An unusual die has six faces labelled 1,2,3,5,7 9. If two of these dice are rolled, and the numbers showing on the upper faces are added, what is the number of possible different sums?

14?

Flora had an average of 56% on her first 7 exams. What would she have to make on her eighth exam to obtain an average of 60% on 8 exams?
88%?

The Band Committee of 100 people wishes to set up a telephone call system. The initial contact person calls three other people, each of whom call three others and so on, until all the people in the Band Committee have been contacted. What is the maximum number of people they need to make the calls?
66?
I added 27+27+6+6

To find the number of possible different sums when two unusual dice are rolled, we need to first determine all the possible combinations of numbers that can be obtained from each die. The given die has six faces labelled 1, 2, 3, 5, 7, and 9.

So, to get the possible sums, we can create all possible combinations by pairing the numbers from each die.

The number of possible sums is equal to the total number of combinations we can form. In this case, to get all possible combinations, we multiply the number of options on one die (6) by the number of options on the other die (6), which gives us 6x6 = 36 possible combinations.

Therefore, there are 36 possible different sums when two unusual dice are rolled.

For the second question, to determine the minimum score Flora needs on her eighth exam to obtain an average of 60% on 8 exams, we need to find the total sum of scores she needs to achieve.

If Flora has an average of 56% on her first 7 exams, the sum of those scores would be 7 * 56% = 392%.

To obtain an average of 60% on 8 exams, Flora needs a total sum of 8 * 60% = 480%.

Therefore, on her eighth exam, Flora needs to make a score of 480% - 392% = 88%.

So, Flora needs to score 88% on her eighth exam to obtain an average of 60% on 8 exams.

Regarding the question about the Band Committee and the telephone call system, the problem essentially asks for the maximum number of people needed to make the calls.

The initial contact person calls three other people, and each of those three people calls three others, and so on until all the people in the Band Committee have been contacted.

To find the maximum number of people needed, we can use a geometric progression formula.

The formula for the sum of a geometric series is S = a(1 - r^n) / (1 - r), where a is the first term, r is the common ratio, and n is the number of terms.

In this scenario, the first term (a) is 1 (since it starts with the initial contact person), the common ratio (r) is 3 (since each person calls three others), and the number of terms (n) is the number of levels in the calling tree.

To calculate the maximum number of people needed, we can plug in these values into the formula.

S = 1(1 - 3^5) / (1 - 3) = 1 - 243 / -2 = 1 + 121.5 = 122.5

Therefore, the maximum number of people needed to make the calls is 122.5. Since we can't have half of a person, we round up to the nearest whole number. Hence, the answer is 123.

So, the maximum number of people needed to make the calls is 123.