The trifecta at most racetracks consists of selecting the first-, second-, and third-place finishers in a particular race in their proper order. If there are nine entries in the trifecta race, how many tickets must you purchase to guarantee a win?

The total tickets to be purchased to guarantee the win =  504 tickets

Step-by-step explanation:

Step 1 :

Number of entries in the trifecta race = 9

The win is to select the first finisher, second finisher and third finisher in their proper order.

We need to find the number of tickets to be purchased to guarantee the win

Step 2 :

Number of ways to select the first finisher = 9

Number of ways to select the second finisher = 8 [the first is selected and fixed. So the number of available finishes is reduced by 1]

Number of ways to select the third finisher = 7

Hence the total tickets to be purchased to guarantee the win = 9 × 8 × 7 = 504

Step 3 :

Answer :

The total tickets to be purchased to guarantee the win =  504 tickets

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.