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Tasya [4]
3 years ago
14

A shop has 55 posters for sale. 34 posters show sports. The rest of the posters show animals. How many posters show animals?

Mathematics
2 answers:
sdas [7]3 years ago
5 0

Answer: 21 posters is the answer.

maksim [4K]3 years ago
3 0
21 Posters.


55-34=21
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Take 110 and divide by 2 so 110/2 = 55 = b
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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
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