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Rudiy27
2 years ago
12

Noah planned a parallelogram-shaped sidewalk to lead from the street to the front door of his house. He realized the walkway was

too short and he needs to change the length to 72 feet. He wrote a ratio to find the change in the scale factor. 24 ft 72 ft = The change in the scale factor is . The change of scale means that 1 inch represented feet, but now 1 inch represents feet.

Mathematics
2 answers:
nekit [7.7K]2 years ago
8 0

Answer:

Step-by-step explanation:

8090 [49]2 years ago
8 0

Answer:

part one 1/3 second part is 8 last part is 24

Step-by-step explanation:

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A certain type of thread is manufactured with a mean tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilogram
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Answer:

(a) The variance decreases.

(b) The variance increases.

Step-by-step explanation:

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

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

\mu_{\bar x}=\mu

And the standard deviation of the sample mean is given by,

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

The standard deviation of sample mean is inversely proportional to the sample size, <em>n</em>.

So, if <em>n</em> increases then the standard deviation will decrease and vice-versa.

(a)

The sample size is increased from 64 to 196.

As mentioned above, if the sample size is increased then the standard deviation will decrease.

So, on increasing the value of <em>n</em> from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for <em>n</em> = 64 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7

The standard deviation of the sample mean for <em>n</em> = 196 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4

The standard deviation of the sample mean decreased from 0.7 to 0.4 when <em>n</em> is increased from 64 to 196.

Hence, the variance also decreases.

(b)

If the sample size is decreased then the standard deviation will increase.

So, on decreasing the value of <em>n</em> from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for <em>n</em> = 784 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2

The standard deviation of the sample mean for <em>n</em> = 49 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8

The standard deviation of the sample mean increased from 0.2 to 0.8 when <em>n</em> is decreased from 784 to 49.

Hence, the variance also increases.

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im soory got to gt to class ill make sure i thank u and answer a nother of ur questions sorry ccoykdnt answer.

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