↓can somone answer this for 10 points↓

⇵in how many ways can you fill three different positions by choosing from 15 different people? ⇵

a. 15 · 14 · 13
⇵b. 15 · 14 · 12 · 3⇵
c. 3 · 2 · 1
d. 15 · 3

⇒∵thans∴

15P3 = 15!/ (15-3)! = 15*14*13*12! / 12! = 15*14*13

Step-by-step explanation:

When the seats are empty, there are a total of 15 people that can possibly sit in there.  There will only be 1 person left.

When the 1st seat is taken there are now 14 people who can possibly sit on another chair. Same goes when 2 seaats are taken.  

To find the total number of possibilities, we multiply together the possible number of people for seat 1, seat2, and seat 3.  

Or, if we use stat, nPr = n! / (n-r)!  

Meaning:  

15P3 = 15!/ (15-3)! = 15*14*13*12! / 12! = 15*14*13

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.