Tommy teacher wants to analyze the weights of the students at his school. he knows the weight of the students is normally distributed so he can use the standard normal distribution. he measures the weight of 100 randomly selected students in the school. he finds the mean is 80 pounds and the standard deviation is 8 pounds.

out of 100 students, how many should weigh between 64 and 96 pounds?

68
48
95

 95

Step-by-step explanation:

Let x be the random variable that represents the weight of the students.

Given :   and .

Using formula :

Also, the weight of the students is normally distributed.

Z-score corresponds to x= 64

Z-score corresponds to x= 96

The  probability that the students weigh between 64 and 96 pounds:

[using z-table for right tailed test]

Out of 100 students, the number of students weigh between 64 and 96 pounds :

Hence, the umber of students weigh between 64 and 96 pounds =95

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.

my guess would be b and d, but i didn’t actually get out the pen and paper. lol