brainly won't let me type my answer out so i'll just put it in the comments, sorry if it's inconvenient.
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.
6 divided by 1/4 = 24 + 12 divided by 1/4 = 48, then 48 + 24 = 72
9514 1404 393
Answer:
3. 7
Step-by-step explanation:
The distance formula applies in 3 dimensions as well as 2.
d = √((x2 -x1)² +(y2 -y1)² +(z2 -z1)²)
d = √((-2)² +6² +3²) = √(4 +36 +9) = √49
d = 7
The distance between the two points is 7 units.
Y=4-3/x
y-4=-3/x
(y-4)x=-3
x(y-4)=-3
x=-3/y-4
there's you answer