When grading using “the curve”, the percentages are as follows:
A: Top 10% of scores
B: Scores between P70 and P90
C: Middle 40% (scores between P30 and P70)
D: Scores between P10 and P30
F: Bottom 10% of scores
CLASS #1: Suppose that the exam grades for Class #1 are as follows:
35 44 46 47 48 51 54 55 55 57
57 58 59 60 60 60 60 61 62 64
68 69 70 72 73 75 75 77 82 85
µ = 61.3 standard deviation = 11.5
a) Using the mean and standard deviation above and the invNorm feature of your calculator,
determine “the curve” cutoffs for the PK values listed above. (These are not standard deviations.)
Round these test grade cutoffs to one decimal. Label them underneath the curve.
Test Grades 0100

b) Use the one-decimal place cut-offs in the picture you created in part (a) to fill in the chart below
with the new range of scores for each letter grade under this plan. Then, look at the 30 test grades
and count how many test grades would fall in each letter grade based on the new scale.

Letter Grade Range of Scores Number of Tests
A to 100
B
C
D
F 0 to

Factor x2 – 8x + 15. which pair of numbers has a product of ac and a sum of what is the factored form of the

Abucket of paint has spilled on a tile floor. the paint flow can be expressed with the function p(t) = 6(t), where t represents time in minutes and p represents how far the paint is spreading. the flowing paint is creating a circular pattern on the tile. the area of the pattern can be expressed as a(p) = 3.14(p)^2 part a: find the area of the circle of spilled paint as a function of time, or a[p(t)]. show your work. part b: how large is the area of spilled paint after 8 minutes? you may use 3.14 to approximate pi in this problem.

Determine if the following statement is true or false. the normal curve is symmetric about its​ mean, mu. choose the best answer below. a. the statement is false. the normal curve is not symmetric about its​ mean, because the mean is the balancing point of the graph of the distribution. the median is the point where​ 50% of the area under the distribution is to the left and​ 50% to the right.​ therefore, the normal curve could only be symmetric about its​ median, not about its mean. b. the statement is true. the normal curve is a symmetric distribution with one​ peak, which means the​ mean, median, and mode are all equal.​ therefore, the normal curve is symmetric about the​ mean, mu. c. the statement is false. the mean is the balancing point for the graph of a​ distribution, and​ therefore, it is impossible for any distribution to be symmetric about the mean. d. the statement is true. the mean is the balancing point for the graph of a​ distribution, and​ therefore, all distributions are symmetric about the mean.