3 years ago

Please help, this is due soon!

1 answer:

Oliga [24]3 years ago

6 0

Before outlier : 7.25
After outlier : 4

7.25 - 4 = 3.25

So the mean decreases by 3.25 when the outlier is removed.

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Ksju [112]

Answer:converge at $I=\frac{1}{3}$

Step-by-step explanation:

Given

Improper Integral I is given as

$I=\int^{\infty}_{3}\frac{1}{x^2}dx$

integration of $\frac{1}{x^2}$ is - $\frac{1}{x}$

$I=\left [ -\frac{1}{x}\right ]^{\infty}_3$

substituting value

$I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]$

$I=-\left [ 0-\frac{1}{3}\right ]$

$I=\frac{1}{3}$

so the value of integral converges at $\frac{1}{3}$

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4 years ago