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3 years ago

What's 63,889 rounded to the nearest hundred

Mathematics

2 answers:

levacccp [35]3 years ago

4 0

63,889 rounded to the nearest hundred is 63,900

Jlenok [28]3 years ago

4 0

The hundred's place here is 8, if you look at the number before 8 it's also 8 which is larger than 5, so you round the hundred's place 8 up to a 9. Now you have your answer. 63,900

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Answer:

$\tt{}1.440.000 \\$

Step-by-step explanation:

$\tt{}1.200 \times 1.200 \: \: \: \: \: \: \: \: \: \: \\ \\ \tt{}((12 \times 12) + (0000)) \\ \\ 1.440.000 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \\$

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2 years ago

Write money amounts using a leading zero if necessary & with 2 decimal places (For example: 0.80 or 0.72 or 1.23)

goldenfox [79]

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Answer:

63 mph
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Step-by-step explanation:

In this context, "per" and "for" mean "divided by."

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3 years ago

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timofeeve [1]

.........................................................

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3 years ago

At Burnt Mesa Pueblo, archaeological studies have used the method of tree-ring dating in an effort to determine when prehistoric

Zanzabum

Answer:

a) (1215, 1297)

b) (1174, 1338)

c) (1133, 1379)

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 1256

Standard Deviation, σ = 41

Empirical Rule:

Also known as 68-95-99.7 rule.
It states that almost all the data lies within three standard deviation for a normal data.
About 68% of data lies within 1 standard deviation of mean.
About 95% of data lies within two standard deviation of mean.
About 99.7% of data lies within 3 standard deviation of mean.

a) range of years centered about the mean in which about 68% of the data lies

$\mu \pm 1(\sigma)\\=1256 \pm 41\\=(1215, 1297)$

68% of data will be found between 1215 years and 1297 years.

b) range of years centered about the mean in which about 95% of the data lies

$\mu \pm 2(\sigma)\\=1256 \pm 2(41)\\=1256 \pm 82\\=(1174, 1338)$

95% of data will be found between 1174 years and 1338 years.

c) range of years centered about the mean in which about all of the data lies

$\mu \pm 3(\sigma)\\=1256 \pm 3(41)\\=1256 \pm 123\\=(1133, 1379)$

All of data will be found between 1133 years and 1379 years.

3 0

3 years ago