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Kisachek [45]
3 years ago
13

Is 0.83 equal to 83/100

Mathematics
2 answers:
scoundrel [369]3 years ago
5 0

Answer:

Yes

Step-by-step explanation:

0.83 means that it's out of 100, if it was 1.0 that means that it would be 100, a full number.

83/100 means that it's out of 100 as well. If it were to be a whole number it would be 100/100

ipn [44]3 years ago
4 0

Answer:

Yes.

Step-by-step explanation:

83/100=0.83 : )

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Write -0.37 as a quotient of integers to show that it is rational
Zina [86]

Answer:

-37/100

Step-by-step explanation:

Here, we want to write -0.37 as a quotient of integers

What we are saying is that we want to write it as a fraction of two numbers

That would be;

-0.37 = -37/100

4 0
2 years ago
Suppose that in one region of the country the mean amount of credit card debt perhousehold in households having credit card debt
kvv77 [185]

Answer:

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 15250, \sigma = 7125, n = 1600, s = \frac{7125}{\sqrt{1600}} = 178.125

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly

This probability is the pvalue of Z when X = 1600 + 300 = 1900 subtracted by the pvalue of Z when X = 1600 - 300 = 1300. So

X = 1900

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

By the Central Limit Theorem

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

Z = \frac{1900 - 1600}{178.125}

Z = 1.68

Z = 1.68 has a pvalue of 0.9535.

X = 1300

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

Z = \frac{1300 - 1600}{178.125}

Z = -1.68

Z = -1.68 has a pvalue of 0.0465.

0.9535 - 0.0465 = 0.907.

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.

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3 years ago
{(3,4)(5,8)(3,7)} is a function. True or false
Dmitry_Shevchenko [17]
True, it is a function because mainly the x and y values are the same;]
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A restaurant has 2 manager, 5 waiters, and 3 chefs. 1 person is randomly selected to be transferred to a nearby store. What is t
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Answer:

80%

Step-by-step explanation:

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3 years ago
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