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noname [10]
3 years ago
11

Which logarithmic equation is equivalent to 8^2=64

Mathematics
2 answers:
NISA [10]3 years ago
7 0

Answer:

2=log_{8}64

Step-by-step explanation:

We are given a equation:

   8²=64

We have to convert this into a logarithmic equation:

 Taking log base 8 on both sides, we get

    log_{8}8^{2}=log_{8}64

   2log_{8}8=log_{8}64   (since, log_{a}m^n=nlog_{a}m)

    2=log_{8}64  (since,log_{a}a=1 )

Hence, option first is true

adell [148]3 years ago
5 0

Answer:

The answer will be the first one...


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The mean number of words per minute (WPM) read by sixth graders is 84 with a standard deviation of 15 WPM. If 188 sixth graders
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0.1836 = 18.36% probability that the sample mean would differ from the population mean by greater than 1.46 WPM

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean number of words per minute (WPM) read by sixth graders is 84 with a standard deviation of 15 WPM.

This means that \mu = 84, \sigma = 15

Sample of 188

This means that n = 188, s = \frac{15}{\sqrt{188}}

What is the probability that the sample mean would differ from the population mean by greater than 1.46 WPM?

Greater than 84 + 1.46 = 85.46 or less than 84 - 1.46 = 82.54. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Probability is is less than 82.54.

P-value of Z when X = 82.54. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{82.54 - 84}{\frac{15}{\sqrt{188}}}

Z = -1.33

Z = -1.33 has a p-value of 0.0918

2*0.0918 = 0.1836

0.1836 = 18.36% probability that the sample mean would differ from the population mean by greater than 1.46 WPM

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