1)In your last 23 basketball games, you attempted 101 free throws and made 66. Find the experimental probability that you make a free throw. Write the probability as a percent, to the nearest tenth of a percent.

A65.3%

B69.8%

C69.7%

D65.7%

2)A number cube is rolled with these results: 64 ones, 67 twos, 73 threes, 59 fours, 72 fives, and 71 sixes. What is the experimental probability of rolling an even number?

Write your answer as a percent, to the nearest tenth of a percent.

The answer is to number 1 is A. 65.3% because she has 101 tries and made 66 of it.
=> 66 / 101 = 0.653
=> 0.653 * 100%
=> 65.3%

The answer for number 2 is:
P(even) = [67+59+71] / [67+59+71+64+73+72] = 48.5%

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virtuematane virtuematane · Answer 2 · 2019-01-07T06:02:21-05:00

Answer:

1) option:A

2)48.52%

Step-by-step solution:

1)

The Probability of an event is defined as:

\(Probability=\frac{\text{Number of favourable events}}{\text{total number of events}}\)

now we are given that we have made 66 free throws out of total attempted 101 throws.

Hence, \(Probability=\frac{66}{101}=0.65346\)

Hence, the experimental probability that you make a free throw in percent=65.346%

Hence, option A is correct.

2)

The results of a number cube are: 64 ones,67 twos, 73 threes,59 fours,72 fives,and 71 sixes.

Total outcomes=64+67+73+59+72+71=406

the outcomes of an even number=67+59+71=197

hence, probability of rolling an even number is:

\(\dfrac{197}{406}=0.48522\)

hence, the experimental probability of rolling an even number in percent is: 48.52%.