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Advocard [28]
2 years ago
5

Betty has 427 oranges and needs to pack them up equally in 23 boxes. How

Mathematics
1 answer:
stellarik [79]2 years ago
3 0

it's 18 with a remainder of 13. I attached a photo that shows the work of it

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Canceled cheques are negotiable at the bank for the face value.<br> True<br> False
balandron [24]

Canceled cheques are negotiable at the bank for the face value is False

4 0
3 years ago
Find the median of: 1, 3, 4, 6, 2, 4, 5, 6, 2, 3, 1, 4, 0, 4, 4, 4, 8, 9, 7, 4
Alborosie

Solution,

Arranging the data in ascending order:

0,1,1,2,2,3,3,4,4,4,4,4,4,4,5,6,6,7,8,9

N(total number of items)= 20

Now,

Median:

(\frac{n + 1}{2)} ) ^{th \: item}  \\  =  (\frac{20 + 1}{2} ) ^{th \: item}  \\  =  \frac{21}{2}  \\  = 10.5 \: th \: \: item

Again,

Median:

\frac{10  \: th \: item + 11 \: th \: item}{2}  \\  =  \frac{4 + 4}{2}  \\  =  \frac{8}{2}  \\  = 4

5 0
3 years ago
Read 2 more answers
To decrease the impact on the environment, factory chimneys must be high enough to allow pollutants to dissipate over a larger a
Komok [63]

Answer:

The probability hat the sample mean height for the 40 chimneys is greater than 102 meters is 0.1469.

Step-by-step explanation:

Let the random variable <em>X</em> be defined as the height of chimneys in factories.

The mean height is, <em>μ</em> = 100 meters.

The standard deviation of heights is, <em>σ</em> = 12 meters.

It is provided that a random sample of <em>n</em> = 40 chimney heights is obtained.

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

Since the sample selected is quite large, i.e. <em>n</em> = 40 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean heights of chimneys.

\bar X\sim N(\mu_{\bar x},\ \sigma^{2}_{\bar x})

Compute the probability hat the sample mean height for the 40 chimneys is greater than 102 meters as follows:

P(\bar X>102)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}})>\frac{102-100}{12/\sqrt{40}})

                    =P(Z>1.05)\\=1-P(Z

*Use a <em>z</em>-table fr the probability.

Thus, the probability hat the sample mean height for the 40 chimneys is greater than 102 meters is 0.1469.

8 0
3 years ago
Which is the graph of g(x) = [X +3]?
Daniel [21]

Answer:

b

Step-by-step explanation:

just took it i got 100%

4 0
3 years ago
an outlier may be defined as a data point that is more than 1.5 times the interquartile range below the lower quartile or is mor
Ira Lisetskai [31]

Supposing a normal distribution, we find that:

The diameter of the smallest tree that is an outlier is of 16.36 inches.

--------------------------------

We suppose that tree diameters are normally distributed with <u>mean 8.8 inches and standard deviation 2.8 inches.</u>

<u />

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}

  • The Z-score 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.<u> </u>

<u />

In this problem:

  • Mean of 8.8 inches, thus \mu = 8.8.
  • Standard deviation of 2.8 inches, thus \sigma = 2.8.

<u />

The interquartile range(IQR) is the difference between the 75th and the 25th percentile.

<u />

25th percentile:

  • X when Z has a p-value of 0.25, so X when Z = -0.675.

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

-0.675 = \frac{X - 8.8}{2.8}

X - 8.8 = -0.675(2.8)

X = 6.91

75th percentile:

  • X when Z has a p-value of 0.75, so X when Z = 0.675.

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

0.675 = \frac{X - 8.8}{2.8}

X - 8.8 = 0.675(2.8)

X = 10.69

The IQR is:

IQR = 10.69 - 6.91 = 3.78

What is the diameter, in inches, of the smallest tree that is an outlier?

  • The diameter is <u>1.5IQR above the 75th percentile</u>, thus:

10.69 + 1.5(3.78) = 16.36

The diameter of the smallest tree that is an outlier is of 16.36 inches.

<u />

A similar problem is given at brainly.com/question/15683591

3 0
2 years ago
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