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Kisachek [45]
3 years ago
6

Data Entry - No Scoring During your procedure to determine the uncalibrated volume near the stopcock of your buret, what was the

mass in grams of water that drained out when the meniscus started just above the 25 mL mark and ended just above the stopcock? Report to the thousandth of a gram.
Mathematics
1 answer:
DaniilM [7]3 years ago
4 0

Answer:

0.025 grams

Step-by-step explanation:

The water in the stopcock has a volume of 25 mL initially, After that, the whole water was drained out. So we have:

Volume of drained water = (25 mL)(1 x 10⁻⁶ m³/1 mL)

Volume of drained water = 25 x 10⁻⁶ m³

Density of drained water = 1000 kg/m³

So, for the mass of drained water:

Density of drained water = Mass of drained water/Volume of drained water

Mass of drained water = (Density of drained water)(Volume of drained water)

Mass of drained water = (1000 kg/m³)(25 x 10⁻⁶ m³)

<u>Mass of drained water = 0.025 gram</u>

Density

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3 years ago
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ary Stevens earns $6 an hour at her job and is entitled to time-and-a-half for overtime and double time on holidays. Last week s
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I will give brainliest!
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3 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
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Answer:

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}}

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The margin of error is:

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

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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}}

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