A teacher wants to award prizes for 1st, 2nd, 3rd, 4th and 5th in the class of
30. In how many ways can the prizes be awarded (assume no two students
tie)?

2,052,086,400.

Step-by-step explanation:

Permutation

Pr = n! / (n-r)!

30P5 = 30! / (30-5)

30P5 = 30x29x28x27x26x25! / 25!

(Just cancel the 25! on the numerator and denominator for easier solving.)

30P5 = 30x29x28x27x26

30P5 = 17,100,720.It is not the final answer because we can still arrange the 5 winners.

5! = 5×4×3×2×1 = 12017,100,720 × 120 = 2,052,086,400

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.