2x^2-7=4
subtract 4 from both sdies
2x^2-11=0
therefor
2x^2=11
divide by 2
x^2=5.5
square root both sdies
x=-2.345 or 2.345
75/5-(4-1)^2
75 / 5 = 15
15-(4-1)^2
15-3^2
answer: 6
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.
Depending on the age of the kitten they can weigh anywhere between 1 to 2 1/2 pounds or even 3 pounds if its a big kitten. Hope this answers your question!:)
Teo fifty 250 easy divide