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ipn [44]
4 years ago
11

Help me please. 30 points :)

Mathematics
1 answer:
Nataly [62]4 years ago
3 0

\frac{(5 \:  -  \: 8x)}{( \sqrt{5 \:  -  \: 8x} )}  \:  =  \:  \sqrt{5 \:  -  \: 8x}
For the result to be Real,
5 \:  -  \: 8x \:  \geqslant  \: 0
8x \:  \leqslant  \: 5
x \:  \leqslant  \:  \frac{5}{8}
( \frac{(5 - 8x)}{( \sqrt{5 - 8x} )} ) \:  \times  \:  (\frac{ \sqrt{5 - 8x} }{ \sqrt{5 - 8x} } ) \:  =  \:  \frac{ {(5 \:  -  \: 8x)}^{ \frac{3}{2} } }{(5 \:  -  \: 8x)}
Hence,
\frac{{(5 \:  -  \: 8x)}^{ \frac{3}{2} } }{(5 \:  -  \: 8x)}
is the required form.

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The statement below is fallacious. Indicate which fallacy the speaker has committed:
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Regression to the mean fallacy

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It assumes that something has returned to normal because of corrective actions taken while it was abnormal. This fails to account for natural fluctuations. It is frequently a special kind of the post hoc fallacy.

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3 years ago
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Two balls are drawn in succession out of a box containing 4 red and 3 white balls. Find the probability that at least 1 ball was
kifflom [539]

Answer:

A) 40/49

B) 36/42

Step-by-step explanation:

Given:

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White Balls, W : 3

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A) If balls were replaced before the 2nd draw

P (White) = 3/7; P(Red) = 4

P(at least 1 red ball),

= 1 - P(no red balls)

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= 1-(9/49)

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A) If balls were replaced before the 2nd draw

P(1st White Ball) = 3/7 ; P(2nd White Ball) = 2/6

P(at least 1 red ball),

= 1 - P(no red balls)

= 1  - P(1st White, 2nd White)

= 1 - (3/7)(2/6)

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= 36/42

7 0
4 years ago
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vladimir2022 [97]
The answer to your problem is B.
7 0
3 years ago
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This means that for every group of 92 cats at the city compound there is a group of 144 dogs. So it also means that for every group of 23 cats there is a group of 36 dogs.

92 to 144 or 92/144 = 23/26  or 23 to 36

<h2>Hope This Helps Out</h2>
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2 years ago
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