Answer:
136 in²
Step-by-step explanation:
8x2=16
6x2=12
6x2=12
6x2=12
10x2=20
10x2=20
10x2=20
8x2=16
2x2=4
2x2=4
136 in²
Answer:
4
hours
Step-by-step explanation:
he was gone for 5
hours, he spent
an hour driving.
5
-
= 4
hours at work.
Answer:
319 is the term
Step-by-step explanation:
Please give me brainliest :)
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.