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kakasveta [241]

3 years ago

6

Solve for x ~

%20" id="TexFormula1" title="2(x - 3) + 5(3x + 1) - 16x = x \\ " alt="2(x - 3) + 5(3x + 1) - 16x = x \\ " align="absmiddle" class="latex-formula">
thankyou~

1 answer:

Mars2501 [29]3 years ago

3 0

Answer:

0

Step-by-step explanation:

<h2><em>2</em><em>x</em><em>-</em><em>6</em><em>+</em><em>1</em><em>5</em><em>x</em><em>+</em><em>5</em><em>-</em><em>1</em><em>6</em><em>x</em><em>=</em><em>x</em></h2>

<h2><em>2</em><em>x</em><em>+</em><em>1</em><em>5</em><em>x</em><em>-</em><em>1</em><em>6</em><em>x</em><em>-</em><em>x</em><em>=</em><em>6</em><em>-</em><em>5</em></h2><h2 /><h2 /><h2><em>0</em><em>×</em><em>x</em><em>=</em><em>1</em></h2><h2 /><h2 /><h2 /><h2><em>x</em><em>=</em><em>1</em><em>÷</em><em>0</em></h2><h2 /><h2 /><h2><em>x</em><em>=</em><em>0</em></h2>

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Which of the following statements are true?

Fantom [35]

Answer:

option A and B are correct.

Step-by-step explanation:

Given: $p(t) = \frac{64}{1 + 11e^{-.08t} }$

Option A: $\lim_{t \to \infty} \frac{64}{1 + 11e^{-.08(\infty)} } = \frac{64}{1 + 0} = 64$

Option A is true,

Option B: $P(0) = \frac{64}{1 + 11} = 5.33$

Option B is also true.

Option C:

$P(t + 1) = \frac{64}{1 + 11e^{-.08(t + 1)} } = \frac{64}{1 + 10.15e^{-.08t} } \\1.08 · P(t) = \frac{64 · 1.08 }{1 + 11e^{-.08t} } = \frac{69.12}{1 + 11e^{-.08t} }$

P(t + 1) ≠ 1.08 · P(t)

Option C is incorrect.

Option D: It is also incorrect, because according to option 2 earth's population will not grow exponentially without Bound.

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3 years ago

Tom rolls 2 fair dice and adds the results from each. Work out the probability of getting a total that is a factor of 21.

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Step-by-step explanation:

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3 years ago

A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at

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Answer:

$P(Same)=\frac{61}{190}$

Step-by-step explanation:

Given

$Red = 5$

$White = 6$

$Black = 9$

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

$Total = 5 + 6 + 9$

$Total = 20$

This is calculated as:

$P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)$

So, we have:

$P(Same)=\frac{n(Red)}{Total} * \frac{n(Red) - 1}{Total - 1} + \frac{n(White)}{Total} * \frac{n(White) - 1}{Total - 1} + \frac{n(Black)}{Total} * \frac{n(Black) - 1}{Total - 1}$

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

$P(Same)=\frac{5}{20} * \frac{5 - 1}{20- 1} + \frac{6}{20} * \frac{6 - 1}{20- 1} + \frac{9}{20} * \frac{9- 1}{20- 1}$

$P(Same)=\frac{5}{20} * \frac{4}{19} + \frac{6}{20} * \frac{5}{19} + \frac{9}{20} * \frac{8}{19}$

$P(Same)=\frac{20}{380} + \frac{30}{380} + \frac{72}{380}$

$P(Same)=\frac{20+30+72}{380}$

$P(Same)=\frac{122}{380}$

$P(Same)=\frac{61}{190}$

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