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forsale [732]
3 years ago
7

11.25 for 2/3 pound how much per pound?

Mathematics
1 answer:
Artist 52 [7]3 years ago
6 0
Divide 11.25 by (2/3). The answer is 16.875
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A player of the National Basketball Association’s Portland Trail Blazers is the best free-throw shooter on the team, making 94%
g100num [7]

Answer:

The data for the probabilities are shown in the table below.

- A represents the probability of making the two shots for each of the best and worst shooter on the Portland Trail Blazers' team

- B represents the probability of making at least one shot for each of the best and worst shooter on the Portland Trail Blazers' team.

- C represents the probability of not making any of the two shots for each of the best and worst shooter on the Portland Trail Blazers' team.

N | Best ||| Worst

A | 0.8836 | 0.3136

B | 0.9964 | 0.8064

C | 0.0036 | 0.1936

It becomes evident why fouling the worst shooter on the team is a better tactic. The probabilities of the best shooter making the basket over the range of those two free shots are way better than the chances for the worst shooter.

Step-by-step explanation:

Part 1

Probability of the best shooter of the National Basketball Association’s Portland Trail Blazers making a shot = P(B) = 94% = 0.94

Probability that he doesn't make a shot = P(B') = 1 - 0.94 = 0.06

a) Probability that the best shooter on the team makes the two shots awarded = P(B) × P(B) = 0.94 × 0.94 = 0.8836

b) Probability that the best shooter on the team makes at least one shot.

This is a sum of probabilities that he makes only one shot and that he makes two shots.

Probability that he makes only one shot

= P(B) × P(B') + P(B') + P(B)

= (0.94 × 0.06) + (0.06 × 0.94) = 0.1128

Probability that he makes two shots = 0.8836 (already calculated in part a)

Probability that he makes at least one shot = 0.1128 + 0.8836 = 0.9964

c) Probability that the best shooter on the team makes none of the two shots = P(B') × P(B') = 0.06 × 0.06 = 0.0036

d) If the worst shooter on the team, whose success rate is 56% is now fouled to take the two shots.

Probability of the worst shooter on the team making a shot = P(W) = 56% = 0.56

Probability that the worst shooter on the team misses a shot = P(W') = 1 - 0.56 = 0.44

Part 2

a) Probability that the worst shooter on the team makes the two shots = P(W) × P(W)

= 0.56 × 0.56 = 0.3136

b) Probability that the worst shooter on the team makes at least one shot.

This is a sum of probabilities that he makes only one shot and that he makes two shots.

Probability that he makes only one shot

= P(W) × P(W') + P(W') + P(W)

= (0.56 × 0.44) + (0.44 × 0.56) = 0.4928

Probability that he makes two shots = 0.3136 (already calculated in part a)

Probability that he makes at least one shot = 0.4928 + 0.3136 = 0.8064

c) Probability that the worst shooter makes none of the two shots = P(W') × P(W') = 0.06 × 0.06 = 0.1936

From the probabilities obtained

N | Best ||| Worst

A | 0.8836 | 0.3136

B | 0.9964 | 0.8064

C | 0.0036 | 0.1936

It becomes evident why fouling the worst shooter on the team is a better tactic. The probabilities of the best shooter making the basket over the range of those two free shots are way better than the chances for the worst shooter.

Hope this Helps!!!

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