Which does not apply?
1 answer:
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I'm assuming you only need help with part B. If that's the case, then all we do is add up the values outside of the table to get the inner cell results. For example, in row 1, column 1, we have 1's outside. They add to 1+1 = 2. So we'll have 2 in the first row and first column of the table. The other cells are computed in this way. In figure 1 (see attached image), I show the intermediate step of adding. Then in figure 2, I actually add up the values and show the final result of what the table will look like when all said and done. Take note of the color coding to help see how the values match up for each inner cell of the table. The color coding is optional.
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Edit:
Part C -- Part 1) Probability of rolling double sixes
There are five copies of "6" in the table (see figure 2 of the attached image) out of 36 total possibilities. How did I get 36? It's the result of multiplying 6*6. This is because there are 6 faces per die. As the table shows, there are 6 rows and 6 columns giving 36 cells total.
So we divide the number of times "6" shows up (5 times) out of the number of total possibilities (36 total) to get: 5/36 = 0.1389 = 13.89%
Answer as a fraction: 5/36
Answer as a decimal: 0.1389
Answer as a percent: 13.89%
The decimal and percent forms are approximate
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Part C -- Part 2) Probability of rolling doubles
To roll doubles, all we need to do is roll the same number twice on each die. In other words, we have "twin values" so to speak. Back in part 1 above, we dealt with just '6's showing up twice. There are other doubles as well
The doubles possibilities are:
1 and 1
2 and 2
3 and 3
4 and 4
5 and 5
6 and 6
Each combo listed above only shows up once. So we have 6 instances where we have doubles (one instance per side of the die). This is out of 36 total
Divide the values mentioned: 6/36 = 1/6 = 0.1667 = 16.67%
Answer as a fraction: 1/6
Answer as a decimal: 0.1667
<span>Answer as a percent: 16.67%</span>
The decimal and percent forms are approximate
Answer:
x=1
Step-by-step explanation:
if it is truly 9^(x-1)=27^(x-1), then you dont really need to use the quadratic fromula.
