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8_murik_8 [283]
3 years ago
13

Please help and show all work so I can fully understand it, thanks.

Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

\frac{1}{m-4}

Step-by-step explanation:

\frac{\frac{4m-5}{m^4 -7m^3 +12m^2}}{\frac{4m-5}{m^3 -3m^2}}

Factor the equation:

\frac{\frac{4m-5}{m^2(m^2 -7m+12)}}{\frac{4m-5}{m^2(m-3)}}

\frac{\frac{4m-5}{m^2(m-3)(m-4)}}{\frac{4m-5}{m^2(m-3)}}

Rewrite to suit the format of multiplying two fractions. Remember, dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second. A reciprocal of a fraction is when one switches the place of the numerator and the denominator, that is, the value on top (numerator), and the value on the bottom (denominator).

\frac{4m-5}{m^2(m-3)(m-4)}*\frac{m^2(m-3)}{4m-5}

Simplify, take out common terms that are found on both the numerator and denominator

\frac{4m-5}{m^2(m-3)(m-4)}*\frac{m^2(m-3)}{4m-5}

\frac{1}{m-4}

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On a standard IQ test, the standard deviation is 15. How many random IQ scores must be obtained if we want to find the true popu
igor_vitrenko [27]

Answer:

We need at least 4238 scores.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.97}{2} = 0.015

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.015 = 0.985, so z = 2.17

Now, find the margin of error M

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How many random IQ scores must be obtained if we want to find the true population mean (with an allowable error of 0.5) and we want 97 percent confidence in the results

We need at least n scores, in which N is found when M = 0.5. So

M = z*\frac{\sigma}{\sqrt{n}}

0.5 = 2.17*\frac{15}{\sqrt{n}}

0.5\sqrt{n} = 2.17*15

0.5\sqrt{n} = 32.55

\sqrt{n} = \frac{32.55}{0.5}

\sqrt{n} = 65.1

\sqrt{n}^{2} = (65.1)^{2}

n = 4238

We need at least 4238 scores.

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