Answer:
4032 different tickets are possible.
Step-by-step explanation:
Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.
To find : How many different tickets are possible ?
Solution :
In the first race there are 9 ways to pick the winner for first and second place.
Number of ways for first place - 
Number of ways for second place - 
In the second race there are 8 ways to pick the winner for first and second place.
Number of ways for first place -
Number of ways for second place - 
Total number of different tickets are possible is
Therefore, 4032 different tickets are possible.
Answer:
C
Step-by-step explanation:
This problem is analogous to the extraction of 6 elements from a total of 10 elements. It's the same if they are marbles, chips, or in this case, people, as here we don't care about the order of the selection as we only are drawing a sample.
Thus, the problem implies solving the amount of possible combinations of 10 people if we take by 6. There is a formula for this and is:
10 C 6 = 10!(6!4!)
If we operate, knowing that for any number x, x!=x*(x-1)*(x-2)*...*1
10 C 6 = 10!(6!4!) = 10*9*8*7*6*...*1 / [(6*5*...*1) * (4*3*2*1)]
10 C 6 = 10*9*8*7*6! / [(6*5*...*1) * (4*3*2*1)]
We have a 6! multiplying and another dividing, so they get eliminated, and as 4*2=8 and 9=3*3
10 C 6 = 10*9*8*7*/ [(4*3*2*1)] = 10*3*3*8*7*/ [(8*3*1)]
We can eliminate the 8s and one of the 3s on the numerator with the one on the denominator:
10 C 6 = 10*3*7*/1 = 210/1= 210
So, option C
Answer:
AD= 5.82 units (3 s.f.)
Step-by-step explanation:
Please see attached picture for full solution.
*skipped these steps in the beginning:
∠ABC +∠ABD= 180° (adj. ∠s on a str. line)
90° +∠ABD= 180°
Answer:
b
Step-by-step explanation:
