Example Question - geometric mean
Here are examples of questions we've helped users solve.
Calculating Specific Expansion Terms and Investigating Geometric Mean Relation
For the 2nd question (problem 8), since it is cut off, we are unable to provide a complete solution. However, for the 3rd question (problem 9), we can proceed: <p>The sum of the two numbers is $6$ times their geometric mean, then </p> <p>Let the two numbers be $a$ and $b$. Given that $a+b = 6\sqrt{ab}$ and we are to show that $a$ and $b$ are in the ratio $(3 + 2\sqrt{2}):(3 - 2\sqrt{2})$.</p> <p>Since we have a ratio of sums to a geometric mean, let's assume $a = (3 + 2\sqrt{2})k$ and $b = (3 - 2\sqrt{2})k$, where $k$ is some positive constant.</p> <p>\[ a + b = (3 + 2\sqrt{2})k + (3 - 2\sqrt{2})k\]</p> <p>\[ a + b = 3k + 2\sqrt{2}k + 3k - 2\sqrt{2}k\]</p> <p>\[ a + b = 6k\]</p> <p>The geometric mean of $a$ and $b$ is $\sqrt{ab}$:</p> <p>\[\sqrt{ab} = \sqrt{(3 + 2\sqrt{2})k \cdot (3 - 2\sqrt{2})k}\]</p> <p>\[\sqrt{ab} = \sqrt{(9 - 8)k^2}\]</p> <p>\[\sqrt{ab} = k\]</p> <p>Now, we compare the sum and geometric mean:</p> <p>\[6k = 6\sqrt{ab}\]</p> <p>\[k = \sqrt{ab}\]</p> <p>This confirms that the sum of $a$ and $b$ is indeed $6$ times their geometric mean, and thus the numbers are in the desired ratio.</p> For question 9, which discusses mean and standard deviation, no specific mathematical equations are shown, and thus a solution cannot be provided based on the information given in the image.
Comparing Arithmetic and Geometric Mean for Positive Numbers
这个问题是关于算术平均数和几何平均数的比较的。根据题意，算术平均数为80，几何平均数是某个正数的平方，要求我们找出这个正数的可能值。 假设这个正数为x，因为几何平均数是x的平方，所以我们有： 算术平均数 = (x + x²) / 2 = 80 这是一个二次方程，我们可以将其改写为： x² + x - 160 = 0 我们使用求根公式（负b加减根号下b平方减4ac除以2a），得： x = [-1 ± √(1 + 4 * 160)] / 2 接下来，我们来计算： x = [-1 ± √641] / 2 显然，因为求的是正数的可能值，我们只需考虑带正号的解： x = [-1 + √641] / 2 ≈ (25.2 - 1) / 2 ≈ 24.2 / 2 ≈ 12.1 所以符合题目条件的正数x的大致值是12.1。然而，查看选项，我们看到最接近的整数值是11，所以最恰当的选项是 (A) 11。
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