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2 years ago

Enter the number that 95.746 rounds to when rounded to the nearest hundredth.

Mathematics

1 answer:

vladimir1956 [14]2 years ago

5 0

Answer:

95.75

Step-by-step explanation:

Note the place values:

9 (tens place value)

5 (ones place value)

7 (tenths place value)

4 (hundredths place value)

6 (thousandths place value)

In this case, you want to round to the nearest hundredth place value. to round, look at the place value directly right of the place value you are solving for (in this case, the thousandths place value). The value is 6.

Note:

When the value of the digit is 5 or greater, you round up.

When the value of the digit is 4 or less, you round down.

In this case, the value is 6, so you will round up:

95.746 rounded to the nearest hundredth is 95.75

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<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

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z is the critical value.

n is the sample size.

In this problem, the parameters are:

$z = 2.575, n = 2000, \pi = 0.77$

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7458$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7942$

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More can be learned about the z-distribution at brainly.com/question/25890103

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2 years ago

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viktelen [127]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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