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3 years ago

Evaluate the integral ∫_0^1{36dx/((2x+1)^3)}

1 answer:

german3 years ago

3 0

$36 \int\limits^1_0 { \frac{1}{(2 x + 1) ^{3} } } \, dx =$
u = 2 x + 1, d u = 2 d x, d x = d u / 2
$36*1/2 \int\limits^1_0 {u ^{-3} } \, du = 18 * t ^{-2}/(-2) = \\ - 9 * t^{-2}= \\ -9 * ( 2 x + 1 ) ^{-2}$
- 9 / ( 2 + 1 )² - ( - 9 / ( 0 + 1 )² =
- 9 / 9 + 9 = - 1 + 9 = 8

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The standard deviation increased but there was no change in the interquantile range

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

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a) Formula:

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where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

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13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

$S.D = \sqrt{\frac{75924.2}{19}} = 63.21$

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

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IQR = $482 - 395 = 87$

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

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