You want to draft a four-player tennis team. There are eight players to choose from. How many different teams can you form?
(1 point)
•70****
•64
•48
•336
Elizabeth has two identical number cubes. Both cubes have faces numbered 1 through 6. If Elizabeth rolls each cube once, what is the probability that the sum of the two numbers on the top faces will be 10?
•1/36
•1/12***
•1/10
•1/9
Suppose 10% of the flights arriving at an airport arrive early, 60% arrive on time, and 30% arrive late. Valerie used the random-number table to find the experimental probability that of 5 flights, at least 2 will arrive late. The digit 0 represents flights arriving early. The digits 1, 2, 3, 4, 5, and 6 represent flights arriving on time. The digits 7, 8, and 9 represent flights arriving late.
65926, 31459, 31986, 65809, 80462
27387, 39075, 46738, 21986, 59837
91384, 10987, 26491, 68498, 98796
32596, 26448, 31235, 63256, 53121
Find the experimental probability that of 5 flights, at least 2 will arrive late.
(1 point)
•3/10
•2/5****
•9/20
•11/20
The first 2 are correct,
don't quite follow the 3rd
To solve the first question, we need to find the number of different teams that can be formed from 8 players. Since we need to choose 4 players out of the 8, we can use the combination formula, which is given by:
C(n,r) = n! / (r!(n-r)!)
In this case, n = 8 (number of players) and r = 4 (number of players to choose). Substituting the values into the formula, we get:
C(8,4) = 8! / (4!(8-4)!)
= 8! / (4!4!)
Now, let's calculate the value:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320
4! = 4 x 3 x 2 x 1 = 24
Substituting the values back into the formula:
C(8,4) = 40,320 / (24 x 24) = 70
Therefore, the answer is 70.
To calculate the number of different teams that can be formed from a pool of eight players to create a four-player tennis team, you can use the combination formula. The formula for calculating combinations is nCr, where n is the number of items to choose from and r is the number of items to choose.
In this case, you have eight players to choose from and need to select four players for the team. The formula for combinations is:
nCr = n! / (r!(n-r)!)
Plugging in the values for this problem:
8C4 = 8! / (4!(8-4)!)
Evaluating the factorial terms:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
4! = 4 x 3 x 2 x 1
8-4 = 4
Now substitute the factorial terms into the combination formula:
8C4 = (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / ((4 x 3 x 2 x 1) x (4 x 3 x 2 x 1))
= 70
Therefore, you can form 70 different teams from the eight players available.
For the second question, to find the probability of getting a sum of 10 when rolling two identical number cubes, you first need to determine the total number of possible outcomes. Each cube has six faces with values from 1 to 6, so the total number of possible outcomes is 6 x 6 = 36.
Next, count the number of outcomes that result in a sum of 10:
(4, 6), (5, 5), (6, 4)
There are three such outcomes.
Finally, divide the number of favorable outcomes by the total number of possible outcomes:
Probability = number of favorable outcomes / total number of outcomes
= 3 / 36
= 1/12
Therefore, the probability that the sum of the two numbers on the top faces will be 10 is 1/12.
For the third question, you are given the results of five flights and need to find the experimental probability that at least two of them will arrive late. To do this, you need to count the number of outcomes that satisfy this condition and divide it by the total number of flights.
Analyzing the given flight results:
65926 (no late flights)
31459 (no late flights)
31986 (no late flights)
65809 (no late flights)
80462 (no late flights)
27387 (one late flight)
39075 (no late flights)
46738 (no late flights)
21986 (one late flight)
59837 (no late flights)
91384 (no late flights)
10987 (no late flights)
26491 (no late flights)
68498 (no late flights)
98796 (no late flights)
32596 (no late flights)
26448 (no late flights)
31235 (no late flights)
63256 (no late flights)
53121 (no late flights)
Out of the given flights, only two flights have arrived late (27387 and 21986). Therefore, the number of outcomes where at least two flights arrive late is 2.
The total number of flights is 5.
Finally, divide the number of favorable outcomes by the total number of outcomes:
Probability = number of favorable outcomes / total number of outcomes
= 2 / 5
= 2/5
Therefore, the experimental probability that of 5 flights, at least 2 will arrive late is 2/5.