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2 years ago

Estimate the unit rate if 12 pairs of socks sell for $5.79.

1 answer:

3 0

If 12 pairs of socks sell for $5.79, then the unit rate is 12 : 5.79. If we want to find a unit rate for one pair of socks, we need to divide 5.79 by 12:
5.79 / 12 = 0,4825
So the unit rate is 1 : 0.4825
Hope this was helpful, and if so, please mark as brainliest. ((: Thank you!

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H(x)=3x-4 what is h(6)

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H(x)=3x-4 what is h(6)
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answer
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2 years ago

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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(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 μ and standard deviation σ and we take appropriately huge random samples (n ≥ 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,

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

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

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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 n from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for n = 64 is:

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

The standard deviation of the sample mean for n = 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 n 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 n from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for n = 784 is:

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

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

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

Hence, the variance also increases.

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