Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that
P(Ch) = 88/520
P(C) = 76/520
P(Ch and C) = 31/520
and we want to find P(Ch or C).
Using the inclusion/exclusion principle, we have
P(Ch or C) = P(Ch) + P(C) - P(Ch and C)
P(Ch or C) = 88/520 + 76/520 - 31/520
P(Ch or C) = 133/520
X = 13
( x + 4) / 51 = (2x - 7) / 57
585 = 45x
585/45 = 45x/45
x = 13
Answer:
7.9
Step-by-step explanation:
Answer:
C. 34.15 oz
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:

How heavy does a box have to be for it to be labeled overweight?
Top 5%, so X when Z has a pvalue of 1-0.05 = 0.95.
So X when Z = 1.645.




So the correct answer is:
C. 34.15 oz