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OleMash [197]
2 years ago
10

Select the expression that has a value of 62.45. (2 points) Group of answer choices 0.006245 × 105 0.6245 × 102 6.245 ÷ 101 624.

5 × 103
Mathematics
1 answer:
Free_Kalibri [48]2 years ago
3 0

The expression that has a value of 62.45 is the second answer choice 0.6245 × 10²

To determine the expression that has a value of 62.45 among the given answer choices, we will evaluate each of the answer choices.

  • For the first answer choice - 0.006245 × 105 = 0.006245 × 10⁵

0.006245 × 10⁵ = 624.5

  • For the second answer choice - 0.6245 × 102 = 0.6245 × 10²

0.6245 × 10² = 62.45

  • For the third answer choice - 6.245 ÷ 101 = 6.245 ÷ 10¹

6.245 ÷ 10¹ = 0.6245

  • For the fourth answer choice - 624.5 × 103 = 624.5 × 10³

624.5 × 10³ = 624500

From above, the second answer choice is the expression that has a value of 62.45

Hence, the expression that has a value of 62.45 is 0.6245 × 10²

Learn more here: brainly.com/question/17633165

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2 years ago
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro
Gnom [1K]

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A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of \pi, so we use the worst case scenario, which is \pi = 0.5

Then

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}

n = 1067.11

Rounding up

A sample of 1068 is needed.

8 0
2 years ago
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