Find the average of 320 , 330 , 340 .
(319 + 329 + 339 )
To find the average of a set of numbers, we sum all the numbers and then divide by the total count of numbers. In this question, we need to find the average of three numbers: \(3^{20}\), \(3^{30}\), and \(3^{40}\).
The numbers are:
The total count of numbers is 3.
The average is given by:
$$ \text{Average} = \frac{\text{Sum of numbers}}{\text{Count of numbers}} $$
Substituting the given numbers:
$$ \text{Average} = \frac{3^{20} + 3^{30} + 3^{40}}{3} $$
We can rewrite the denominator \(3\) as \(3^1\). To simplify the expression, we can divide each term in the numerator by \(3^1\). Recall the exponent rule: \( \frac{a^m}{a^n} = a^{m-n} \).
Let's apply this rule to each term:
So, the average becomes the sum of these simplified terms:
$$ \text{Average} = 3^{19} + 3^{29} + 3^{39} $$
Comparing this result with the given options, we find that it matches one of them.
The calculated average of \(3^{20}\), \(3^{30}\), and \(3^{40}\) is \(3^{19} + 3^{29} + 3^{39}\).
| Concept | Description | Example |
|---|---|---|
| Base | The number being multiplied by itself. | In \(3^5\), 3 is the base. |
| Exponent (Power) | The number of times the base is multiplied by itself. | In \(3^5\), 5 is the exponent. \(3^5 = 3 \times 3 \times 3 \times 3 \times 3\) |
| Division Rule | When dividing powers with the same base, subtract the exponents. \( \frac{a^m}{a^n} = a^{m-n} \) | \( \frac{5^7}{5^3} = 5^{7-3} = 5^4 \) |
Calculating the average is a fundamental concept in mathematics used to find a central value in a set of numbers. When dealing with numbers expressed as powers, like \(3^{20}\), \(3^{30}\), and \(3^{40}\), it's important to apply the rules of exponents correctly. These numbers are very large, so calculating their exact values before finding the average would be impractical. Simplifying the expression algebraically using exponent rules, as shown above, is the correct approach.
The expression \(3^{19} + 3^{29} + 3^{39}\) represents the simplified form of the average \( \frac{3^{20} + 3^{30} + 3^{40}}{3} \).
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