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blagie [28]
3 years ago
15

If

n="absmiddle" class="latex-formula"> and \sf b = log_54, find \sf log_615 in terms of a and b.​
Mathematics
1 answer:
Norma-Jean [14]3 years ago
7 0

Answer:

  • (4a + b)/(2ab + b)

Step-by-step explanation:

<h3>Given</h3>
  • a = log9 2
  • b = log5 4
<h3>To find</h3>
  • log6 15 in terms of a and b
<h3>Solution</h3>

<u>Rewrite the given</u>

  • log9 2 = a ⇒ log2/log9 = a ⇒ 1/2*log2/log3 = a ⇒  log3 = 1/2*log2/a
  • log5 4 = b ⇒ log4/log5 = b ⇒ log5 = 2*log2/b

<u>Rewriting the log6 15:</u>

  • log6 15 = log15/log6= (log 5 + log3)/(log2 + log3)

<u>Substitute as follows:</u>

  • (log 5 + log3)/(log2 + log3) =
  • (2*log2/b + 1/2*log2/a)/(log2 + 1/2*log2/a) =
  • (2/b + 1/(2a))/(1 + 1/(2a)) =
  • (4a + b)/2ab ÷ (2a + 1)/2a =
  • (4a + b)/(2ab + b)
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1 year ago
The population of scores on a nationally standardized test forms a normal distribution with μ = 300 and σ = 50. If you take a ra
madreJ [45]

Answer:

P(\bar X

And using a calculator, excel or the normal standard table we have that:

P(Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

X \sim N(300,50)  

Where \mu=300 and \sigma=50

Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

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P(\bar X

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P(Z

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