Let's calculate the total weight of the luggage first.
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Total Weight of the Suitcases:
- Convert the weights to an improper fraction if necessary:
- 24 1/2 lb = 24.5 = 49/2 lb
- 13 3/8 lb = 13.375 = 107/8 lb
- 3 1/2 lb = 3.5 = 7/2 lb
- 17 3/4 lb = 17.75 = 71/4 lb
Finding a common denominator: The denominators are 2, 8, 2, and 4. The least common multiple (LCM) is 8.
- Convert to eighths:
- 49/2 = 196/8
- 107/8 = 107/8
- 7/2 = 28/8
- 71/4 = 142/8
Add them up: \[ \frac{196}{8} + \frac{107}{8} + \frac{28}{8} + \frac{142}{8} = \frac{196 + 107 + 28 + 142}{8} = \frac{473}{8} \]
Now, convert \(\frac{473}{8}\) to a mixed number: \[ \frac{473}{8} = 59 \frac{1}{8} \text{ lb} \]
Total weight is \(59 \frac{1}{8}\) lb, but for accurate selection from given options, let’s check.
- Correct Options for Weight Selection: None of the provided options is accurate. The calculation holds, so it seems like an error in worthiness.
- Convert the weights to an improper fraction if necessary:
-
Reducing weight by 12.2 lb:
- Since \(59 \frac{1}{8}\) lb = 59.125 lb.
- Subtract 12.2 lb: \[ 59.125 - 12.2 = 46.925 \text{ lb} \]
Closest option from the choices: None seem to match the amount depending but theoretically displays an error.
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Total height of the suitcases:
- Each suitcase is \(14 \frac{1}{6}\) inches tall.
- Total height = \(4 \times 14 \frac{1}{6}\) inches = \(4 \times \frac{85}{6} = \frac{340}{6} = \frac{170}{3} = 56 \frac{2}{3}\) inches.
How They Fit in Trunk of 50 Inches
- Total height of the luggage is \(56 \frac{2}{3}\) inches, meaning No, they will not fit.
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Area of the bumper sticker:
- Area \(A = \text{length} \times \text{width} = 4.56 \times 2.3\). \[ A = 4.56 \times 2.3 = 10.488 \text{ in}^2 \]
The area calculation shows 10.488 in² which is valid.
Final Answers:
- Total weight of the luggage: Doesn't match known estimates, calculate manually.
- Total weight after reduction is specifically theorized within all false ends.
- No, the height with all will exceed trunk space.
- Total area of the bumper sticker is \(10.488 \text{ in}²\).