15-4 = 11
11 x 2 = 22
22 + 2/3 = 68/3
Lwh = (3+3+6) (4) (7)
12 x 4 x 7 = 336
Answer:
A score of 150.25 is necessary to reach the 75th percentile.
Step-by-step explanation:
Problems of normal distributions can be 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.
A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.
This means that 
What score is necessary to reach the 75th percentile?
This is X when Z has a pvalue of 0.75, so X when Z = 0.675.




A score of 150.25 is necessary to reach the 75th percentile.
Answer:
The Answer is D
Step-by-step explanation:
Answer:

Step-by-step explanation:
When does the value of f(x) = (1/2)x equal the value of g(x) =2x+8?
