The following are the scores of a 23-student class on Machine Learning, out of 100. As can be seen, based on the grading scale listed in your syllabus sheet, most of the students will fail. The professor therefore decides to grade on the (normal distribution) curve with the mean score receiving a B- grade, the grade going up or down one step from there with every 1/3rd of the standard deviation change in the score. Assume SJSU letter grading [F to A+]. Write a program in python to determine each student’s grade. 47, 63, 71, 39, 47, 49, 43, 37, 81, 69, 38, 13, 29, 61, 49, 53, 57, 23, 58, 17, 73, 33, 29

x equals 1

step-by-step explanation:

f(4)=f(×)

x=1

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.