$\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}\times \dfrac{1\ remaining\ prize}{27\ remaining\ people}=\dfrac{6}{21,924}\\\\\\.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =\large\boxed{\dfrac{1}{3,654}}$

2) First and Second

$\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}=\dfrac{6}{812}=\large\boxed{\dfrac{13}{406}}$

$3)\quad \dfrac{29!}{(29-10)!}=\large\boxed{72,684,900,288,000}$

$4)\quad _{10}C_3=\dfrac{10!}{3!(10-3)!}=\large\boxed{120}$

8 0

3 years ago

Pre Calculus:

Gwar [14]

Let n be a number of friends at the perty. If dinner costs $30 per person, then it cost $30n for all friends.

Robert has booked a banquet hall which costs $200, then his total expenses are $200+30n.

If the final bill is p, then p=200+30n.

Answer: p=200+30n.

3 0

3 years ago

Read 2 more answers

X+y=8 and 3x+2y=14 for soliving system

kumpel [21]

X+y=8
multiply the first equation by 3 to get rid of x
3x+3y=24
3x+2y=14
subtract equation 1 from 2
y=10
substitute to find the value of x
x+10=8
x=-2

7 0

3 years ago