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umka21 [38]
3 years ago
6

Round 483 to the nearest ten

Mathematics
2 answers:
madam [21]3 years ago
7 0
To round to the nearest ten, we will look at the number in the ones place, and determine if it is higher, lower, or equal to 5. In this case, 3 is less than 5, so we round down. This means that 483 rounded to the nearest ten is 480. :) I hope this helps you in the future! A common phrase you can think about is:
Four or less, give it a rest. 5 and above, give it a shove.
Luden [163]3 years ago
6 0

Answer:

480

We round the number up to the nearest ten if the last digit in the number is 5, 6, 7, 8, or 9.

We round the number down to the nearest ten if the last digit in the number is 1, 2, 3, or 4.

If the last digit is 0, then we do not have to do any rounding, because it is already to the ten.


You might be interested in
According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe
Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1 - p)}{n}}, as long as np \geq 10 and n(1 - p) \geq 10.

In this problem:

  • 22% of couples meet online, hence p = 0.22.
  • A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

\mu = p = 0.22

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem:

Z = \frac{X - \mu}{s}

Z = \frac{0.25 - 0.22}{0.0338}

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

Z = \frac{X - \mu}{s}

Z = \frac{0.2 - 0.22}{0.0338}

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

Z = \frac{X - \mu}{s}

Z = \frac{0.15 - 0.22}{0.0338}

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

4 0
2 years ago
Can someone pls help out
weqwewe [10]

Answer:

A. 99.87%

Step-by-step explanation:

i don't explain things good

6 0
3 years ago
Pleease help me with this math work
OleMash [197]
The first one because the left bound is included [ and the right one is not )
6 0
4 years ago
What is 184.24 to 4 significant figures
siniylev [52]
184.2 is the answer to 4 significant figure
4 0
3 years ago
Im so confused all i know is that it isn't C
hodyreva [135]
Well I’m not 100 percent sure but I believe it’s b
3 0
3 years ago
Read 2 more answers
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