I've worked out the first part of the question, just need help with the second part which is:
The spinner has three sections. While no two are the same size, one of the sections is half the size of another.
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You are more likely to get 2/3 than one if you sum two spins, but 5/6 is the most likely sum of all.
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The sum of the three numbers on the spinner is one.
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The largest number you can get in two spins of this spinner is one.
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If you spin the spinner twice and add, you get a sum of one about a quarter of the time.
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If you spun the spinner a hundred times and added up all the numbers, you'd probably get somewhere near 40.
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DRAW THE SPINNER.
I don't think the spinner is possible given the requirements you provided.
The largest number in two spins is 1 so, one section must have a value of 1/2 (or 6/12)
The most likely value for two spins is 5/6 so, the largest area must have a value of 5/12.
Which means the value for the last area must be 1/12 -- which means there is no way to get a value of 2/3 on two spins.
I would like to see the correct answer.
Based on the given information, let's break down the problem and solve it step by step.
1. The spinner has three sections, and no two sections are the same size. One section is half the size of another:
- Let's assume the three sections are represented by A, B, and C.
- If we assign A as the smallest section, B will be twice the size of A, and C will be larger than A and B.
2. You are more likely to get 2/3 than one if you sum two spins, but 5/6 is the most likely sum of all:
- This suggests that the largest section, C, should have a probability of 2/3 when spun individually.
- If you sum two spins, the probability of getting 2/3 is higher than the probability of getting 1.
- The most likely sum of all is 5/6, which suggests that the sum of B and C should be 5/6.
3. The sum of the three numbers on the spinner is one:
- If we let A + B + C = 1, we can use this equation to determine the values of A, B, and C.
4. The largest number you can get in two spins of this spinner is one:
- This implies that the sum of the two largest sections (B and C) should not exceed 1.
5. If you spin the spinner twice and add, you get a sum of one about a quarter of the time:
- This suggests that the sum of the three numbers on the spinner should be distributed in a way that gives a sum close to 1 about 25% of the time.
Based on the given information, we have derived the following conditions:
- A is the smallest section.
- B is twice the size of A.
- The sum of B and C is 5/6.
- A + B + C = 1.
- The sum of B and C does not exceed 1.
- The sum of A + B or A + C should be close to 1 about 25% of the time.
Without further information, it is difficult to determine the exact values for A, B, and C or draw the spinner accurately. However, with the given conditions, you can try different combinations of values that satisfy the conditions and draw the spinner accordingly. Remember to test if the conditions hold true for each combination you try.