Answer:
The minimum score required for an A grade is 83.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 72.3 and a standard deviation of 8.
This means that 
Find the minimum score required for an A grade.
This is the 100 - 9 = 91th percentile, which is X when Z has a pvalue of 0.91, so X when Z = 1.34.




The minimum score required for an A grade is 83.
Answer:

Step-by-step explanation:
The picture of the question in the attached figure
step 1
Let
r ---> the radius of the sector
s ---> the arc length of sector
Find the radius r
we know that



solve for r

step 2
Find the value of s

substitute the value of r

step 3
we know that
The area of complete circle is equal to

The complete circle subtends a central angle of 2π radians
so
using proportion find the area of the sector by a central angle of angle theta
Let
A ---> the area of sector with central angle theta

substitute the value of r


Convert to function notation

Assuming a d-heap means the order of the tree representing the heap is d.
Most of the computer applications use binary trees, so they are 2-heaps.
A heap is a complete tree where each level is filled (complete) except the last one (leaves) which may or may not be filled.
The height of the heap is the number of levels. Hence the height of a binary tree is Ceiling(log_2(n)), for example, for 48 elements, log_2(48)=5.58.
Ceiling(5.58)=6. Thus a binary tree of 6 levels contains from 2^5+1=33 to 2^6=64 elements, and 48 is one of the possibilities. So the height of a binary-heap with 48 elements is 6.
Similarly, for a d-heap, the height is ceiling(log_d(n)).