 - 0.2358 = 0.5284
52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct