14.07 is the answer to the question above
Correct Ans:Option A. 0.0100
Solution:We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.
First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.
Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.
Given the points A and B
The coordinates of point A = ( 3 , 1 )
The coordinates of point B = (-1 , -1)
The midpoint of AB, is the point C
C will be calculated as following :

so, the midpoint of AB = (1 , 0 )
This question means the than (8 + 10) x 2 = red markers. She has 32 red markers, 8 blue markers, and 10 green markers. If you add them all up, you get 50 markers.