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Olenka [21]
2 years ago
7

Which expression is equivalent to the following complex fraction? StartFraction 3 Over x minus 1 EndFraction minus 4 divided by

2 minus StartFraction 2 Over x minus 1 EndFraction.
Mathematics
1 answer:
zimovet [89]2 years ago
5 0

The expression which is equivalent to the provided complex fraction is equal to the expression,

{\dfrac{-4x+7}{2(x-2)}}

<h3>What is the equivalent expression?</h3>

Equivalent expressions are the expression whose result is equal to the original expression, but the way of representation is different.

The expression given in the problem is,

\dfrac{\dfrac{3}{x-1}-4}{2-\dfrac{2}{x-1}}

Let the expression which is equivalent to the above expression is f(<em>x</em>). Therefore,

f(x)=\dfrac{\dfrac{3}{x-1}-4}{2-\dfrac{2}{x-1}}

Take the LCM (x-1) in the numerator and denominator as,

f(x)=\dfrac{\dfrac{3-4(x-1)}{x-1}}{\dfrac{2(x-1)-2}{x-1}}

Simplify the expression and cancel out the (x-1) from both the fraction as,

f(x)={\dfrac{3-4(x-1)}{2(x-1)-2}}\\f(x)={\dfrac{3-4x+4}{2x-2-2}}\\f(x)={\dfrac{-4x+7}{2(x-2)}}

Hence, the expression which is equivalent to the provided complex fraction is equal to the expression,

{\dfrac{-4x+7}{2(x-2)}}

Learn more about the equivalent expression here;

brainly.com/question/2972832

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In a test of the effectiveness of garlic for lowering​ cholesterol, 8181 subjects were treated with raw garlic. Cholesterol leve
xz_007 [3.2K]

Answer:

With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

Step-by-step explanation:

The dependent <em>t</em>-test (also known as the paired <em>t</em>-test or paired samples <em>t</em>-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

In this case a paired <em>t</em>-test is used to determine the effectiveness of garlic for lowering​ cholesterol.

A random sample of 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment.

The hypothesis for the test can be defined as follows:

<em>H₀</em>: With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0, i.e. <em>d</em> ≤ 0.

<em>Hₐ</em>: With garlic​ treatment, the mean change in LDL cholesterol is greater than 0, i.e. <em>d</em> > 0.

The information provided is:

\bar d=0.40\\SD_{d}=16.2\\\alpha =0.01

Compute the test statistic value as follows:

t=\frac{\bar d}{SD_{d}/\sqrt{n}}\\\\=\frac{0.40}{16.2/\sqrt{81}}\\\\=0.22

The test statistic value is 0.22.

Decision rule:

If the <em>p</em>-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the <em>p</em>-value of the test as follows:

p-value=P(t_{n-1}>0.22)\\=P(t_{80}>0.22)\\=0.4132

*Use a <em>t</em>-table.

The <em>p</em>-value of the test is 0.4132.

<em>p-</em>value= 0.4132 > <em>α</em> = 0.01

The null hypothesis was failed to be rejected.

Thus, it can be concluded that with garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

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Your probability is a 30% chance of picking up a red marble then when getting a yellow one you have a 33.3% and so on of 3s chance of picking a yellow one 

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Ivenika [448]
Here’s the full solution:

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