There were 16 students running in a race. How many different arrangements of first, second, and third place are possible?

the expected value of the random variable : 0, 1 and 2

step-by-step explanation:

6x = 48

i hope this . 6 cards each and a total of $48. dvide each side by 6 and you will have your answer.

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0