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ahrayia [7]
2 years ago
11

How would logx (49) = x look like in exponential form

Mathematics
1 answer:
Ivanshal [37]2 years ago
5 0

I'm assuming the given log equation is \log_{\text{x}}(49) = \text{x}

If so, then the exponential form is \text{x}^{\text{x}} = 49

This is because the general form \log_{\text{b}}(\text{x}) = \text{y} transforms into \text{y} = \text{b}^{\text{x}}

For both equations, the 'b' is the base.

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Step-by-step explanation:

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Suppose a continuous probability distribution has an average of μ=35 and a standard deviation of σ=16. Draw 100 times at random
yulyashka [42]

Answer:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

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 normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

For sums, we have that the mean is \mu*n and the standard deviation is s = \sigma \sqrt{n}

In this problem, we have that:

\mu = 100*35 = 3500, \sigma = \sqrt{100}*16 = 160

This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.

X = 4000

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

By the Central Limit Theorem

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

Z = \frac{4000 - 3500}{160}

Z = 3.13

Z = 3.13 has a pvalue of 0.9991

X = 3000

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

Z = \frac{3000 - 3500}{160}

Z = -3.13

Z = -3.13 has a pvalue of 0.0009

0.9991 - 0.0009 = 0.9982

So the correct answer is:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

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Answer:

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Step-by-step explanation:

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