To solve the problem, we start by calculating the current expected value of the slips of paper in the bag.
We have the following slips in the bag:
- 8 slips with the number 7
- 2 slips with the number 9
The total number of slips is \(10\).
First, we calculate the total contribution of the slips to the expected value.
The total value contributed by the slips with 7 is:
\[ 8 \times 7 = 56 \]
The total value contributed by the slips with 9 is:
\[ 2 \times 9 = 18 \]
Adding these, the total value of all the slips is:
\[ 56 + 18 = 74 \]
Now, to find the current expected value, we divide the total value by the total number of slips:
\[ \text{Current expected value} = \frac{74}{10} = 7.4 \]
Next, we introduce \(x\) slips of paper with the number 2. The total number of slips now becomes:
\[ 10 + x \]
The total value of the slips corresponding to the new slips with 2s will be:
\[ 2x \]
So, the new total value becomes:
\[ 74 + 2x \]
We need the new expected value to equal 10:
\[ \frac{74 + 2x}{10 + x} = 10 \]
To eliminate the fraction, we multiply both sides by \(10 + x\):
\[ 74 + 2x = 10(10 + x) \]
Expanding the right side:
\[ 74 + 2x = 100 + 10x \]
Next, we isolate \(x\) by first moving all the \(x\) terms to one side and constants to the other:
\[ 74 - 100 = 10x - 2x \]
This simplifies to:
\[ -26 = 8x \]
Now, we solve for \(x\):
\[ x = -\frac{26}{8} = -\frac{13}{4} = -3.25 \]
Since \(x\) must be a non-negative integer and we obtained a negative result, it indicates that adding slips with a value of 2 will never achieve an expected value of 10. In fact, since the maximum expected value obtainable from the slips (with increasing counts of 2s) can only decrease, we conclude that it is impossible to achieve an expected value of 10 with the current setup.
Thus, we have:
\[ \boxed{0} \]
In conclusion, it is impossible to achieve an expected value of 10 by adding slips with the value of 2; therefore, the answer is 0 slips.
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