Jeff has ten packages that he wants to mail. Nine identical packages weigh \( 2 \frac{7}{8} \) pounds each. A tenth package weighs two times as much as one of the other packages. How many pounds do all ten packages weigh?

First, we need to calculate the weight of the ninth package. Each of the nine identical packages weighs \( 2 \frac{7}{8} \) pounds. To make our calculations simpler, let's convert \( 2 \frac{7}{8} \) into an improper fraction: \[ 2 \frac{7}{8} = \frac{16}{8} + \frac{7}{8} = \frac{23}{8} \text{ pounds} \] Next, we multiply this weight by 9 to find the total weight of the nine identical packages: \[ 9 \times \frac{23}{8} = \frac{207}{8} \text{ pounds} \] Now, for the tenth package, which weighs two times as much as one of the other packages: \[ 2 \times \frac{23}{8} = \frac{46}{8} = \frac{23}{4} \text{ pounds} \] Now we add the weight of the ninth package to the weight of the tenth package: \[ \text{Total weight} = \frac{207}{8} + \frac{23}{4} \] To add these, we need a common denominator. The common denominator of 8 and 4 is 8, so we can convert \( \frac{23}{4} \) into eighths: \[ \frac{23}{4} = \frac{23 \times 2}{4 \times 2} = \frac{46}{8} \] Now, we can add the two fractions: \[ \frac{207}{8} + \frac{46}{8} = \frac{253}{8} \text{ pounds} \] Finally, let's convert \( \frac{253}{8} \) back into a mixed number: \[ 253 \div 8 = 31 \text{ R } 5 \] So, \( \frac{253}{8} = 31 \frac{5}{8} \). Therefore, the total weight of all ten packages is \( \boxed{31 \frac{5}{8}} \) pounds.