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oksano4ka [1.4K]
3 years ago
12

Tamar is thinking of a number in the hundredths. Her number is greater than 0.8 and less than 0.9. The greatest digit in the num

ber is in the hundredths place. What number is Tamar thinking of? Explain.
Mathematics
2 answers:
Yanka [14]3 years ago
5 0

Answer:

Therefore, Tamar is thinking of the number 0.89.

Step-by-step explanation:

Let x be the number Tamar is thinking of.

0.8 < x < 0.9

Since the greatest digits in the hundreths is greater digit, and there's an "8" in the tenths, the number can described as:

0.88< x < 0.9

There is only a single number in the hundredths that would fit the expression above, and that is 0.89.

Therefore, Tamar is thinking of the number 0.89.

Ivahew [28]3 years ago
3 0
0.89 i think its closest to 0.9
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Find f(2) and f(3).<br> f(1) = 3<br> f(n) = 7f(n = 1)
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Write 0.96 as a fraction in simplest form.<br> 0.96 =
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The average amount of water in randomly selected 16-ounce bottles of water is 16.15 ounces with a standard deviation of 0.45 oun
vekshin1

Answer:

0.0179 = 1.79% probability that the mean of this sample is less than 15.99 ounces of water.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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.

In this question, we have that:

\mu = 16.15, \sigma = 0.45, n = 35, s = \frac{0.45}{\sqrt{35}} = 0.0761

What is the probability that the mean of this sample is less than 15.99 ounces of water?

This is the pvalue of Z when X = 15.99. So

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

By the Central Limit Theorem

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

Z = \frac{15.99 - 16.15}{0.0761}

Z = -2.1

Z = -2.1 has a pvalue of 0.0179

0.0179 = 1.79% probability that the mean of this sample is less than 15.99 ounces of water.

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