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3 years ago

What is the relationship between the values of the given digits. 4s in 44,000

2 answers:

8 0

The 44 in 44,000 is indicating the value of the thousand (,000). Because it is a 44, it would be 44 thousand. The first 4 is the 10,000 digit, and the second 4 is the 1,000 digit. All together, it is 4 ten thousands, plus 4 thousands, making 44,000

Softa [21]3 years ago

3 0

Answer:

'The 4 in the ten thousand is 10 times the 4 in the thousands place'.

Step-by-step explanation:

Given : Number 4s in 44,000.

To find : What is the relationship between the values of the given digits ?

Solution :

The place value system is

Ten Thousand Thousand Hundred Tens Ones

4 4 0 0 0

The most-significant 4 has a value of 40,000.

The least-significant 4 has a value of 4,000.

So, $\frac{40000}{4000}=10$

i.e. The most -significant 4 value is 10 times the least-significant 4 value.

Therefore, The relationship between them is

'The 4 in the ten thousand is 10 times the 4 in the thousands place'.

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Read 2 more answers

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kobusy [5.1K]

Using the normal distribution, it is found that:

a) The pilot is at the 72th percentile.

b) 19.13% of pilots are unable to fly.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 72.6, \sigma = 2.7$ .

Item a:

The percentile is the <u>p-value of Z when X = 74.2</u>, hence:

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

$Z = \frac{74.2 - 72.6}{2.7}$

Z = 0.59

Z = 0.59 has a p-value of 0.7224.

72th percentile.

Item b:

The proportion that is able to fly is the <u>p-value of Z when X = 78 subtracted by the p-value of Z when X = 70</u>, hence:

X = 78:

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

$Z = \frac{78 - 72.6}{2.7}$

Z = 2

Z = 2 has a p-value of 0.9772.

X = 70:

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

$Z = \frac{70 - 72.6}{2.7}$

Z = -0.96

Z = -0.96 has a p-value of 0.1685.

0.9772 - 0.1685 = 0.8087 = 80.87%.

Hence the percentage that is unable to fly is:

100 - 80.87 = 19.13%.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

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