MATHEMATICS HIGH SCHOOL

1. Each of 9 friends chooses her favorite positive integera. The median of the chosen number is 91, what is the smallest the average of the 9 chosen numbers could be?

b. The median of the chosen number is 91, is there an limit to how large the aerge of the chosen numbers can be? If so, what is the largest the average can be?

c. The average of the chosen number is 91, what is the smallest the median of the 9 chosen numbers could be?

d. The average of the chosen numbers is 91. What is the largest the median of the chosen numbers could be?

Answers

Answer 1
Answer:

Answer:

a) 1

b) There is no limit to which the largest number can be because we are only given information about the median.

c) 1

d) 90
Answer 2
Answer:

Final answer:

The smallest average is 49 and the largest average is 91. The smallest median is 91 and the largest median is also 91.

Explanation:

a. Since the median is 91, at least 5 friends must choose numbers greater than or equal to 91, and at most 4 friends can choose numbers less than 91. To minimize the average, let's assume the four friends choose the smallest possible numbers less than 91 (1, 2, 3, and 4). The remaining five friends can then choose 91, 91, 91, 91, and 91. The average of the nine chosen numbers is (1 + 2 + 3 + 4 + 91 + 91 + 91 + 91 + 91)/9 = 49.

b. There is no limit to how large the average of the chosen numbers can be. The nine friends can all choose the same number, such as 91, which would make the average 91.

c. Since the average is 91, let's assume the eight friends choose the smallest possible numbers less than 91 (1, 2, 3, ..., 8). The remaining friend can then choose a number greater than or equal to 91. To minimize the median, the friend can choose the smallest possible number greater than or equal to 91, which is 91. So, the smallest median would be 91.

d. Since the average is 91, let's assume the eight friends choose the largest possible numbers less than 91 (84, 85, ..., 91). The remaining friend can then choose a number greater than or equal to 91. To maximize the median, the friend can choose the largest possible number greater than or equal to 91, which is 91. So, the largest median would also be 91.

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In △ABC, points M and P are points on sides AC and BC respectively. Find the area of △MPC, if BM∩AP=O, AAOM=45 dm2, ABOP=15 dm2, and AAOB=75 dm2. I WILL GIVE YOU BRAINLIEST PLS HELP.

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so i believe that the answer is C.) Smaller.

I really hope that this helps

  
i think it is c because it says that the is larger but its not the volume its the ratio so change the ratio to a volume and the volume is still bigger than the ratio given 

Hopes this helps

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