Super easy all you do is grab 3500*0.7306 which equals 2557.1 2557.1 is correct hope this helps
Answer:
i think but I am not sure so maybe ask somebody else but I think it is A.
Step-by-step explanation:
I just think it is
Answer:
2.28% of tests has scores over 90.
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:

What proportion of tests has scores over 90?
This proportion is 1 subtracted by the pvalue of Z when X = 90. So



has a pvalue of 0.9772.
So 1-0.9772 = 0.0228 = 2.28% of tests has scores over 90.
Answer:
see attached
Step-by-step explanation:
For x < -6, the function has a slope of -1 and an x-intercept of -6.
For x > -6, the function has a slope of 2 and an x-intercept of -6.
The function given here is not defined at x=6, so there is a hole at (-6, 0).
First divide by 8:_
38 / 8 = 4 remainder 6
so its 46