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The average distance between the Sun and a certain planet is approximately \(2.3 \times 10^{14}\) inches. Which of the following is closest to the average distance between the Sun and the planet, in kilometers? (1 kilometer is approximately \(3.9 \times 10^{4}\) inches.)
Let's start by understanding what we're being asked to do. We have a distance measured in inches, and we want to know that same distance measured in kilometers instead.
Think of it like this: imagine you measured your room in inches and got a really big number. Now someone asks "how many feet is that?" You'd need to convert by dividing by 12 (since there are 12 inches in a foot).
Here, we have:
• Distance in inches: \(2.3 \times 10^{14}\) inches
• Conversion factor: 1 kilometer = \(3.9 \times 10^4\) inches
• We want: Distance in kilometers
The question we're answering is: "How many kilometers fit into \(2.3 \times 10^{14}\) inches?"
Process Skill: TRANSLATE - Converting the word problem into a clear mathematical relationship
Now let's think about the conversion logically. If 1 kilometer equals \(3.9 \times 10^4\) inches, then to find how many kilometers we have, we need to see how many "chunks" of \(3.9 \times 10^4\) inches fit into our total of \(2.3 \times 10^{14}\) inches.
This is division: we divide the total inches by the inches per kilometer.
In plain English: "How many groups of \(3.9 \times 10^4\) inches can we make from \(2.3 \times 10^{14}\) inches?"
Mathematically: Distance in kilometers = (\(2.3 \times 10^{14}\) inches) ÷ (\(3.9 \times 10^4\) inches per kilometer)
When we divide numbers in scientific notation, we can split this into two parts:
• Divide the regular numbers: \(2.3 \div 3.9\)
• Divide the powers of 10: \(10^{14} \div 10^4\)
Let's handle each part:
Step 1: Divide the coefficients
\(2.3 \div 3.9 \approx 0.59\) (we can approximate since 2.3 is a bit more than half of 3.9)
Step 2: Divide the powers of 10
\(10^{14} \div 10^4 = 10^{(14-4)} = 10^{10}\)
Step 3: Combine the results
\(0.59 \times 10^{10}\)
To express this in proper scientific notation (where the first number is between 1 and 10):
\(0.59 \times 10^{10} = 5.9 \times 10^9\)
Our calculated result is \(5.9 \times 10^9\) kilometers.
Looking at the answer choices:
(A) \(7.1 \times 10^8\) - This is about 10 times smaller than our answer
(B) \(5.9 \times 10^9\) - This matches our calculation exactly!
(C) \(1.6 \times 10^{10}\) - This is about 3 times larger than our answer
(D) \(1.6 \times 10^{11}\) - This is about 30 times larger than our answer
(E) \(5.9 \times 10^{11}\) - This is about 100 times larger than our answer
The average distance between the Sun and the planet is approximately \(5.9 \times 10^9\) kilometers.
The correct answer is (B) \(5.9 \times 10^9\).
Students often confuse which number should be divided by which. Since 1 kilometer = \(3.9 \times 10^4\) inches, some students might incorrectly multiply \(2.3 \times 10^{14}\) by \(3.9 \times 10^4\) instead of dividing. They think "if 1 km equals a certain number of inches, then more inches means more kilometers," missing that they need to find how many groups of \(3.9 \times 10^4\) fit into the total.
Students might flip the conversion factor and think that 1 inch = \(3.9 \times 10^4\) kilometers instead of 1 kilometer = \(3.9 \times 10^4\) inches. This fundamental misreading of the given information would lead them to divide \(2.3 \times 10^{14}\) by \(3.9 \times 10^4\) when they should actually be multiplying, resulting in a vastly incorrect answer.
When dividing \(10^{14}\) by \(10^4\), students might add the exponents (getting \(10^{18}\)) instead of subtracting them (to get \(10^{10}\)). This is a common mistake when students confuse the rules for multiplication (where you add exponents) with division (where you subtract exponents).
Students might struggle with dividing 2.3 by 3.9, potentially getting values like 0.6, 0.7, or other approximations that are significantly off from the correct 0.59. Some might even get confused and calculate \(3.9 \div 2.3 \approx 1.7\) instead.
After getting \(0.59 \times 10^{10}\), students might forget to convert this to proper scientific notation. They could leave the answer as \(0.59 \times 10^{10}\) or incorrectly convert it to \(59 \times 10^8\) instead of the correct \(5.9 \times 10^9\).
Even if students get the coefficient approximately right (around 5.9), they might select an answer choice with the wrong power of 10. For instance, they might choose (E) \(5.9 \times 10^{11}\) because the coefficient matches, without carefully checking that their calculation yielded \(10^9\), not \(10^{11}\).