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.
Perimeter of rectangle is 2L + 2W
perimeter is 296
length of field is 79
Solve
2L + 2W = 296
2(79) + 2W = 296
158 + 2W = 296
2W = 296 - 158 = 138
W = 138/2 = 69 yards
Answer:
the correct answer is shown below
Step-by-step explanation:
hope this helps