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3 years ago

What is 26.50 rounded to the nearest hundredth

1 answer:

6 0

Well,

26.50

In this case, the hundredths place is the 0.

Since there are no more digits after that, 26.50 must already be rounded to the nearest hundredth.

26.50 is the final answer.

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Plz help! Giving a lot of points! :)

monitta

(B) carlos: 920 - 300 = 680
Rita: 980 - 375 = 680
(Sorry i could only answer B :(

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3 years ago

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Jeremy wrote this equivalent equation to solve for x=-11y-140 Which of these equations might have been Jeremy’s original equatio

Alex73 [517]

The correct answers are 3x + 33y = -420 AND
X + 140 = -11 y

Please see the work attached for the steps for getting these answers.

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3 years ago

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Pleas answer the question in the attached file.

ch4aika [34]

A. My best friends, whose name is Laura, is in law school.

You know this because, if you were to take out the part in between the commas, the sentence would still make sense. It just adds additional detail.

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3 years ago

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Which of the following is a linear function?

klasskru [66]

I hope this helps you

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4 years ago

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A company receives shipments of a component used in the manufacture of a component for a high-end acoustic speaker system. When

Tanya [424]

Answer:

$ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434$

Step-by-step explanation:

For this case we have a sample size of n = 250 units and in this sample they found that 24 units failed one or more of the tests.

We are interested in the proportion of units that fail to meet the company's specifications, and we can estimate this with:

$\hat p = \frac{24}{250}= 0.096$

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval for a proportion is given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

And the margin of error would be:

$ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434$

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4 years ago